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Chapter 8 Performance analysis and design of Bernoulli lines

Chapter 8 Performance analysis and design of Bernoulli 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 :

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Chapter 8 Performance analysis and design of Bernoulli lines

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  1. Chapter 8 Performance analysis and design of Bernoulli 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 : J. Li and S.M. Meerkov, Production Systems Engineering

  2. Plan • Definitions and justifications • Two-machine Bernoulli lines • Long Bernoulli Lines • Continuous Improvement of Bernoulli Lines • Constrained Improvability • Unconstrained Improvability

  3. Definitions and justifications

  4. Bernoulli lines Definition M1 B1 M2 B2 M3 B3 M4 • A Bernoulli line is a synchronuous line with all machines having identical cycle time. • It is a slotted time model with time indexed t = 0, 1, 2, ... Justification: appropriate for high volume assembly lines.

  5. Bernoulli lines Definition M1 B1 M2 B2 M3 B3 M4 • Machines are subject to Time DependentFailures(TDF). • Justifications: • For mostpractical cases, the difference of performance measureswith TDF and ODFmodelsiswithin 1% - 3% (especiallywhen buffers are not toosmall). • The errorresultingfrom the selection of failure model issmallwith respect to usualerrors in identification of reliabilityparameters (rarellyknownwithaccuracybetterthan 5% - 10%. • The TDF model issimpler for analysis

  6. Bernoulli lines Definition M1 B1 M2 B2 M3 B3 M4 • Each machine is characterized by a Bernoulli reliability model. • At the beginning of each time slot, • the status of each machine Mi - UP or DOWN - is determined by a chance experiment. • It is UP with proba pi and DOWN with proba 1-pi, independent of its status in all previous time slots and independent of the status of remaining system. • Justification: • It ispractical for describingassemblyoperationswhere the downtimeistypicallyvery short and comparable with the cycle time of the machine.

  7. Bernoulli lines Operating rules M1 B1 M2 B2 M3 B3 M4 • A Bernoulli line can be represented by a vector • (p1, ..., pM, N1, ..., NM-1) • of machine reliability parameters and buffer capacities. • The time is slotted with the cycle time t of the machines. • The status of each machine is determined at the beginning and the state of the buffers at the end of each time slot. • The status of a machine is UP with proba pi and DOWN with proba (1-pi) and it is independent of past history and the status of the remaining system

  8. Bernoulli lines Operating rules M1 B1 M2 B2 M3 B3 M4 • Blocking Before Service: • an UP machine is blocked if its downstream buffer is full at the end of previous time slot and the downstream machine cannot produce. • It is starved if its upstream buffer is empty at the end of the previous time slot. • At the end of a time slot, an UP machine that is neither blocked nor starved removes one part from its upstream buffer and adds one part in its downstream buffer. • The first machine is never starved; the last machine is never blocked.

  9. Transformation of a failure-prone line into a Bernoulli line M1 B1 M2 B2 M3 B3 M4 • A failure-prone line with parameters : • ti = 1/Ui, li, mi, hi • Bernoulli Line transformation • t = min{ti, "i} • pi = tei/ti, with ei = 1/(1+ li/mi) • Ni = min{himiti+1, himi+1ti} + 1 • Justifications: • Fromnumericalresultswith real data, the errorbetween the twomodelsisquitesmall (lessthan 4%) for the case Ni ≥ 2 and is up to 7% - 8% for the case Ni < 2. • The theory and resultswork for fractional buffer sizes as well.

  10. Transformation of a failure-prone line into a Bernoulli line M1 B1 M2 B2 M3 B3 M4 • Why Ni = Ni = min{himiti+1, himi+1ti} +1: • A Bernoulli buffer canpreventstarvation of the downstream machine and the blockage of upstream machine for a number of time slots atmostequal to Ni • hiti+1 = largest time duringwhich the downstream machine isprotectedfromfailure of upstream machine • himiti+1= fraction of averagedowntime of the upstream machine thatcanbeaccommodated by the buffer. • himi+1ti= fraction of averagedowntime of the downstream machine thatcanbeaccommodated by the buffer. • Fractional buffer sizes are allowedin thischapter

