Scheduling || A Solvable Case of the One-Machine Scheduling Problem with Ready and Due Times

A Solvable Case of the One-Machine Scheduling Problem with Ready and Due TimesAuthor(s): Hiroshi Kise, Toshihide Ibaraki and Hisashi MineSource: Operations Research, Vol. 26, No. 1, Scheduling (Jan. - Feb., 1978), pp. 121-126Published by: INFORMSStable URL: http://www.jstor.org/stable/169895 .

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OPERATIONS RESEARCH 0030-364X/78/2601-0121 $01.25 Vol. 26, No. 1, January-February 1978 ? 1978 Operations Research Society of America

A Solvable Case of the One-Machine Scheduling Problem with Ready and Due Times

HIROSHI KISE

Kyoto Institute of Technology, Kyoto, Japan

TOSHIHIDE IBARAKI and HISASHI MINE

Kyoto University, Kyoto, Japan

(Received April 1976; accepted June 1977)

We consider a class of n-job one-machine scheduling problems with ready time r(i), processing time p(i), and due time d(i) for each job i. Preemption is not allowed, and precedence constraints among jobs are not assumed. For this problem we show that there is a 0(n2)-time algorithm to find a schedule that minimizes the number of tardy jobs, under the assumption that r(i) <r(j) implies d(i) ?d(j).

THIS PAPER gives an 0(n2) -time algorithm for solving the n-job one machine scheduling problem defined as follows.

(i) Let J={, , n} be a set of n jobs. Each job i has ready time r(i), processing time p(i), and due time d(i), which are all non-negative real numbers. It is assumed that r(i) <r(j) implies d(i) <d(j) and jobs are ordered by

r(l) _** <r(n) and d(1)< ** <?d(n). (1)

(ii) Jobs i are processed on the machine one at a time when they are ready (i.e., time> r(i)), and preemption is not allowed. No precedence relation among jobs is assumed.

If the processing of job i finishes before or at d(i), it is called early; otherwise, tardy. Our objective is to find a schedule that minimizes the number of tardy jobs. Such a schedule is called optimal.

Once an order of jobs to be processed on the machine is specified in this problem, we can assume without loss of generality that the next job j is processed as soon as the machine becomes idle and job j is ready. Thus we can completely define a schedule by specifying a permutation (an order) of n jobs.

Although the constraint (1) seems to be very restrictive, this includes Moore's well-known problem [6] as a special case (i.e., the case of r(l) - ... =r(n)). Furthermore, if (1) is removed, the problem becomes

121



122 Kise, Ibaraki, and Mine

NP-complete [2, 5] even if all the jobs except one have the same ready time and the same due time. (Thus it is unlikely that this problem has a polynomial time algorithm [1].)

There are also other n-job, one-machine scheduling problems with due times that can be solved in polynomial time under appropriate additional assumptions [2-4, 7]. In particular, Lawler [4] generalizes Moore's result to an O(nlogn) algorithm for the case of agreeable weights, where each job i has weight w(i) and p(i) <p(j) implies w(i) ?w(j). He introduces the concept of "j-optimality" to prove the correctness of his algorithm. Our algorithm is quite similar to his in structure, and the correctness will also be proved in a fashion parallel to his by employing the concept of j-optimality.

1. THE ALGORITHM

A schedule on a subset S of J is feasible if all jobs in S are early. The next properties are obvious characteristics of our problem.

(i) A subset S of J has a feasible schedule if and only if the schedule that orders the jobs in S according to their indices (see (1)) is feasible.

(ii) It is possible to assume without loss of generality that an optimal schedule is given in the form (S, J-S), SCJ, such that (a) jobs in S are all early, while jobs in J- S are all tardy, and (b) jobs in S are ordered according to their indices.

Therefore, a schedule is completely specified by simply giving a feasible subset S of J. Assume that a subset S= (k1, ... , km)(ki< * <km) is given. Then, by definition, the finishing time Fki(S) of job ki is given by

Fki(S) =max {Fki._..(S), r(k)} +p(ki), (2)

where Fk0 (S) =0 is assumed for convenience. The finishing time of all jobs in S is defined by F(S) =Fkm(S).

