1 Chapter 7 Performance analysis of failure- prone production lines Learning objectives :...

Chapter 7

Performance analysis of failure-prone production lines

Learning objectives :Understanding the mathematical models of production lines

Understanding the impact of machine failures

Understanding the role of buffers

Able to correctly dimension buffer capacities

Textbook :S.B. Gershwin, Manufacturing Systems Engineering, Prentice Hall,

J. Li and S.M. Meerkov, Production Systems Engineering

PlanPlan

• Basic concepts

• Failure-prone single-machine systems

• Production lines with unlimited buffers

• Production lines without buffers

• Aggregation of parallel machines and consecutive dependent machines

• Two-machine production lines with intermediate buffer

• Long failure-prone production lines

Basic concepts

Production lines or tranfer linesProduction lines or tranfer lines

Frequent production disruption by machine failures

Buffers are of finite capacity

Production often varies wildly

M1 B1 M2 B2 M3 B3 M4

machine Buffer

Production lines or tranfer linesProduction lines or tranfer lines

Production capability of a machineProduction capability of a machine

Cycle time or processing time () :

•time necessary to process a part by a machine

•Constant cycle times (large assembly systems)

•Variable or random cycle time (job-shop environment).

Maximal production rate or max. capacity (U) :

•U = 1/ (parts per time unit)

Machine reliability modelMachine reliability model

• TBF = Time Between Failure (Tup)

• TTR = Time To Repair (Tdown)

• MTBF = Mean TBF

• MTTR = Mean TTR

• Failure rate = 1/MTBF

• Repair rate = 1/MTTR

UP DOWN

Models of machine failuresModels of machine failures

ODF – Operation-Dependent Failure: the state of the machine degrades ony when it produces.

Implication : an ODF cannot fail when it is not producing.

TDF – Time-Dependent Failure: the state of the machine degrades all the time even if it is not producing.

Implication : a TDF machine can fail even if it is not producing.

Models of machine failuresModels of machine failures

Cause of failures:

ODF: tool wear

TDF: electricity supply, electronic components, …

ODF failures account for 80% of disruptions in manufacturing systems (Hanifin & Buzacott)

In this chapter, we mainly focus on ODF failures.

Example of a two machine production line to explain the difference

Mathematical models of buffersMathematical models of buffers

Buffer capacity : N

State of a buffer : number of parts in it varying from 0 to N

Assumptions:

•A part, produced by a machine, is immediately placed in the downstream buffer, if it is not full.

•A part is immediately available for processing by a machine, if the upstream buffer is not empty.

Mathematical models of buffersMathematical models of buffers

Buffering capacity of a moving convey

travel

travel0

where l = length of the convey v = speed of the convey K = maximum number of carriers in the convey

Interaction between machines and buffersInteraction between machines and buffers

Blocking Before Service (BBS):

A machine cannot operate and is blocked if

•it is up

•its downstream buffer is full

•no part can be removed from that buffer

Blocking After Service (BAS):

•A machine continue to produce even if the downstream buffer is full.

•The machine is blocked at the completion of the part if the buffer remains full.

Buffer capacity convetion: NBBS = NBAS +1

Starvation : an idle up machine is starved if its upstream buffer is empty/

Performance measuresPerformance measures

Throughput rate also called productivity (TH):

•Number of parts produced per time unit.

Production rate (PR):

•Number of parts produced per cycle time.

•Concept appropriate for synchronuous production systems with all machines having identical cycle times

TH = U×PR

Performance measuresPerformance measures

Work-in-process of the i-th buffer (WIPi)

•Average number of parts contained in the i-th buffer.

Total work-in-process (WIP):

•Average number of parts) in the system

•WIP = WIP1 + WIP2 + ...

Probability of blocking (BLi)

Probability of starvation (STi)

Failure-prone single-machine Failure-prone single-machine systemssystems

Throughput rateThroughput rate

Proof:

•Average length of an UP period = MTBF

•Average length of a DOWN period = MTTR

•Production of an UP period = U. MTBF

•Length of an UP-DOWN cycle = MTBF + MTTR

•Throughput rate : TH = (U.MTBF) / (MTBF + MTTR).