  11. Transformation of a failure-prone line into a Bernoulli line Examples to work out: Line 1 e = {0.867; 0.852; 0.925; 0.895; 0.943; 0.897; 0.892; 0.935; 0.903; 0.870}; Tdown = {14.23; 16.89; 18.83; 16.08; 7.65; 11.09; 19.05; 18.76; 11.15; 18.42}; N = {7.026; 17.350; 33.461; 5.345; 9.861; 12.097; 11.955; 26.133; 14.527}; U = {1.950; 1.231; 1.607; 1.486; 1.891; 1.762; 1.457; 1.019; 1.822; 1.445}. Line 2 e = {0.945; 0.873; 0.911; 0.899; 0.939; 0.926; 0.896; 0.852; 0.932; 0.895}; Tdown = {14.22; 16.89; 18.83; 16.08; 7.65; 11.09; 19.05; 18.76; 11.15; 18.42}; N = {5.535; 31.138; 20.578; 37.614; 21.310; 19.653; 34.618; 23.380; 12.093}; U = {1.672; 1.838; 1.020; 1.681; 1.380; 1.832; 1.503; 1.709; 1.429; 1.305}. Line 3 e = {0.869; 0.869; 0.918; 0.880; 0.904; 0.865; 0.920; 0.888; 0.936; 0.935}; Tdown = {13.91; 12.45; 18.48; 17.33; 14.68; 17.27; 14.90; 10.13; 9.35; 10.12}; N = {26.746; 32.819; 38.490; 23.291; 35.805; 11.054; 39.291; 14.501; 13.832}; U = {1.534; 1.727; 1.309; 1.839; 1.568; 1.370; 1.703; 1.547; 1.445; 1.695}.

  12. Two-machine Bernoulli lines

  13. p01 N-1 N 0 1 … DTMC model p1 N > 0 p2 M1 B M2 • States of the system: • Bernoulli machines are memoryless • System state = Buffer state xn at the end of time slot n • xn is a discrete time Markov chain • State transition diagram pN-1,N-1 p11 pN-2,N-1 pNN p00 p12 pN-1,N pN,N-1 pN-1,N-2 p10 p21

  14. p01 N-1 N 0 1 … DTMC model p1 N p2 M1 B M2 • Blockage of M1 in period n+1 • xn = N • M1 is UP • M2 is DOWN • Starvation of M2 in period n+1 • xn = 0 • M2 is UP pN-1,N-1 p11 pN-2,N-1 pNN p00 p12 pN-1,N pN,N-1 pN-1,N-2 p10 p21

  15. p01 N-1 N 0 1 … DTMC model Transition probabilities p1 N p2 M1 B M2 p00 = 1 - p1 p01 = p1 p10 = (1 - p1)p2 pii = p1p2 + (1 - p1) (1 - p2) pi,i+1 = p1(1 - p2), i = 1, ..., N-1 pi+1,i = (1 - p1)p2 pNN = p1p2 + (1 - p1) (1 - p2) + p1(1 - p2) = p1p2 +1 - p2 pN-1,N-1 p11 pN-2,N-1 pNN p00 p12 pN-1,N pN,N-1 pN-1,N-2 p10 p21

  16. p01 N-1 N 0 1 … DTMC model Steady state distribution p1 N p2 M1 B M2 • Equilibrium equation • states {0,1, ..., i} : pi+1pi+1,i = pipi,i+1, "i < N • Normalization equation • p0+ p1 + ... + pN = 1 pN-1,N-1 p11 pN-2,N-1 pNN p00 p12 pN-1,N pN,N-1 pN-1,N-2 p10 p21

  17. p01 N-1 N 0 1 … DTMC model Steady state distribution p1 N p2 M1 B M2 To be shown : pN-1,N-1 p11 pN-2,N-1 pNN p00 p12 pN-1,N pN,N-1 pN-1,N-2 p10 p21

  18. p01 N-1 N 0 1 … DTMC model Steady state distribution p1 N p2 M1 B M2 Case of identical machines, p1 = p2 = p For practical case with p  1, p0 0 pi 1/N, "i > 0 pN-1,N-1 p11 pN-2,N-1 pNN p00 p12 pN-1,N pN,N-1 pN-1,N-2 p10 p21

  19. p01 N-1 N 0 1 … DTMC model Steady state distribution p1 N p2 M1 B M2 Case of nonidentical machines, i.e. p1 ≠p2 pN-1,N-1 p11 pN-2,N-1 pNN p00 p12 pN-1,N pN,N-1 pN-1,N-2 p10 p21

  20. DTMC model Steady state distribution p1 = 0.8, p2 = 0.82, N = 5 p1 = 0.82, p2 = 0.8, N = 5 p1 = 0.6, p2 = 0.9, N = 5 p1 = 0.9, p2 = 0.6, N = 5

