Chapter 7 Performance analysis of failureprone 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, 1994. J. Li and S.M. Meerkov, Production Systems Engineering 1 Plan • 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 2 Basic concepts 3 Production lines or tranfer lines M1 B1 M2 B2 machine M3 B3 M4 Buffer Frequent production disruption by machine failures Buffers are of finite capacity Production often varies wildly 4 Production lines or tranfer lines 5 Production capability of a machine Cycle time or processing time (t) : • 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/t (parts per time unit) 6 Machine reliability model • TBF = Time Between Failure (Tup) • TTR = Time To Repair (Tdown) • MTBF = Mean TBF • MTTR = Mean TTR • Failure rate l = 1/MTBF • Repair rate m = 1/MTTR l UP DOWN m 7 Models 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. 8 Models 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 9 Mathematical 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. 10 Mathematical models of buffers Buffering capacity of a moving convey l v Ttravel N0 t N K N0 Ttravel where l = length of the convey v = speed of the convey K = maximum number of carriers in the convey 11 Interaction 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/ 12 Performance 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 13 Performance 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) 14 Failure-prone single-machine systems 15 Throughput rate MTBF m TH U U MTBF MTTR lm 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). UP DOWN 16 A machine operating at a reduced speed U' < U Operation Dependent Failure case TH U ' m U' l m U Time Dependent Failure case: TH U ' m lm 17 Production lines with unlimited buffers 18 Assumptions M1 B1 M2 B2 M3 B3 M4 • The first machine M1 is never starved • The last machine is never blocked • Each machine produces if its upstream buffer is not empty 19 Bottleneck machine M1 B1 M2 B2 M3 B3 M4 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. Ui mi li mi Uj mj lj m j , j i 20 Throughput rate M1 B1 M2 B2 M3 B3 M4 The throughput rate of a production line with unlimited buffers is equal to that of the bottleneck machines, i.e. (Why?) TH MIN i U i MTBFi Um MIN i i i MTBFi MTTRi li mi The throughput rate of a machine Mi is equal to that of the slowest upstream machine, i.e. TH j MIN i j U i MTBFi Um MIN i i i j li mi MTBFi MTTRi 21 Case of a single bottleneck machine M1 B1 M2 B2 B3 M3 M4 The level of the input buffer of the bottleck machine grows without limit (at which slope?) All downstream buffers remain limited. B2 B3 22 Case of two bottleneck machines M1 B1 M2 B2 B3 M3 M4 The level of the input buffer of each bottleneck machines grows without limit. All other buffers downstream of the first bottleneck remain limited. B1 B3 B2 Master GI2007 23 Production lines without buffers 24 Assumptions M1 B1 M2 B2 M3 B3 M4 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. 25 Case of reliable machines with different cycle times t1 1 t2 2 M1 M2 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 i MIN i 2 i 2 1 ti M1 wait 2 26 Failure-prone lines with identical cycle times M1 B1 M2 B2 M3 B3 M4 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) 27 Failure-prone lines with identical cycle times M1 B1 M2 B2 M3 B3 M4 Productivity: TH U li 1 i mi 1 li t 1 i mi 28 Proof 1) Each time interval can be decomposed as follows: UP M4 All UP UP M2 UP M1 some machine DOWN 2) tn : instant when the line produces n parts. The time interval [0, tn) includes : 3) • a total duration of nt of all UP periods, • for each machine Mi, nt/MTBFi failures requiring with total repair time of (nt/MTBFi) MTTRi. tn nt i nt MTTRi MTBFi l nt 1 i i mi n 1 x tn l t 1 i i mi TH lim 29 Impact of the length of the