In the following, we use the concept of a "j-optimal" set [4]: A set Ej is j-optimal if it is the set of early jobs in an optimal schedule for the set of jobs I 1, . , j} cJ. It is clear by definition that an n-optimal set En yields an optimal schedule (Es, J-E.) for J. Our algorithm obtains j-optimal sets Ej in the order of j =1, ... , n by

E= fE1UAj} if F(Ej_1U{j})<d(j) (3) Ej-1U } - {I l otherwise,

where Eo=0 is assumed for convenience and 1 is a job satisfying

F(Ej1Ufj} -{l}) <F(E5,1iUj}-{i})- (4)

for all iEEj U{j}. Inequality (4) is a key property for Ej to be j-optimal and will be proved in the next section.



One-Machine Scheduling Problem 123

Main Algorithm to Compute an Optimal Schedule

Step 1. Eo0-4O, j*-O and go to Step 2. (a- stands for the assignment operation represented by = in Algol.)

Step 2. If j=n, then go to Step 3. Otherwise, j<-j+1; compute Ej by (3) and (4) and return to Step 2.

Step S. Halt. (Es J-En) is an optimal schedule.

Ej in Step 2 can be computed in at most 0(n) steps. If F(Ej_1Utj1 ) ? d(j), Ej is obviously computed in constant time; otherwise, 1 of (4) is computed in at most 0(n) steps by using the following subalgorithm. For a given Ej-1 k1, i , km-il and j= ki, the subalgorithm computes F(Si) and F (Si') for i-1, * , m, where St-{ki , * *i , kh-1, k+1*, ... ki, Si'={kI, * , kit1 and l=kh satisfy F(Si)<F(Si'), i1=, * , m. Upon termination (i.e., i=m), l=kh excluded from Sm becomes equal to 1 of (4).

Subalgorithm to Compute I of (4)

Step 1. If m= 1, then Siz-, 1<-km and halt; else, i -2, So-< k2l, F(S,) <-r(k2)+p(k2), Si'<-{k1}, F(Si')<-r(k1)+p(ki), 1*-k1 and go to Step 2.

Step 2. If F(S.)>F(S.'), then Si--Si', F(Si)<-F(St'), and le-k,. If i=m, then halt; else, go to Step 3.

Step 3. S+1i-SiU{i+ 1i, F(Si+1)<-max fF(Si),r(ki+1)1 +p(ki+?), S'+&-Si'U{il, F(S'+i)<-max {F(Si'), r(k)) +p(k,), i<-i+l and return to Step 2.

THEOREM. The proposed algorithm is valid (i.e., yields an optimal schedule) and requires 0(n2) time.

Proof. The validity of the main algorithm immediately follows if Ej defined in (3) and (4) is j-optimal, which will be proved in the next section. It is obvious that the subalgorithm correctly computes 1 of (4) in at most 0(m) ( ?0(n) ) steps. Thus Ej is computed in at most 0(n) steps, and the main algorithm computes Ej exactly n times. Hence at most 0(n2) steps are required in total.

2. VALIDITY OF THE PROPOSED ALGORITHM

To prove the validity of the algorithm, we show by a series of lemmas that Ej given by (3) and (4) is j-optimal.

LEMMA 1. Ei is feasible for all J.

Proof. By induction on j. Obviously, Eo=4o is feasible. Assume that Ei-1 is feasible. If F(Ej_1Uj}) <d(j), then Ej=Ej_1U~j}. Therefore, Ej is feasible. If Ej_1U{j >d(j), then Ej=Ej_1Utjl-Ill. Note first that Ej-1-{l) is feasible since so is Ej-1. In addition, d(j)?d(j-1)>F(Ej_1) ?F(E_1Utj} -{l}) follows from (1) and (4); job j is early. Therefore, Ej= Ej_1Utj} -{l} is feasible.