MTBFTH U U

MTBF MTTR

UP DOWN

A machine operating at a reduced speed U' < UA machine operating at a reduced speed U' < U

Operation Dependent Failure case

Time Dependent Failure case:

Production lines with unlimited Production lines with unlimited buffersbuffers

Assumptions

• The first machine M1 is never starved

• The last machine is never blocked

• Each machine produces if its upstream buffer is not empty

M1 B1 M2 B2 M3 B3 M4

Bottleneck machine

A machine Mi is said to be a bottleneck if it proper productivity (or isolated productivity) is smaller than that of other machines, i.e.

M1 B1 M2 B2 M3 B3 M4

,jii j

i i j j

U U j i

Throughput rate

The throughput rate of a production line with unlimited buffers is equal to that of the bottleneck machines, i.e. (Why?)

The throughput rate of a machine Mi is equal to that of the slowest upstream machine, i.e.

M1 B1 M2 B2 M3 B3 M4

i i i i

i ii i i i

U MTBF UTH MIN MIN

MTBF MTTR

i i i ij

i j i ji i i i

U MTBF UTH MIN MIN

MTBF MTTR

Case of a single bottleneck machineCase of a single bottleneck machine

The level of the input buffer of the bottleck machine grows without limit (at which slope?)

All downstream buffers remain limited.

M1 B1 M2 B2 M3 B3 M4

Master GI2007 23

Case of two bottleneck machinesCase of two bottleneck machines

The level of the input buffer of each bottleneck machines grows without limit.

All other buffers downstream of the first bottleneck remain limited.

M1 B1 M2 B2 M3 B3 M4

Production lines without buffersProduction lines without buffers

Assumptions

If a machine breaks down or it takes longer time for an operation,

then all other machines must wait.

(immedicate propagration of disruptions)

Impact : the productivity of the line is usually smaller than that of the bottleneck machine.

M1 B1 M2 B2 M3 B3 M4

Case of reliable machines with different cycle times

The progress of products in the line is synchronized to allow the completion of all on-going operations .

The cycle time of the line is that of the slowest machine, i.e.

TH MIN U MIN

M1 wait

Failure-prone lines with identical cycle times

Assumptions:

When a machine breaks down, all other machines must wait.

The probability of two machines failed at the same time is small enough and can be neglected. (true in practice)

M1 B1 M2 B2 M3 B3 M4

Failure-prone lines with identical cycle times

Productivity:

M1 B1 M2 B2 M3 B3 M4

1 1i i

i i i i

1) Each time interval can be decomposed as follows:

2) tn : instant when the line produces n parts. The time interval [0, tn) includes :

• a total duration of n of all UP periods,

• for each machine Mi, n/MTBFi failures requiring with total repair time of (n/MTBFi) MTTRi.

UP M4 UP M2 UP M1

All UP some machine DOWN

n ii i

nt n MTTR

1x n i

0 5 10 15 20

Nb machines

Impact of the length of the line

Unlimited buffer

Zero-buffers Un

lost capacity

The longer the line is, the higher the capacity loss is.

Aggregation of parallel machines and consecutive dependent machines

Aggregation

M1 B1 M2 M4 B4M5

M1 B1 M234 B4 M56

consecutive dependent machines parallel

machines

Aggregation of parallel machines

Identical parallel machines : i = = 1/U, i = , i =

Ueq = S×Ueq = eq =

Aggregation of parallel machines

Non Identical parallel machines : i = 1/Ui, i, i, ei = 1/(1+i/i)

eq i iieq

e U eU

i iieq eq

eqeq eq

av failure frequency

availability of Meq

Aggregration of consecutive dependent machines

M1 MSM2

eqSeq i

ieq eq i

Failure rate equivalence

... Meq

Machines of identical cycle time : i = = 1/U, i, i

Flow rate equivalence

Average stoppage time equivalence

Aggregration of consecutive dependent machines

1 1S i

ieq eq i

Machines of nonidentical cycle time : i = 1/Ui, i, i

• All machines slowed down to slowest one : U = min{Ui, i= 1, ..., S}

• Reduced failure rate : i = Ui /Ui

• Equivalent machine cycle time : Ueq = U

• Failure rate equivalence : eq = i i

• Flow rate equivalence : THeq = TH(L)

• Average stoppage time equivalence

Two-machine production lines with intermediate buffer

Motivation & cost of intermediate buffersMotivation & cost of intermediate buffers

Motivation: • Avoid loss of production capacity

Costs:• Increasing WIP and production delay• Larger factory space• More complicated material handling