line The longer the line is, the higher the capacity loss is. Unlimited buffer U/(1l/m) 0,6 0,5 Throughput 0,4 lost capacity 0,3 Zero-buffers U/(1nl/m) 0,2 0,1 0 0 5 10 15 20 Nb machines 30 Aggregation of parallel machines and consecutive dependent machines 31 Aggregation M5 M1 B1 M2 M3 M4 B6 B4 M7 M6 consecutive dependent machines M1 B1 M234 parallel machines B4 M56 B6 M7 32 Aggregation of parallel machines M1 M2 Meq MS Identical parallel machines : ti = t = 1/U, li = l, mi = m Ueq = S×U leq = l meq = m 33 Aggregation of parallel machines M1 M2 Meq MS Non Identical parallel machines : ti = 1/Ui, li, mi, ei = 1/(1+ li/mi) S U eq U i leq eeq i 1 eeq 1 U eq S i 1 U i ei 1 1 leq meq S i 1 li ei S eeq av failure frequency availability of Meq 34 Aggregration of consecutive dependent machines M2 M1 ... Meq MS Machines of identical cycle time : ti = t = 1/U, li, mi leq i 1 li S eeq 1 1 E ( L) leq meq 1 S li 1 i 1 meq leq mi Failure rate equivalence 1 S li 1 i 1 mi Flow rate equivalence Average stoppage time equivalence 35 Aggregration of consecutive dependent machines Machines of nonidentical cycle time : ti = 1/Ui, li, mi • All machines slowed down to slowest one : U = min{Ui, i= 1, ..., S} • Reduced failure rate : li = Uli /Ui • Equivalent machine cycle time : Ueq = U • Failure rate equivalence : leq = Si li • Flow rate equivalence : THeq = TH(L) 1 S li 1 i 1 • Average stoppage time equivalence meq leq mi 36 Two-machine production lines with intermediate buffer 37 Motivation & cost of intermediate buffers M1 B1 M2 B2 M3 B3 M4 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 38 Motivation & cost of intermediate buffers New phenomena: •Blocking Failure of M3 •Starvation M1 M2 M3 M4 M3 M4 3t time units after the failure where t is the cycle time M1 M2 39 CTMC model of reliable line with exponential processing times M1 B M2 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 40 CTMC model of reliable line with exponential processing times B M1 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 p1 0 p1 … 1 p2 p1 p2 p1 K+1 p2 K+2 p2 41 CTMC model of reliable line with exponential processing times M1 B M2 CTMC model equivalent to M/M/1/(K+2). 0 1 r n , r 0 , if r 1 n K 3 1 r 1 n , if r 1 K 3 with r = p1/p2, corresponding to the traffic intensity. p1 0 p1 … 1 p2 p1 p2 p1 K p2 K+2 p2 42 CTMC model of reliable line with exponential processing times Performance measures (case r ≠ 1) Starving probability of M2 : 0 = (1-r)/(1-rK+3) Blocking probability of M1: K+2 = rK+2(1-r)/(1-rK+3) Throughput rate : TH = p2(1-0) = p1(1-K+2) = p1(1- rK+2)/(1-rK+3) Mean WIP K 2 r E X n n K 3 1 r n 0 1 r K 2 K 2 ( K 2) r 1 r 43 CTMC model of reliable line with exponential processing times Example : p1 = 10, p2 = 9, r = 10/9 0,4 0,35 0,3 0,25 Blocking prob. 0,2 9,5 Unlimited buffer 9 0,15 0,1 starving prob 8,5 0,05 0 0 7,5 5 10 15 20 25 30 35 Buffer capacity 7 30 Zero-buffer 6,5 25 6 20 5,5 5 0 5 10 15 20 Buffer capacity 25 30 35 Encours Throughput 8 15 10 5 0 0 5 10 15 20 Buffer capacity 25 30 35 44 Exponential model of Failure-prone lines M1 B M2 Assumptions: • Machines can break down. • Exponentially distributed times to failures and time to repair with TBFi = EXP(li) and TTRi = EXP (mi) • 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 45 Exponential model of Failure-prone lines 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 ai = 1 if Mi is UP and 0 if Mi is DOWN The state vector (a1, a2, x) is a continuous time Markov chain 46 Exponential model of Failure-prone lines 0, 0, 0 0, 0, 1 0, 0, 2 1, 0, 0 1, 0, 1 1, 0, 2 m2 0, 1, 0 m1 0, 1, 1 l1 1, 1, 0 0, 1, 2 l2 p2 1, 1, 1 1, 1, 2 p1 Exponential model of the case K = 0 • Analytical expressions of steady-state probabilities available in the book of SB Gershwin • Can be used to evaluate the performance measures 47 Slotted time model of a failure-prone line Assumptions M1 B M2 • Synchronized line, i.e. ti = t, 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. 