Hopp-Iravani-Shou v53.5 - Institute for Operations Research and the

Online Companion for

“Serial Agile Production Systems with Automation”

Operations Research Volume 53, Number 5

September-October 2005

Wallace J. Hopp Northeastern University

Seyed M.R. Iravani Northeastern University

and

Biying Shou

Northeastern University

©2005

ON-LINE APPENDIX

for

Serial Agile Production Systems with Automation

Wallace J. Hopp, Seyed M.R. Iravani and Biying Shou

Department of Industrial Engineering and Management SciencesNorthwestern University, Evanston, IL 60208, USA

Abstract

To gain insights into design and control of manufacturing cells with automation, we studysimple models of serial production systems where one ßexible worker attends a set of automatedstations. We (a) characterize the operational beneÞts of automation, (b) determine the mostdesirable placement of automation within a line, and (c) investigate how best to allocate labordynamically in a line with manual and automatic equipment. We do this by Þrst considering2-station MDP models and then studying 3-station simulations. Our results show that thecapacity of production lines with automatic machines can be signiÞcantly lower than the rate ofthe bottleneck. We also show that automating a manual machine can have a dramatic effect onthe average WIP level, provided that labor is the system bottleneck. Once a machine becomesthe bottleneck, the beneÞts from further automation are dramatically reduced. In general, weÞnd that automation is more effective when placed toward the end of the line rather than towardthe front. Finally, we show that automation level increases the priority workers should give toa station when selecting a work location.

1

ON-LINE APPENDIX

The MDP Model for the Capacity of a Two-Station AAP Line

Consider a two-station AAP line with one cross-trained worker and automatic machines at both stations.Assuming inÞnite arrival rate at station 1, which results in inÞnite WIP at that station, the capacity of theline can be obtained by solving the following MDP model. In the MDP model, the state of the system is(n2, s1, s2), where n2 is the WIP at station 2, and si = 0 indicates that the machine at station i is waiting tobe loaded, si = 1 indicates that the machine at station i is automatically processing a job, and si = 2 indicatesthat the machine at station i is waiting to be unloaded (i = 1, 2). Setting Λ = l1 + µ1 + u1 + l2 + µ2 + u2,the optimality equations are:

g + V (n2, 0, 0) = max

V (n2, 0, 0) ; Idlingl1ΛV (n2, 1, 0) +

Λ−l1Λ V (n2, 0, 0) ; Loading station 1

l2ΛV (n2, 0, 1) +


g + V (n2, 0, 1) = max

½ µ2Λ V (n2, 0, 2) +

Λ−µ2Λ V (n2, 0, 1) ; Idling

l1ΛV (n2, 1, 1) +

µ2Λ V (n2, 0, 2) +

Λ−l1−µ2Λ V (n2, 0, 1) ; Loading station 1

g + V (n2, 0, 2) = max

V (n2, 0, 2) ; Idlingl1ΛV (n2, 1, 2) +


u2Λ [V (n2 − 1, 0, 0) + 1] + Λ−u2

Λ V (n2, 0, 2) ; Unloading station 2

g + V (n2, 1, 0) = max

½µ1Λ V (n2, 2, 0) +


µ1Λ V (n2, 2, 0) +

l2ΛV (n2, 1, 1) +

Λ−µ1−l2Λ V (n2, 1, 0) ; Loading station 2

g + V (n2, 1, 1) =µ1ΛV (n2, 2, 1) +

µ2ΛV (n2, 1, 2) +

Λ− µ1 − µ2Λ

V (n2, 1, 1) ; Idling

g + V (n2, 1, 2) = max

½µ1Λ V (n2, 2, 2) +


µ1Λ V (n2, 2, 2) +

u2Λ [V (n2 − 1, 1, 0) + 1] + Λ−µ1−u2


g + V (n2, 2, 0) = max

V (n2, 2, 0) ; Idlingu1Λ V (n2 + 1, 0, 0) +

Λ−u1Λ V (n2, 2, 0) ; Unloading station 1

l2ΛV (n2, 2, 1) +


g + V (n2, 2, 1) = max

½µ2Λ V (n2, 2, 2) +


u1Λ V (n2 + 1, 0, 1) +

µ2Λ V (n2, 2, 2) +

Λ−u1−µ2Λ V (n2, 2, 1) ; Unloading station 1

g + V (n2, 2, 2) = max

V (n2, 2, 2) ; Idlingu1Λ V (n2 + 1, 0, 2) +

Λ−u1Λ V (n2, 2, 2) ; Unloading station 1

u2Λ [V (n2 − 1, 2, 0) + 1] + Λ−u2


1

ON-LINE APPENDIX

PROOFS OF ANALYTICAL RESULTS

PROOF OF LEMMA 1:

In order to prove part (i) of Lemma 1, it is sufficient to show that under an inÞnite job arrival rate, theutilization of the worker (the bottleneck) is less than 100%, regardless of the worker�s policy. It is obviousthat policies that idle the worker when there are jobs in the system that can be loaded or unloaded froman idle machine do not achieve the maximum capacity of the line. Therefore, we only focus on the class ofnon-idling policies Π. We prove Lemma 1 by showing that under any policy π ∈ Π there is always a positiveprobability that the worker becomes idle.

Suppose that policy π is an effective policy that does not result in worker idle time due to a lack of jobs atwork station 2. If such a policy does not exist, then the worker will become idle, and the proof of part (i)is completed. However, if such a policy exists, it implies that there are enough jobs at both stations 1 and2 to keep the worker busy. Now consider the time that under policy π the worker Þnishes loading a job on

the machine at station 1. At that time, with probabilities p(π)µ , p

(π)u , and p

(π)l the machine at station 2 is

automatically processing a job, waiting to be unloaded, and waiting to be loaded, respectively. Note that

p(π)µ + p

(π)u + p

(π)l = 1. We consider each of the above cases as follows:

Case I: If the machine at station 2 is automatically processing a job, then the worker will become idle untileither the machine at station 1 or the machine at station 2 Þnishes its automatic processing time.

Case II: If the machine at station 2 has Þnished its processing and therefore requires unloading, thenthe worker starts unloading the machine. The probability the worker becomes idle after completingthis unloading is Pr{Xu2 +Xl2 < Xµ1}. This probability represents situations where the automaticprocessing time at station 1 takes longer than unloading and reloading station 2. Since condition E1does not hold, then this probability is strictly positive (i.e, a positive probability for worker idleness).

Case III: If the machine at station 2 requires loading, then the worker starts loading the machine. Withprobability Pr{Xl2 < Xµ1} the worker Þnishes loading the machine at station 2 before the machineat station 1 Þnishes its processing. It is clear that if condition E1 does not hold, then probabilityPr{Xl2 < Xµ1} is strictly positive. This means that there exists a positive probability that the workerbecomes idle, after s/he Þnishes loading the machine at station 2 (since both machines at stations 1and 2 are automatically processing jobs).

Therefore, since there are positive probabilities of worker idleness in all of the above cases, and p(π)µ + p

(π)u +

p(π)l = 1, we can conclude that worker utilization is less than 100%. This implies that even if the two-stationline has an inÞnite arrival rate, the worker, which is the bottleneck, will have some idle time. Therefore,s/he cannot produce µb = 1/t0 items per unit time. The same argument applies when E2 does not hold,which completes the proof for part (i).

In order to prove part (ii), we need to show that under the cyclic policy the worker is always busy. It isclear that if condition E1 (condition E2) holds, then the time required to unload and reload a job at station2 (station 1) is always shorter than the automatic processing time at station 1 (station 2). This impliesthat, every time the worker Þnishes unloading and reloading station 2 and switches to station 1, the machineat station 1 is ready to be unloaded. Furthermore, when the worker Þnishes unloading and reloading thatmachine and switches to station 2, the machine at that station is ready for unloading. Therefore, if theworker repeats this cycle (i.e., follows the cyclic policy), s/he will never become idle. This guarantees thatthe line produces items at its bottleneck rate µb = 1/t0.

In order to prove part (iii), we show that if jobs are always available at station 1, the utilization of theworker (the bottleneck) can be 100%. For part (iii) we must consider two cases where: (a) the automatedstation is the Þrst station, and (b) the automated station is the second station.

2

(a) If the Þrst station is automated, then the maximum capacity of the line is achieved under any classof policies that guarantees inÞnite WIP at the second station. This is because, since station 2 is manual,inÞnite WIP at the second station guarantees 100% worker utilization under any non-idling policy. Considerthe class of non-idling policies Γ1 that give preemptive priority to the automatic machine at station 1. Underany policy γ1 ∈ Γ1 and unlimited jobs at station 1, station 1 completes a job every t1 units of time. In otherwords, station 2 faces an arrival rate of 1/t1 per unit time. Furthermore, since the worker is the bottleneck,

t0 > t11

l1+1

u1+ t2 >

1

l1+1

µ1+1

u1=⇒ t2 >

1

µ1. (1)

If we deÞne cycle time C(γ1)cy as the time elapsed between two consecutive loading completions of a job at

station 1 when policy γ1 is used, then in every cycle time C(γ1)cy , the worker has, on average, 1/µ1 units of

time to spend at station 2. However, according to inequality (1), this time is less than the average timerequired to Þnish a job at station 2. Therefore, under policy γ1 station 2 will have inÞnite WIP, whichguarantees 100% utilization for the worker.

(b) If the second station is automated, it is easy to show that any non-idling policy that gives preemptivepriority to the automatic machine at station 2 guarantees 100% utilization of the worker. The proof is similarto part (a) and is therefore omitted.

PROOF OF LEMMA 2:

(i) If both automated machines at stations 1 and 2 are bottlenecks, then the bottleneck rate is µb = 1/t1 =1/t2. In order to achieve the capacity equal to the bottleneck rate, both machines at stations 1 and 2 mustremain 100% utilized. However, it is easy to see that this cannot happen when at least one operation (i.e.,loading, unloading, automatic processing) is stochastic.

When both machine are bottlenecks, then both machines are 100% utilized only if none of the machines willever be idle because the worker is working on the other machine. This requires that unloading and loadingoperations on machine 1 (machine 2) always start and end during the automatic processing of machine 2(machine 1). However, when loading, unloading or automatic processing times are stochastic, this does notalways happen. Any randomness in loading, unloading or processing will result in a positive probability thatunloading and loading operation on machine 1 (machine 2) does not start and end during the automaticprocessing of machine 2 (machine 1). Therfore, there exists times that one of the machines are idle andwaiting for unloading, and the worker is loading (or unloading) the other machine. This implies that oneof the machines (or both machines) cannot remain 100% utilized. Thus, the capacity of the line becomesstrictly less than its bottleneck rate µb = 1/ti (i = 1 or 2).

(ii) If station 1 is automated and the machine in that station is the bottleneck, then the maximum capacityof the line is achieved under the class of policies Γ1 that gives preemptive priority to the bottleneck machineat station 1. This is because, under policy γ1 ∈ Γ1, station 1 completes jobs at the bottleneck rate µb = 1/t1.Since the machine is the bottleneck, we have

t0 < t11

l1+1

u1+ t2 <

1

l1+1

µ1+1

u1=⇒ t2 <

1

µ1. (2)

Note that in every cycle time C(γ1)cy , the worker has, on average, 1/µ1 units of time to spend at station 2.

On the other hand, according to inequality (2), this average time is more than the average time requiredto Þnish a job at station 2. Therefore, under policy γ1, station 2 will be able to process all jobs that arecompleted at station 1 and maintain a Þnite WIP level at station 2. Hence, the throughput of the line willequal the rate of station 1, which is the bottleneck rate.

3

PROOF OF THEOREM 1:

Before we start the proof of Theorem 1, we Þrst introduce Π0 as the class of stationary policies that neveridle when n1 = 0, n2 > 0, and s = 0 (i.e., when there is no job at the Þrst station, there is at least one job atthe second station, and the automatic machine in the Þrst station is not processing). We show that idlingunder such situation is not optimal. Therefore, without loss of generality, we revise our MDP by excludingthe non-optimal action idling from the corresponding states (i.e., states in which n1 = 0, n2 > 0, and s = 0).We call the new MDP model, the revised MDP.

Proposition: Policies that idle at states (0, n2 > 0, s = 0) are not optimal.

Proof: We use a sample path argument to do that. Suppose policy ζ is a policy that idles when n1 = 0, n2 >0, and s = 0. Consider t0 as the Þrst time under policy ζ that the server starts idling for some n1 = 0, n2 > 0,and s = 0. Then after some idle time I, the server may decide to: (i) serve the second station, or (ii) loadthe Þrst station, if there has been an arrival(s), and an arriving job(s) is waiting at station 1.

� (i) Consider the case that after idle time I the server decides to serve the next job in station 2 withservice time S2. A new service policy κ can be constructed as follows: κ follows ζ in [0, t0), serves acustomer at station 2, then idles for time I, and then follows what ζ does in [t0 + I + S2,∞). Since ζand κ behave exactly the same in [0, t0), reach the same state at time t0+ I +S2, and behave exactlythe same after time t0+I+S2, these two policies can be compared only during interval [t0, t0+I+S2].It is easy to see that during interval [t0, t0 + S2) the total WIP under both policy ζ and policy κ isthe same, but during interval [t0 + S2, t0 + I + S2) the system under policy κ carries one unit lessWIP than that under policy ζ. This is because in the former the second station Þnishes the automaticprocessing and releases the job out of the system at time t0 + S2 while in the latter the job is notreleased until time t0 + I + S2. Therefore, policy ζ cannot be optimal.

� (ii) Consider the case that after idle time I a new job has arrived, and the server decides to load theÞrst station which takes S1 units of time. Suppose t0 + I + S1 +M is the Þrst time under policy ζthat the server starts loading the second station after t0, which takes S2 units of time.

A new service policy χ can be constructed as follows: χ follows ζ in [0, t0), then starts to serve acustomer at station 2.

� If the service time S2 ≤ I, then χ idles for I−S2 units of time, follows what ζ does in [t0+I, t0+I +S1+M), idles again during [t0+ I +S1+M, t0+ I +S1+M +S2), and then follows ζ againin [t0 + I + S1 +M + S2,∞). Since ζ and χ behave exactly the same in [0, t0), reach the samestate at time t0+ I +S1+M +S2, and behave exactly the same after time t0+ I +S1+M +S2,the WIP under ζ and χ can be compared by focusing only on interval [t0, t0 + I + S1+M + S2].It is easy to see that during interval [t0, t0+S2) the total WIP under both policy ζ and policy χis the same, but during interval [t0+S2, t0+ I+S1+M +S2), the system under policy χ carriesone unit less WIP than that under policy ζ. This is because in the former the second stationÞnishes unloading and releases the job out of the system at time t0 + S2, while in the latter thejob is not released until time t0 + I + S1 +M + S2. Therefore, policy ζ cannot be optimal.

� If the service time S2 > I, policy χ stops serving the second station at time t0 + I and followswhat ζ does in [t0 + I,∞). Obviously, the performance of ζ will be as good as that of policy χ.

Combining the analysis of cases (i) and (ii), we can conclude that policies that idle when n1 = 0, n2 > 0,and s = 0 are not optimal. This completes the proof of our Proposition.

We now show that every stationary policy in our revised MDP model generates a Markov chain with asingle irreducible class. First, because the arrivals to the system are Poisson, state (0, 0, 0) can reach anystate (n1, 0, 0) with a positive probability. Second, state (n1, n2, 1) can reach state (n1 − 1, n2 + 1, 0) witha positive probability where n1 ≥ 1, n2 ≥ 0, as the Þrst station Þnishes automatic processing. Finally, sincethe stationary policies in Π0 never idles when n1 = 0, n2 > 0, and s = 0, state (0, 0, 0) is reachable by any

4

state (0, n2, 0) with a positive probability. Therefore, the Markov chain corresponding to every stationarypolicy in our revised MDP model is irreducible, and thus we can use Theorem 8.10.7 in Puterman (1994)to prove that there exists an average-cost optimal stationary policy with a constant average cost g. On theother hand, since the cost r(s, a) = −(n1+n2)/Λ ≤ 0 for any state s = (n1, n2, 0), s = (n1, n2, 1) and actiona, it suffices to show that Theorem 8.10.9 in Puterman (1994) holds.

In order to show that Theorem 8.10.9 in Puterman (1994) holds, Þrst we need to prove that when λ < µb,there exists a stationary policy that induces a positive recurrent Markov chain on the whole state space ofour revised MDP model, and also produces a stable system with Þnite mean queue lengths. For a serialtwo-station AAP system with one fully cross-trained worker and station 1 automated, we must discuss thefollowing two cases: (a) worker is the bottleneck, and (b) machine is the bottleneck.

(a)When the worker is the bottleneck, consider the class of non-idling policies Γ1 ⊂ Π0 that give preemptivepriority to the automatic machine at station 1. Since the worker is the bottleneck, i.e., µb = 1/t0,

1

µb= t0 > t1 =⇒ 1

λ> t1 since λ < µb.

Hence, under any policy γ1 ∈ Γ1 and an arrival rate λ < 1/t1, station 1 becomes a standard M/G/1queue with ρ < 1. The M/G/1 system has a Þnite queue length and completes a job, on average,every 1/λ units of time. Station 2, on the other hand, faces an arrival rate of λ per unit time. If we

deÞne cycle time C(γ1)cy as the time elapsed between two job arrivals at station 2 when policy γ1 is

used, then in every cycle time C(γ1)cy the worker has, on average, 1/λ− 1/l1 units of time to spend at

station 2. Since

1

λ> t1 =

1

l1+1

µ2=⇒ 1

λ− 1

l1>1

µ2,

this means that, on average, the worker has enough time to complete a job at station 2. Therefore,under policy γ1 station 2 will also have Þnite average queue length.

(b) When the automated station 1 is the bottleneck (i.e., 1/µb = t1), consider any policy γ1 ∈ Γ1 ⊂ Π0

which is a non-idling policy and gives preemptive priority to the automatic machine at station 1. Underthe arrival rate λ < µb = 1/t1, station 1 becomes a standard M/G/1 queue with ρ < 1. Therefore, ithas Þnite queue length, and it completes a job, on average, every 1/λ units of time. Station 2, on the

other hand, faces an arrival rate of λ per unit time. If we deÞne cycle time C(γ1)cy as the time elapsed

between two job arrivals at station 2 when policy γ1 is used, then in every cycle time C(γ1)cy the worker

has, on average, 1/λ − 1/l1 units of time to spend at station 2. Similar to the previous case, we canshow that (1/λ− 1/l1) > 1/µ2, which means that, on average, worker has enough time to complete ajob at station 2. Therefore, under policy γ1 station 2 will also have Þnite average queue length.

Thus, in both cases (a) and (b), the system involves a Þnite average cost (i.e., Þnite average queue lengths). Onthe other hand, since under policy γ1 all states can reach state (0,0,0) within a Þnite time, the derived Markovchain is positive recurrent. Furthermore, it can be shown that set {s ∈ S : C(s, a) < gγ1 for some a ∈ As}is nonempty and Þnite. Since gγ1 is Þnite and C(s, a) = (n1 + n2)/Λ (for all a ∈ As) is increasing in n1 andn2, then it is clear that only a Þnite number of states s = (n1, n2) satisfy C(s, a) = (n1 + n2)/Λ < g

γ1 . Onthe other hand, considering the fact that gγ1 > 0, for state so in which the system is empty (i.e., n1 = 0 andn2 = 0), we will have C(so, a) = 0 < g

γ1 . This implies that set {s ∈ S : C(s, a) < gγ1 for some a ∈ As} isa non-empty set. Thus, the assumptions of Theorem 8.10.9 in Puterman (1994) are veriÞed. Hence, thereexists an average-cost optimal stationary policy, and the average cost g is constant.

Finally, we must show that the value iteration algorithm converges. Proposition 4.3 of Sennott [1996] statesthat, if (i) there exists a stationary policy e inducing an irreducible and positive recurrent Markov chainwith the Þnite average cost Je < ∞, and (ii) there exists ² > 0 such that B = {s | there exists a such thatC(s, a) < Je+²} is Þnite, then the value iteration algorithm converges. Considering policy γ1 as policy e, thenwe have already shown (i). Furthermore, we have also shown that {s ∈ S : C(s, a) < gγ1 for some a ∈ As}

5

is a Þnite set. Using a similar argument, it can be proven that set B is also a Þnite set (assume ² = 0.1 forexample). Thus, both conditions of Proposition 4.3 of Sennott [1996] is satisÞed, and therefore the valueiteration algorithm converges.

PROOF OF PROPOSITION 1

Conditions V1, V2, V3 and C3 can be rewritten as follows:

V1 V (n1, n2 − 1, 1) ≤ V (n1, n2, 1) for n1 ≥ 1, n2 ≥ 1.V2 V (n1, n2, 1) ≤ V (n1, n2, 0) for n1 ≥ 1V3 V (n1, n2 − 1, 0) ≤ V (n1, n2, 0) for n2 ≥ 1.C1 V (n1 − 1, n2 + 1, 0) ≤ V (n1, n2, 1) for n1 ≥ 1.

We use induction and the value iteration algorithm to prove the above conditions. The optimality equationsfor the value iteration algorithm are:

Vk(n1, n2, 0) =n1 + n2Λ

+λ

ΛVk−1(n1 + 1, n2, 0) +

µ1ΛVk−1(n1, n2, 0)

+1

Λmin

(l1 + µ2)Vk−1(n1, n2, 0)l1Vk−1(n1, n2, 1) + µ2Vk−1(n1, n2, 0)µ2Vk−1(n1, n2 − 1, 0) + l1Vk−1(n1, n2, 0)

Vk(n1, n2, 1) =n1 + n2Λ

+λ

ΛVk−1(n1 + 1, n2, 1) +

µ1ΛVk−1(n1 − 1, n2 + 1, 0) + l1

ΛVk−1(n1, n2, 1)

+µ2Λmin

½Vk−1(n1, n2, 1)Vk−1(n1, n2 − 1, 1).

(V1,V2,V3,C1) Iteration 0: At iteration 0, all V0(n1, n2, s) = 0 for all (n1, n2, s). Therefore, V1, V2,V3 and C1 hold at iteration 0.

(V1,V2,V3,C1) Iteration k-1: We assume that at iteration k − 1, V1, V2,V3 and C1 hold. That is,

V1 Vk−1(n1, n2 − 1, 1) ≤ Vk−1(n1, n2, 1) for n1 ≥ 1, n2 ≥ 1V2 Vk−1(n1, n2, 1) ≤ Vk−1(n1, n2, 0) for n1 ≥ 1V3 Vk−1(n1, n2 − 1, 0) ≤ Vk−1(n1, n2, 0) for n2 ≥ 1C1 Vk−1(n1 − 1, n2 + 1, 0) ≤ Vk−1(n1, n2, 1) for n1 ≥ 1.

(V1,V2,V3,C1) Iteration k: Now, we show that (V1,V2,V3,C1) also hold at iteration k.

Proof for V1: Vk(n1, n2 − 1, 1) ≤ Vk(n1, n2, 1) for n1 ≥ 1, n2 ≥ 1.We discuss the following two cases:

� Case I: n1 ≥ 1 and n2 = 1;� Case II: n1 ≥ 1 and n2 ≥ 2.