LEMMA 2. Let b satisfy F (J-{a) ) _ F(J-{b} ) for each a E J. Then F(J- U-{a3) > F(J-U-{b3) holds for any UC{c I c>a, c~bl.

Proof. We may assume that U is a singleton { c3; the general case then immediately follows by induction. Note that for any three subsets of J) X, Y and Z such that (ViE Z)(VijEXUY)(j<i) holds, F(X)<F(Y) implies F(XUZ) <F(YUZ) by (2). Using this fact, we can show the lemma holds if c>max {a, b}. Therefore, assume that a<c<b. Furthermore, we assume for simplicity that a =1 and b = n. (The extension to the general case is straightforward.) Now it is sufficient to prove that F(J- { 1} ) > F(J-{n}) implies F(J-{1, c})_F(J-{n, c}). We may assume that F(J-{c, n}) >r(n), for otherwise F(J-{1,c}) = r(n) + p(n) > F(J-{c, n}) (see (2)). Let A(c; x) =F(J) -F(J- {x}) and also A(x; y) = F(J-{x})-F(J-{x,y}). Now from F(J-{c,n})>r(n) and (2) A(4; n)= A(c; n)=p(n) follows. Furthermore, F(J-{1})_?F(J-{n}) implies A (c; 1) < A (c; n) = p (n).

Note that A(4; c)+A(c; 1)=A(4; 1)+A(1; c)5A (c; n)+A(1; c), and hence F(J-{1, c})-F(J-{c, n})= A(c; n)- A(c; 1)>A(4; c)-A(1; c). Therefore, the lemma is proved if

A *; c) >- t(1; c) (5)

always holds. To prove (5), first assume that n=c+1. Then F(J-{c, n})=F(S,-i)>

r(n) > r(c) by assumption, and hence

A(c; c) =max {F(Sci1)+p(c), r(n)} -max {F(Sc-1), r(n)} A (1; c) = max f max [F(Sc-, - f1 ), r(c) ]+p (c), r(n)}I

-max {F(Sc-l-{ 1} ), r(n)}

=max {max [F(Sc1-{1}), r(c)]+p(c), r(n)}

-max {max [F(S01-{1}), r(c)], r(n)}

holds by (2), where Sj={l, * , j}. Then A(c; c)>A(1; c) immediately follows since A(s) =max {s+p, r(n)} -max {s, r(n)} is obviously non- decreasing in s and F(Sc-1) _ max [F(S01-{ 1} ), r(c) ] holds.

The general case n ? c+1 is then proved by induction. Let An-i(x; y) denote A (x; y) for J = {1, *,n-1}. In this case we have

A(c; c) =max {F(Snl), r(n)} -max {F(Sn~-{c}), r(n)}

=max {F(Sn1), r(n)} -max {F(S.-,)-An- (c; c), r(n)}

=max {O, min [Ain-, (; c), F(Sn-1)-r(n)]}

A(1; c)=max {O, min [A 1(; c), F(S,,,-{1})-r(n)]}.

Now A(c; c)>A(1; c) follows from An-1(40; c)->A-n1(1; c) (induction hypothesis) and F( S,,-) > F(Sn-1- { 1} )



One-Machine Scheduling Problem 125

LEMMA 3. For all k, j with 1? k <j < n, there is a j-optimal set contained in EkU{k+1, * * *, j}, where Ek is obtained by (3) and (4).

Proof. Proof is by induction on k. Clearly, EoU{I 1, * , j} contains a j-optimal set. Assuming that EklU{ k, k+ 1, *., j} contains a j-optimal set denoted by E, we show that EkU{k+1, *., j} contains a j-optimal set denoted by E'. If F(Ek-lU{k})) <d(k), let E'=E. Then E' is j-optimal by definition and is contained in EkU{k+l, ... , j} since Ek =Ek-lU{ k} (by (3) and (4)) and hence EkU{k+1, ..., j} =Ek1lUlk, k+1, . ., j}. If F(Ek-lU{k})>d(k) and 1 obtained by (4) in the computation of Ek is not contained in E, again let E' =E. Since Ek-lU{k, k+1, ..., j}= EkU(k+1,. ,j}U{I} by Ek=EklU~k}-{l1}, lE implies ECEkU k+1, 2 {,j}. Finally, consider the case F(Ek-lU{k} ) >d(k) and l E E.