48 Slotted time model of a failure-prone line Discrete Time Markov chain M1 B M2 The state vector (a1, a2, x) with • ai(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) + a1(t)×1{x(t-1)<N} - a2(t)×1{x(t-1)>0} • 0≤x(t) ≤ N 49 Slotted time model of a failure-prone line Discrete Time Markov chain 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 (a1, a2, 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. 50 Slotted time model of a failure-prone line Discrete Time Markov chain M1 B M2 Performance measures : • Efficiency of M1 : E1 = Sa1 = 1, x < N (a1, a2, x) • Efficiency of M2 : E2 = Sa2 = 1, x > 0 ( a1, a2, x) • Throughput rate : TH = E1×U = E2×U • WIP : S x( a1, a2, x) • Probability of starvation : (0, 1, 0) • Probability of blocking : (1, 0, N) 51 Slotted time model of a failure-prone line Analytical results Efficiency of the line E(N) = probability a machine is producing 1 r * N , if I1 I 2 * N 1 I1 (1 I 2 ) r E(N) r1 r2 r1r2 r1r2 (1 I ) N ,if I1 I 2 I ( r r )(1 2 I ) r r (1 I )2 ( N 1) 1 2 1 2 a1 p1 p2 p1 p2 r1 p2 a 2 p1 p2 p1 p2 r2 p1 1 r1 r2 r1r2 p1r2 2 r1 r2 r1r2 p2 r1 p a rp 1 2 , r* 1 2 1 , Ii i 2a1 r2 p1 2 ri WIP to be determined with expressions of steady-state probabilities available in the book of S.B. Gerswhin 52 Continuous flow models of failure-prone lines Assumptions 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 53 Continuous flow models of failure-prone lines How good is continuous flow approximation 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. 54 Continuous flow models of failure-prone lines Dynamic behavior M1 B M2 Model parameters: li: failure rate of Mi mi: repair rate of Mi • U: maximum production rate of Mi • h : buffer capacity State variables ai(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 55 Continuous flow models of failure-prone lines Dynamic behavior u1(t) h u2(t) M1 B M2 a1(t) x(t) a2(t) u1 = U u1 = U u1 = 0 u1 = U u1 = 0 u1 = 0 a1 = 1 a1 = 1 a1 = 1 a1 = 1 a1 = 0 a1 = 0 u2 = U u2 = 0 u2 = 0 u2 = U u2 = U u2 = 0 a2 = 1 a2 = 0 a2 = 0 a2 = 1 a2 = 1 a2 = 1 x(t) F2 blocking R2 F1 Starving R1 F2 56 Continuous flow models of failure-prone lines Dynamic behavior u1(t) M1 a1(t) u2(t) B x(t), h M2 a2(t) Blockage of M1 : a1(t) = 1, x(t) = h, a2(t) = 0 Starvation of M2 : a1(t) = 0, x(t) = 0, a2(t) = 1 In all other case : u1(t) = a1(t) U, u2(t) = a2(t) U 57 Continuous flow models of failure-prone lines Performance measures online proof 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) = mi/(li+mi) Mean buffer level: Q(h) 58 Continuous flow models of failure-prone lines Analytical solution Case I1 = l1/m1 I2 = l2/m2 E (h) Q (h) a I 2 eah I1 I 2 (1 I 2 ) e ah I1 (1 I1 ) I1 I 2 m1 m2 U 1 e ah I 2 (1 I 2 ) he ah I 2 I1 m1m2 ( ) I 2 (1 I 2 ) eah I1 (1 I1 ) l2 m1 l1m2 l1 l2 m1 m2 U ( l1 l2 )( m1 m2 ) 59 Continuous flow models of failure-prone lines Analytical solution Case I1 = l1/m1 I2 = l2/m2 = I E (h) l l 1 I h 1 2 U I l1l2 (1 I )2 h I (1 2 I ) l1 l2 U l1l2 h I l1 l2 (1 I ) 1 U 2 l1 m2 Q (h) h l1 l2 2 1 I h I 1 2 I U ( ) ( ) 2 l1l2 60 Continuous flow models of failure-prone lines Analytical solution Case U=1, l1 m1 l2 m2 = 0,1 0,55 infinite buffer 0,5 Throughput 0,45 In this case, 0,4 Q(h) = 0,5h zero-buffer 0,35 0,3 0,25 0,2 0 50 100 150 200 250 300 350 Buffer capacity h 61 Numerical results U = 1, l1 = 0.1, m2 = 0.1, l2 = 0.1 Throughput m1 = 0,14 m1 = 0,12 m1 = 0,10 m1 = 0,08 m1 = 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 m1 lower? 