V1-Case I:When n1 ≥ 1 and n2 = 1, conditionV1 becomes Vk(n1, 0, 1) ≤ Vk(n1, 1, 1). Since for Vk(n1, 0, 1)there�s no job at the second station, idling is the only option. Therefore,

Vk(n1, 0, 1) =n1Λ+λ

ΛVk−1(n1 + 1, 0, 1) +

µ1ΛVk−1(n1 − 1, 1, 0) + l1

ΛVk−1(n1, 0, 1) +

µ2ΛVk−1(n1, 0, 1)

6

For Vk(n1, 1, 1), we have:

Vk(n1, 1, 1) =n1 + 1

Λ+λ

ΛVk−1(n1 + 1, 1, 1) +

µ1ΛVk−1(n1 − 1, 2, 0) + l1

ΛVk−1(n1, 1, 1)

+µ2Λmin{Vk−1(n1, 1, 1) , Vk(n1, 0, 1)}.

Therefore,we have

Vk(n1, 0, 1)− Vk(n1, 1, 1) =−1Λ+λ

Λ[Vk−1(n1 + 1, 0, 1)− Vk−1(n1 + 1, 1, 1)]

+µ1Λ[Vk−1(n1 − 1, 1, 0)− Vk−1(n1 − 1, 2, 0)]

+l1Λ[Vk−1(n1, 0, 1)− Vk−1(n1, 1, 1)]

+µ2Λ[Vk−1(n1, 0, 1)−min{Vk−1(n1, 1, 1), Vk(n1, 0, 1)}]. (3)

According to induction assumption V1, we have

Vk−1(n1 + 1, 0, 1)− Vk−1(n1 + 1, 1, 1) ≤ 0Vk−1(n1, 0, 1)− Vk−1(n1, 1, 1) ≤ 0.

Thus, the second and the forth terms in (3) are nonpositive, and

min{Vk−1(n1, 1, 1)− Vk−1(n1, 0, 1)} = Vk−1(n1, 0, 1),which implies that the Þfth term on the right-hand side of (3) is equal to:

µ2Λ{Vk−1(n1, 0, 1)− Vk(n1, 0, 1)} = 0.

Finally, according to induction assumption V3

Vk−1(n1 − 1, 1, 0)− Vk−1(n1 − 1, 2, 0) ≤ 0,which implies that the third term on the right-hand side of (3) is nonpositive. Since all terms on theright-hand side of (3) are nonpositive, the proof for V1-Case I is complete.

V1-Case II: When n1 ≥ 1 and n2 ≥ 2, we will have

Vk(n1, n2 − 1, 1) =n1 + n2 − 1

Λ+λ

ΛVk−1(n1 + 1, n2 − 1, 1) + µ1

ΛVk−1(n1 − 1, n2, 0)

+l1ΛVk−1(n1, n2 − 1, 1) + µ2

Λmin{Vk−1(n1, n2 − 1, 1), Vk(n1, n2 − 2, 1)},

Vk(n1, n2, 1) =n1 + n2Λ

+λ

ΛVk−1(n1 + 1, n2, 1) +

µ1ΛVk−1(n1 − 1, n2 + 1, 0) + l1

ΛVk−1(n1, n2, 1)

+µ2Λmin{Vk−1(n1, n2, 1), Vk(n1, n2 − 1, 1)}.

Hence,

Vk(n1, n2 − 1, 1)− Vk(n1, n2, 1) =−1Λ+λ

Λ[Vk−1(n1 + 1, n2 − 1, 1)− Vk−1(n1 + 1, n2, 1)]

+µ1Λ[Vk−1(n1 − 1, n2, 0)− Vk−1(n1 − 1, n2 + 1, 0)]

+l1Λ[Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 1)]

+µ2Λ[min{Vk(n1, n2 − 1, 1), Vk(n1, n2 − 2, 1)}−min{Vk(n1, n2, 1)− Vk(n1, n2 − 1, 1)}]. (4)

7

According to induction assumption V1, we have

Vk−1(n1 + 1, n2 − 1, 1)− Vk−1(n1 + 1, n2, 1) ≤ 0,Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 1) ≤ 0.

Therefore, the second and the forth terms on the right-hand side of (4) are nonpositive. Also, due to V1

min{Vk−1(n1, n2 − 1, 1), Vk−1(n1, n2 − 2, 1)} = Vk−1(n1, n2 − 2, 1),min{Vk−1(n1, n2, 1), Vk−1(n1, n2 − 1, 1)} = Vk−1(n1, n2 − 1, 1),

which imply that the Þfth term in (4) is equal to the following nonpositive term:

µ2Λ[Vk−1(n1, n2 − 2, 1)− Vk−1(n1, n2 − 1, 1)].

Finally, the third term in (4) is nonpositive since according to induction assumption V3 we have

Vk−1(n1 − 1, n2, 0)− Vk−1(n1 − 1, n2 + 1, 0) ≤ 0.

This completes the proof for V1-Case II.

Proof for V2: Vk(n1, n2, 1) ≤ Vk(n1, n2, 0) for n1 ≥ 1.We discuss the following two cases:

� Case I: n1 ≥ 1 and n2 = 0;� Case II: n1 ≥ 1 and n2 ≥ 1.

V2-Case I:When n1 ≥ 1 and n2 = 0, conditionV2 becomes Vk(n1, 0, 1) ≤ Vk(n1, 0, 0). Since for Vk(n1, 0, 1)there�s no job at the second station, idling is the only option. Therefore,

Vk(n1, 0, 1) =n1Λ+λ

ΛVk−1(n1 + 1, 0, 1) +

µ1ΛVk−1(n1 − 1, 1, 0) + l1

ΛVk−1(n1, 0, 1) +

µ2ΛVk−1(n1, 0, 1).

On the other hand, since at state (n1, 0, 0) there�s no job at the second station, to process a job at station 2is not feasible. Therefore,

Vk(n1, 0, 0) =n1Λ+λ

ΛVk−1(n1 + 1, 0, 0) +

µ1ΛVk−1(n1, 0, 0) +

µ2ΛVk−1(n1, 0, 0)

+l1Λmin{Vk−1(n1, 0, 0) , Vk−1(n1, 0, 1)}.

Hence,

Vk(n1, 0, 1)− Vk(n1, 0, 0) =λ

Λ[Vk−1(n1 + 1, 0, 1)− Vk−1(n1 + 1, 0, 0)]

+µ1Λ[Vk−1(n1 − 1, 1, 0)− Vk−1(n1, 0, 0)]

+µ2Λ[Vk−1(n1, 0, 1)− Vk−1(n1, 0, 0)]

+l1Λ[Vk−1(n1, 0, 1)−min{Vk−1(n1, 0, 0)− Vk−1(n1, 0, 1)}]. (5)

According to induction assumption V2

Vk−1(n1 + 1, 0, 1)− Vk−1(n1 + 1, 0, 0) ≤ 0,Vk−1(n1, 0, 1)− Vk−1(n1, 0, 0) ≤ 0.

8

Therefore, the Þrst and the third terms on the right-hand side of (5) are nonpositive, and

min{Vk−1(n1, 0, 0), Vk−1(n1, 0, 1)} = Vk−1(n1, 0, 1),

which implies that the forth term on the right-hand side of (5) is equal to zero.Finally, according to induction assumption C1 and V2,

Vk−1(n1 − 1, 1, 0) ≤ Vk−1(n1, 0, 1) ≤ Vk−1(n1, 0, 0),

which implies that the second term on the right-hand side of (5) is nonpositive. Since all terms on theright-hand side of (5) are nonpositive, the proof for V2-Case I is completed.

V2-Case II: When n1 ≥ 1 and n2 ≥ 1, we will have

Vk(n1, n2, 1) =n1 + n2Λ

+λ

ΛVk−1(n1 + 1, n2, 1) +

µ1ΛVk−1(n1 − 1, n2 + 1, 0)

+l1ΛVk−1(n1, n2, 1) +

µ2Λmin

½Vk−1(n1, n2, 1)Vk(n1, n2 − 1, 1),

Vk(n1, n2, 0) =n1 + n2Λ

+λ

ΛVk−1(n1 + 1, n2, 0) +

µ1ΛVk−1(n1, n2, 0)

+1

Λmin

(l1 + µ2)Vk−1(n1, n2, 0)l1Vk−1(n1, n2, 1) + µ2Vk−1(n1, n2, 0)µ2Vk−1(n1, n2 − 1, 0) + l1Vk−1(n1, n2, 0).

Hence,

Vk(n1, n2, 1)− Vk(n1, n2, 0) =λ

Λ[Vk−1(n1 + 1, n2, 1)− Vk−1(n1 + 1, n2, 0)]

+µ1Λ[Vk−1(n1 − 1, n2 + 1, 0)− Vk−1(n1, n2, 0)]

+l1ΛVk−1(n1, n2, 1) +

µ2Λmin{Vk−1(n1, n2, 1), Vk(n1, n2 − 1, 1)}

− 1Λmin


(6)


Vk−1(n1 + 1, n2, 1)− Vk−1(n1 + 1, n2, 0) ≤ 0,

which implies that the Þrst term on the right-hand side of (6) is nonpositive. On the other hand, accordingto induction assumption C1 and V2

Vk−1(n1 − 1, n2 + 1, 0) ≤ Vk−1(n1, n2, 1) ≤ Vk−1(n1, n2, 0),

which implies that the second term on the right-hand side of (6) is nonpositive. Therefor, we only need toshow that the remaining terms on the right-hand side of (6) is nonpositive. We call the remaining term π.Therefore,

π =l1ΛVk−1(n1, n2, 1) +

µ2Λmin{Vk−1(n1, n2, 1), Vk(n1, n2 − 1, 1)}

− 1Λmin


9


Vk−1(n1, n2 − 1, 1) ≤ Vk−1(n1, n2, 1).Hence, the second term of π will be

µ2Λmin{Vk−1(n1, n2 − 1, 1), Vk−1(n1, n2, 1)} = µ2

ΛVk−1(n1, n2 − 1, 1).

Also, according to induction assumption V2 and V3

Vk−1(n1, n2, 0) ≥ Vk−1(n1, n2, 1),Vk−1(n1, n2, 0) ≥ Vk−1(n1, n2 − 1, 0).

Therefore,(l1 + µ2)Vk−1(n1, n2, 0) ≥ l1Vk−1(n1, n2, 1) + µ2Vk−1(n1, n2, 0)(l1 + µ2)Vk−1(n1, n2, 0) ≥ µ2Vk−1(n1, n2 − 1, 0) + l1Vk−1(n1, n2, 0),

which imply:

min

(l1 + µ2)Vk−1(n1, n2, 0)l1Vk−1(n1, n2, 1) + µ2Vk−1(n1, n2, 0)µ2Vk−1(n1, n2 − 1, 0) + l1Vk−1(n1, n2, 0)

= min

½l1Vk−1(n1, n2, 1) + µ2Vk−1(n1, n2, 0)µ2Vk−1(n1, n2 − 1, 0) + l1Vk−1(n1, n2, 0).

Thus,

π =l1ΛVk−1(n1, n2, 1) +

µ2ΛVk(n1, n2 − 1, 1)− 1

Λmin


To show that π ≤ 0, we discuss two cases:π ≤ 0: Case 1. In Case 1 we assume that

min

½l1Vk−1(n1, n2, 1) + µ2Vk−1(n1, n2, 0)µ2Vk−1(n1, n2 − 1, 0) + l1Vk−1(n1, n2, 0) = µ2Vk−1(n1, n2 − 1, 0) + l1Vk−1(n1, n2, 0),

then,

π =l1ΛVk−1(n1, n2, 1) +

µ2ΛVk(n1, n2 − 1, 1)− µ2

ΛVk−1(n1, n2 − 1, 0)− l1

ΛVk−1(n1, n2, 0)

=l1Λ[Vk−1(n1, n2, 1)− Vk−1(n1, n2, 0)] + µ2

Λ[Vk(n1, n2 − 1, 1)− Vk−1(n1, n2 − 1, 0)].


Vk−1(n1, n2, 1) ≤ Vk−1(n1, n2, 0),Vk−1(n1, n2 − 1, 1) ≤ Vk−1(n1, n2 − 1, 0).

This implies that both terms of π are nonpositive. Therefore, π ≤ 0.π ≤ 0: Case 2. In Case 2 we assume that

min

½l1Vk−1(n1, n2, 1) + µ2Vk−1(n1, n2, 0)µ2Vk−1(n1, n2 − 1, 0) + l1Vk−1(n1, n2, 0) = l1Vk−1(n1, n2, 1) + µ2Vk−1(n1, n2, 0),

then,

π =l1ΛVk−1(n1, n2, 1) +

µ2ΛVk(n1, n2 − 1, 1)− l1

ΛVk−1(n1, n2, 1)− µ2

ΛVk−1(n1, n2, 0)

=µ2Λ[Vk(n1, n2 − 1, 1)− Vk−1(n1, n2, 0)].

10

According to induction assumption V2 and V3

Vk−1(n1, n2 − 1, 1) ≤ Vk−1(n1, n2 − 1, 0) ≤ Vk−1(n1, n2, 0).

Therefore, π ≤ 0. This completes the proof for V2-Case II.Proof for V3: Vk(n1, n2 − 1, 0) ≤ Vk(n1, n2, 0) for n2 ≥ 1.We dicuss the following four cases:

� Case I: n1 = 0 and n2 = 1;� Case II: n1 = 0 and n2 ≥ 2;� Case III: n1 ≥ 1 and n2 = 1;� Case IV: n1 ≥ 1 and n2 ≥ 2;

V3-Case I:When n1 = 0 and n2 = 1, condition V3 becomes Vk(0, 0, 0) ≤ Vk(0, 1, 0). Since at state (0, 0, 0)there�s no job at the Þrst and second stations, idling is the only option. Therefore,

Vk(0, 0, 0) =λ

ΛVk−1(1, 0, 0) +

l1 + µ1 + µ2Λ

Vk−1(0, 0, 0).

Since at state (0, 1, 0) there�s no job at the Þrst station, the worker can either idle or process a job at station2, which means:

Vk(0, 1, 0) =1

Λ+λ

ΛVk−1(1, 1, 0) +

µ1ΛVk−1(0, 1, 0) +

l1ΛVk−1(0, 1, 0) +

µ2Λmin{Vk−1(0, 1, 0) , Vk−1(0, 0, 0)}.

Hence,

Vk(0, 0, 0)− Vk(0, 1, 0) = − 1Λ+λ

Λ[Vk−1(1, 0, 0)− Vk−1(1, 1, 0)] + µ1 + l1

Λ[Vk−1(0, 0, 0)− Vk−1(0, 1, 0)]

+µ2Λ[Vk−1(0, 0, 0)−min{Vk−1(0, 1, 0), Vk−1(0, 0, 0)}]. (7)


Vk−1(1, 0, 0)− Vk−1(1, 1, 0) ≤ 0,Vk−1(0, 0, 0)− Vk−1(0, 1, 0) ≤ 0.

Therefore, the second and the third terms on the right-hand side of (7) are nonpositive, and

min{Vk−1(0, 1, 0), Vk−1(0, 0, 0)} = Vk−1(0, 0, 0),

which implies that the forth term is equal to:

µ2Λ{Vk−1(0, 0, 0)− Vk−1(0, 0, 0)} = 0.

Since all terms on the right-hand side of (7) are nonpositive, the proof for V3-Case I is completed.

V3-Case II: When n1 = 0 and n2 ≥ 2, condition V3 becomes Vk(0, n2− 1, 0) ≤ Vk(0, n2, 0). Since for bothVk(0, n2 − 1, 0) and Vk(0, n2, 0) there�s no job at the Þrst station, the worker can either idle or process a jobat station 2. Thus,

Vk(0, n2 − 1, 0) =n2 − 1Λ

+λ

ΛVk−1(1, n2 − 1, 0) + µ1

ΛVk−1(0, n2 − 1, 0) + l1

ΛVk−1(0, n2 − 1, 0)

+µ2Λmin{Vk−1(0, n2 − 1, 0) , Vk−1(0, n2 − 2, 0)},

11

Vk(0, n2, 0) =n2Λ+λ

ΛVk−1(1, n2, 0) +

µ1ΛVk−1(0, n2, 0) +

l1ΛVk−1(0, n2, 0)

+µ2Λmin{Vk−1(0, n2, 0) , Vk−1(0, n2 − 1, 0)}.

Therefore, we have

Vk(0, n2 − 1, 0)− Vk(0, n2, 0) = − 1Λ+λ

Λ[Vk−1(1, n2 − 1, 0)− Vk−1(1, n2, 0)]

+µ1 + l1Λ

[Vk−1(0, n2 − 1, 0)− Vk−1(0, n2, 0)]

+µ2Λ[min{Vk−1(0, n2 − 1, 0), Vk−1(0, n2 − 2, 0)}−min{Vk−1(0, n2, 0), Vk−1(0, n2 − 1, 0)}]. (8)


Vk−1(1, n2 − 1, 0)− Vk−1(1, n2, 0) ≤ 0,Vk−1(0, n2 − 1, 0)− Vk−1(0, n2, 0) ≤ 0.

Thus, the second and the third terms on the right-hand side of (8) are nonpositive, and,

min{Vk−1(0, n2 − 1, 0), Vk−1(0, n2 − 2, 0)} = Vk−1(0, n2 − 2, 0)min{Vk−1(0, n2, 0), Vk−1(0, n2 − 1, 0)} = Vk−1(0, n2 − 1, 0),

which imply that the forth term on the right-hand side of (8) is equal to:

Vk−1(0, n2 − 2, 0)− Vk−1(0, n2 − 1, 0),which, according to induction assumption V3, is nonpositive. Hence, all terms on the right-hand side of (8)are nonpositive. This completes the proof for V3-Case II.

V3-Case III: When n1 ≥ 1 and n2 = 1, condition V3 becomes Vk(n1, 0, 0) ≤ Vk(n1, 1, 0). Since forVk(n1, 0, 0) there�s no job at the second station, serving 2 is not feasible. Therefore,

Vk(n1, 0, 0) =n1Λ+λ

ΛVk−1(n1 + 1, 0, 0) +

µ1ΛVk−1(n1, 0, 0) +

µ2ΛVk−1(n1, 0, 0)

+l1Λmin{Vk−1(n1, 0, 0) , Vk−1(n1, 0, 1)}.

On the other hand,

Vk(n1, 1, 0) =n1 + 1

Λ+λ

ΛVk−1(n1 + 1, 1, 0) +

µ1ΛVk−1(n1, 1, 0)

+1

Λmin

(l1 + µ2)Vk−1(n1, 1, 0)l1Vk−1(n1, 1, 1) + µ2Vk−1(n1, 1, 0)µ2Vk−1(n1, 0, 0) + l1Vk−1(n1, 1, 0).

Hence,

Vk(n1, 0, 0)− Vk(n1, 1, 0) = − 1Λ+λ

Λ[Vk−1(n1 + 1, 0, 0)− Vk−1(n1 + 1, 1, 0)]

+µ1Λ[Vk−1(n1, 0, 0)− Vk−1(n1, 1, 0)]

+µ2ΛVk−1(n1, 0, 0) +

l1Λmin{Vk−1(n1, 0, 0), Vk−1(n1, 0, 1)}

− 1Λmin


(9)

12


Vk−1(n1 + 1, 0, 0) ≤ Vk−1(n1 + 1, 1, 0),Vk−1(n1, 0, 0) ≤ Vk−1(n1, 1, 0).

Thus, the second and the third terms on the right-hand side of (9) are nonpositive. Therefore, we only needto show that the remaining terms in (9) are nonpositive. DeÞne φ as the remaining part, we will have

φ =µ2ΛVk−1(n1, 0, 0) +

l1Λmin{Vk−1(n1, 0, 0), Vk−1(n1, 0, 1)}− 1

Λmin



Vk−1(n1, 1, 1) ≤ Vk−1(n1, 1, 0),Vk−1(n1, 0, 0) ≤ Vk−1(n1, 1, 0).

Therefore,

φ =µ2ΛVk−1(n1, 0, 0) +

l1ΛVk−1(n1, 0, 1)− 1

Λmin

½l1Vk−1(n1, 1, 1) + µ2Vk−1(n1, 1, 0)µ2Vk−1(n1, 0, 0) + l1Vk−1(n1, 1, 0).

To show that φ ≤ 0, we discuss two cases:φ ≤ 0: Case 1.

min

½l1Vk−1(n1, 1, 1) + µ2Vk−1(n1, 1, 0)µ2Vk−1(n1, 0, 0) + l1Vk−1(n1, 1, 0)

= µ2Vk−1(n1, 0, 0) + l1Vk−1(n1, 1, 0),

then,

φ =µ2ΛVk−1(n1, 0, 0) +

l1ΛVk−1(n1, 0, 1)− µ2

ΛVk−1(n1, 0, 0)− l1

ΛVk−1(n1, 1, 0)

=l1Λ[Vk−1(n1, 0, 1)− Vk−1(n1, 1, 0)].


Vk−1(n1, 0, 1) ≤ Vk−1(n1, 0, 0) ≤ Vk−1(n1, 1, 0).Therefore, φ ≤ 0.

φ ≤ 0: Case 2.

min


= l1Vk−1(n1, 1, 1) + µ2Vk−1(n1, 1, 0),

then,

φ =µ2ΛVk−1(n1, 0, 0) +

l1ΛVk−1(n1, 0, 1)− l1

ΛVk−1(n1, 1, 1)− µ2

ΛVk−1(n1, 1, 0)

=l1Λ[Vk−1(n1, 0, 1)− Vk−1(n1, 1, 1)] + µ2

Λ[Vk−1(n1, 0, 0)− Vk−1(n1, 1, 0)].

According to induction assumption V1 and V3, we have

Vk−1(n1, 0, 1) ≤ Vk−1(n1, 1, 1),Vk−1(n1, 0, 0) ≤ Vk−1(n1, 1, 0).

Therefore, φ ≤ 0.

13

This completes the proof for V3-Case III.

V3-Case IV: When n1 ≥ 1 and n2 ≥ 2:

Vk(n1, n2 − 1, 0) =n1 + n2 − 1

Λ+λ

ΛVk−1(n1 + 1, n2 − 1, 0) + µ1

ΛVk−1(n1, n2 − 1, 0)

+1

Λmin

(l1 + µ2)Vk−1(n1, n2 − 1, 0)l1Vk−1(n1, n2 − 1, 1) + µ2Vk−1(n1, n2 − 1, 0)µ2Vk−1(n1, n2 − 2, 0) + l1Vk−1(n1, n2 − 1, 0),

Vk(n1, n2, 0) =n1 + n2Λ

+λ

ΛVk−1(n1 + 1, n2, 0) +

µ1ΛVk−1(n1, n2, 0)

+1

Λmin


Therefore,

Vk(n1, n2 − 1, 0)− Vk(n1, n2, 0) = − 1Λ+λ

Λ[Vk−1(n1 + 1, n2 − 1, 0)− Vk−1(n1 + 1, n2, 0)]

+µ1Λ[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)]

+1

Λmin

(l1 + µ2)Vk−1(n1, n2 − 1, 0)l1Vk−1(n1, n2 − 1, 1) + µ2Vk−1(n1, n2 − 1, 0)µ2Vk−1(n1, n2 − 2, 0) + l1Vk−1(n1, n2 − 1, 0)

− 1Λmin


(10)


Vk−1(n1 + 1, n2 − 1, 0) ≤ Vk−1(n1 + 1, n2, 0),Vk−1(n1, n2 − 1, 0) ≤ Vk−1(n1, n2, 0),

which imply that the second and the third terms on the right-hand side of (10) are nonpositive. Therefore,we only need to show that the remaining terms are nonpositive. DeÞning σ as the remaining terms, we willhave

σ = min

(l1 + µ2)Vk−1(n1, n2 − 1, 0)l1Vk−1(n1, n2 − 1, 1) + µ2Vk−1(n1, n2 − 1, 0)µ2Vk−1(n1, n2 − 2, 0) + l1Vk−1(n1, n2 − 1, 0)

−min (l1 + µ2)Vk−1(n1, n2, 0)l1Vk−1(n1, n2, 1) + µ2Vk−1(n1, n2, 0)µ2Vk−1(n1, n2 − 1, 0) + l1Vk−1(n1, n2, 0).