Effect of failures: • Unlimited buffer : no upward propagation of disruptions• No buffer: Instantaneous propagation • Finite buffers : delayed and partial propagations

M1 B1 M2 B2 M3 B3 M4

Motivation & cost of intermediate buffersMotivation & cost of intermediate buffers

New phenomena:

•Blocking

•Starvation

M1 M3M2 M4

Failure of M3

M1 M3M2 M4

3 time units after the failure where is the cycle time

CTMC model of CTMC model of reliable line with exponential processing timesreliable line with exponential processing times

Assumptions:

• The two machines are reliable and never fail

• The processing times are exponentially distributed random variables with mean 1/p1 on M1 and 1/p2 on M2

• The buffer capacity is K

• Each machine can hold a part on it for processing.

• M1 is never starved and M2 is never blocked

M1 B M2

The following state variable

X(t) = number of parts in B

+ the part on M2 if any

+ the finished part blocked on M1 if any

is a continuous time Markov chain

M1 B M2

0 1 K+1 K+2p1

CTMC model equivalent to M/M/1/(K+2).

with = p1/p2, corresponding to the traffic intensity.

M1 B M2

0 1 K K+2p1

1, , if 1

1, if 1

Performance measures (case ≠ 1)

Starving probability of M2 : 0 = (1-)/(1-K+3)

Blocking probability of M1: K+2 = K+2(1-)/(1-K+3)

Throughput rate :

TH = p2(1-0) = p1(1-K+2) = p1(1- K+2)/(1-K+3)

Mean WIP

E X n K

0 5 10 15 20 25 30 35

Buffer capacity

0 5 10 15 20 25 30 35

Buffer capacity

0 5 10 15 20 25 30 35

Buffer capacity

Blocking prob.

starving prob

Zero-buffer

Unlimited buffer

Example : p1 = 10, p2 = 9, = 10/9

Exponential model Exponential model of Failure-prone linesof Failure-prone lines

Assumptions:

• Machines can break down.

• Exponentially distributed times to failures and time to repair with TBFi = EXP(i) and TTRi = EXP (i)

• Exponential processing times with T1 = EXP(p1) and T2 = EXP(p2)

• Buffer of capacity K

• Each machines holds the part in process.

• M1 never starved and M2 never blocked

M1 B M2

The system can be described by the following state variables:

x = number of parts in B + part on M2 if any + finished part blocked on M1 if any

i = 1 if Mi is UP and 0 if Mi is DOWN

The state vector (1, 2, x) is a continuous time Markov chain

Exponential model of the case K = 0

0, 0, 1

1, 0, 1

0, 1, 1

1, 1, 1

0, 0, 2

1, 0, 2

0, 1, 2

1, 1, 2

0, 0, 0

1, 0, 0

0, 1, 0

1, 1, 0

• Analytical expressions of steady-state probabilities available in the book of SB Gershwin

• Can be used to evaluate the performance measures

Slotted time model of a failure-prone lineSlotted time model of a failure-prone lineAssumptionsAssumptions

• Synchronized line, i.e. i = , with a buffer of capacity N.

• All parts remain in buffers and machines do not hold parts.

• Slotted time indexed t = 1, 2, 3, …

• Machine state change at the begining of a period: machine working (W), under repair (R), blocked (B), starved (I).

• Buffer state change at the end of a period.

• Blocking Before Service: M1 blocked if B is full, M2 starved if B is empty

• A machine Mi in state W in t breaks down in period t+1 with proba pi and, with proba 1 - pi, moves to state W or B or I.

• A machine Mi in state R in t moves to state W in t+1 with proba ri and, with proba 1 - ri, remains in R.

• A machine in B or I in t moves to W in t+1 if the other machine is repaired.

M1 B M2

Slotted time model of a failure-prone lineSlotted time model of a failure-prone lineDiscrete Time Markov chainDiscrete Time Markov chain

The state vector (1, 2, x) with

• i(t) = 1/0 depending on whether Mi is UP or DOWN at the begining of t

• x(t) = number of parts in B at the end of t

is a discrete time Markov chain.