62 Numerical results U = 1, l1 = 0.1, m2 = 0.1, l2 = 0.1 WIP m1 = 0,14 m1 = 0,12 m1 = 0,10 m1 = 0,08 m1 = 0,06 Buffer capacity Discussions: • Why are the curves increasing? • Why different asymptotes? • What is the limit of WIP as N→? • Why are the curves with smaller m1 lower? 63 Numerical results U = 1, l1 = 0.1, m2 = 0.1, l2 = 0.1 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? 64 Numerical results U = 1, l1 = 0.1, m1 = 0.1, m2 = 0.1, l2 = 0.1 Throughput Improvement to nonbottleneck machine. Same graph for improvement of machine 2 Buffer capacity 65 Numerical results U = 1, l1 = 0.1 , m1 = 0.1, m2 = 0.1, l2 = 0.1 Average inventory Inventory increases as the (non-bottleneck upstream machine is improved and as the buffer space is increased. Buffer capacity 66 Numerical results U = 1, l1 = 0.1 , m1 = 0.1, m2 = 0.1, l2 = 0.1 Average inventory • Inventory decreases as the (non-bottleneck) downstream machine is improved •Inventory increases as the buffer space isincreased. Buffer capacity 67 Numerical results U = 1, m2 = 0.8, l2 = 0.09, h = 10 Should we prefer short and frequent disruptions or long and infrequent disruptions? Throughput • l1 and m1 vary together and m1/(l1 + m1) = 0.9 • Answer: short and frequent failures. • Why? Repair rate m1 68 Continuous flow models of failure-prone lines Reversiblity M1 B M2 M1 B M2 L L' 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') Proof for the continuous L2 line 69 Continuous flow models of failure-prone lines Dynamic behavior u1(t) M1 a1(t) u2(t) B x(t), h M2 a2(t) A continuous time Markov process with hybrid state space characterized by Internal state distribution (0 < x < h): Fa1a2(x) = P{a1(t) = a1, a2(t) = a2, 0 < x(t) x} fa1a2(x) = d Fa1a2(x) /dx Boundary distribution: Pa1a2(0), Pa1a2(h) 70 Case I1 = l1/m1 I2 = l2/m2 a l2 m1 l1m2 l1 l2 m1 m2 U ( l1 l2 )( m1 m2 ) f10 ( x ) f 01 ( x ) Ce ax f 00 ( x ) l1 l2 ax m m2 ax Ce , f11 ( x ) 1 Ce m1 m2 l1 l2 P11 ( h ) U l1 Ce ah , P11 ( 0 ) U l2 C l1 l2 ah l1 l2 P10 ( h ) U Ce , P01 ( 0 ) U C l1m2 l2 m1 P01 ( h ) P00 ( h ) P10 ( 0 ) P00 ( 0 ) 0 1 U ( l1 l2 ) 1 I 2 ah 1 I1 e C l2 m1 l1m2 I1 I2 71 Case I1 = l1/m1 I2 = l2/m2=I f10 ( x ) f 01 ( x ) C0 C0 I U U P11 ( h ) C0 , P11 ( 0 ) C0 f 00 ( x ) IC0 , f11 ( x ) l1 l2 l1 l2 l1 l2 P10 ( h ) U C0 , P01 ( 0 ) U C0 l1m2 l2 m1 P01 ( h ) P00 ( h ) P10 ( 0 ) P00 ( 0 ) 0 l l 1 U 1 2 (1 2 I ) C0 l1l2 (1 I ) I 2 h 72 Long failure-prone production lines 73 Introduction M1 B1 M2 B2 M3 B3 M4 • 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 74 Notation M1 B1 M2 B2 M3 B3 M4 Given isolated machine performances: Ii = li/mi 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 75 Aggregation method Equivalent machine M1 B1 M2 B2 M1 B1 M2 L12 M3 B3 M4 Replace L12 by a machine M12 of equivalent isolation throughput rate (flow equivalence), i.e. 1 m12 eM E ( L12 ) l12 m12 1 l12 / m12 E ( L12 ) 12 Repair time of M12 = Average stoppage time of M2 in L12: P(a 2 0)m2 ps( L12 )m1 1 1 1 m12 P(a 2 0)m2 ps( L12 )m1 m2 P(a 2 0)m2 ps( L12 )m1 m1 P(a 2 0) 1 E ( L12 ) ps( L12 ) 76 Aggregation method Equivalent machine M1 B1 M2 M12 B2 B2 M3 B3 M4 M3 B3 M4 M123 B3 M4 M1234 Repeating the aggregation process leads to an approximated estimation of the throughput of the line. 77 Decomposition method Properties of a continuous line M1 B1 M2 B2 M3 B3 M4 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{ai(t) = 1 & Mi not blocked & Mi not starved} = P{ai(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. 