Vk−1(n1 + 1, n2 − 2, 0) ≤ Vk−1(n1 + 1, n2 − 1, 0),Vk−1(n1 + 1, n2 − 1, 0) ≤ Vk−1(n1 + 1, n2, 0),

On the other hand, according to induction assumption V2

Vk−1(n1, n2 − 1, 1) ≤ Vk−1(n1, n2 − 1, 0),Vk−1(n1, n2, 1) ≤ Vk−1(n1, n2, 0).

Therefore,

σ = min

½l1Vk−1(n1, n2 − 1, 1) + µ2Vk−1(n1, n2 − 1, 0)µ2Vk−1(n1, n2 − 2, 0) + l1Vk−1(n1, n2 − 1, 0) −min


To show that σ ≤ 0, we discuss two cases:

14

σ ≤ 0: Case 1.

min

½l1Vk−1(n1, n2, 1) + µ2Vk−1(n1, n2, 0)µ2Vk−1(n1, n2 − 1, 0) + l1Vk−1(n1, n2, 0) = l1Vk−1(n1, n2, 1) + µ2Vk−1(n1, n2, 0),

then,

σ = min

½l1Vk−1(n1, n2 − 1, 1) + µ2Vk−1(n1, n2 − 1, 0)µ2Vk−1(n1, n2 − 2, 0) + l1Vk−1(n1, n2 − 1, 0) − l1Vk−1(n1, n2, 1)− µ2Vk−1(n1, n2, 0)

≤ l1Vk−1(n1, n2 − 1, 1) + µ2Vk−1(n1, n2 − 1, 0)− l1Vk−1(n1, n2, 1)− µ2Vk−1(n1, n2, 0)= l1[Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 1)] + µ2[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)].

According to induction assumption V1 and V3, we have

Vk−1(n1, n2 − 1, 1) ≤ Vk−1(n1, n2, 1),Vk−1(n1, n2 − 1, 0) ≤ Vk−1(n1, n2, 0).

Therefore, σ ≤ 0.σ ≤ 0: Case 2.

min

½l1Vk−1(n1, n2, 1) + µ2Vk−1(n1, n2, 0)µ2Vk−1(n1, n2 − 1, 0) + l1Vk−1(n1, n2, 0) = µ2Vk−1(n1, n2 − 1, 0) + l1Vk−1(n1, n2, 0),

then,

σ = min

½l1Vk−1(n1, n2 − 1, 1) + µ2Vk−1(n1, n2 − 1, 0)µ2Vk−1(n1, n2 − 2, 0) + l1Vk−1(n1, n2 − 1, 0) − µ2Vk−1(n1, n2 − 1, 0)− l1Vk−1(n1, n2, 0)

≤ l1Vk−1(n1, n2 − 1, 1) + µ2Vk−1(n1, n2 − 1, 0)− µ2Vk−1(n1, n2 − 1, 0)− l1Vk−1(n1, n2, 0)= l1[Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 0)].


Vk−1(n1, n2 − 1, 1) ≤ Vk−1(n1, n2 − 1, 0) ≤ Vk−1(n1, n2, 0).

Therefore, σ ≤ 0.This completes the proof for V3-Case IV.

Proof for C1: Vk(n1 − 1, n2 + 1, 0) ≤ Vk(n1, n2, 1) for n1 ≥ 1.We discuss the following four cases

� Case I: n1 = 1 and n2 = 0,� Case II: n1 ≥ 2 and n2 = 0,� Case III: n1 = 1 and n2 ≥ 1,� Case IV: n1 ≥ 2 and n2 ≥ 1.

C1-Case I: When n1 = 1 and n2 = 0, C1 becomes Vk(0, 1, 0) ≤ Vk(1, 0, 1). Since at state (0, 1, 0) there�sno job at the Þrst station, the worker can either idle or process a job at the second station. Therefore,

Vk(0, 1, 0) =1

Λ+λ

ΛVk−1(1, 1, 0) +

µ1ΛVk−1(0, 1, 0) +

l1ΛVk−1(0, 1, 0)

+µ2Λmin{Vk−1(0, 1, 0) , Vk−1(0, 0, 0)}.

15

On the other hand, at state (1, 0, 1) there�s no job at the second station and the Þrst station is underautomatic processing, so idling is the only option, and therefore,

Vk(1, 0, 1) =1

Λ+λ

ΛVk−1(2, 0, 1) +

µ1ΛVk−1(0, 1, 0) +

l1 + µ2Λ

Vk−1(1, 0, 1).

Hence,

Vk(0, 1, 0)− Vk(1, 0, 1) =λ

Λ[Vk−1(1, 1, 0)− Vk−1(2, 0, 1)] + l1

Λ[Vk−1(0, 1, 0)− Vk−1(1, 0, 1)]

+µ2Λ{min{Vk−1(0, 1, 0), Vk−1(0, 0, 0)}− Vk−1(1, 0, 1)]. (11)

According to induction assumption C1

Vk−1(1, 1, 0)− Vk−1(2, 0, 1) ≤ 0,Vk−1(0, 1, 0)− Vk−1(1, 0, 1) ≤ 0.

Thus, the Þst and the second terms on the right-hand side of (11) are nonpositive. Also, according toinduction assumption V3

min{Vk−1(0, 1, 0), Vk−1(0, 0, 0)} = Vk−1(0, 0, 0),which implies that the last term on the right-hand side of (11) is equal to

µ2Λ{Vk−1(0, 0, 0)− Vk−1(1, 0, 1)}.

Furthermore, according to induction assumption V3 and C1

Vk−1(0, 0, 0) ≤ Vk−1(0, 1, 0) ≤ Vk−1(1, 0, 1).Thus, the last term on the right-hand side of (11) is nonpositive. This completes the proof for C1-Case I.

C1-Case II: When n1 ≥ 2, n2 = 0, C1 becomes Vk(n1 − 1, 1, 0) ≤ Vk(n1, 0, 1), and

Vk(n1 − 1, 1, 0) =n1 − 1+ 1

Λ+λ

ΛVk−1(n1, 1, 0) +

µ1ΛVk−1(n1 − 1, 1, 0)

+1

Λmin

(l1 + µ2)Vk−1(n1 − 1, 1, 0)l1Vk−1(n1 − 1, 1, 1) + µ2Vk−1(n1 − 1, 1, 0)µ2Vk−1(n1 − 1, 0, 0) + l1Vk−1(n1 − 1, 1, 0).

Since at state (n1, 0, 1), there�s no job at the second station and the Þrst station is under automatic processing,idling is the only option. Thus,

Vk(n1, 0, 1) =n1Λ+λ

ΛVk−1(n1 + 1, 0, 1) +

µ1ΛVk−1(n1 − 1, 1, 0) + l1 + µ2

ΛVk(n1, 0, 1).

Therefore,

Vk(n1 − 1, 1, 0)− Vk(n1, 0, 1) =λ

Λ[Vk−1(n1, 1, 0)− Vk−1(n1 + 1, 0, 1)]

+1

Λmin

(l1 + µ2)Vk−1(n1 − 1, 1, 0)l1Vk−1(n1 − 1, 1, 1) + µ2Vk−1(n1 − 1, 1, 0)µ2Vk−1(n1 − 1, 0, 0) + l1Vk−1(n1 − 1, 1, 0)

− l1 + µ2Λ

Vk(n1, 0, 1). (12)


Vk−1(n1, 1, 0) ≤ Vk−1(n1 + 1, 0, 1),

16

which implies that the Þrst term on the right-hand side of (12) is nonpositive. Therefore,

Vk(n1 − 1, 1, 0)− Vk(n1, 0, 1) ≤ 1

Λmin

(l1 + µ2)Vk−1(n1 − 1, 1, 0)l1Vk−1(n1 − 1, 1, 1) + µ2Vk−1(n1 − 1, 1, 0)µ2Vk−1(n1 − 1, 0, 0) + l1Vk−1(n1 − 1, 1, 0)

− l1 + µ2Λ

Vk(n1, 0, 1)

≤ µ2ΛVk−1(n1 − 1, 0, 0) + l1

ΛVk−1(n1 − 1, 1, 0)− l1 + µ2

ΛVk(n1, 0, 1)

=µ2Λ[Vk−1(n1 − 1, 0, 0)− Vk(n1, 0, 1)] + l1

Λ[Vk−1(n1 − 1, 1, 0)− Vk(n1, 0, 1)].

According to induction assumption V3 and C1

Vk−1(n1 − 1, 0, 0) ≤ Vk−1(n1 − 1, 1, 0) ≤ Vk−1(n1, 0, 1).

Therefore,µ2Λ[Vk−1(n1 − 1, 0, 0)− Vk(n1, 0, 1)] + l1

Λ[Vk−1(n1 − 1, 1, 0)− Vk(n1, 0, 1)] ≤ 0,

which implies Vk(n1 − 1, 1, 0) ≤ Vk(n1, 0, 1). This completes the proof for C1-Case II.C1-Case III: When n1 = 1, n2 ≥ 1, C1 becomes Vk(0, n2 + 1, 0) ≤ Vk(1, n2, 1). Since at state (0, n2 + 1, 0)there�s no job at the Þrst station, the worker can either idle or process a job at the second station. Therefore,

Vk(0, n2 + 1, 0) =n2 + 1

Λ+λ

ΛVk−1(1, n2 + 1, 0) +

µ1ΛVk−1(0, n2 + 1, 0) +

l1ΛVk−1(0, n2 + 1, 0)

+µ2Λmin{Vk−1(0, n2 + 1, 0) , Vk−1(0, n2, 0)},

Vk(1, n2, 1) =1+ n2Λ

+λ

ΛVk−1(2, n2, 1) +

µ1ΛVk−1(0, n2 + 1, 0) +

l1ΛVk−1(1, n2, 1)

+µ2Λmin{Vk−1(1, n2, 1) , Vk−1(1, n2 − 1, 1)}.

Hence,

Vk(0, n2 + 1, 0)− Vk(1, n2, 1) =λ

Λ[Vk−1(1, n2 + 1, 0)− Vk−1(2, n2, 1)] + l1

Λ[Vk−1(0, n2 + 1, 0)− Vk−1(1, n2, 1)]

+µ2Λ[min{Vk−1(0, n2 + 1, 0) , Vk−1(0, n2, 0)}−min{Vk−1(1, n2, 1) , Vk−1(1, n2 − 1, 1)}]. (13)


Vk−1(1, n2 + 1, 0) ≤ Vk−1(2, n2, 1),Vk−1(0, n2 + 1, 0) ≤ Vk−1(1, n2, 1).

Therefore, the Þrst and second terms on the right-hand side of (13) are negative. We only need to show thatthe remaining terms are nonpositive. DeÞning γ as the remaining terms, we have:

γ = min{Vk−1(0, n2 + 1, 0) , Vk−1(0, n2, 0)}−min{Vk−1(1, n2, 1) , Vk−1(1, n2 − 1, 1)}.


Vk−1(0, n2, 0) ≤ Vk−1(0, n2 + 1, 0),Vk−1(1, n2 − 1, 1) ≤ Vk−1(1, n2, 1).

17

Therefore,

γ =µ2Λ[Vk−1(0, n2, 0)− Vk−1(1, n2 − 1, 1)].


Vk−1(0, n2, 0) ≤ Vk−1(1, n2 − 1, 1).

Therefore, γ ≤ 0. This completes the proof for C1-Case III.C1-Case IV: When n1 ≥ 1 and n2 ≥ 2, we will have

Vk(n1 − 1, n2 + 1, 0) =n1 − 1+ n2 + 1

Λ+λ

ΛVk−1(n1, n2 + 1, 0) +

µ1ΛVk−1(n1 − 1, n2 + 1, 0)

+1

Λmin

(l1 + µ2)Vk−1(n1 − 1, n2 + 1, 0)l1Vk−1(n1 − 1, n2 + 1, 1) + µ2Vk−1(n1 − 1, n2 + 1, 0)µ2Vk−1(n1 − 1, n2, 0) + l1Vk−1(n1 − 1, n2 + 1, 0),

Vk(n1, n2, 1) =n1 + n2Λ

+λ

ΛVk−1(n1 + 1, n2, 1) +

µ1ΛVk−1(n1 − 1, n2 + 1, 0) + l1

ΛVk−1(n1, n2, 1)

+µ2Λmin{Vk−1(n1, n2, 1) , Vk(n1, n2 − 1, 1)}.

Hence,

Vk(n1 − 1, n2 + 1, 0)− Vk(n1, n2, 1) =λ

Λ[Vk−1(n1, n2 + 1, 0)− Vk−1(n1 + 1, n2, 1)]

+1

Λmin

(l1 + µ2)Vk−1(n1 − 1, n2 + 1, 0)l1Vk−1(n1 − 1, n2 + 1, 1) + µ2Vk−1(n1 − 1, n2 + 1, 0)µ2Vk−1(n1 − 1, n2, 0) + l1Vk−1(n1 − 1, n2 + 1, 0)

− l1ΛVk−1(n1, n2, 1)

−µ2Λmin{Vk−1(n1, n2, 1) , Vk(n1, n2 − 1, 1)}. (14)


Vk−1(n1, n2 + 1, 0) ≤ Vk−1(n1 + 1, n2, 0).

Therefore, the Þrst term on the right-hand side of (14) is nonpositive. We only need to show that theremaining terms are nonpositive. DeÞne δ as the remaining terms, then

δ =1

Λmin

(l1 + µ2)Vk−1(n1 − 1, n2 + 1, 0)l1Vk−1(n1 − 1, n2 + 1, 1) + µ2Vk−1(n1 − 1, n2 + 1, 0)µ2Vk−1(n1 − 1, n2, 0) + l1Vk−1(n1 − 1, n2 + 1, 0)

− l1ΛVk−1(n1, n2, 1)− µ2

Λmin{Vk−1(n1, n2, 1) , Vk(n1, n2 − 1, 1)}.

According to induction assumption V2,V3, and V1, we have

Vk−1(n1 − 1, n2 + 1, 1) ≤ Vk−1(n1 − 1, n2 + 1, 0),Vk−1(n1 − 1, n2, 0) ≤ Vk−1(n1 − 1, n2 + 1, 0),Vk−1(n1, n2 − 1, 1) ≤ Vk−1(n1, n2, 1).

18

Therefore,

δ =1

Λmin

½l1Vk−1(n1 − 1, n2 + 1, 1) + µ2Vk−1(n1 − 1, n2 + 1, 0)µ2Vk−1(n1 − 1, n2, 0) + l1Vk−1(n1 − 1, n2 + 1, 0)

− l1ΛVk−1(n1, n2, 1)− µ2

ΛVk(n1, n2 − 1, 1)

≤ µ2ΛVk−1(n1 − 1, n2, 0) + l1

ΛVk−1(n1 − 1, n2 + 1, 0)− l1

ΛVk−1(n1, n2, 1)− µ2

ΛVk(n1, n2 − 1, 1)

=µ2Λ[Vk−1(n1 − 1, n2, 0)− Vk(n1, n2 − 1, 1)] + l1

Λ[Vk−1(n1 − 1, n2 + 1, 0)− Vk−1(n1, n2, 1)].


Vk−1(n1 − 1, n2, 0) ≤ Vk−1(n1, n2 − 1, 1),Vk−1(n1 − 1, n2 + 1, 0) ≤ Vk−1(n1, n2, 1).

Therefore, δ ≤ 0. This completes the proof for C1-Case IV.


ConditionsW1, W2, D1, D2, M1, M2, M3 and M4 can be rewritten as follows:

W1 µ2[V (n1, n2 − 1, 0)− V (n1, n2, 0)] + l1[V (n1, n2, 0)− V (n1, n2, 1)]≥ µ2[V (n1, n2, 0)− V (n1, n2 + 1, 0)] + l1[V (n1, n2 + 1, 0)− V (n1, n2 + 1, 1)] for n1 ≥ 1, n2 ≥ 1

W2 µ2[V (n1, n2 − 1, 0)− V (n1, n2, 0)] + l1[V (n1, n2, 0)− V (n1, n2, 1)]≤ µ2[V (n1 + 1, n2 − 1, 0)− V (n1 + 1, n2, 0)] + l1[V (n1 + 1, n2, 0)− V (n1 + 1, n2, 1)] for n1 ≥ 1, n2 ≥ 1

D1 V (n1, n2 + 1, 1)− V (n1 − 1, n2 + 2, 0)− [V (n1, n2, 1)− V (n1 − 1, n2 + 1, 0)] ≤ 0 for n1 ≥ 1, n2 ≥ 0D2 V (n1, n2, 1)− V (n1 − 1, n2 + 1, 0)− [V (n1 + 1, n2, 1)− V (n1, n2 + 1, 0)] ≤ 0 for n1 ≥ 1, n2 ≥ 0M1 V (n1, n2, 0)− V (n1, n2, 1)− [V (n1 + 1, n2, 0)− V (n1 + 1, n2, 1)] ≤ 0 for n1 ≥ 1, n2 ≥ 1M2 V (n1, n2, 0)− V (n1, n2 − 1, 0)− [V n1 + 1, n2, 0)− V (n1 + 1, n2 − 1, 0)] ≤ 0 for n1 ≥ 0, n2 ≥ 1M3 V (n1, n2, 1)− V (n1, n2 − 1, 1)− [V (n1 + 1, n2, 1)− V (n1 + 1, n2 − 1, 1)] ≤ 0 for n1 ≥ 1, n2 ≥ 1M4 l1[V (n1, 1, 0)− V (n1, 1, 1)− V (n1, 0, 0) + V (n1, 0, 1)] + µ2[V (n1, 0, 0)− V (n1, 1, 0)] ≤ 0 for n1 ≥ 1.We use induction and the value iteration algorithm to prove the above conditions.

(W1,W2,D1,D2,M1,M2,M3,M4) Iteration 0: At iteration 0, all V0(n1, n2, s) = 0 for all (n1, n2, s).Therefore, (W1,W2,D1,D2,M1,M2,M3,M4) hold at iteration 0.

(W1,W2,D1,D2,M1,M2,M3,M4) Iteration k-1: Assume at iteration k−1 (W1,W2,D1,D2,M1,M2,M3,M4)hold. That is,

W1 µ2[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)] + l1[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)]≥ µ2[Vk−1(n1, n2, 0)− Vk−1(n1, n2 + 1, 0)] + l1[Vk−1(n1, n2 + 1, 0)− Vk−1(n1, n2 + 1, 1)];

W2 µ2[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)] + l1[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)]≤ µ2[Vk−1(n1 + 1, n2 − 1, 0)− Vk−1(n1 + 1, n2, 0)] + l1[Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2, 1)];

D1 Vk−1(n1, n2 + 1, 1)− Vk−1(n1 − 1, n2 + 2, 0)− [Vk−1(n1, n2, 1)− Vk−1(n1 − 1, n2 + 1, 0)] ≤ 0;D2 Vk−1(n1, n2, 1)− Vk−1(n1 − 1, n2 + 1, 0)− [Vk−1(n1 + 1, n2, 1)− Vk−1(n1, n2 + 1, 0)] ≤ 0M1 Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)− [Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2, 1)] ≤ 0M2 Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 0)− [V n1 + 1, n2, 0)− Vk−1(n1 + 1, n2 − 1, 0)] ≤ 0M3 Vk−1(n1, n2, 1)− Vk−1(n1, n2 − 1, 1)− [Vk−1(n1 + 1, n2, 1)− Vk−1(n1 + 1, n2 − 1, 1)] ≤ 0M4 l1[Vk−1(n1, 1, 0)− Vk−1(n1, 1, 1)− Vk−1(n1, 0, 0) + Vk−1(n1, 0, 1)] + µ2[Vk−1(n1, 0, 0)− Vk−1(n1, 1, 0)] ≤ 0.

(W1,W2,D1,D2,M1,M2,M3,M4) Iteration k: Now, we show that (W1,W2,D1,D2,M1,M2,M3,M4)also hold at iteration k.

19

Proof for W1: At iteration k, for n1 ≥ 1 and n2 ≥ 1,W1 is

µ2[Vk(n1, n2 − 1, 0)− Vk(n1, n2, 0)] + l1[Vk(n1, n2, 0)− Vk(n1, n2, 1)]≥ µ2[Vk(n1, n2, 0)− Vk(n1, n2 + 1, 0)] + l1[Vk(n1, n2 + 1, 0)− Vk(n1, n2 + 1, 1)].

For convenience, we deÞne:

Ak = Vk(n1, n2 − 1, 0)− Vk(n1, n2, 0),Bk = Vk(n1, n2, 0)− Vk(n1, n2, 1),Ck = Vk(n1, n2, 0)− Vk(n1, n2 + 1, 0),Dk = Vk(n1, n2 + 1, 0)− Vk(n1, n2 + 1, 1).

We need to show thatµ2Ck + l1Dk − µ2Ak − l1Bk ≤ 0.

Since we have already proven that idling is not optimal, using the optimality equations, we will have

Ak = − 1Λ+λ

Λ[Vk−1(n1 + 1, n2 − 1, 0)− Vk−1(n1 + 1, n2, 0)] + µ1

Λ[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)]

+1

Λmin

½l1Vk−1(n1, n2 − 1, 1) + µ2Vk−1(n1, n2 − 1, 0)µ2Vk−1(n1, n2 − 2, 0) + l1Vk−1(n1, n2 − 1, 0)

− 1Λmin

½l1Vk−1(n1, n2, 1) + µ2Vk−1(n1, n2, 0)µ2Vk−1(n1, n2 − 1, 0) + l1Vk−1(n1, n2, 0),

Bk =λ

Λ[Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2, 1)] + µ1

Λ[Vk−1(n1, n2, 0)− Vk−1(n1 − 1, n2 + 1, 0)]

+1

Λmin

½µ2[Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 1)]l1[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)] + µ2[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2 − 1, 1)],

Ck = − 1Λ+λ

Λ[Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2 + 1, 0)] + µ1

Λ[Vk−1(n1, n2, 0)− Vk−1(n1, n2 + 1, 0)]

+1

Λmin

½l1Vk−1(n1, n2, 1) + µ2Vk−1(n1, n2, 0)µ2Vk−1(n1, n2 − 1, 0) + l1Vk−1(n1, n2, 0)

− 1Λmin

½l1Vk−1(n1, n2 + 1, 1) + µ2Vk−1(n1, n2 + 1, 0)µ2Vk−1(n1, n2, 0) + l1Vk−1(n1, n2 + 1, 0),

Dk =λ

Λ[Vk−1(n1 + 1, n2 + 1, 0)− Vk−1(n1 + 1, n2 + 1, 1)] + µ1

Λ[Vk−1(n1, n2 + 1, 0)− Vk−1(n1 − 1, n2 + 2, 0)]

+1

Λmin

½µ2[Vk−1(n1, n2 + 1, 0)− Vk−1(n1, n2, 1)]l1[Vk−1(n1, n2 + 1, 0)− Vk−1(n1, n2 + 1, 1)] + µ2[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)].