Buffer state change :

• x(t) = x(t-1) + 1(t)×1{x(t-1)<N} - 2(t)×1{x(t-1)>0}

• 0≤x(t) ≤ N

M1 B M2

Transient states :

• (1, 0, 0), (1, 1, 0), (0, 0, 0), (1, 0, 1)

• (0,0,N), (0,1,N), (1,1,N), (0,1,N-1)

Flow balance equations for states (1, 2, x) with 2 ≤ x ≤ N-2

(1,1,x) = (1-p1)(1-p2)(1,1,x) + r1(1-p2)(0,1,x) + (1-p1)r2(1,0,x) + r1r2(0,0,x)

(0,0,x) = (1-r1)(1-r2)(0,0,x) + p1(1-r2)(1,0,x) + (1-r1)p2(0,1,x) + p1p2(0,0,x)

(1,0,x) = (1-p1)p2(1,1,x-1) + (1- p1)(1-r2)(1,0,x-1) + r1p2(0,1,x-1) + r1(1-r2)(0,0,x-1)

(0,1,x) = p1(1-p2)(1,1,x+1) + (1-r1)(1-p2)(0,1,x+1) + p1r2(1,0,x+1) + (1-r1)r2(0,0,x+1)

Other boundary equations can be derived similarly.

M1 B M2

Performance measures :

• Efficiency of M1 : E1 = 1 = 1, x < N 1, 2, x)

• Efficiency of M2 : E2 = 2 = 1, x > 0 1, 2, x)

• Throughput rate : TH = E1×U = E2×U

• WIP : x1, 2, x)

• Probability of starvation : (0, 1, 0)

• Probability of blocking : (1, 0, N)

M1 B M2

Efficiency of the line E(N) = probability a machine is producing

1 2*1 2

1 2 1 2 1 21 22

1 2 1 2

1 1 2 1 2 1 2

2 1 2 1 2 2 1

1 1 2 1 2 1 2

2 1 2 1 2 2 1

*1 2 1 2 1

2 1 2 1 2

1 2 1 1

I I rE N

r r r r r r I NI I I

r r I r r I N

p p p p r p

r r r r p r

pr pr I

Slotted time model of a failure-prone lineSlotted time model of a failure-prone lineAnalytical resultsAnalytical results

WIP to be determined with expressions of steady-state probabilities available in the book of S.B. Gerswhin

Continuous flow models of failure-prone linesContinuous flow models of failure-prone linesAssumptionsAssumptions

M1 B M2

Only synchronuous lines are considered, i.e. Ui = U.

Each machine produces continuously.

When a machine Mi produces,

•a flow moves out of its upstream buffer at rate U

•a flow is injected in its downstream buffer at rate U

M1 B M2

The continuous flow model is a good approximation for a high volume production line with large enough buffer capacity.

Theorem (David, Xie, Dallery):

THContinuous(h) < THDiscret(h) < THContinuous(h+2)

where THDiscret (h) is a discrete flow line similar to the Exponential model but with constant processing times and buffer capacity h.

Result holds for longer lines.

Continuous flow models of failure-prone linesContinuous flow models of failure-prone linesHow good is continuous flow approximationHow good is continuous flow approximation

M1 B M2

Model parameters:

i: failure rate of Mi

i: repair rate of Mi

• U: maximum production rate of Mi

• h : buffer capacity

State variables

i(t) =1/0 : state of machine Mi at time t

• x(t): buffer level at t (a real variable)

Auxillary variable:

•ui(t) : production rate of Mi at t

Continuous flow models of failure-prone linesContinuous flow models of failure-prone linesDynamic behaviorDynamic behavior

M1 B M2

u1(t) u2(t)

1(t) 2(t)x(t)

u1 = U

u2 = U

u1 = U

u2 = 0

F2 blocking R2 F1 R1Starving

u1 = 0

u2 = 0

u1 = U

u2 = U

u1 = 0

u2 = U

u1 = 0

u2 = 0

Blockage of M1 : 1(t) = 1, x(t) = h, 2(t) = 0

Starvation of M2 : 1(t) = 0, x(t) = 0, 2(t) = 1

In all other case : u1(t) = 1(t) U, u2(t) = 2(t) U

M1 B M2

u1(t) u2(t)

1(t) 2(t)x(t), h

Efficiency of the line, E(h) that is the probability a machine is producing

Throughput rate : TH(h) = E(h)U

Probability of starving of M2: ps(h) = 1 – E(h)/e2

Probability of blocking of M1: pb(h) = 1 – E(h)/e1

Isolated efficiency of Mi : ei = 1/(1+Ii) = i/(i+i)

Mean buffer level: Q(h)