78 Decomposition method Decomposition Decompose a K-machine line into K-1 lines of two-machines L: L(1) M1 B1 M2 l1 m1 h1 l2 m2 Mu(1) B(1) Md(1) lu(1) mu(1) h1 ld(1) md(1) B2 h2 B3 M4 l3 m3 h3 l4 m4 Mu(i) = upstream subline of Bi Md(i) = downstream subline of Bi Mu(2) B(2) lu(2) mu(2) h2 Objective: the input/output flow of B(i) is similar to that of Bi in L M3 Md(2) L(2) ld(2) md(2) Mu(3) B(3) Md(3) lu(3) mu(3) h3 ld(3) md(3) L(3) 79 Decomposition method Decomposition 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) 80 Decomposition method Decomposition 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) 81 Decomposition method Decomposition Repair time of Mu(i) : A) Failure of Mu(i) = failure of Mi, with prob. 1- a = failure of Mu(i-1), with prob. a B) Repair time of Mu(i) MTTRu(i) = aMTTRu(i-1) + (1-a) MTTRi 1/mu(i) = a/mu(i-1) + (1-a) /mi (9) where a = percentage of stoppages of Mu(i) caused by a failure of Mu(i-1) 82 Decomposition method Decomposition C) nb of flows interruptions of B(i-1) = a ps ( i 1) mu ( i 1) E ( i ) lu ( i ) nb of flow resumptions of B(i-1) (10) Nb of failures of Mu(i) State-transition of Mu(i) lu(i) E(i) working DOWN mu(i) idle 83 Decomposition method Decomposition Combine (9)-(10), ps ( i 1) mu ( i 1) 1 1 mu ( i ) E ( i ) I u ( i ) mu ( i ) E ( i ) I u ( i ) mu ( i ) mi ps ( i 1) ps ( i 1) mu ( i 1) 1 mi mu ( i ) E i I i E ( i ) Iu ( i ) ( ) u( ) 1 ps ( i 1) 1 mu(i) = X. mu(i-1) + (1-X) mi (III) with X = ps(i-1) / (Iu(i).E(i)). 84 Decomposition method Decomposition Repair time of Md(i) md(i) = Y. mu(i+1) + (1-Y) mi+1 (IV) with Y = pb(i+1) / (Id(i).E(i)). Boundary equations: lu(1) = l1, mu(1) = m1, ld(K-1) = lK, mu(K-1) = mK, (V) 85 Decomposition method Decomposition 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 (III) mu(i) = Xmu(i-1) + (1-X) mi with X ps ( i 1) Iu (i ) E (i ) pb ( i 1) (IV) md(i) = Ymu(i+1) + (1-Y) mi+1 with Y I ( i ) E ( i ) d (V) lu(1) = l1, mu(1) = m1, ld(K-1) = lK, mu(K-1) = mK • 4(K-1) equation • 4(K-1) unknowns : lu(i), mu(i), ld(i), md(i) • E(i), ps(i), pb(i) are functions of lu(i), mu(i), ld(i), md(i) 86 Decomposition method Decomposition DDX Algorithm: Step 1: Initialisation lu(i) = li, mu(i) = mi, ld(i) = li+1, mu(i) = mi+1 Step 2: Forward update lu(i), mu(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), mu(i) = Xmu(i-1) + (1-X) mi with X = ps(i-1) / (Iu(i).E(i-1)). Step 3: Backward update ld(i), md(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), md(i) = Ymd(i+1) + (1-Y) mi+1 with Y = pb(i+1) / (Id(i).E(i+1)) Step 4: Repeat (2) – (3) till convergence, i.e. E(i) = E(1). 87 Distribution of material in a line with average buffer 50 identical machines l = 0.01, m = 0.1, U = 1, hi = 20 88 Effect of bottleneck average buffer 50 identical machines : l = 0.01, m = 0.1, U = 1, hi = 20 except bottleneck at M10 with l10 =0.0375 89 Increase one buffer capacity 8-machines with l = 0.09, m = 0.75, U = 1.2, hi = 30 except h6 Why buffer increases and which buffer decreases? buffer capacity h6 90 Distribution of buffer capacity Which has a higher throughput rate? 9-machine line with two buffering options: • 8 buffers equally sized M1 B1 M2 B2 M3 B3 M4 B4 M5 B5 M6 B6 M7 B7 M8 B8 M9 • 2 buffers equally sized M1 M2 M3 B3 M4 M5 M6 B6 M7 M8 M9 91 Distribution of buffer capacity Throughput All machines have l = 0.001, m = 0.019, U = 1 What are the asymptotes Is 8 buffers always faster? Total buffer space 92 Distribution of buffer capacity Throughput Is 8 buffers always faster? Perhaps not, but the difference is not significant in systems with very small buffers. Total buffer space 93 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 inprocess inventory space so that the line meets a production rate target. 94 Design buffer space distribution The common operation time is one operation per minute. The target production rate is 0.88 parts per minute. 95 Design buffer space distribution 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) 96 Design buffer space distribution 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) 97 Design buffer space distribution Optimal buffer space distribution Observation: Buffer space is needed most where buffer level variability is greatest! 98