Furthermore, we let

A0k =1

Λmin


− 1Λmin


B0k =1

Λmin

½µ2[Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 1)]l1[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)] + µ2[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2 − 1, 1)]

20

C 0k =1

Λmin


− 1Λmin

½l1Vk−1(n1, n2 + 1, 1) + µ2Vk−1(n1, n2 + 1, 0)µ2Vk−1(n1, n2, 0) + l1Vk−1(n1, n2 + 1, 0)

D0k =

1

Λmin

½µ2[Vk−1(n1, n2 + 1, 0)− Vk−1(n1, n2, 1)]l1[Vk−1(n1, n2 + 1, 0)− Vk−1(n1, n2 + 1, 1)] + µ2[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)].

Then,

µ2Ck + l1Dk − µ2Ak − l1Bk=

λ

Λ{µ2[Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2 + 1, 0)] + l1[Vk−1(n1 + 1, n2 + 1, 0)− Vk−1(n1 + 1, n2 + 1, 1)]−µ2[Vk−1(n1 + 1, n2 − 1, 0)− Vk−1(n1 + 1, n2, 0)]− l1[Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2, 1)]}

+µ2µ1Ck−1 + l1µ1[Dk−1 + Vk−1(n1, n2 + 1, 1)− Vk−1(n1 − 1, n2 + 2, 0)]−µ2µ1Ak−1 − l1µ1[Bk−1 + Vk−1(n1, n2, 1)− Vk−1(n1 − 1, n2 + 1, 0)]+µ2C

0k + l1D

0k − µ2A0k − l1B0k

=λ

Λ{µ2[Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2 + 1, 0)] + l1[Vk−1(n1 + 1, n2 + 1, 0)− Vk(n1 + 1, n2 + 1, 1)]−µ2[Vk−1(n1 + 1, n2 − 1, 0)− Vk−1(n1 + 1, n2, 0)]− l1[Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2, 1)]}

+µ1[µ2Ck−1 + l1Dk−1 − µ2Ak−1 − l1Bk−1]+l1µ1{[Vk−1(n1, n2 + 1, 1)− Vk−1(n1 − 1, n2 + 2, 0)]− [Vk−1(n1, n2, 1)− Vk−1(n1 − 1, n2 + 1, 0)]}+µ2C

0k + l1D

0k − µ2A0k − l1B0k.

According to induction assumption W1, the Þrst and the second terms are nonpositive. Furthermore,according to induction assumption D1, the third term is nonpositive. Therefore, we only need to show that

µ2C0k + l1D

0k − µ2A0k − l1B0k ≤ 0.

Based on induction assumption W1, we know that if at state (n1, n2 − 1, 0) serving station 2 is optimal,then at states (n1, n2, 0) and (n1, n2+ 1, 0), serving station 2 is also optimal. Hence, we only need to discussthe following four cases:

Case Vk(n1, n2 − 1, 0) Vk(n1, n2, 0) Vk(n1, n2 + 1, 0)

1 Serve 2 Serve 2 Serve 2




W1-Case 1: When serve 2 is optimal at states (n1, n2 − 1, 0), (n1, n2, 0) and (n1, n2 + 1, 0), then

A0k =µ2Λ[Vk−1(n1, n2 − 2, 0)− Vk−1(n1, n2 − 1, 0)] + l1

Λ[Vk−1(n1, n2 − 1, 0)]− Vk−1(n1, n2, 0)]

B0k =l1Λ[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)] + µ2

Λ[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2 − 1, 1)]

C 0k =µ2Λ[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)] + l1

Λ[Vk−1(n1, n2, 0)]− Vk−1(n1, n2 + 1, 0)]

D0k =

l1Λ[Vk−1(n1, n2 + 1, 0)− Vk−1(n1, n2 + 1, 1)] + µ2

Λ[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)].

Therefore,

Λ[C 0k −A0k] = µ2[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)] + l1[Vk−1(n1, n2, 0)− Vk−1(n1, n2 + 1, 0)]

21

−µ2[Vk−1(n1, n2 − 2, 0)− Vk−1(n1, n2 − 1, 0)]− l1[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)]= µ2[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)] + l1[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)]

−µ2[Vk−1(n1, n2 − 2, 0)− Vk−1(n1, n2 − 1, 0)]− l1[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2 − 1, 1)]+l1{[Vk−1(n1, n2, 1)− Vk−1(n1, n2 + 1, 0)]− [Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 0)]}.

According to induction assumptionW1

µ2[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)] + l1[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)]−µ2[Vk−1(n1, n2 − 2, 0)− Vk−1(n1, n2 − 1, 0)]− l1[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2 − 1, 1)] ≤ 0.

Hence,

Λ[C 0k −A0k] ≤ l1{[Vk−1(n1, n2, 1)− Vk−1(n1, n2 + 1, 0)]− [Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 0)]}.Similarly,

Λ[D0k −B0k] = l1[Vk−1(n1, n2 + 1, 0)− Vk−1(n1, n2 + 1, 1)] + µ2[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)]

−l1[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)]− µ2[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2 − 1, 1)]= l1[Vk−1(n1, n2 + 1, 0)− Vk−1(n1, n2 + 1, 1)] + µ2[Vk−1(n1, n2, 0)− Vk−1(n1, n2 + 1, 0)]

−l1[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)]− µ2[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)]+µ2{[Vk−1(n1, n2 + 1, 0)− Vk−1(n1, n2, 1)]− [Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 1)]}.


l1[Vk−1(n1, n2 + 1, 0)− Vk−1(n1, n2 + 1, 1)] + µ2[Vk−1(n1, n2, 0)− Vk−1(n1, n2 + 1, 0)]−l1[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)]− µ2[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)] ≤ 0.

Hence,

Λ[D0k −B0k] ≤ µ2{[Vk−1(n1, n2 + 1, 0)− Vk−1(n1, n2, 1)]− [Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 1)]}.

Therefore, we have

Λ[µ2C0k + l1D

0k − µ2A0k − l1B0k]

= Λµ2[C0k −A0k] + Λl1[D0

k −B0k]≤ µ2l1{[Vk−1(n1, n2, 1)− Vk−1(n1, n2 + 1, 0)]− [Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 0)]}+l1µ2{[Vk−1(n1, n2 + 1, 0)− Vk−1(n1, n2, 1)]− [Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 1)]}= 0,

which implies that µ2C0k + l1D

0k − µ2A0k − l1B0k ≤ 0. This completes the proof forW1-Case 1.

W1-Case 2: When serving station 1 is optimal at state (n1, n2 − 1, 0), but serving station 2 is optimal atstates (n1, n2, 0) and (n1, n2 + 1, 0), we will have

A0k =1

Λ[l1Vk−1(n1, n2 − 1, 1) + µ2Vk−1(n1, n2 − 1, 0)]− 1

Λ[µ2Vk−1(n1, n2 − 1, 0) + l1Vk−1(n1, n2, 0)]

=l1Λ[Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 0)],

B0k, C0k, D

0k are same as inW1-Case 1. Therefore,

Λ[C 0k −A0k] = µ2[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)] + l1[Vk−1(n1, n2, 0)− Vk−1(n1, n2 + 1, 0)]−l1[Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 0)]

= µ2[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)] + l1[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)]+l1{[Vk−1(n1, n2, 1)− Vk−1(n1, n2 + 1, 0)]− [Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 0)]}.

22

Since inW1-Case 2 serving station 2 is optimal at state (n1, n2, 0), that is,

µ2[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)] + l1[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)] ≤ 0,then,

Λ[C 0k −A0k] ≤ l1{[Vk−1(n1, n2, 1)− Vk−1(n1, n2 + 1, 0)]− [Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 0)]}.Since D0

k and B0k are the same as those in Case 1, we know from Case 1 that

Λ[D0k −B0k] ≤ µ2{[Vk−1(n1, n2 + 1, 0)− Vk−1(n1, n2, 1)]− [Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 1)]}.

Therefore, we have

Λ[µ2C0k + l1D

0k − µ2A0k − l1B0k]

= Λµ2[C0k −A0k] + Λl1[D0

k −B0k]≤ µ2l1{[Vk−1(n1, n2, 1)− Vk−1(n1, n2 + 1, 0)]− [Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 0)]}+l1µ2{[Vk−1(n1, n2 + 1, 0)− Vk−1(n1, n2, 1)]− [Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 1)]}= 0,



W1-Case 3: When serving station 1 is optimal at states (n1, n2 − 1, 0) and (n1, n2, 0), but serving station2 is optimal at state (n1, n2 + 1, 0), we will have

A0k =l1Λ[Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 1)] + µ2

Λ[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)]

B0k =µ2Λ[Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 1)]

C 0k =l1Λ[Vk−1(n1, n2, 1)− Vk−1(n1, n2 + 1, 0)]

D0k =

l1Λ[Vk−1(n1, n2 + 1, 0)− Vk−1(n1, n2 + 1, 1)] + µ2

Λ[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)].

Therefore,

Λ[C 0k −A0k] = l1{[Vk−1(n1, n2, 1)− Vk−1(n1, n2 + 1, 0)]− [Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 1)]}+µ2[Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 0)]

= l1Vk−1[(n1, n2, 1)− Vk−1(n1, n2, 0)] + µ2Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 0)+l1{[Vk−1(n1, n2, 0)− Vk−1(n1, n2 + 1, 0)]− [Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 1)]}.

Since inW1-Case 3, serving station 1 is optimal at state (n1, n2, 0), that is,

l1[Vk−1(n1, n2, 1)− Vk−1(n1, n2, 0)] + µ2[Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 0)] ≤ 0,then

Λ[C 0k −A0k] ≤ l1{[Vk−1(n1, n2, 0)− Vk−1(n1, n2 + 1, 0)]− [Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 1)]}.Similarly,

Λ[D0k −B0k] = l1[Vk−1(n1, n2 + 1, 0)− Vk−1(n1, n2 + 1, 1)] + µ2[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)]

−µ2[Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 1)]= l1[Vk−1(n1, n2 + 1, 0)− Vk−1(n1, n2 + 1, 1)] + µ2[Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 1)]= l1[Vk−1(n1, n2 + 1, 0)− Vk−1(n1, n2 + 1, 1)] + µ2[Vk−1(n1, n2, 0)− Vk−1(n1, n2 + 1, 0)]

+µ2{[Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 1)]− [Vk−1(n1, n2, 0)− Vk−1(n1, n2 + 1, 0)]}.

23

Since inW1-Case 3, serving station 2 is optimal at state (n1, n2 + 1, 0), that is,

l1[Vk−1(n1, n2 + 1, 0)− Vk−1(n1, n2 + 1, 1)] + µ2[Vk−1(n1, n2, 0)− Vk−1(n1, n2 + 1, 0)] ≤ 0,then

Λ[D0k −B0k] ≤ µ2{[Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 1)]− [Vk−1(n1, n2, 0)− Vk−1(n1, n2 + 1, 0)]}.

Therefore, we have

Λ[µ2C0k + l1D

0k − µ2A0k − l1B0k]

= Λµ2[C0k −A0k] + Λl1[D0

k −B0k]≤ µ2l1{[Vk−1(n1, n2, 0)− Vk−1(n1, n2 + 1, 0)]− [Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 1)]}+l1µ2{[Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 1)]− [Vk−1(n1, n2, 0)− Vk−1(n1, n2 + 1, 0)]}= 0,



W1-Case 4: When serving station 1 is optimal at states (n1, n2 − 1, 0), (n1, n2, 0) and (n1, n2 + 1, 0), wewill have

A0k =l1Λ[Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 1)] + µ2

Λ[Vk−1(n1, n2 − 1, 0)]− Vk−1(n1, n2, 0)]

B0k =µ2Λ[Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 1)]

C 0k =l1Λ[Vk−1(n1, n2, 1)− Vk−1(n1, n2 + 1, 1)] + µ2

Λ[Vk−1(n1, n2, 0)− Vk−1(n1, n2 + 1, 0)]

D0k =

µ2Λ[Vk−1(n1, n2 + 1, 0)− Vk−1(n1, n2, 1)].

Therefore,

Λ[C 0k −A0k] = l1[Vk−1(n1, n2, 1)− Vk−1(n1, n2 + 1, 1)] + µ2[Vk−1(n1, n2, 0)− Vk−1(n1, n2 + 1, 0)]−l1[Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 1)]− µ2[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)]

= l1[Vk−1(n1, n2 + 1, 0)− Vk−1(n1, n2 + 1, 1)] + µ2[Vk−1(n1, n2, 0)− Vk−1(n1, n2 + 1, 0)]−l1[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)]− µ2[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)]+l1{[Vk−1(n1, n2, 1)− Vk−1(n1, n2 + 1, 0)] + [Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 1)]}.

According to induction assumptionW1,

l1[Vk−1(n1, n2 + 1, 0)− Vk−1(n1, n2 + 1, 1)] + µ2[Vk−1(n1, n2, 0)− Vk−1(n1, n2 + 1, 0)]−l1[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)]− µ2[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)] ≤ 0.

Therefore,

Λ[C 0k −A0k] ≤ l1{[Vk−1(n1, n2, 1)− Vk−1(n1, n2 + 1, 0)] + [Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 1)]}.On the other hand,

Λ[D0k −B0k] = µ2{[Vk−1(n1, n2 + 1, 0)− Vk−1(n1, n2, 1)]− [Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 1)]}.

Therefore, we have

Λ[µ2C0k + l1D

0k − µ2A0k − l1B0k]

= Λµ2[C0k −A0k] + Λl1[D0

k −B0k]≤ µ2l1{[Vk−1(n1, n2, 1)− Vk−1(n1, n2 + 1, 0)] + [Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 1)]}+l1µ2{[Vk−1(n1, n2 + 1, 0)− Vk−1(n1, n2, 1)]− [Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 1)]}= 0,

24



Proof for W2: At iteration k for n1 ≥ 1 and n2 ≥ 1, conditionW2 is

µ2[Vk(n1, n2 − 1, 0)− Vk(n1, n2, 0)] + l1[Vk(n1, n2, 0)− Vk(n1, n2, 1)]≤ µ2[Vk(n1 + 1, n2 − 1, 0)− Vk(n1 + 1, n2, 0)] + l1[Vk(n1 + 1, n2, 0)− Vk(n1 + 1, n2, 1)].

For convenience, we deÞne:

Ik = Vk(n1, n2 − 1, 0)− Vk(n1, n2, 0),Jk = Vk(n1, n2, 0)− Vk(n1, n2, 1),Mk = Vk(n1 + 1, n2 − 1, 0)− Vk(n1 + 1, n2, 0),Nk = Vk(n1 + 1, n2, 0)− Vk(n1 + 1, n2, 1).

Therefore,W2 is equivalent toµ2Ik + l1Jk − µ2Mk − l1Nk ≤ 0.

Since idling is not optimal, we will have

Ik = − 1Λ+λ

Λ[Vk−1(n1 + 1, n2 − 1, 0)− Vk−1(n1 + 1, n2, 0)] + µ1

Λ[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)]

+1

Λmin


− 1Λmin


Jk =λ

Λ[Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2, 1)] + µ1

Λ[Vk−1(n1, n2, 0)− Vk−1(n1 − 1, n2 + 1, 0)]

+1

Λmin


Mk = − 1Λ+λ

Λ[Vk−1(n1 + 2, n2 − 1, 0)− Vk−1(n1 + 2, n2, 0)]

+µ1Λ[Vk−1(n1 + 1, n2 − 1, 0)− Vk−1(n1 + 1, n2, 0)]

+1

Λmin

½l1Vk−1(n1 + 1, n2 − 1, 1) + µ2Vk−1(n1 + 1, n2 − 1, 0)µ2Vk−1(n1 + 1, n2 − 2, 0) + l1Vk−1(n1 + 1, n2 − 1, 0)

− 1Λmin

½l1Vk−1(n1 + 1, n2, 1) + µ2Vk−1(n1 + 1, n2, 0)µ2Vk−1(n1 + 1, n2 − 1, 0) + l1Vk−1(n1 + 1, n2, 0)

Nk =λ

Λ[Vk−1(n1 + 2, n2, 0)− Vk−1(n1 + 2, n2, 1)] + µ1

Λ[Vk−1(n1 + 1, n2, 0)− Vk−1(n1, n2 + 1, 0)]

+1

Λmin

µ2[Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2 − 1, 1)]l1[Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2, 1)]

+µ2[Vk−1(n1 + 1, n2 − 1, 0)− Vk−1(n1 + 1, n2 − 1, 1)].Furthermore, we let

I 0k =1

Λmin


− 1Λmin


25

J 0k =1

Λmin


M 0k =

1

Λmin

½l1Vk−1(n1 + 1, n2 − 1, 1) + µ2Vk−1(n1 + 1, n2 − 1, 0)µ2Vk−1(n1 + 1, n2 − 2, 0) + l1Vk−1(n1 + 1, n2 − 1, 0)

− 1Λmin


N 0k =

1

Λmin

µ2[Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2 − 1, 1)]l1[Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2, 1)]

+µ2[Vk−1(n1 + 1, n2 − 1, 0)− Vk−1(n1 + 1, n2 − 1, 1)].Then,

µ2Ik + l1Jk − µ2Mk − l1Nk =λ

Λ{µ2[Vk(n1 + 1, n2 − 1, 0)− Vk(n1 + 1, n2, 0)] + l1[Vk(n1 + 1, n2, 0)− Vk(n1 + 1, n2, 1)]−µ2[Vk(n1 + 2, n2 − 1, 0)− Vk(n1 + 2, n2, 0)]− l1[Vk(n1 + 2, n2, 0)− Vk(n1 + 2, n2, 1)]}

+µ1[µ2Ik−1 + l1Jk−1 − µ2Mk−1 − l1Nk−1]+l1µ1{[Vk−1(n1, n2, 1)− Vk−1(n1 − 1, n2 + 1, 0)]− [Vk−1(n1 + 1, n2, 1)− Vk−1(n1, n2 + 1, 0)]}+µ2I

0k + l1J

0k − µ2M 0

k − l1N 0k.

According to induction assumption W2, the Þrst and the second terms are nonpositive. Furthermore,according to induction assumption D2, the third term is also nonpositive. Therefore, we only need to provethat

µ2I0k + l1J

0k − µ2M 0

k − l1N 0k ≤ 0.

We discuss the following six cases:

Case Vk(n1, n2 − 1, 0) Vk(n1, n2, 0) Vk(n1 + 1, n2 − 1, 0) Vk(n1 + 1, n2, 0)

1 Serve 2 Serve 2 Serve 2 Serve 2






W2-Case 1: For Case 1 of the above table, we will have

I 0k =µ2Λ[Vk−1(n1, n2 − 2, 0)− Vk−1(n1, n2 − 1, 0)] + l1

Λ[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)]

J 0k =l1Λ[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)] + µ2

Λ[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2 − 1, 1)]

M 0k =

µ2Λ[Vk−1(n1 + 1, n2 − 2, 0)− Vk−1(n1 + 1, n2 − 1, 0)] + l1

Λ[Vk−1(n1 + 1, n2 − 1, 0)− Vk−1(n1 + 1, n2, 0)]

N 0k =

l1Λ[Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2, 1)] + µ2

Λ[Vk−1(n1 + 1, n2 − 1, 0)− Vk−1(n1 + 1, n2 − 1, 1)].

Therefore,

Λ[I 0k −M 0k] = µ2[Vk−1(n1, n2 − 2, 0)− Vk−1(n1, n2 − 1, 0)] + l1[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)]

−µ2[Vk−1(n1 + 1, n2 − 2, 0)− Vk−1(n1 + 1, n2 − 1, 0)]

26

−l1[Vk−1(n1 + 1, n2 − 1, 0)− Vk−1(n1 + 1, n2, 0)]= {µ2[Vk−1(n1, n2 − 2, 0)− Vk−1(n1, n2 − 1, 0)] + l1[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2 − 1, 1)]

−µ2[Vk−1(n1 + 1, n2 − 2, 0)− Vk−1(n1 + 1, n2 − 1, 0)]−l1[Vk−1(n1 + 1, n2 − 1, 0)− Vk−1(n1 + 1, n2 − 1, 1)]}

+l1{[Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 0)]− [Vk−1(n1 + 1, n2 − 1, 1)− Vk−1(n1 + 1, n2, 0)]}.According to induction assumptionW2, the Þrst term is nonpositive. Thus,

Λ[I 0k −M 0k] ≤ l1{[Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 0)]− [Vk−1(n1 + 1, n2 − 1, 1)− Vk−1(n1 + 1, n2, 0)]}.

Similarly,

Λ[J 0k −N 0k] = µ2Ik−1 + l1Jk−1 − µ2Mk−1 − l1Nk−1

+µ2{[Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 1)]− [Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2 − 1, 1)]}≤ µ2{[Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 1)]− [Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2 − 1, 1)]}.

Therefore, we have

Λ[µ2I0k + l1J

0k − µ2M 0

k − l1N 0k]

≤ µ2l1{[Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 0)]− [Vk−1(n1 + 1, n2 − 1, 1)− Vk−1(n1 + 1, n2, 0)]}+µ2l1{[Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 1)]− [Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2 − 1, 1)]}= 0,

which implies that µ2I0k + l1J

0k − µ2M 0

k − l1N 0k ≤ 0. This completes the proof forW2-Case 1.

W2-Case 2: I 0k, J0k, N

0k are the same as inW2-Case 1. However, we have

M 0k =

l1Λ[Vk−1(n1 + 1, n2 − 1, 1)− Vk−1(n1 + 1, n2, 0)].

Therefore,

Λ[I 0k −M 0k] = µ2[Vk−1(n1, n2 − 2, 0)− Vk−1(n1, n2 − 1, 0)] + l1[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2 − 1, 1)]

+l1{[Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 0)]− [Vk−1(n1 + 1, n2 − 1, 1)− Vk−1(n1 + 1, n2, 0)]}.

Since inW2-Case 2 serving station 2 is optimal at state (n1, n2 − 1, 0), thenµ2[Vk−1(n1, n2 − 2, 0)− Vk−1(n1, n2 − 1, 0)] + l1[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2 − 1, 1)] ≤ 0.

Hence,


Since J 0k and N0k are the same as those in Case 1, we know from Case 1 that

Λ[J 0k −N 0k] ≤ µ2{[Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 1)]− [Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2 − 1, 1)]}.