LetSk={1,* ,k}and

EnSk = EklU{k} -X. (6)

Then X(C Sk) Al) since Ek-lU{k} is not feasible by assumption. We show that

E -E U fij- I1}, (7)

where i is the smallest element in X, is also j-optimal. Clearly, E and E' have the same cardinality and E nSkCEklU{k}-{l} (by ECEklU {k, k+1, *, j} and iEEk1U{k}) =Ek. These imply that E'fnSk is feasible (since Ek is feasible by Lemma 1). Hence showing

F(E'nSk) <F(EnSk) (8)

amounts to proving that E' is j-optimal since E'- Sk = E- Sk. Let X'= X-{i} . Since EnSk=EklU{k}-X'-{i} (see (6)), it follows from (7) that

E'nSk = (EU{i} -{l} )nSk= (Ek1U{k} -X'-{i} )U{i} -{l}

=EklU{k} -X'-{l}.

Furthermore, from the definition of l in (4)

F(Ek) =F(EklU{k}-{l} ) ?F(EklU{k}-{i} ) (9)

follows. Now since i is the smallest element in X, Lemma 2 asserts that (9) implies (8) (take J=Ekl1Ufk}, b-1, a=i and U=X').

LEMMA 4. Ej satisfying (3) and (4) is j-optimal.

Proof. Proof is by induction on j. Eo=4 is trivially j-optimal. Assume that Ej-1 is (j-1)-optimal. If F(Ej_1U{j}) d(j), then Ej=Ej11U{j} is j-optimal since Ej is feasible and jEjE > JEjE11. Now assume that F(Ej_1Ut j} ) >d (j). Then a j-optimal set is a proper subset of Ej-1U I j}




by infeasibility of Ej_1U{j} and by Lemma 3 with k=j-1. Hence Ej= Ej_1Uj}j-{l} is j-optimal since Ej is feasible by Lemma 1, and IE'j= Ej_jU~j} I-1.

ACKNOWLEDGMENT

The authors wish to thank Professor T. Hasegawa of Kyoto University for his comments. They are also grateful to one of the anonymous referees for his many suggestions, which were quite helpful in improving read- ability and shortening the proofs, and for his having informed us of references [4, 5].

REFERENCES

1. R. M. KARP, "Reducibility among Combinatorial Problems," in Complexity of Computer Computations, pp. 85-103, R. E. Miller and J. W. Thatcher (Eds.). Plenum Press, New York, 1972.

2. H. KISE, T. IBARAKI, AND H. MINE, "n-Job One-Machine Scheduling Problems with Ready and Due Times," Working Paper, Dept. of Applied Math. and Physics, Kyoto University, Kyoto Japan, 1975; also presented at the XXII International TIMS Meeting, Kyoto, July 1975.

3. E. L. LAWLER, "Sequencing Problems with Deferral Costs," Management Sci. 11, 280-288 (1964).

4. E. L. LAWLER, "Sequencing to Minimize the Weighted Number of Tardy Jobs," Rev. Franz. Automat. Informa. Rech. Opnelle. 10.5 Suppl., 27-33 (1976).

5. J. K. LENSTRA, A. H. G. RINNooY KAN, AND P. BRUCKER, "Complexity of Machine Scheduling Problems," Ann. Discrete Math. 1, 343-362 (1977.)

6. J. M. MOORE, "Sequencing n Jobs on One Machine to Minimize the Number of Tardy Jobs," Management Sci. 15, 102-109 (1968).

7. J. B. SIDNEY, "An Extension of Moore's Due Date Algorithm," in Symposium on the Theory of Scheduling and Its Applications, pp. 393-298, S. E. Elmagh- raby (Ed.). Springer-Verlag, Berlin, 1973.



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Scheduling || A Solvable Case of the One-Machine Scheduling Problem with Ready and Due Times