Continuous flow models of failure-prone linesContinuous flow models of failure-prone linesPerformance measuresPerformance measures

online proof

2 2 1 1

1 2 1 22 2

2 1 1 2

2 2 1 1

2 1 1 2 1 2 1 2

1 2 1 2

I e IE h

I I e I I

I IU e I I he

I IQ h

I I e I I

Case I1 = 1/1 I2 = 2/2

Continuous flow models of failure-prone linesContinuous flow models of failure-prone linesAnalytical solutionAnalytical solution

Ih I U

h II U

Q h hI h I I U

Case I1 = 1/1 I2 = 2/2 = I

0 50 100 150 200 250 300 350

Buffer capacity h

tCase U=1, = 0,1

In this case,

Q(h) = 0,5hzero-buffer

infinite buffer

Numerical resultsU = 1, 1 = 0.1, 2 = 0.1, 2 = 0.1

Throughput

1 = 0,14

1 = 0,12

1 = 0,10

1 = 0,08

1 = 0,06

Buffer capacity

Discussions:

Why are the curves increasing? Why do there reach an asymptote?

What is TH when N= 0? What is the limit of TH as N tends to infinity?

Why are the curves with smaller 1 lower?

Discussions:

•Why are the curves increasing?

•Why different asymptotes?

•What is the limit of WIP as N→?

•Why are the curves with smaller 1 lower?

1 = 0,14

1 = 0,12

1 = 0,10

1 = 0,08

1 = 0,06

Buffer capacity

Questions :

•If we want to increas production rate, which machine should we improve?

•What would happen to production rate if we improved any other machine?

Improvement to non-bottleneck machine.

Same graph for improvement of machine 2

Numerical resultsU = 1, 1 = 0.1, 1 = 0.1, 2 = 0.1, 2 = 0.1

Throughput

Buffer capacity

Inventory increases as the (non-bottleneck upstream machine is improved and as the buffer space is increased.

Numerical resultsU = 1, 1 = 0.1 , 1 = 0.1, 2 = 0.1, 2 = 0.1

Average inventory

Buffer capacity

• Inventory decreases as the (non-bottleneck) downstream machine is improved

•Inventory increases as the buffer space isincreased.

Numerical resultsU = 1, 1 = 0.1 , 1 = 0.1, 2 = 0.1, 2 = 0.1

Average inventory

Buffer capacity

• 1 and 1 vary together and 1/(1 +1) = 0.9

• Answer: short and frequent failures.

• Why?

Numerical resultsU = 1, 2 = 0.8, 2 = 0.09, h = 10

Throughput

Repair rate 1

Should we prefer short and frequent disruptions or long and infrequent disruptions?

Reversibility Theorem

(hold for any nb of machine and for all models)

E(L) = E(L')

Q(L) = h - Q(L')

ps(L) = pb(L')

pb(L) = ps(L')

Continuous flow models of failure-prone linesContinuous flow models of failure-prone linesReversiblityReversiblity

M1 B M2

Proof for the continuous L2 line

A continuous time Markov process with hybrid state space characterized by

Internal state distribution (0 < x < h):F12(x) = P{1(t) = 1, 2(t) = 2, 0 < x(t) x}

f12(x) = d F12(x) /dx

Boundary distribution:P12(0), P12(h)

M1 B M2

u1(t) u2(t)

1(t) 2(t)x(t), h

Case I1 = 1/1 I2 = 2/2

2 1 1 2 1 2 1 2

1 2 1 2

1 2 1 200 11

1 2 1 2

11 111 2

1 2 1 210 01

1 2 2 1

01 00 10 00

1 2 2 1

2 1 1 2 1 2

f x f x Ce

f x Ce f x Ce

U UP h Ce P C

P h U Ce P U C

P h P h P P

U I Ie

Case I1 = 1/1 I2 = 2/2=I

10 01 0

000 0 11

11 0 11 01 2

1 2 1 210 0 01 0

1 2 2 1

01 00 10 00

f x f x C

Cf x IC f x

P h C P C

P h U C P U C

P h P h P P

IU I h

Long failure-prone production linesLong failure-prone production lines

Introduction

• The performance evaluation of a general failure-prone line is difficult due to the lack of analytical solution and the state space explosion.

• The number of states for a M machines lines with buffers of capacity N is about 2M(N+1)M-1. For M = 10 and N = 100, there are over 1021 states.