Therefore, we have

Λ[µ2I0k + l1J

0k − µ2M 0

k − l1N 0k]

≤ µ2l1{[Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 0)]− [Vk−1(n1 + 1, n2 − 1, 1)− Vk−1(n1 + 1, n2, 0)]}+l1µ2{[Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 1)]− [Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2 − 1, 1)]}= 0,

which implies that µ2I0k + l1J

0k − µ2M 0

k − l1N 0k ≤ 0. This completes the proof forW2-Case 2.

27

W2-Case 3: I 0k, J0k are the same as those in Case 1. However, we have

M 0k =

l1Λ[Vk−1(n1 + 1, n2 − 1, 1)− Vk−1(n1 + 1, n2, 1)] + µ2

Λ[Vk−1(n1 + 1, n2 − 1, 0)− Vk−1(n1 + 1, n2, 0)],

N 0k =

µ2Λ[Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2 − 1, 1)].

Therefore,

Λ[I 0k −M 0k] = µ2[Vk−1(n1, n2 − 2, 0)− Vk−1(n1, n2 − 1, 0)] + l1[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)]

−l1[Vk−1(n1 + 1, n2 − 1, 1)− Vk−1(n1 + 1, n2, 1)]−µ2[Vk−1(n1 + 1, n2 − 1, 0)− Vk−1(n1 + 1, n2, 0)]

= µ2[Vk−1(n1, n2 − 2, 0)− Vk−1(n1, n2 − 1, 0)] + l1[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2 − 1, 1)]}−µ2[Vk−1(n1 + 1, n2 − 1, 0)− Vk−1(n1 + 1, n2, 0)]− l1[Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2, 1)]+l1{[Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 0)]− [Vk−1(n1 + 1, n2 − 1, 1)− Vk−1(n1 + 1, n2, 0)]}.

Since inW2-Case 3, serving station 2 is optimal at state (n1, n2 − 1, 0), while serving 1 is optimal at state(n1 + 1, n2, 0), then we will have

µ2[Vk−1(n1, n2 − 2, 0)− Vk−1(n1, n2 − 1, 0)] + l1[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2 − 1, 1)]} ≤ 0,µ2[Vk−1(n1 + 1, n2 − 1, 0)− Vk−1(n1 + 1, n2, 0)] + l1[Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2, 1)] ≥ 0.

Hence,


Similarly,

Λ[J 0k −N 0k] = µ2[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2 − 1, 1)] + l1[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)]

−µ2[Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2 − 1, 1)]= µ2[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)] + l1[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)]

+µ2{[Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 1)]− [Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2 − 1, 1)]}.Since inW2-Case 3, serving station 2 is optimal at state (n1, n2, 0), that is,

µ2[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)] + l1[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)] ≤ 0,then,


Therefore, we have

Λ[µ2I0k + l1J

0k − µ2M 0

k − l1N 0k]

≤ µ2l1{[Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 0)]− [Vk−1(n1 + 1, n2 − 1, 1)− Vk−1(n1 + 1, n2, 0)]}+l1µ2{[Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 1)]− [Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2 − 1, 1)]}= 0.

This completes the proof forW2-Case 3.

W2-Case 4: J 0k, N0k are the same as those in Case 1. However, we have

I 0k =l1Λ[Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 0)],

M 0k =

l1Λ[Vk−1(n1 + 1, n2 − 1, 1)− Vk−1(n1 + 1, n2, 0)].

28

Therefore,

Λ[I 0k −M 0k] = l1{[Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 0)]− [Vk−1(n1 + 1, n2 − 1, 1)− Vk−1(n1 + 1, n2, 0)]}.

On the other hand,

Λ[J 0k −N 0k] ≤ µ2{[Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 1)]− [Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2 − 1, 1)].

Hence,

Λ[µ2I0k + l1J

0k − µ2M 0

k − l1N 0k]

≤ µ2l1{[Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 0)]− [Vk−1(n1 + 1, n2 − 1, 1)− Vk−1(n1 + 1, n2, 0)]}+l1µ2{[Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 1)]− [Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2 − 1, 1)]}

= 0.


W2-Case 5: For Case 5, we have

I 0k =l1Λ[Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 0)]

M 0k =

l1Λ[Vk−1(n1 + 1, n2 − 1, 1)− Vk−1(n1 + 1, n2, 1)] + µ2

Λ[Vk−1(n1 + 1, n2 − 1, 0)− Vk−1(n1 + 1, n2, 0)]

J 0k =l1Λ[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)] + µ2

Λ[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2 − 1, 1)]

N 0k =

µ2Λ[Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2 − 1, 1)].

Therefore,

Λ[I 0k −M 0k] = l1{[Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 0)]− [Vk−1(n1 + 1, n2 − 1, 1)− Vk−1(n1 + 1, n2, 1)]}

−µ2[Vk−1(n1 + 1, n2 − 1, 0)− Vk−1(n1 + 1, n2, 0)]= l1{[Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 0)]− [Vk−1(n1 + 1, n2 − 1, 1)− Vk−1(n1 + 1, n2, 0)]}

−{l1[Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2, 1)]+µ2[Vk−1(n1 + 1, n2 − 1, 0)− Vk−1(n1 + 1, n2, 0)]}.

Since inW2-Case 5, serving station 1 is optimal at state (n1 + 1, n2, 0), which implies,

l1[Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2, 1)] + µ2[Vk−1(n1 + 1, n2 − 1, 0)− Vk−1(n1 + 1, n2, 0)] ≥ 0,

then


On the other hand,

Λ[J 0k −N 0k] = l1[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)] + µ2[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2 − 1, 1)]

−µ2[Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2 − 1, 1)]= l1[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)] + µ2[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)]

+µ2{[Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 1)]− [Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2 − 1, 1)]}.

Since inW2-Case 5, serving station 2 is optimal at state (n1, n2, 0), which implies

l1[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)] + µ2[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)] ≤ 0,

29

then, we will have


Thus,

Λ[µ2I0k + l1J

0k − µ2M 0

k − l1N 0k]


= 0.


W2-Case 6: In Case 6, M 0k, N

0k are the same as those in Case 5. However,

I 0k =l1Λ[Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 1)] + µ2

Λ[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)],

J 0k =µ2Λ[Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 1)].

Therefore,

Λ[I 0k −M 0k] = l1[Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 1)] + µ2[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)]

−l1[Vk−1(n1 + 1, n2 − 1, 1)− Vk−1(n1 + 1, n2, 1)]−µ2[Vk−1(n1 + 1, n2 − 1, 0)− Vk−1(n1 + 1, n2, 0)]

= l1{[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)] + µ2[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)]}−l1[Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2, 1)]− µ2[Vk−1(n1 + 1, n2 − 1, 0)− Vk−1(n1 + 1, n2, 0)]+l1{[Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 0)]− [Vk−1(n1 + 1, n2 − 1, 1)− Vk−1(n1 + 1, n2, 0)]}.


l1[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)] + µ2[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)]−l1[Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2, 1)]− µ2[Vk−1(n1 + 1, n2 − 1, 0)− Vk−1(n1 + 1, n2, 0)] ≤ 0.

Thus,


On the other hand,

Λ[J 0k −N 0k] = µ2{[Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 1)]− [Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2 − 1, 1)]}.

Therefore, we have

Λ[µ2I0k + l1J

0k − µ2M 0

k − l1N 0k]


= 0.


Proof for D1: We must prove that for n1 ≥ 1 and n2 ≥ 0,[Vk(n1, n2 + 1, 1)− Vk(n1 − 1, n2 + 2, 0)]− [Vk(n1, n2, 1)− Vk(n1 − 1, n2 + 1, 0)] ≤ 0.

30

We Þrst discuss the cases when n2 > 0, then we focus on cases when n2 = 0.

D1-Case I: Since idling is not optimal, we have

Vk(n1, n2 + 1, 1) =n1 + n2 + 1

Λ+λ

ΛVk−1(n1 + 1, n2 + 1, 1) +

µ1ΛVk−1(n1 − 1, n2 + 2, 0)

+l1ΛVk−1(n1, n2 + 1, 1) +

µ2ΛVk−1(n1, n2, 1)

Vk(n1 − 1, n2 + 2, 0) =n1 + n2 + 1

Λ+λ

ΛVk−1(n1, n2 + 2, 0) +

µ1ΛVk−1(n1 − 1, n2 + 2, 0)

+1

Λmin

½l1Vk−1(n1 − 1, n2 + 2, 1) + µ2Vk−1(n1 − 1, n2 + 2, 0)µ2Vk−1(n1 − 1, n2 + 1, 0) + l1Vk−1(n1 − 1, n2 + 2, 0)

Vk(n1, n2, 1) =n1 + n2Λ

+λ

ΛVk−1(n1 + 1, n2, 1) +

µ1ΛVk−1(n1 − 1, n2 + 1, 0)

+l1ΛVk−1(n1, n2, 1) + µ2Vk−1(n1, n2 − 1, 1)

Vk(n1 − 1, n2 + 1, 0) =n1 + n2Λ

+λ

ΛVk−1(n1, n2 + 1, 0) +

µ1ΛVk−1(n1 − 1, n2 + 2, 0)

+1

Λmin

½l1Vk−1(n1 − 1, n2 + 1, 1) + µ2Vk−1(n1 − 1, n2 + 1, 0)µ2Vk−1(n1 − 1, n2, 0) + l1Vk−1(n1 − 1, n2 + 1, 0).

Therefore,

[Vk(n1, n2 + 1, 1)− Vk(n1 − 1, n2 + 2, 0)]− [Vk(n1, n2, 1)− Vk(n1 − 1, n2 + 1, 0)]=

λ

Λ{[Vk(n1 + 1, n2 + 1, 1)− Vk(n1, n2 + 2, 0)]− [Vk(n1 + 1, n2, 1)− Vk(n1, n2 + 1, 0)]}

+l1ΛVk−1(n1, n2 + 1, 1) +

µ2ΛVk−1(n1, n2, 1)

− 1Λmin


− l1ΛVk−1(n1, n2, 1)− µ2

ΛVk−1(n1, n2 − 1, 1)

+1

Λmin


According to induction assumption D1, the Þrst term is nonpositive. Therefore, we only need to show thatthe remaining terms, which we deÞne as α, are nonpositive.

α =l1ΛVk−1(n1, n2 + 1, 1) +

µ2ΛVk−1(n1, n2, 1)

− 1Λmin


− l1ΛVk−1(n1, n2, 1)− µ2

ΛVk−1(n1, n2 − 1, 1)

+1

Λmin


We discuss the following three cases:

Case Vk(n1 − 1, n2 + 1, 0) Vk(n1 − 1, n2 + 2, 0)1 Serve 2 Serve 2

2 Serve 1 Serve 2

3 Serve 1 Serve 1

31

D1-Case I-1: When serving station 2 is optimal at states (n1 − 1, n2 + 1, 0) and (n1 − 1, n2 + 2, 0), then

α =l1ΛVk−1(n1, n2 + 1, 1) +

µ2ΛVk−1(n1, n2, 1)− 1

Λ[µ2Vk−1(n1 − 1, n2 + 1, 0) + l1Vk−1(n1 − 1, n2 + 2, 0)]

− l1ΛVk−1(n1, n2, 1)− µ2

ΛVk−1(n1, n2 − 1, 1) + 1

Λ[µ2Vk−1(n1 − 1, n2, 0) + l1Vk−1(n1 − 1, n2 + 1, 0)]

=l1Λ{[Vk−1(n1, n2 + 1, 1)− Vk−1(n1 − 1, n2 + 2, 0)]− [Vk−1(n1, n2, 1)− Vk−1(n1 − 1, n2 + 1, 0)]}

+µ2{[Vk−1(n1, n2, 1)− Vk−1(n1 − 1, n2 + 1, 0)]− [Vk−1(n1, n2 − 1, 1)− Vk−1(n1 − 1, n2, 0)]}.According to D1,

[Vk−1(n1, n2 + 1, 1)− Vk−1(n1 − 1, n2 + 2, 0)]− [Vk−1(n1, n2, 1)− Vk−1(n1 − 1, n2 + 1, 0)] ≤ 0,[Vk−1(n1, n2, 1)− Vk−1(n1 − 1, n2 + 1, 0)]− [Vk−1(n1, n2 − 1, 1)− Vk−1(n1 − 1, n2, 0)] ≤ 0.

Hence, α ≤ 0, which completes the proof for D1-Case I-1.D1-Case I-2: When serving station 1 and 2 are optimal at states (n1 − 1, n2 + 1, 0) and (n1 − 1, n2 + 2, 0),respectively, then

α =1

Λ[l1Vk−1(n1, n2 + 1, 1) + µ2Vk−1(n1, n2, 1)]− 1

Λ[µ2Vk−1(n1 − 1, n2 + 1, 0) + l1Vk−1(n1 − 1, n2 + 2, 0)]

− 1Λ[l1Vk−1(n1, n2, 1) + µ2Vk−1(n1, n2 − 1, 1)] + 1

Λ[l1Vk−1(n1 − 1, n2 + 1, 1) + µ2Vk−1(n1 − 1, n2 + 1, 0)]


+l1Λ[Vk−1(n1 − 1, n2 + 1, 1)− Vk−1(n1 − 1, n2 + 1, 0)]

+µ2Λ{[Vk−1(n1, n2, 1)− Vk−1(n1 − 1, n2 + 1, 0)]− [Vk−1(n1, n2 − 1, 1)− Vk−1(n1 − 1, n2, 0)]}

+µ2Λ[Vk−1(n1 − 1, n2 + 1, 0)− Vk−1(n1 − 1, n2, 0)].

According to D1,

[Vk−1(n1, n2 + 1, 1)− Vk−1(n1 − 1, n2 + 2, 0)]− [Vk−1(n1, n2, 1)− Vk−1(n1 − 1, n2 + 1, 0)] ≤ 0[Vk−1(n1, n2, 1)− Vk−1(n1 − 1, n2 + 1, 0)]− [Vk−1(n1, n2 − 1, 1)− Vk−1(n1 − 1, n2, 0)] ≤ 0.

Hence,

α ≤ l1Λ[Vk−1(n1 − 1, n2 + 1, 1)− Vk−1(n1 − 1, n2 + 1, 0)] + µ2

Λ[Vk−1(n1 − 1, n2 + 1, 0)− Vk−1(n1 − 1, n2, 0)].

Since in D1-Case I-2 serving station 1 is optimal at state (n1 − 1, n2 + 1, 0), thenl1[Vk−1(n1 − 1, n2 + 1, 1)− Vk−1(n1 − 1, n2 + 1, 0)] + µ2[Vk−1(n1 − 1, n2 + 1, 0)− Vk−1(n1 − 1, n2, 0)] ≤ 0.Hence, α ≤ 0, which completes the proof for D1-Case I-2.D1-Case I-3: When serving station 1 is optimal at states (n1 − 1, n2 + 1, 0) and (n1 − 1, n2 + 2, 0), then

α =1

Λ[l1Vk−1(n1, n2 + 1, 1) + µ2Vk−1(n1, n2, 1)]− 1

Λ[l1Vk−1(n1 − 1, n2 + 2, 1) + µ2Vk−1(n1 − 1, n2 + 2, 0)]

− 1Λ[l1Vk−1(n1, n2, 1) + µ2Vk−1(n1, n2 − 1, 1)] + 1

Λ[l1Vk−1(n1 − 1, n2 + 1, 1) + µ2Vk−1(n1 − 1, n2 + 1, 0)]


32

+l1Λ{[Vk−1(n1 − 1, n2 + 2, 0)− Vk−1(n1 − 1, n2 + 2, 1)]− [Vk−1(n1 − 1, n2 + 1, 0)− Vk−1(n1 − 1, n2 + 1, 1)]}

+µ2Λ{[Vk−1(n1, n2, 1)− Vk−1(n1 − 1, n2 + 1, 0)]− [Vk−1(n1, n2 − 1, 1)− Vk−1(n1 − 1, n2, 0)]}

+µ2Λ{[Vk−1(n1 − 1, n2 + 1, 0)− Vk−1(n1 − 1, n2 + 2, 0)]− [Vk−1(n1 − 1, n2, 0)− Vk−1(n1 − 1, n2 + 1, 0)]}.

According to D1,

[Vk−1(n1, n2 + 1, 1)− Vk−1(n1 − 1, n2 + 2, 0)]− [Vk−1(n1, n2, 1)− Vk−1(n1 − 1, n2 + 1, 0)] ≤ 0[Vk−1(n1, n2, 1)− Vk−1(n1 − 1, n2 + 1, 0)]− [Vk−1(n1, n2 − 1, 1)− Vk−1(n1 − 1, n2, 0)] ≤ 0.

Hence,

α ≤ l1Λ{[Vk−1(n1 − 1, n2 + 2, 0)− Vk−1(n1 − 1, n2 + 2, 1)]− [Vk−1(n1 − 1, n2 + 1, 0)− Vk−1(n1 − 1, n2 + 1, 1)]}

+µ2Λ{[Vk−1(n1 − 1, n2 + 1, 0)− Vk−1(n1 − 1, n2 + 2, 0)]− [Vk−1(n1 − 1, n2, 0)− Vk−1(n1 − 1, n2 + 1, 0)]}.

According to induction assumptionW1,

l1[Vk−1(n1 − 1, n2 + 2, 0)− Vk−1(n1 − 1, n2 + 2, 1)] + µ2[Vk−1(n1 − 1, n2 + 1, 0)− Vk−1(n1 − 1, n2 + 2, 0)]−{l1[Vk−1(n1 − 1, n2 + 1, 0)− Vk−1(n1 − 1, n2 + 1, 1)] + µ2[Vk−1(n1 − 1, n2, 0)− Vk−1(n1 − 1, n2 + 1, 0)]} ≤ 0.

Therefore, α ≤ 0, which completes the proof for D1-Case I-3.D1-Case II: When n2 = 0, D1 becomes:

[Vk(n1, 1, 0)− Vk(n1 − 1, 2, 0)]− [Vk(n1, 0, 1)− Vk(n1 − 1, 1, 0)] ≤ 0.At iteration k of the value iteration algorithm we have

Vk(n1, 1, 1) =n1 + 1

Λ+λ

ΛVk−1(n1 + 1, 1, 1)

+µ1ΛVk−1(n1 − 1, 2, 0) + l1

ΛVk−1(n1, 1, 1) +

µ2ΛVk−1(n1, 0, 1)

Vk(n1 − 1, 2, 0) =n1 + 1

Λ+λ

ΛVk−1(n1, 2, 0) +

µ1ΛVk−1(n1 − 1, 2, 0)

+1

Λmin

½l1Vk−1(n1 − 1, 2, 1) + µ2Vk−1(n1 − 1, 2, 0)µ2Vk−1(n1 − 1, 1, 0) + l1Vk−1(n1 − 1, 2, 0)

Vk(n1, 0, 1) =n1Λ+λ

ΛVk−1(n1 + 1, 0, 1)

+µ1ΛVk−1(n1 − 1, 1, 0) + l1

ΛVk−1(n1, 0, 1) + µ2Vk−1(n1, 0, 1)

Vk(n1 − 1, 1, 0) =n1Λ+λ

ΛVk−1(n1, 1, 0)

+µ1ΛVk−1(n1 − 1, 2, 0)

+1

Λmin

½l1Vk−1(n1 − 1, 1, 1) + µ2Vk−1(n1 − 1, 1, 0)µ2Vk−1(n1 − 1, 0, 0) + l1Vk−1(n1 − 1, 1, 0).

33

Therefore,

[Vk(n1, 1, 1)− Vk(n1 − 1, 2, 0)]− [Vk(n1, 0, 1)− Vk(n1 − 1, 1, 0)]=

λ

Λ{[Vk−1(n1 + 1, 1, 1)− Vk−1(n1, 2, 0)]− [Vk−1(n1 + 1, 0, 1)− Vk−1(n1, 1, 0)]}

+l1ΛVk−1(n1, 1, 1) +

µ2ΛVk−1(n1, 0, 1)

− 1Λmin


− l1ΛVk−1(n1, 0, 1)− µ2

ΛVk−1(n1, 0, 1)

+1

Λmin


Since D1 holds at iteration k − 1, the Þrst term on the right hand side is non-positive. Therefore, we onlyneed to show that the remaining terms, which we deÞne as η, is nonpositive.

η =l1ΛVk−1(n1, 1, 1) +

µ2ΛVk−1(n1, 0, 1)

− 1Λmin


− l1ΛVk−1(n1, 0, 1)− µ2

ΛVk−1(n1, 0, 1)

+1

Λmin


We discuss the following three cases:

Case Vk(n1 − 1, 1, 0) Vk(n1 − 1, 2, 0)1 Serve 2 Serve 2

2 Serve 1 Serve 2

3 Serve 1 Serve 1

D1-Case II-1: When serving 2 is optimal at states (n1 − 1, 1, 0) and (n1 − 1, 2, 0), then

η =l1ΛVk−1(n1, 1, 1) +

µ2ΛVk−1(n1, 0, 1)

− 1Λ[µ2Vk−1(n1 − 1, 1, 0) + l1Vk−1(n1 − 1, 2, 0)]

− l1ΛVk−1(n1, 0, 1)− µ2

ΛVk−1(n1, 0, 1)

+1

Λ[µ2Vk−1(n1 − 1, 0, 0) + l1Vk−1(n1 − 1, 1, 0)]

=l1Λ{[Vk−1(n1, 1, 1)− Vk−1(n1 − 1, 2, 0)]− [Vk−1(n1, 0, 1)− Vk−1(n1 − 1, 1, 0)]}

+µ2Λ[Vk−1(n1 − 1, 0, 0)− Vk−1(n1 − 1, 1, 0)].

According to D1,

[Vk−1(n1, 1, 1)− Vk−1(n1 − 1, 2, 0)]− [Vk−1(n1, 0, 1)− Vk−1(n1 − 1, 1, 0)] ≤ 0.On the other hand, according to V3, we have

Vk−1(n1 − 1, 0, 0)− Vk−1(n1 − 1, 1, 0) ≤ 0.

34

Hence, η ≤ 0, which completes the proof for D1-Case II-1.D1-Case II-2:When serving station 1 and 2 are optimal at states (n1−1, 1, 0) and (n1−1, 2, 0), respectively,then

η =1

Λ[l1Vk−1(n1, 1, 1) + µ2Vk−1(n1, 0, 1)]

− 1Λ[µ2Vk−1(n1 − 1, 1, 0) + l1Vk−1(n1 − 1, 2, 0)]

− 1Λ[l1Vk−1(n1, 0, 1) + µ2Vk−1(n1, 0, 1)]

+1

Λ[l1Vk−1(n1 − 1, 1, 1) + µ2Vk−1(n1 − 1, 1, 0)]

=l1Λ{[Vk−1(n1, 1, 1)− Vk−1(n1 − 1, 2, 0)]− [Vk−1(n1, 0, 1)− Vk−1(n1 − 1, 1, 0)]}

+l1Λ{[Vk−1(n1 − 1, 1, 1)− Vk−1(n1 − 1, 1, 0)]

According to D1,

[Vk−1(n1, 1, 1)− Vk−1(n1 − 1, 2, 0)]− [Vk−1(n1, 0, 1)− Vk−1(n1 − 1, 1, 0)] ≤ 0.

Furthermore, according to V2, we have

Vk−1(n1 − 1, 1, 1)− Vk−1(n1 − 1, 1, 0) ≤ 0.

Hence, η ≤ 0, which completes the proof for D1-Case II-2.D1-Case II-3: When serving station 1 is optimal at states (n1 − 1, 1, 0) and (n1 − 1, 2, 0), then

η =1

Λ[l1Vk−1(n1, 1, 1) + µ2Vk−1(n1, 0, 1)]

− 1Λ[µ2Vk−1(n1 − 1, 2, 0) + l1Vk−1(n1 − 1, 2, 1)]

− 1Λ[l1Vk−1(n1, 0, 1) + µ2Vk−1(n1, 0, 1)]

+1

Λ[l1Vk−1(n1 − 1, 1, 1) + µ2Vk−1(n1 − 1, 1, 0)]

=l1Λ{[Vk−1(n1, 1, 1)− Vk−1(n1 − 1, 2, 1)]− [Vk−1(n1, 0, 1)− Vk−1(n1 − 1, 1, 1)]}

+µ2Λ[Vk−1(n1 − 1, 1, 0)− Vk−1(n1 − 1, 2, 0)]

=l1Λ{[Vk−1(n1, 1, 1)− Vk−1(n1 − 1, 2, 0)]− [Vk−1(n1, 0, 1)− Vk−1(n1 − 1, 1, 0)]}

+l1Λ{[Vk−1(n1 − 1, 2, 0)− Vk−1(n1 − 1, 2, 1)]− [Vk−1(n1 − 1, 1, 0)− Vk−1(n1 − 1, 1, 1)]}

+µ2Λ{[Vk−1(n1 − 1, 1, 0)− Vk−1(n1 − 1, 2, 1)]− [Vk−1(n1 − 1, 0, 0)− Vk−1(n1 − 1, 1, 0)]}

+µ2Λ[Vk−1(n1 − 1, 0, 0)− Vk−1(n1 − 1, 1, 0)].

According to D1,

[Vk−1(n1, 1, 1)− Vk−1(n1 − 1, 2, 0)]− [Vk−1(n1, 0, 1)− Vk−1(n1 − 1, 1, 0)] ≤ 0.

35

Also, according to induction assumptionW1,

l1{[Vk−1(n1 − 1, 2, 0)− Vk−1(n1 − 1, 1, 1)]− [Vk−1(n1 − 1, 1, 0)− Vk−1(n1 − 1, 1, 1)]}+µ2{[Vk−1(n1 − 1, 1, 0)− Vk−1(n1 − 1, 2, 0)]− [Vk−1(n1 − 1, 0, 0)− Vk−1(n1 − 1, 1, 0)]} ≤ 0.

Finally, according to V3,Vk−1(n1 − 1, 0, 0)− Vk−1(n1 − 1, 1, 0) ≤ 0.

Hence, η ≤ 0, which completes the proof for D1-Case II-3.

Proof for D2: For n1 ≥ 1 and n2 ≥ 0 we must prove that

Vk(n1, n2, 1)− Vk(n1 − 1, n2 + 1, 0)− Vk(n1 + 1, n2, 1) + Vk(n1, n2 + 1, 0) ≤ 0.

We discuss two cases:

� Case 1: n1 ≥ 1 and n2 = 0,� Case 2: n1 ≥ 1 and n2 ≥ 1.

D2-Case I: We need to show that for n1 ≥ 1,

Vk(n1, 0, 1)− Vk(n1 − 1, 1, 0)− Vk(n1 + 1, 0, 1) + Vk(n1, 1, 0) ≤ 0.

We have

Vk(n1, 0, 1) =n1Λ+λ

ΛVk−1(n1 + 1, 0, 1) +

µ1ΛVk−1(n1 − 1, 1, 0) + l1 + µ2

ΛVk−1(n1, 0, 1)

Vk(n1 − 1, 1, 0) =n1Λ+λ

ΛVk−1(n1, 1, 0) +

µ1ΛVk−1(n1 − 1, 1, 0)

+1

Λmin


Vk(n1 + 1, 0, 1) =n1 + 1

Λ+λ

ΛVk−1(n1 + 2, 0, 1) +

µ1ΛVk−1(n1, 1, 0) +

l1 + µ2Λ

Vk−1(n1 + 1, 0, 1)

Vk(n1, 1, 0) =n1 + 1

Λ+λ

ΛVk−1(n1 + 1, 1, 0) +

µ1ΛVk−1(n1, 1, 0)

+1

Λmin


Vk(n1, 0, 1)− Vk(n1 − 1, 1, 0)− Vk(n1 + 1, 0, 1) + Vk(n1, 1, 0)=

λ

Λ[Vk(n1 + 1, 0, 1)− Vk(n1, 1, 0)− Vk(n1 + 2, 0, 1) + Vk(n1 + 1, 1, 0)]

+l1 + µ2Λ

[Vk−1(n1, 0, 1)− Vk−1(n1 + 1, 0, 1)]

− 1Λmin


+1

Λmin

½l1Vk−1(n1, 1, 1) + µ2Vk−1(n1, 1, 0)µ2Vk−1(n1, 0, 0) + l1Vk−1(n1, 1, 0).

36

According to D2, the Þrst term on the right-hand side is nonpositive. In the following, we show that theremaining terms, which we deÞne as κ, are nonpositive. We discuss three cases:

Case Vk(n1 − 1, 1, 0) Vk(n1, 1, 0)

1 Serve 1 Serve 1

2 Serve 2 Serve 1

3 Serve 2 Serve 2

D2-Case I-1: We have

Λ× κ = l1[Vk−1(n1, 0, 1)− Vk−1(n1 + 1, 0, 1)− Vk−1(n1 − 1, 1, 1) + Vk−1(n1, 1, 1)]+µ2[Vk−1(n1, 0, 1)− Vk−1(n1 + 1, 0, 1)− Vk−1(n1 − 1, 1, 0) + Vk−1(n1, 1, 0)]

= (l1 + µ2)[Vk−1(n1, 0, 1)− Vk−1(n1 + 1, 0, 1)− Vk−1(n1 − 1, 1, 0) + Vk−1(n1, 1, 0)]+l1[Vk−1(n1 − 1, 1, 0)− Vk−1(n1 − 1, 1, 1)− Vk−1(n1, 1, 0) + Vk−1(n1, 1, 1)].

According to D2, the Þrst term is nonpositive. Furthermore, according to M1, the second term is alsononpositive. This completes the proof for D2-Case I-1.

D2-Case I-2: For this case we have


= (l1 + µ2)[Vk−1(n1, 0, 1)− Vk−1(n1 + 1, 0, 1)− Vk−1(n1 − 1, 1, 0) + Vk−1(n1, 1, 0)]+{l1[Vk−1(n1, 1, 1)− Vk−1(n1, 1, 0)] + µ2[Vk−1(n1, 1, 0)− Vk−1(n1, 0, 0)]}+µ2[Vk−1(n1 − 1, 1, 0)− Vk−1(n1 − 1, 0, 0)− Vk−1(n1, 1, 0) + Vk−1(n1, 0, 0)].

According to D2, the Þrst term is nonpositive. Furthermore, since it is optimal to serve station 1 at state(n1, 1, 0), the second term is also nonpositive. Finally, according to M2, the last term is nonpositive. Thiscompletes the proof for D2-Case I-2.

D2-Case I-3: We have


= (l1 + µ2)[Vk−1(n1, 0, 1)− Vk−1(n1 + 1, 0, 1)− Vk−1(n1 − 1, 1, 0) + Vk−1(n1, 1, 0)]+µ2[Vk−1(n1 − 1, 1, 0)− Vk−1(n1 − 1, 0, 0)− Vk−1(n1, 1, 0) + Vk−1(n1, 0, 0)].

According to D2, the Þrst term is nonpositive. Also, according toM2, the second term is nonpositive. Thiscompletes the proof for D2-Case I-3.

D2-Case II: Since idling is not optimal, we will have

Vk(n1, n2, 1) =n1 + n2Λ

+λ

ΛVk−1(n1 + 1, n2, 1) +

µ1ΛVk−1(n1 − 1, n2 + 1, 0)

+l1ΛVk−1(n1, n2, 1) +

µ2ΛVk−1(n1, n2 − 1, 1)

Vk(n1 − 1, n2 + 1, 0) =n1 + n2Λ

+λ

ΛVk−1(n1, n2 + 1, 0) +

µ1ΛVk−1(n1 − 1, n2 + 1, 0)

+1

Λmin


Vk(n1 + 1, n2, 1) =n1 + 1+ n2

Λ+λ

ΛVk−1(n1 + 2, n2, 1) +

µ1ΛVk−1(n1, n2 + 1, 0)

+l1ΛVk−1(n1 + 1, n2, 1) +

µ2ΛVk−1(n1 + 1, n2 − 1, 1)

37

Vk(n1, n2 + 1, 0) =n1 + n2 + 1

Λ+λ

ΛVk−1(n1 + 1, n2 + 1, 0) +

µ1ΛVk−1(n1, n2 + 1, 0)

+1

Λmin

½l1Vk−1(n1, n2 + 1, 1) + µ2Vk−1(n1, n2 + 1, 0)µ2Vk−1(n1, n2, 0) + l1Vk−1(n1, n2 + 1, 0).

Therefore,

[Vk(n1, n2, 1)− Vk(n1 − 1, n2 + 1, 0)]− [Vk(n1 + 1, n2, 1)− Vk(n1, n2 + 1, 0)]=λ

Λ{[Vk(n1 + 1, n2, 1)− Vk(n1, n2 + 1, 0)]− [Vk(n1 + 2, n2, 1)− Vk(n1 + 1, n2 + 1, 0)]}

+l1ΛVk−1(n1, n2, 1) +

µ2ΛVk−1(n1, n2 − 1, 1)

− 1Λmin


− l1ΛVk−1(n1 + 1, n2, 1)− µ2

ΛVk−1(n1 + 1, n2 − 1, 1)

+1

Λmin


According to induction assumption D2, the Þrst term on the right-hand side is nonpositive. Thus, we onlyneed to show that the remaining terms, which we deÞne as β, is also nonpositive.

β =l1ΛVk−1(n1, n2, 1) +

µ2ΛVk−1(n1, n2 − 1, 1)

− 1Λmin


− l1ΛVk−1(n1 + 1, n2, 1)− µ2

ΛVk−1(n1 + 1, n2 − 1, 1)

+1

Λmin


Based on the induction assumption of W2, we know that: if to serving station 1 is optimal at state (n1 −1, n2 + 1, 0), then it is also optimal at state (n1, n2 + 1, 0). Therefore, we only need to discuss the followingthree cases:

Case Vk(n1 − 1, n2 + 1, 0) Vk(n1, n2 + 1, 0)

1 Serve 1 Serve 1

2 Serve 2 Serve 1

3 Serve 2 Serve 2

D2-Case II-1: When serving station 1 is optimal at states (n1 − 1, n2 + 1, 0) and (n1, n2 + 1, 0), then

β =l1Λ{[Vk−1(n1, n2, 1)− Vk−1(n1 − 1, n2 + 1, 1)]− [Vk−1(n1 + 1, n2, 1)− Vk−1(n1, n2 + 1, 1)]}

+µ2Λ{[Vk−1(n1, n2 − 1, 1)− Vk−1(n1 − 1, n2 + 1, 0)]− [Vk−1(n1 + 1, n2 − 1, 1)− Vk−1(n1, n2 + 1, 0)]}

=l1Λ{[Vk−1(n1, n2, 1)− Vk−1(n1 − 1, n2 + 1, 0)]− [Vk−1(n1 + 1, n2, 1)− Vk−1(n1, n2 + 1, 0)]}

+l1Λ{[Vk−1(n1 − 1, n2 + 1, 0)− Vk−1(n1 − 1, n2 + 1, 1)]− [Vk−1(n1, n2 + 1, 0)− Vk−1(n1, n2 + 1, 1)]}

+µ2Λ{[Vk−1(n1, n2 − 1, 1)− Vk−1(n1 − 1, n2, 0)]− [Vk−1(n1 + 1, n2 − 1, 1)− Vk−1(n1, n2, 0)]}

+µ2Λ{[Vk−1(n1 − 1, n2, 0)− Vk−1(n1 − 1, n2 + 1, 0)]− [Vk−1(n1, n2, 0)− Vk−1(n1, n2 + 1, 0)]}.

According to D2,

[Vk−1(n1, n2, 1)− Vk−1(n1 − 1, n2 + 1, 0)]− [Vk−1(n1 + 1, n2, 1)− Vk−1(n1, n2 + 1, 0)] ≤ 0,

38

[Vk−1(n1, n2 − 1, 1)− Vk−1(n1 − 1, n2, 0)]− [Vk−1(n1 + 1, n2 − 1, 1)− Vk−1(n1, n2, 0)] ≤ 0.Hence,

β ≤ l1Λ{[Vk−1(n1 − 1, n2 + 1, 0)− Vk−1(n1 − 1, n2 + 1, 1)]− [Vk−1(n1, n2 + 1, 0)− Vk−1(n1, n2 + 1, 1)]}

+µ2Λ{[Vk−1(n1 − 1, n2, 0)− Vk−1(n1 − 1, n2 + 1, 0)]− [Vk−1(n1, n2, 0)− Vk−1(n1, n2 + 1, 0)]}.

According to induction assumptionW2, the above right hand side is nonpositive. This completes the prooffor D2-Case II-1.

D2-Case II-2: When serving station 2 and station 1 are optimal, respectively, at states (n1 − 1, n2 + 1, 0)and (n1, n2 + 1, 0), then


+µ2Λ{[Vk−1(n1, n2 − 1, 1)− Vk−1(n1 − 1, n2, 0)]− [Vk−1(n1 + 1, n2 − 1, 1)− Vk−1(n1, n2 + 1, 0)]}

=l1Λ{[Vk−1(n1, n2, 1)− Vk−1(n1 − 1, n2 + 1, 0)]− [Vk−1(n1 + 1, n2, 1)− Vk−1(n1, n2 + 1, 0)]}

+l1Λ[Vk−1(n1, n2 + 1, 1)− Vk−1(n1, n2 + 1, 0)]

+µ2Λ{[Vk−1(n1, n2 − 1, 1)− Vk−1(n1 − 1, n2, 0)]− [Vk−1(n1 + 1, n2 − 1, 1)− Vk−1(n1, n2, 0)]}

+µ2Λ[Vk−1(n1, n2 + 1, 0)− Vk−1(n1, n2, 0)].

According to D2,

[Vk−1(n1, n2, 1)− Vk−1(n1 − 1, n2 + 1, 0)]− [Vk−1(n1 + 1, n2, 1)− Vk−1(n1, n2 + 1, 0)] ≤ 0,

[Vk−1(n1, n2 − 1, 1)− Vk−1(n1 − 1, n2, 0)]− [Vk−1(n1 + 1, n2 − 1, 1)− Vk−1(n1, n2, 0)] ≤ 0.Hence,

β ≤ l1Λ[Vk−1(n1, n2 + 1, 1)− Vk−1(n1, n2 + 1, 0)] + µ2

Λ[Vk−1(n1, n2 + 1, 0)− Vk−1(n1, n2, 0)].

Since in D2-Case II-2 serving station 1 is optimal at state (n1, n2 + 1, 0), then

l1[Vk−1(n1, n2 + 1, 1)− Vk−1(n1, n2 + 1, 0)] + µ2[Vk−1(n1, n2 + 1, 0)− Vk−1(n1, n2, 0)] ≤ 0.

Therefore, β ≤ 0, which completes the proof for D2-Case II-2.D2-Case II-3: When serving station 2 is optimal at states (n1 − 1, n2 + 1, 0) and (n1, n2 + 1, 0), then


+µ2Λ{[Vk−1(n1, n2 − 1, 1)− Vk−1(n1 − 1, n2, 0)]− [Vk−1(n1 + 1, n2 − 1, 1)− Vk−1(n1, n2, 0)]}.

According to D2, both terms of β are nonpositive. Hence, β ≤ 0, which completes the proof for D2-CaseII-3.

Proof for M1: At iteration k, for n1 ≥ 1 and n2 ≥ 1, M1 is

M1 Vk(n1, n2, 0)− Vk(n1, n2, 1)− [Vk(n1 + 1, n2, 0)− Vk(n1 + 1, n2, 1)] ≤ 0.

39

On the other hand,

Vk(n1, n2, 0) =n1 + n2Λ

+λ

ΛVk−1(n1 + 1, n2, 0) +

µ1ΛVk−1(n1, n2, 0)

+1

Λmin


Vk(n1, n2, 1) =n1 + n2Λ

+λ

ΛVk−1(n1 + 1, n2, 1) +

µ1ΛVk−1(n1 − 1, n2 + 1, 0) + l1

ΛVk−1(n1, n2, 1)

+µ2ΛVk(n1, n2 − 1, 1)

Vk(n1 + 1, n2, 0) =n1 + n2 + 1

Λ+λ

ΛVk−1(n1 + 2, n2, 0) +

µ1ΛVk−1(n1 + 1, n2, 0)

+1

Λmin


Vk(n1 + 1, n2, 1) =n1 + n2 + 1

Λ+λ

ΛVk−1(n1 + 2, n2, 1) +

µ1ΛVk−1(n1, n2 + 1, 0) +

l1ΛVk−1(n1 + 1, n2, 1)

+µ2ΛVk(n1 + 1, n2 − 1, 1).

Hence,

Vk(n1, n2, 0)− Vk(n1, n2, 1)− Vk(n1 + 1, n2, 0) + Vk(n1 + 1, n2, 1)=

λ

Λ[Vk(n1 + 1, n2, 0)− Vk(n1 + 1, n2, 1)− Vk(n1 + 2, n2, 0) + Vk(n1 + 2, n2, 1)]

+µ1Λ[Vk(n1, n2, 0)− Vk(n1 − 1, n2 + 1, 0)− Vk(n1 + 1, n2, 0) + Vk(n1, n2 + 1, 0)]

+1

Λmin


− l1ΛVk−1(n1, n2, 1)− µ2

ΛVk(n1, n2 − 1, 1)

− 1Λmin


+l1ΛVk−1(n1 + 1, n2, 1) +

µ2ΛVk(n1 + 1, n2 − 1, 1). (15)

According to induction assumption M1, the Þrst term on the right-hand side of (15) is nonpositive.Furthermore, according to M1 and D2, the second term is also nonpositive. Therefore, we only need toshow that the remaining terms on the right-hand side of (15) are nonpositive. We deÞne the remaining termsas ², and we discuss the following three cases:

Case Vk(n1, n2, 0) Vk(n1 + 1, n2, 0)

1 Serve 1 Serve 1

2 Serve 2 Serve 1

3 Serve 2 Serve 2

² ≤ 0: Case 1.Λ× ² = µ2[Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 1)− Vk−1(n1 + 1, n2, 0) + Vk−1(n1 + 1, n2 − 1, 1)]

= µ2[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)− Vk−1(n1 + 1, n2, 0) + Vk−1(n1 + 1, n2, 1)]+µ2[Vk−1(n1, n2, 1)− Vk−1(n1, n2 − 1, 1)− Vk−1(n1 + 1, n2, 1) + Vk−1(n1 + 1, n2 − 1, 1)].

According to induction assumptionM1 and M3, both terms are nonpositive. Therefore, ² ≤ 0.

40

² ≤ 0: Case 2.Λ× ² = {µ2[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)] + l1[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)]}

+µ2[Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 1)− Vk−1(n1 + 1, n2, 0) + Vk−1(n1 + 1, n2 − 1, 1)].Since in case 2 it is optimal to serve station 2 at state (n1, n2, 0), the Þrst term is nonpositive.The second term is the same as what we had in Case 1; therefore, according to M1 and M3, it isnonpositive. Thus, ² ≤ 0.

² ≤ 0: Case 3.Λ× ² = l1[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)]− [Vk−1(n1 + 1, n2, 0) + Vk−1(n1 + 1, n2, 1)]

+µ2[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2 − 1, 1)− Vk−1(n1 + 1, n2 − 1, 0) + Vk−1(n1 + 1, n2 − 1, 1)]= {l1[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)]− [Vk−1(n1 + 1, n2, 0) + Vk−1(n1 + 1, n2, 1)]

+µ2[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)− Vk−1(n1 + 1, n2 − 1, 0) + Vk−1(n1 + 1, n2, 0)]}+µ2[Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 1)− Vk−1(n1 + 1, n2, 0) + Vk−1(n1 + 1, n2 − 1, 1)].

According toW2, the Þrst term is nonpositive. Furthermore, according to M1 and M3, the secondterm is also nonpositive. Therefore, ² ≤ 0.

This completes the proof for M1.

Proof for M2: At iteration k, for n1 ≥ 0 and n2 ≥ 1,M2 Vk(n1, n2, 0)− Vk(n1, n2 − 1, 0)− [Vk(n1 + 1, n2, 0)− Vk(n1 + 1, n2 − 1, 0)] ≤ 0.

We Þrst show the boundary case, and then discuss the general case.

M2-Case I: When n1 = 0 and n2 = 1, M2 becomes:

Vk(0, 1, 0)− Vk(0, 0, 0)− [Vk(1, 1, 0)− Vk(1, 0, 0)] ≤ 0.We have,

Vk(0, 1, 0) =1

Λ+λ

ΛVk−1(1, 1, 0) +

µ1 + l1Λ

Vk−1(0, 1, 0) +µ2ΛVk−1(0, 0, 0),

Vk(0, 0, 0) =λ

ΛVk−1(1, 0, 0) +

µ1 + l1 + µ2Λ

Vk−1(0, 0, 0),

Vk(1, 1, 0) =2

Λ+λ

ΛVk−1(2, 1, 0) +

µ1ΛVk−1(1, 1, 0) +

1

Λmin

½l1Vk−1(1, 1, 1) + µ2Vk−1(1, 1, 0)µ2Vk−1(1, 0, 0) + l1Vk−1(1, 1, 0),

Vk(1, 0, 0) =1

Λ+λ

ΛVk−1(2, 0, 0) +

µ1 + µ2Λ

Vk−1(1, 0, 0) +l1ΛVk−1(1, 0, 1).

Hence,

Vk(0, 1, 0)− Vk(0, 0, 0)− Vk(1, 1, 0) + Vk(1, 0, 0) (16)

=λ

Λ[Vk(1, 1, 0)− Vk(1, 0, 0)− Vk(2, 1, 0) + Vk(2, 0, 0)]

+µ1Λ[Vk(0, 1, 0)− Vk(0, 0, 0)− Vk(1, 1, 0) + Vk(1, 0, 0)]

+l1Λ[Vk(0, 1, 0)− Vk(0, 0, 0) + Vk(1, 0, 1)] + µ2

ΛVk−1(1, 0, 0)

− 1Λmin

½l1Vk−1(1, 1, 1) + µ2Vk−1(1, 1, 0)µ2Vk−1(1, 0, 0) + l1Vk−1(1, 1, 0).