• A so-called DDX decomposition method is capable of obtaining an approximative but precise enough analytical estimation.

• Other approximation methods exist but the DDX method is considered as one of the most efficient ones and can be extended to other systems such as assembly lines.

• Focus on Continuous flow model but all results can be extended to discrete flow models

M1 B1 M2 B2 M3 B3 M4

Notation

Given isolated machine performances:

Ii = i/i

ei = 1/(1+Ii) : isolated efficiency of Mi

eiU : isolated productivity of Mi

Unknown system performance measures:

Ei : Probability that Mi is producing

THi = Ei U: throughput rate of Mi

psi : probability of starvation of Mi

pbi : proba of blockage of Mi

M1 B1 M2 B2 M3 B3 M4

Aggregation methodEquivalent machine

Replace L12 by a machine M12 of equivalent isolation throughput rate (flow equivalence), i.e.

M1 B1 M2 B2 M3 B3 M4

M1 B1 M2 L12

1 /Me E L

Repair time of M12 = Average stoppage time of M2 in L12:

12 1212E L

2 2 12 1

12 2 2 12 1 2 2 2 12 1 1

2 12 12

( 0) ( )1 1 1

( 0) ( ) ( 0) ( )

( 0) 1 ( ) ( )

P ps L

P ps L P ps L

P E L ps L

Aggregation methodEquivalent machine

Repeating the aggregation process leads to an approximated estimation of the throughput of the line.

M12 B2 M3 B3 M4

M1 B1 M2 B2 M3 B3 M4

M123 B3 M4

Decomposition methodProperties of a continuous line

Flow conservation:

THi = TH1, i = 2, …, K (1)

Ei = E1, i = 2, …, K (2)

Flow-idle time relation:

Ei = ei (1- psi –pbi) , i = 2, …, K (3)

Proof:

Ei = P{i(t) = 1 & Mi not blocked & Mi not starved}

= P{i(t) = 1 | Mi not blocked & Mi not starved}. P{Mi not blocked & Mi not starved}

= ei (1- psi –pbi)

since the proba that Mi is blocked and starved simultaneously is null in continuous flow model.

M1 B1 M2 B2 M3 B3 M4

Decomposition methodDecomposition

Decompose a K-machine line into K-1 lines of two-machines

M1 B1 M2 B2 M3 B3 M4L:

Mu(1) B(1) Md(1)L(1)

Mu(2) B(2) Md(2) L(2)

Mu(3) B(3) Md(3) L(3)

Mu(i) = upstream subline of Bi

Md(i) = downstream subline of Bi

Objective: the input/output flow of B(i) is similar to that of Bi in L

Notation :

Iu(i), eu(i), Id(i), ed(i)

E(i) : proba that Md(i) is producing

ps(i) : proba of starvation of Md(i)

pb(i) : proba. of blockage of Mu(i)

From the objective of decomposition:

E(i) = Ei+1, i = 1, …, K-1 (4)

ps(i) = psi+1 , i = 1, …, K-1 (5)

pb(i) = pbi , i = 1, …, K-1 (6)

Apply(3) to L(i):

E(i) = eu(i)(1- pb(i)) , i = 1, …, K-1 (7)

E(i) = ed(i)(1- ps(i)) , i = 1, …, K-1 ( 8)

Combine (2) & (4)

E(i) = E(1) , i = 1, …, K-1 (I)

Combine (3), (4), (5), (6),

E(i-1) = ei (1 - ps(i-1) - pb(i))

Combining with (7) & (8)

Id(i-1) + Iu(i) = 1/E(i-1) + Ii –1 (II)

Repair time of Mu(i) :

A) Failure of Mu(i) = failure of Mi, with prob. 1-

= failure of Mu(i-1), with prob.

Repair time of Mu(i)

MTTRu(i) = MTTRu(i-1) + (1-) MTTRi

1/u(i) = /u(i-1) + (1-) /i (9)

where = percentage of stoppages of Mu(i) caused by a failure of Mu(i-1)

ps i i

nb of flows interruptions of B(i-1)

nb of flow resumptions of B(i-1)

Nb of failures of Mu(i)State-transition of Mu(i)

working DOWN

Combine (9)-(10),

1 1 11 1 11

1 1 11

u u u u u i

ps i ps i i

i E i I i i E i I i i

ps i ps i ii

E i I i E i I i

u(i) = X. u(i-1) + (1-X) i (III)

with X = ps(i-1) / (Iu(i).E(i)).

Repair time of Md(i)

d(i) = Y. u(i+1) + (1-Y) i+1 (IV)

with Y = pb(i+1) / (Id(i).E(i)).

Boundary equations:

u(1) = 1, u(1) = 1, d(K-1) = K, u(K-1) = K, (V)

Equation system (I) – (V),

(I) E(i) = E(1) , i = 1, …, K-1

(II) Id(i-1) + Iu(i) = 1/E(i-1) + Ii –1

u(i) = Xu(i-1) + (1-X) i with

(IV)d(i) = Yu(i+1) + (1-Y) i+1 with

(V) u(1) = 1, u(1) = 1, d(K-1) = K, u(K-1) = K

•4(K-1) equation

•4(K-1) unknowns : u(i), u(i), d(i), d(i)

•E(i), ps(i), pb(i) are functions of u(i), u(i), d(i), d(i)

I i E i

DDX Algorithm:

Step 1: Initialisation u(i) = i, u(i) = i, d(i) = i+1, u(i) = i+1

Step 2: Forward update u(i), u(i) by equation (I)-(II)-(III)

For i = 2 to K-1, do

2.1 Evaluate the line L(i-1) to obtain E(i-1), ps(i-1), pb(i-1)

2.2 From (II), Iu(i) = 1/E(i-1) + Ii –1 - Id(i-1)

2.3 From (III)-(I), u(i) = Xu(i-1) + (1-X) i with X = ps(i-1) / (Iu(i).E(i-1)).

Step 3: Backward update d(i), d(i) with equations (I)-(II)-(IV)

For i = K-2 to 1, do

3.1 Evaluate the line L(i+1) to obtain E(i+1), ps(i+1), pb(i+1)

3.2 From (II), Id(i) = 1/E(i+1) + Ii+1 –1 - Iu(i+1)

3.3 From (IV)-(I), d(i) = Yd(i+1) + (1-Y) i+1 with Y = pb(i+1) / (Id(i).E(i+1))

Step 4: Repeat (2) – (3) till convergence, i.e. E(i) = E(1).

Distribution of material in a line with

er50 identical machines = 0.01, = 0.1, U = 1, hi = 20

Effect of bottleneck av

50 identical machines : = 0.01, = 0.1, U = 1, hi = 20except bottleneck at M10 with 10 =0.0375

Increase one buffer capacity

buffer capacity h6

Why buffer increases and which buffer decreases?

8-machines with = 0.09, = 0.75, U = 1.2, hi = 30 except h6

Distribution of buffer capacity

Which has a higher throughput rate?

9-machine line with two buffering options:

•8 buffers equally sized

•2 buffers equally sized

M1 B1 M2 B2 M3 B3 M4 B4 M5 B5 M6 B6 M7 B7 M8 B8 M9

M1 M2 M3 B3 M4 M5 M6 B6 M7 M8 M9

Throughput

Total buffer space

All machines have = 0.001, = 0.019, U = 1

What are the asymptotes

Is 8 buffers always faster?

Throughput

Total buffer space

Is 8 buffers always faster?

Perhaps not, but the difference is not significant in systems with very small buffers.

Design buffer space distribution

Design the buffers for a 20-machine production line

The machines have been selected, and the only decision remaining is the amount of space to allocate for in-process inventory.

The goal is to determine the smallest amount of in-process inventory space so that the line meets a production rate target.

The common operation time is one operation per minute.

The target production rate is 0.88 parts per minute.

Case 1 : MTBF = 200 minutes and MTTR = 10.5 minutes for all machines (ei = 0.95 parts per minute)

Case 2 : Like Case 1 except Machine 5. For Machine 5, MTBF = 100 and MTTR = 10.5 minutes (ei = 0.905 parts per minute)

Case 3 : Like Case 1 except Machine 5. For Machine 5, MTBF = 200 and MTTR = 21 minutes (ei = 0.905 parts per minute)

Are buffers really needed?

Line Production rate with no buffer

Case 1 0.487

Case 2 0.475

Case 3 0.475

Yes. How to compute these numbers? (homework)

Optimal buffer space distribution

Observation:

Buffer space is needed most where buffer level variability is greatest!

1 Chapter 7 Performance analysis of failure- prone production lines Learning objectives :...

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