41

According to M2, the Þrst and the second terms on the right-hand side are nonpositive; therefore weonly need to show that the remaining terms, which we deÞne as ζ, are nonpositive. We consider two cases:

ζ ≤ 0: Case 1. When l1Vk−1(1, 1, 1) + µ2Vk−1(1, 1, 0) ≤ µ2Vk−1(1, 0, 0) + l1Vk−1(1, 1, 0),

ζ =l1Λ[Vk(0, 1, 0)− Vk(0, 0, 0)− Vk(1, 1, 1) + Vk(1, 0, 1)] + µ2

Λ[Vk(1, 0, 0)− Vk(1, 1, 0)]

=l1Λ[Vk(0, 1, 0)− Vk(0, 0, 0)− Vk(1, 1, 0) + Vk(1, 0, 0)]

+{ l1Λ[Vk(1, 1, 0)− Vk(1, 1, 1)− Vk(1, 0, 0) + Vk(1, 0, 1)] + µ2

Λ[Vk(1, 0, 0)− Vk(1, 1, 0)]}.

According toM2, the Þrst term on the right-hand side is nonpositive. Furthermore, according toM4,the second term is also nonpositive. Therefore, ζ ≤ 0.

ζ ≤ 0: Case 2. When µ2Vk−1(1, 0, 0) + l1Vk−1(1, 1, 0) ≤ l1Vk−1(1, 1, 1) + µ2Vk−1(1, 1, 0),

ζ =l1Λ[Vk(0, 1, 0)− Vk(0, 0, 0)− Vk(1, 1, 0) + Vk(1, 0, 1)]

=l1Λ[Vk(0, 1, 0)− Vk(0, 0, 0)− Vk(1, 1, 0) + Vk(1, 0, 0)] + l1

Λ[Vk(1, 0, 1)− Vk(1, 0, 0)].

According toM2, the Þrst term is nonpositive, and according to V2, the second term is nonpositive.Therefore, ζ ≤ 0.

This completes the proof for M2-Case I.

M2-Case II: For all other cases where (n1 = 0, n2 ≥ 1) or (n1 ≥ 1, n2 ≥ 1), we need to show that

Vk(n1, n2, 0)− Vk(n1, n2 − 1, 0)− [Vk(n1 + 1, n2, 0)− Vk(n1 + 1, n2 − 1, 0)] ≤ 0.

We have,

Vk(n1, n2, 0) =n1 + n2Λ

+λ

ΛVk−1(n1 + 1, n2, 0) +

µ1ΛVk−1(n1, n2, 0)

+1

Λmin


Vk(n1, n2 − 1, 0) =n1 + n2 − 1

Λ+λ

ΛVk−1(n1 + 1, n2 − 1, 0) + µ1

ΛVk−1(n1, n2 − 1, 0)

+1

Λmin


Vk(n1 + 1, n2, 0) =n1 + n2 + 1

Λ+λ

ΛVk−1(n1 + 2, n2, 0) +

µ1ΛVk−1(n1 + 1, n2, 0)

+1

Λmin


Vk(n1 + 1, n2 − 1, 0) =n1 + n2Λ

+λ

ΛVk−1(n1 + 2, n2 − 1, 0) + µ1

ΛVk−1(n1 + 1, n2 − 1, 0)

+1

Λmin

½l1Vk−1(n1 + 1, n2 − 1, 1) + µ2Vk−1(n1 + 1, n2 − 1, 0)µ2Vk−1(n1 + 1, n2 − 2, 0) + l1Vk−1(n1 + 1, n2 − 1, 0).

42

Hence,

Vk(n1, n2, 0)− Vk(n1, n2 − 1, 0)− Vk(n1 + 1, n2, 0) + Vk(n1 + 1, n2 − 1, 0)=

λ

Λ[Vk(n1 + 1, n2, 0)− Vk(n1 + 1, n2 − 1, 0)− Vk(n1 + 2, n2, 0) + Vk(n1 + 2, n2 − 1, 0)]

+µ1Λ[Vk(n1, n2, 0)− Vk(n1, n2 − 1, 0)− Vk(n1 + 1, n2, 0) + Vk(n1 + 1, n2 − 1, 0)]

+1

Λmin


− 1Λmin


− 1Λmin


+1

Λmin

½l1Vk−1(n1 + 1, n2 − 1, 1) + µ2Vk−1(n1 + 1, n2 − 1, 0)µ2Vk−1(n1 + 1, n2 − 2, 0) + l1Vk−1(n1 + 1, n2 − 1, 0).

According to induction assumption M2, the Þrst and the second terms are nonpositive, so we only need toshow that the remaining terms, which we deÞne as χ, is nonpositive. We discuss the following cases:

Case Vk(n1, n2, 0) Vk(n1, n2 − 1, 0) Vk(n1 + 1, n2, 0) Vk(n1 + 1, n2 − 1, 0)1 Serve 1 Serve 1 Serve 1 Serve 1






χ ≤ 0: Case 1.Λ× χ = l1[Vk−1(n1, n2, 1)− Vk−1(n1, n2 − 1, 1)− Vk−1(n1 + 1, n2, 1) + Vk−1(n1 + 1, n2 − 1, 1)]

+µ2[Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 0)− Vk−1(n1 + 1, n2, 0) + Vk−1(n1 + 1, n2 − 1, 0)].According to induction assumptionM3 and M2, both terms are nonpositive. Therefore, χ ≤ 0.


+µ2[Vk−1(n1 + 1, n2 − 1, 0)− Vk−1(n1 + 1, n2, 0)].= l1[Vk−1(n1, n2, 1)− Vk−1(n1, n2 − 1, 1)− Vk−1(n1 + 1, n2, 1) + Vk−1(n1 + 1, n2 − 1, 1)]

+l1[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)] + µ2[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)]+µ2[Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 0)− Vk−1(n1 + 1, n2, 0) + Vk−1(n1 + 1, n2 − 1, 0)].

According to induction assumption M3 and M2, the Þrst and the third terms are nonpositive. Fur-thermore, since in Case 2 it is optimal to serve station 2 at state (n1, n2, 0), the second term is alsononpositive. Therefore, χ ≤ 0.


= l1[Vk−1(n1, n2, 1)− Vk−1(n1, n2 − 1, 1)− Vk−1(n1 + 1, n2, 1) + Vk−1(n1 + 1, n2 − 1, 1)]+l1[Vk−1(n1, n2, 0)− Vk−1(n1, n2, 1)− Vk−1(n1 + 1, n2, 0) + Vk−1(n1 + 1, n2, 1)].

According to M3, the Þrst term is nonpositive. Furthermore, according to M1, the second term isnonpositive. Therefore, χ ≤ 0.

43


+µ2[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2 − 2, 0)− Vk−1(n1 + 1, n2, 0) + Vk−1(n1 + 1, n2 − 1, 0)]= l1[Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 0)− Vk−1(n1 + 1, n2, 0) + Vk−1(n1 + 1, n2 − 1, 0)]

+{l1[Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2, 1)− Vk−1(n1 + 1, n2 − 1, 0)− Vk−1(n1 + 1, n2 − 1, 1)]+µ2[Vk−1(n1 + 1, n2 − 1, 0)− Vk−1(n1 + 1, n2, 0)− Vk−1(n1 + 1, n2 − 2, 0) + Vk−1(n1 + 1, n2 − 1, 0)]}+µ2[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2 − 2, 0)− Vk−1(n1 + 1, n2 − 1, 0) + Vk−1(n1 + 1, n2 − 2, 0)].

According to induction assumption M2, the Þrst and the third terms are nonpositive. Furthermore,according to induction assumptionW1, the second term is also nonpositive. Therefore, χ ≤ 0.


+µ2[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2 − 2, 0)]= l1[Vk−1(n1, n2, 0)− Vk−1(n1, n2 − 1, 0)− Vk−1(n1 + 1, n2, 0) + Vk−1(n1 + 1, n2 − 1, 0)]

+{l1[Vk−1(n1 + 1, n2 − 1, 1)− Vk−1(n1 + 1, n2 − 1, 0)]+ µ2[Vk−1(n1 + 1, n2 − 1, 0)− Vk−1(n1 + 1, n2 − 2, 0)]}

+µ2[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2 − 2, 0)− Vk−1(n1 + 1, n2 − 1, 0) + Vk−1(n1 + 1, n2 − 2, 0)].According to induction assumption M2, the Þrst and the third terms are nonpositive. Furthermore,since in Case 5, it is optimal to serve station 1 at (n1+1, n2−1, 0), the second term is also nonpositive.Therefore, χ ≤ 0.


+µ2[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2 − 2, 0)− Vk−1(n1 + 1, n2 − 1, 0) + Vk−1(n1 + 1, n2 − 2, 0)].According to induction assumptionM2, both terms are nonpositive. Therefore, χ ≤ 0.

This completes the proof for M2-Case II.

Proof for M3: We must prove that for n1 ≥ 1 and n2 ≥ 1,[Vk(n1, n2, 1)− Vk(n1, n2 − 1, 1)]− [Vk(n1 + 1, n2, 1)− Vk(n1 + 1, n2 − 1, 1)] ≤ 0.

We Þrst discuss the case when n1 ≥ 1 and n2 = 1, then we focus on the case when n1 ≥ 1 and n2 ≥ 2.M3-Case I: When n1 ≥ 1 and n2 = 1, M3 becomes

[Vk(n1, 1, 1)− Vk(n1, 0, 1)]− [Vk(n1 + 1, 1, 1)− Vk(n1 + 1, 0, 1)] ≤ 0.Since

Vk(n1, 1, 1) =n1 + 1

Λ+λ

ΛVk−1(n1 + 1, 1, 1) +

µ1ΛVk−1(n1 − 1, 2, 0) + l1

ΛVk−1(n1, 1, 1) +

µ2ΛVk−1(n1, 0, 1),

Vk(n1, 0, 1) =n1Λ+λ

ΛVk−1(n1 + 1, 0, 1) +

µ1ΛVk−1(n1 − 1, 1, 0) + l1 + µ2

ΛVk−1(n1, 0, 1),

Vk(n1 + 1, 1, 1) =n1 + 2

Λ+λ

ΛVk−1(n1 + 2, 1, 1) +

µ1ΛVk−1(n1, 2, 0) +

l1ΛVk−1(n1 + 1, 1, 1) +

µ2ΛVk−1(n1 + 1, 0, 1),

44

Vk(n1 + 1, 0, 1) =n1 + 1

Λ+λ

ΛVk−1(n1 + 2, 0, 1) +

µ1ΛVk−1(n1, 1, 0) +

l1 + µ2Λ

Vk−1(n1 + 1, 0, 1),

then, we will have

Vk(n1, 1, 1)− Vk(n1, 0, 1)− Vk(n1 + 1, 1, 1) + Vk(n1 + 1, 0, 1)=

λ

Λ[Vk(n1 + 1, 1, 1)− Vk(n1 + 1, 0, 1)− Vk(n1 + 2, 1, 1) + Vk(n1 + 2, 0, 1)]

+l1Λ[Vk(n1, 1, 1)− Vk(n1, 0, 1)− Vk(n1 + 1, 1, 1) + Vk(n1 + 1, 0, 1)]

+µ1Λ[Vk(n1 − 1, 2, 0)− Vk(n1 − 1, 1, 0)− Vk(n1, 2, 0) + Vk(n1, 1, 0)].

According to M3, the Þrst and the second terms are nonpositive. Furthermore, according to M2, the thirdterm is also nonpositive. Therefore, the proof for M3-Case I is completed.

M3-Case II: When n1 ≥ 1 and n2 ≥ 2, we have

Vk(n1, n2, 1)− Vk(n1, n2 − 1, 1)− Vk(n1 + 1, n2, 1) + Vk(n1 + 1, n2 − 1, 1)=

λ

Λ[Vk(n1 + 1, n2, 1)− Vk(n1 + 1, n2 − 1, 1)− Vk(n1 + 2, n2, 1) + Vk(n1 + 2, n2 − 1, 1)]

+l1Λ[Vk(n1, n2, 1)− Vk(n1, n2 − 1, 1)− Vk(n1 + 1, n2, 1) + Vk(n1 + 1, n2 − 1, 1)]

+µ1Λ[Vk(n1 − 1, n2 + 1, 0)− Vk(n1 − 1, n2, 0)− Vk(n1, n2 + 1, 0) + Vk(n1, n2, 0)]

+µ2Λ[Vk(n1, n2 − 1, 1)− Vk(n1, n2 − 2, 1)− Vk(n1 + 1, n2 − 1, 1) + Vk(n1 + 1, n2 − 2, 1)].

According toM3, the Þrst, the second, and the forth terms are nonpositive. The third term is also nonpositiveaccording to M2. This completes the proof for M3-Case II.

Proof for M4: We must prove that for n1 ≥ 1,

l1[Vk(n1, 1, 0)− Vk(n1, 1, 1)− Vk(n1, 0, 0) + Vk(n1, 0, 1)] + µ2[Vk(n1, 0, 0)− Vk(n1, 1, 0)] ≤ 0.

On the other hand, we have

Vk(n1, 1, 0) =n1 + 1

Λ+λ

ΛVk−1(n1 + 1, 1, 0) +

µ1ΛVk−1(n1, 1, 0)

+1

Λmin

½l1Vk−1(n1, 1, 1) + µ2Vk−1(n1, 1, 0)µ2Vk−1(n1, 0, 0) + l1Vk−1(n1, 1, 0),

Vk(n1, 1, 1) =n1 + 1

Λ+λ

ΛVk−1(n1 + 1, 1, 1) +

µ1ΛVk−1(n1 − 1, 2, 0) + l1

ΛVk−1(n1, 1, 1) +

µ2ΛVk−1(n1, 0, 1),

Vk(n1, 0, 0) =n1Λ+λ

ΛVk−1(n1 + 1, 0, 0) +

µ1 + µ2Λ

Vk−1(n1, 0, 0) +l1ΛVk−1(n1, 0, 1),

Vk(n1, 0, 1) =n1Λ+λ

ΛVk−1(n1 + 1, 0, 1) +

µ1ΛVk−1(n1 − 1, 1, 0) + l1 + µ2

ΛVk−1(n1, 0, 1).

Hence,

l1[Vk(n1, 1, 0)− Vk(n1, 1, 1)− Vk(n1, 0, 0) + Vk(n1, 0, 1)] + µ2[Vk(n1, 0, 0)− Vk(n1, 1, 0)]

45

=λ

Λ{l1[Vk(n1 + 1, 1, 0)− Vk(n1 + 1, 1, 1)− Vk(n1 + 1, 0, 0) + Vk(n1 + 1, 0, 1)] + µ2[Vk(n1 + 1, 0, 0)− Vk(n1 + 1, 1, 0)]}

+µ1Λ{[Vk(n1, 1, 0)− Vk(n1 − 1, 2, 0)− Vk(n1, 0, 0) + Vk(n1 − 1, 1, 0)] + µ2[Vk(n1, 0, 0)− Vk(n1, 1, 0)]}

+l1Λ{min


− l1Vk−1(n1, 1, 1)− µ2Vk−1(n1, 0, 0)}

+µ2Λ{l1Vk−1(n1, 0, 1) + µ2Vk−1(n1, 0, 0)−min


}.

According toM4, the Þrst term in nonpositive. Furthermore, according toM4 and D1, the second term isalso nonpositive. Hence, we only need to show that the remaining terms are nonpositive. We deÞne themas τ , and we discuss the following two cases:

τ ≤ 0: Case 1. When it is optimal to serve station 1 at state (n1, 1, 0),

Λ× τ = l1[l1Vk(n1, 1, 1) + µ2Vk(n1, 1, 0)− l1Vk(n1, 1, 1)− µ2Vk(n1, 0, 0)]+µ2[l1Vk(n1, 0, 1) + µ2Vk(n1, 0, 0)− l1Vk(n1, 1, 1) + µ2Vk(n1, 1, 0)]

= l1µ2[Vk(n1, 1, 0)− Vk(n1, 0, 0)− Vk(n1, 1, 1) + Vk(n1, 0, 1)]+µ2µ2[Vk(n1, 0, 0)− Vk(n1, 1, 0)].

This, according to M4, is nonpositive. Therefore, the proof for M4-Case I is completed.

τ ≤ 0: Case 2. When it is optimal to serve station 2 at state (n1, 1, 0), then

Λ × τ = l1[l1Vk(n1, 1, 0) + µ2Vk(n1, 0, 0)− l1Vk(n1, 1, 1)− µ2Vk(n1, 0, 0)]+µ2[l1Vk(n1, 0, 1)− l1Vk(n1, 1, 0)]

= l1{[l1[Vk(n1, 1, 0)− Vk(n1, 1, 1)] + µ2[Vk(n1, 0, 0)− Vk(n1, 1, 0)]}+l1µ2[Vk(n1, 0, 1)− Vk(n1, 0, 0)].

Since it is optimal to serve station 2 at state (n1, 1, 0), the Þrst term is nonpositive. Also, accordingto V2, the second term is nonpositive. Therefore, the proof for M4-Case II is completed.


We use induction and value iteration algorithm to prove conditionsA1, X1, X2, X3 andX4. The optimalityequations for the value iteration algorithm are:

Vk(n1, n2, 0) =n1 + n2Λ

+λ

ΛVk−1(n1 + 1, n2, 0) +

µ2ΛVk−1(n1, n2, 0)

+1

Λmin

(µ1 + l2)Vk−1(n1, n2, 0)µ1Vk−1(n1 − 1, n2 + 1, 0) + l2Vk−1(n1, n2, 0)µ1Vk−1(n1, n2, 0) + l2Vk−1(n1, n2, 1),

Vk(n1, n2, 1) =n1 + n2Λ

+λ

ΛVk−1(n1 + 1, n2, 1) +

l2ΛVk−1(n1, n2, 1) +

µ2ΛVk−1(n1, n2 − 1, 0)

+µ1Λmin

½Vk−1(n1, n2, 1)Vk−1(n1 − 1, n2 + 1, 1).

46

(X1,X2,X3,X4,A1) Iteration 0: At iteration 0, since all V0(n1, n2, s) = 0 for all (n1, n2, s),X1,X2,X3,X4,A1hold at iteration 0.

(X1,X2,X3,X4,A1) Iteration k-1: Assume at iteration k − 1, X1,X2,X3,X4,A1 hold. That is,X1 Vk−1(n1, n2, 1) ≤ Vk−1(n1, n2, 0) for n1 ≥ 0, n2 ≥ 1X2 Vk−1(n1 − 1, n2 + 1, 0) ≤ Vk−1(n1, n2, 0) for n1 ≥ 1, n2 ≥ 0X3 Vk−1(n1 − 1, n2 + 1, 1) ≤ Vk−1(n1, n2, 1) for n1 ≥ 1, n2 ≥ 1X4 l2Vk−1(n1, n2, 1) + µ1Vk−1(n1, n2, 0) ≤ l2Vk−1(n1, n2, 0) + µ1Vk−1(n1 − 1, n2 + 1, 0) for n1 ≥ 1, n2 ≥ 1A1 Vk−1(n1, n2 − 1, 0) ≤ Vk−1(n1, n2, 1) for n1 ≥ 0, n2 ≥ 1.

(X1,X2,X3,X4,A1) Iteration k: Now, we show that (X1,X2,X3,X4,A1) also hold at iteration k.

Proof for X1: Vk(n1, n2, 1) ≤ Vk(n1, n2, 0) for n1 ≥ 0 and n2 ≥ 1.We discuss the following two cases:

� Case 1: n1 = 0 and n2 ≥ 1,� Case 2: n1 ≥ 1 and n2 ≥ 1.

X1-Case 1: Since at states (0, n2, 1) and (0, n2, 0), there�s no job at the Þrst station, processing a job atstation 1 is not feasible. Hence,

Vk(0, n2, 1) =n2Λ+λ

ΛVk−1(1, n2, 1) +

µ1 + l2Λ

Vk−1(0, n2, 1) +µ2ΛVk−1(0, n2 − 1, 0)

Vk(0, n2, 0) =n2Λ+λ

ΛVk−1(1, n2, 0) +

µ1 + µ2Λ

Vk−1(0, n2, 0) +l2Λmin{Vk−1(0, n2, 0) , Vk−1(0, n2, 1)}.

Therefore, we have

Vk(0, n2, 1)− Vk(0, n2, 0) =λ

Λ[Vk−1(1, n2, 1)− Vk−1(1, n2, 0)] + µ1

Λ[Vk−1(0, n2, 1)− Vk−1(0, n2, 0)]

+µ2Λ[Vk−1(0, n2 − 1, 0)− Vk−1(0, n2, 0)]

+l2Λ[Vk−1(0, n2, 1)−min{Vk−1(0, n2, 0) , Vk−1(0, n2, 1)}]. (17)

According to induction assumption X1, the Þrst and the second terms on the right-hand side of (17) arenonpositive. Also, due to X1, we have

min{Vk−1(0, n2, 0) , Vk−1(0, n2, 1)} = Vk−1(0, n2, 1),which implies that the forth terms in (17) is zero.

Finally, according to induction assumption A1 and X1, we have

Vk−1(0, n2 − 1, 0) ≤ Vk−1(0, n2, 1) ≤ Vk−1(0, n2, 0).Therefore, the third term in (17) is nonpositive. Since all terms on the right-hand side of (17) are nonpositive,the proof for X1-Case 1 is completed .

X1-Case 2: When n1 ≥ 1 and n2 ≥ 1, we have

Vk(n1, n2, 1) =n1 + n2Λ

+λ

ΛVk−1(n1 + 1, n2, 1) +

l2ΛVk−1(n1, n2, 1) +

µ2ΛVk−1(n1, n2 − 1, 0)

+µ1Λmin{Vk−1(n1, n2, 1) , Vk−1(n1 − 1, n2 + 1, 1)}

47

Vk(n1, n2, 0) =n1 + n2Λ

+λ

ΛVk−1(n1 + 1, n2, 0) +

µ2ΛVk−1(n1, n2, 0)

+1

Λmin

(µ1 + l2)Vk−1(n1, n2, 0)µ1Vk−1(n1 − 1, n2 + 1, 0) + l2Vk−1(n1, n2, 0)µ1Vk−1(n1, n2, 0) + l2Vk−1(n1, n2, 1).

Hence,

Vk(n1, n2, 1)− Vk(n1, n2, 0) =λ

Λ[Vk−1(n1 + 1, n2, 1)− Vk−1(n1 + 1, n2, 0)]

+µ2Λ[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)]

+l2ΛVk−1(n1, n2, 1) +

µ1Λmin{Vk−1(n1, n2, 1) , Vk−1(n1 − 1, n2 + 1, 1)}

− 1Λmin


(18)

According to induction assumption X1, the Þrst term on the right-hand side of (18) is nonpositive. Also,according to induction assumption A1 and X1, we have

Vk−1(n1, n2 − 1, 0) ≤ Vk−1(n1, n2, 1) ≤ Vk−1(n1, n2, 0).

Therefore, the second term in (18) is nonpositive. Furthermore, due to X1, X2, X3 , and X4

min{Vk−1(n1, n2, 1) , Vk−1(n1 − 1, n2 + 1, 1)} = Vk−1(n1 − 1, n2 + 1, 1),

min

(µ1 + l2)Vk−1(n1, n2, 0)µ1Vk−1(n1 − 1, n2 + 1, 0) + l2Vk−1(n1, n2, 0)µ1Vk−1(n1, n2, 0) + l2Vk−1(n1, n2, 1)

= µ1Vk−1(n1, n2, 0) + l2Vk−1(n1, n2, 1),

which imply that the remaining terms in (18) are equal to:

µ1Λ[Vk−1(n1 − 1, n2 + 1, 1)− Vk−1(n1, n2, 0)].

On the other hand, according to X3 and X1, we have

Vk−1(n1 − 1, n2 + 1, 1) ≤ Vk−1(n1, n2, 1) ≤ Vk−1(n1, n2, 0).

Therefore, all terms on the right-hand side of (18) are nonpositive. This completes the proof for X1-Case2.

Proof for X2: Vk(n1 − 1, n2 + 1, 0) ≤ Vk(n1, n2, 0) for n1 ≥ 1, n2 ≥ 0We discuss the following two cases:

� Case 1: n1 ≥ 1 and n2 = 0,� Case 2: n1 ≥ 1 and n2 ≥ 1.

X2-Case 1: When n1 ≥ 1 and n2 = 0, X2 becomes Vk(n1 − 1, 1, 0) ≤ Vk(n1, 0, 0). We have

Vk(n1 − 1, 1, 0) =n1 − 1+ 1

Λ+λ

ΛVk−1(n1, 1, 0) +

µ2ΛVk−1(n1 − 1, 1, 0)

+1

Λmin

(µ1 + l2)Vk−1(n1 − 1, 1, 0)µ1Vk−1(n1 − 2, 2, 0) + l2Vk−1(n1 − 1, 1, 0)µ1Vk−1(n1 − 1, 1, 0) + l2Vk−1(n1 − 1, 1, 1).

48

On the other hand, since at state (n1, 0, 0) there�s no job at the second station, loading station 2 is notfeasible. Hence,

Vk(n1, 0, 0) =n1Λ+λ

ΛVk−1(n1 + 1, 0, 0) +

l2 + µ2Λ

Vk−1(n1, 0, 0)

+µ1Λmin{Vk−1(n1, 0, 0) , Vk−1(n1 − 1, 1, 0)}.

Therefore, we have

Vk(n1 − 1, 1, 0)− Vk(n1, 0, 0)=

λ

Λ[Vk−1(n1, 1, 0)− Vk−1(n1 + 1, 0, 0)] + µ2

Λ[Vk−1(n1 − 1, 1, 0)− Vk−1(n1, 0, 0)]

+1

Λmin

(µ1 + l2)Vk−1(n1 − 1, 1, 0)µ1Vk−1(n1 − 2, 2, 0) + l2Vk−1(n1 − 1, 1, 0)µ1Vk−1(n1 − 1, 1, 0) + l2Vk−1(n1 − 1, 1, 1)

− l2ΛVk−1(n1, 0, 0)− µ1

Λmin{Vk−1(n1, 0, 0), Vk−1(n1 − 1, 1, 0)}. (19)

According to induction assumption X2, the Þrst and the second terms on the right-hand side of (19) arenonpositive. Also, due to X1, X2, and X4 we have

min

(µ1 + l2)Vk−1(n1 − 1, 1, 0)µ1Vk−1(n1 − 2, 2, 0) + l2Vk−1(n1 − 1, 1, 0)µ1Vk−1(n1 − 1, 1, 0) + l2Vk−1(n1 − 1, 1, 1)

= µ1Vk−1(n1 − 1, 1, 0) + l2Vk−1(n1 − 1, 1, 1),

min{Vk−1(n1, 0, 0) , Vk−1(n1 − 1, 1, 0)} = Vk−1(n1 − 1, 1, 0),which imply that the remaining terms in (19) are equal to the following term:

l2Λ[Vk−1(n1 − 1, 1, 1)− Vk−1(n1, 0, 0)].

According to induction assumption of X1 and X2, we have

Vk−1(n1 − 1, 1, 1) ≤ Vk−1(n1 − 1, 1, 0) ≤ Vk−1(n1, 0, 0),

which implies that the remaining terms on the right-hand side of (19) are nonpositive. Since all terms in(19) are nonpositive, the proof for X2-Case 1 is completed .


Vk(n1 − 1, n2 + 1, 0) =n1 − 1+ n2 + 1

Λ+λ

ΛVk−1(n1, n2 + 1, 0) +

µ2ΛVk−1(n1 − 1, n2 + 1, 0)

+1

Λmin

(µ1 + l2)Vk−1(n1 − 1, n2 + 1, 0)µ1Vk−1(n1 − 2, n2 + 2, 0) + l2Vk−1(n1 − 1, n2 + 1, 0)µ1Vk−1(n1 − 1, n2 + 1, 0) + l2Vk−1(n1 − 1, n2 + 1, 1),

Vk(n1, n2, 0) =n1 + n2Λ

+λ

ΛVk−1(n1 + 1, n2, 0) +

µ2ΛVk−1(n1, n2, 0)

+1

Λmin


49

Hence,

Vk(n1 − 1, n2 + 1, 0)− Vk(n1, n2, 0)=

λ

Λ[Vk−1(n1, n2 + 1, 0)− Vk−1(n1 + 1, n2, 0)] + µ2

Λ[Vk−1(n1 − 1, n2 + 1, 0)− Vk−1(n1, n2, 0)]

+1

Λmin

(µ1 + l2)Vk−1(n1 − 1, n2 + 1, 0)µ1Vk−1(n1 − 2, n2 + 2, 0) + l2Vk−1(n1 − 1, n2 + 1, 0)µ1Vk−1(n1 − 1, n2 + 1, 0) + l2Vk−1(n1 − 1, n2 + 1, 1).

− 1Λmin


(20)

According to induction assumption X2, the Þrst and the second terms on the right-hand side of (20) arenonpositive. Furthermore, due to X1, X2 and X4, we have

min

(µ1 + l2)Vk−1(n1 − 1, n2 + 1, 0)µ1Vk−1(n1 − 2, n2 + 2, 0) + l2Vk−1(n1 − 1, n2 + 1, 0)µ1Vk−1(n1 − 1, n2 + 1, 0) + l2Vk−1(n1 − 1, n2 + 1, 1)

= µ1Vk−1(n1−1, n2+1, 0)+l2Vk−1(n1−1, n2+1, 1),

min


= µ1Vk−1(n1, n2, 0) + l2Vk−1(n1, n2, 1),

which imply that the last two terms on the right-hand side of (20) are equal to:

µ1Λ[Vk−1(n1 − 1, n2 + 1, 0)− Vk−1(n1, n2, 0)] + l2

Λ[Vk−1(n1 − 1, n2 + 1, 1)− Vk−1(n1, n2, 1)]. (21)

According to the induction assumption of X2 and X3, both terms in (21) are nonpositive. Therefore, allterms in (20) are nonpositive. This completes the proof for X2-Case 2.

Proof for X3: Vk(n1 − 1, n2 + 1, 1) ≤ Vk(n1, n2, 1) for n1 ≥ 1, n2 ≥ 1.We discuss the following two cases:

� Case 1: n1 = 1 and n2 ≥ 1,� Case 2: n1 ≥ 2 and n2 ≥ 1.

X3-Case 1:When n1 = 1 and n2 ≥ 1, X3 becomes Vk(0, n2+1, 1) ≤ Vk(1, n2, 1). Since at state (0, n2+1, 1),there�s no job at the Þrst station and the second station is under automatic processing, the worker can onlyidle. Therefore,

Vk(0, n2 + 1, 1) =n2 + 1

Λ+λ

ΛVk−1(1, n2 + 1, 1) +

µ2ΛVk−1(0, n2, 0) +

µ1 + l2Λ

Vk−1(0, n2 + 1, 1)

Vk(1, n2, 1) =1+ n2Λ

+λ

ΛVk−1(2, n2, 1) +

µ2ΛVk−1(1, n2 − 1, 0) + l2

ΛVk−1(1, n2, 1)

+µ1Λmin{Vk−1(1, n2, 1) , Vk−1(0, n2 + 1, 1)}.

Thus, we have

Vk(0, n2 + 1, 1)− Vk(1, n2, 1)=λ

Λ[Vk−1(1, n2 + 1, 1)− Vk−1(2, n2, 1)] + µ1

Λ[Vk−1(0, n2 + 1, 1)−min{Vk−1(1, n2, 1) , Vk−1(0, n2 + 1, 1)}]

+µ2Λ[Vk−1(0, n2, 0)− Vk−1(1, n2 − 1, 0)] + l2

Λ[Vk−1(0, n2 + 1, 1)− Vk−1(1, n2, 1)]. (22)

50

According to induction assumption X3, the Þrst and the forth terms on the right-hand side of (22) arenonpositive. Also, due to X3, we have:

min{Vk−1(1, n2, 1) , Vk−1(0, n2 + 1, 1)} = Vk−1(0, n2 + 1, 1),

which implies that the second term in (22) is zero. Finally, according to induction assumption X2, the thirdterm is nonpositive. Since all terms on the right-hand side of (22) are nonpositive, the proof for X3-Case1 is completed .


Vk(n1 − 1, n2 + 1, 1) =(n1 − 1) + (n2 + 1)

Λ+λ

ΛVk−1(n1, n2 + 1, 1) +

l2ΛVk−1(n1 − 1, n2 + 1, 1)

+µ2ΛVk−1(n1 − 1, n2, 0) + µ1

Λmin{Vk−1(n1 − 1, n2 + 1, 1) , Vk−1(n1 − 2, n2 + 2, 1)}

Vk(n1, n2, 1) =n1 + n2Λ

+λ

ΛVk−1(n1 + 1, n2, 1) +

µ2ΛVk−1(n1, n2 − 1, 0) + l2

ΛVk−1(n1, n2, 1)

+µ1Λmin{Vk−1(n1, n2, 1) , Vk−1(n1 − 1, n2 + 1, 1)}.

Hence,

Vk(n1 − 1, n2 + 1, 1)− Vk(n1, n2, 1)=

λ

Λ[Vk−1(n1, n2 + 1, 1)− Vk−1(n1 + 1, n2, 1)] + µ2

Λ[Vk−1(n1 − 1, n2, 0)− Vk−1(n1, n2 − 1, 0)]

+l2Λ[Vk−1(n1 − 1, n2 + 1, 1)− Vk(n1, n2, 1)]

+µ1Λ[min{Vk−1(n1 − 1, n2 + 1, 1) , Vk−1(n1 − 2, n2 + 2, 1)}−min{Vk−1(n1, n2, 1) , Vk−1(n1 − 1, n2 + 1, 1)}]. (23)

According to induction assumption X3, the Þrst and the third terms on the right-hand side of (23) arenonpositive. Also, due to X3, we have:

min{Vk−1(n1 − 1, n2 + 1, 1) , Vk−1(n1 − 2, n2 + 2, 1)} = Vk−1(n1 − 2, n2 + 2, 1),min{Vk−1(n1, n2, 1) , Vk−1(n1 − 1, n2 + 1, 1)} = Vk−1(n1 − 1, n2 + 1, 1),

which imply that the forth term on the right-hand side of (23) is equal to the nonpositive term:

µ1Λ[Vk−1(n1 − 2, n2 + 2, 1)− Vk−1(n1 − 1, n2 + 1, 1)].

Furthermore, due to X2, the second term on the right-hand side of (23) is nonpositive. Since all terms onthe right-hand side of (23) are nonpositive, the proof for X3-Case 2 is completed.

Proof for X4: We must prove that for n1 ≥ 1 and n2 ≥ 1,

l2[Vk(n1, n2, 1)− Vk(n1, n2, 0)] + µ1[Vk(n1, n2, 0)− Vk(n1 − 1, n2 + 1, 0)] ≤ 0.

At iteration k, we have

Vk(n1, n2, 1) =n1 + n2Λ

+λ

ΛVk−1(n1 + 1, n2, 1) +

l2ΛVk−1(n1, n2, 1) +

µ2ΛVk−1(n1, n2 − 1, 0)

+µ1Λmin{Vk−1(n1, n2, 1) , Vk−1(n1 − 1, n2 + 1, 1)}

51

Vk(n1, n2, 0) =n1 + n2Λ

+λ

ΛVk−1(n1 + 1, n2, 0) +

µ2ΛVk−1(n1, n2, 0)

+1

Λmin


Vk(n1 − 1, n2 + 1, 0) =(n1 − 1) + (n2 + 1)

Λ+λ

ΛVk−1(n1, n2 + 1, 0) +

µ2ΛVk−1(n1 − 1, n2 + 1, 0)

+1

Λmin


Hence,

Vk(n1, n2, 1)− Vk(n1, n2, 0) =λ

Λ[Vk−1(n1 + 1, n2, 1)− Vk−1(n1 + 1, n2, 0)]

+µ2Λ[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 0)]

+l2ΛVk−1(n1, n2, 1) +

µ1Λmin{Vk−1(n1, n2, 1) , Vk−1(n1 − 1, n2 + 1, 1)}

− 1Λmin


On the other hand,

Vk(n1, n2, 0)− Vk(n1 − 1, n2 + 1, 0) =λ

Λ[Vk−1(n1 + 1, n2, 0)− Vk−1(n1, n2 + 1, 0)]

+µ2Λ[Vk−1(n1, n2, 0)− Vk−1(n1 − 1, n2 + 1, 0)]

+1

Λmin


− 1Λmin


According to induction assumption of X1, X2 and X4, we have:

min


= µ1Vk−1(n1, n2, 0) + l2Vk−1(n1, n2, 1),

min

(µ1 + l2)Vk−1(n1 − 1, n2 + 1, 0)µ1Vk−1(n1 − 2, n2 + 2, 0) + l2Vk−1(n1 − 1, n2 + 1, 0)µ1Vk−1(n1 − 1, n2 + 1, 0) + l2Vk−1(n1 − 1, n2 + 1, 1)

= µ1Vk−1(n1−1, n2+1, 0)+l2Vk−1(n1−1, n2+1, 1).

Therefore,

l2[Vk(n1, n2, 1)− Vk(n1, n2, 0)] + µ1[Vk(n1, n2, 0)− Vk(n1 − 1, n2 + 1, 0)]=

λ

Λ{l2[Vk−1(n1 + 1, n2, 1)− Vk−1(n1 + 1, n2, 0)] + µ1[Vk−1(n1 + 1, n2, 0)− Vk−1(n1, n2 + 1, 0)]}

+µ1 + µ2Λ

{l2[Vk−1(n1, n2, 1)− Vk−1(n1, n2, 0)] + µ1[Vk−1(n1, n2, 0)− Vk−1(n1 − 1, n2 + 1, 0)]}

+µ2l2Λ[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 1)]. (24)

52

According to induction assumption X4, the Þrst and the second terms on the right-hand side of (24) arenonpositive. Furthermore, due to induction assumption A1, the last term on the right-hand side of (24) isnonpositive. Therefore, all terms on the right-hand side of (24) are nonpositive. This completes the prooffor X4.

Proof for A1: Vk(n1, n2 − 1, 0) ≤ Vk(n1, n2, 1) for n1 ≥ 0 and n2 ≥ 1We discuss the following four cases:

� Case 1: n1 = 0 and n2 = 1,� Case 2: n1 ≥ 1 and n2 = 1,� Case 3: n1 = 0 and n2 ≥ 2,� Case 4: n1 ≥ 1 and n2 ≥ 2.

A1-Case 1: When n1 = 0 and n2 = 1, A1 becomes Vk(0, 0, 0) ≤ Vk(0, 1, 1), and we have,

Vk(0, 0, 0) =λ

ΛVk−1(1, 0, 0) +

l2 + µ1 + µ2Λ

Vk−1(0, 0, 0)

Vk(0, 1, 1) =1

Λ+λ

ΛVk−1(1, 1, 1) +

µ2ΛVk−1(0, 0, 0) +

l2 + µ1Λ

Vk−1(0, 1, 1).

Therefore,

Vk(0, 0, 0)− Vk(0, 1, 1) = − 1Λ+λ

Λ[Vk−1(1, 0, 0)− Vk−1(1, 1, 1)] + l2 + µ1

Λ[Vk−1(0, 0, 0)− Vk(0, 1, 1)].

According to A1, the last two terms on the right-hand side are nonpositive. The proof for A1-Case 1 istherefore completed.

A1-Case 2: When n1 ≥ 1 and n2 = 1, A1 becomes Vk(n1, 0, 0) ≤ Vk(n1, 1, 1). Since at state (n1, 0, 0)there�s no job at the second station, loading station 2 is not feasible. Hence,

Vk(n1, 0, 0) =n1Λ+λ

ΛVk−1(n1 + 1, 0, 0) +

l2 + µ2Λ

Vk−1(n1, 0, 0) +µ1Λmin{Vk−1(n1, 0, 0) , Vk−1(n1 − 1, 1, 0)}

Vk(n1, 1, 1) =n1 + 1

Λ+λ

ΛVk−1(n1 + 1, 1, 1) +

µ2ΛVk−1(n1, 0, 0) +

l2ΛVk−1(n1, 1, 1)

+µ1Λmin{Vk−1(n1, 1, 1) , Vk−1(n1 − 1, 2, 1)}.

Therefore, we have

Vk(n1, 0, 0)− Vk(n1, 1, 1)= − 1

Λ+λ

Λ[Vk−1(n1 + 1, 0, 0)− Vk−1(n1 + 1, 1, 1)] + l2

Λ[Vk−1(n1, 0, 0)− Vk(n1, 1, 1)]

+µ1Λ[min{Vk−1(n1, 0, 0), Vk−1(n1 − 1, 1, 0)}−min{Vk−1(n1, 1, 1), Vk−1(n1 − 1, 2, 1)}]. (25)

According to induction assumption A1, the second and the third terms on the right-hand side of (25) arenonpositive. Also, due to X2 and X3, we have

min{Vk−1(n1, 0, 0) , Vk−1(n1 − 1, 1, 0)} = Vk−1(n1 − 1, 1, 0),min{Vk−1(n1, 1, 1) , Vk−1(n1 − 1, 2, 1)} = Vk−1(n1 − 1, 2, 1),

53

which imply that the last term on the right-hand side of (25) is equal to:

µ1Λ[Vk−1(n1 − 1, 1, 0)− Vk−1(n1 − 1, 2, 1)].

According to induction assumption A1, we have Vk−1(n1− 1, 1, 0) ≤ Vk−1(n1− 1, 2, 1). Therefore, all termson the right-hand side of (25) are nonpositive. The proof for A1-Case 2 is therefore completed.

A1-Case 3: When n1 = 0 and n2 ≥ 2, we have

Vk(0, n2 − 1, 0) =n2 − 1Λ

+λ

ΛVk−1(1, n2 − 1, 0) + µ2 + µ1

ΛVk−1(0, n2 − 1, 0)

+l2Λmin{Vk−1(0, n2 − 1, 0), Vk−1(0, n2 − 1, 1)},

Vk(0, n2, 1) =n2Λ+λ

ΛVk−1(1, n2, 1) +

l2 + µ1Λ

Vk−1(0, n2, 1) +µ2ΛVk−1(0, n2 − 1, 0).

Hence,

Vk(0, n2 − 1, 0)− Vk(0, n2, 1) = − 1Λ+λ

Λ[Vk−1(1, n2 − 1, 0)− Vk−1(1, n2, 1)]

+µ1Λ[Vk−1(0, n2 − 1, 0)− Vk−1(0, n2, 1)]

+l2Λ[min{Vk−1(0, n2 − 1, 0), Vk−1(0, n2 − 1, 1)}− Vk−1(0, n2, 1)].

According to A1, the second and the third terms are nonpositive. According to A1 and X1, the last termis also nonpositive. This completes the proof for A1-Case 3.

A1-Case 4: When n1 ≥ 1 and n2 ≥ 2, we have

Vk(n1, n2 − 1, 0) =n1 + n2 − 1

Λ+λ

ΛVk−1(n1 + 1, n2 − 1, 0) + µ2

ΛVk−1(n1, n2 − 1, 0)

+1

Λmin

(µ1 + l2)Vk−1(n1, n2 − 1, 0)µ1Vk−1(n1 − 1, n2, 0) + l2Vk−1(n1, n2 − 1, 0)µ1Vk−1(n1, n2 − 1, 0) + l2Vk−1(n1, n2 − 1, 1),

Vk(n1, n2, 1) =n1 + n2Λ

+λ

ΛVk−1(n1 + 1, n2, 1) +

l2ΛVk−1(n1, n2, 1) +

µ2ΛVk−1(n1, n2 − 1, 0)

+µ1Λmin{Vk−1(n1, n2, 1) , Vk−1(n1 − 1, n2 + 1, 1)}.

Hence,

Vk(n1, n2 − 1, 0)− Vk(n1, n2, 1)= − 1

Λ+λ

Λ[Vk−1(n1 + 1, n2 − 1, 0)− Vk−1(n1 + 1, n2, 1)]

+1

Λmin

(µ1 + l2)Vk−1(n1, n2 − 1, 0)µ1Vk−1(n1 − 1, n2, 0) + l2Vk−1(n1, n2 − 1, 0)µ1Vk−1(n1, n2 − 1, 0) + l2Vk−1(n1, n2 − 1, 1).

− l2ΛVk−1(n1, n2, 1)− µ1

Λmin{Vk−1(n1, n2, 1) , Vk−1(n1 − 1, n2 + 1, 1)}. (26)

According to induction assumption A1, the second term on the right-hand side of (26) is nonpositive. Thus,we only need to show that the last three terms on the right-hand side of (26) are nonpositive.

54

Due to X1,X2, X3, and X4

min

(µ1 + l2)Vk−1(n1, n2 − 1, 0)µ1Vk−1(n1 − 1, n2, 0) + l2Vk−1(n1, n2 − 1, 0)µ1Vk−1(n1, n2 − 1, 0) + l2Vk−1(n1, n2 − 1, 1)

= µ1Vk−1(n1, n2 − 1, 0) + l2Vk−1(n1, n2 − 1, 1)

min{Vk−1(n1, n2, 1) , Vk−1(n1 − 1, n2 + 1, 1)} = Vk−1(n1 − 1, n2 + 1, 1),which imply that the last three terms on the right-hand side of (26) are equal to:

µ1Λ[Vk−1(n1, n2 − 1, 0)− Vk−1(n1 − 1, n2 + 1, 1)] + l2

Λ[Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 1)]. (27)

Furthermore,

µ1Λ[Vk−1(n1, n2 − 1, 0)− Vk−1(n1 − 1, n2 + 1, 1)] + l2

Λ[Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2, 1)]

=µ1Λ[Vk−1(n1, n2 − 1, 0)− Vk−1(n1 − 1, n2, 0)] + l2

Λ[Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2 − 1, 0)]

+µ1Λ[Vk−1(n1 − 1, n2, 0)− Vk−1(n1 − 1, n2 + 1, 1)] + l2

Λ[Vk−1(n1, n2 − 1, 0)− Vk−1(n1, n2, 1)].(28)

According to induction assumption X4, we have

µ1Λ[Vk−1(n1, n2 − 1, 0)− Vk−1(n1 − 1, n2, 0)] + l2

Λ[Vk−1(n1, n2 − 1, 1)− Vk−1(n1, n2 − 1, 0)] ≤ 0.

Also, according to A1, the last two terms on the right-hand side of (28) are nonpositive. Hence, (27) isnonpositive, and therefore all terms on the right-hand side of (26) are nonpositive. This completes the prooffor A1-Case 4.

55

Documents

Hopp-Iravani-Shou v53.5 - Institute for Operations Research and the