  21. DTMC model Steady state distribution • Theorem: Function Q(x, y, N) defined below, with 0<x<1, 0<y<1, and N ≥ 1, takes values on (0,1) and is • strictly decreasing in x, • strictly increasing in y • strictly decreasing in N • where

  22. DTMC model Steady state distribution • Theorem: • p0 = Q(p1, p2, N) • pN = Q(p2, p1, N)/(1-p2) • a(y, x) = 1/a(x, y) • Meaning of Q(p1, p2, N) : • The intermediate buffer is empty • Implication : M2 is starved if it is UP • Meaning of Q(p2, p1, N) : • The intermediate buffer is full & its downstream machine does not produce • Implication : M1 is blocked if it is UP

  23. DTMC model Performance measures • Production rate (PR) • PR = p2(1 - p0) • PR = p1(1 - pN(1-p2)) • PR = p2(1 - Q(p1, p2, N)) • PR = p1(1 - Q(p2, p1, N))

  24. DTMC model Performance measures Work In Process (WIP)

  25. DTMC model Blockage and Starvation Blocking probability of M1 (BL1) BL1 = p1pN(1-p2) = p1 Q(p2, p1, N) Starvation probability of M2 (ST2) ST2 = p2p0 = p2 Q(p1, p2, N) Relation with PR PR = p1 - BL1 PR = p2 - ST2

  26. DTMC model L1: p1 = p2 = 0.9 L2: p1 = 0.9, p2 = 0.7 L3: p1 = 0.7, p2 = 0.9

  27. DTMC model Theorem:

  28. Long Bernoulli Lines

  29. DTMC model p1 N1 p2 N2 p3 N3 p4 M1 B1 M2 B2 M3 B3 M4 • The vector of buffer states • (x1(n), x2(n), ..., xM-1(n)) • is a discrete time Markov chain. • Unfortunately, the state space is large with (N1+1) (N2+1)... (NM-1+1) states. • Analytical formula are not available for performance measures of long Bernoulli lines. • Focus on an aggregation approach.

  30. Idea of the aggregation Backward aggregration Mb3 Mb1 Mb2 M1 M1 M1 M2 M2 M3 M4 B1 B1 B1 B2 B2 B3 • pb3 = production rate of the 2-machine line (M3, B3, M4) • Repeating the aggregation process • pbi = production rate of the 2-machine line (Mi, Bi, Mbi+1) • Drawback : is quite different from the production rate of the M-machine line

  31. Idea of the aggregation Forward aggreation Mb2 Mb3 M1 Mf2 Mf3 Mf4 Mb4 B1 B2 B3 • Forward aggreation is introduced to improve the aggregration. • pfi is determine to take into account the starvation of Bi-1 in the 2-machine line (Mfi-1, Bi-1, Mbi) • The whole process repeats to futher improved the aggregation

  32. Aggregation procedure Formal definition The recursive aggregation procedure is as follow (Why?) with initial condition and boundary conditions

  33. Aggregation procedure Example to workout with Excel A 3-machine line L = (0.9, 0.9, 0.9, 2, 2)

  34. Aggregation procedure Convergence Theorem. Both sequence pfi(s) and pbi(s) are converging, i.e. the following limits exist : For each i, the sequence pfi(s) is monotonically decreasing and the sequence pfi(s) is monotonically increasing. Moreover, Interpretation the downstream subline of buffer Bi-1 the upstream subline of buffer Bi

  35. Aggregation procedure Exercice L1 : (0.9, 0.9, 0.9, 0.9, 0.9; 3, 3, 3, 3) L2 : (0.7; 0.75; 0.8; 0.85; 0.9; 3, 3, 3, 3) L3: (0.7; 0.85; 0.9; 0.85; 0.7; 3, 3, 3, 3) L4: (0.9; 0.85; 0.7; 0.85; 0.9; 3, 3, 3, 3) How the production capacity is distributed in above lines?

  36. Aggregation procedure Performance measures Production rate estimation: WIP estimation estimated directly for the corresponding 2-machine line Blockage estimation Starvation estimation

  37. Aggregation procedure Numerical evidence on the accuracy of the estimates • In general, the PR estimate is relatively accurate with the error within 1% for most cases and 3% for the largest error • The accuracy of WIP, ST and BL estimates is typically lower • The highest accuracy of all estimates is for the uniform machine efficiency pattern • The lowest accuracy is for the inverted bowl and "oscillating" pattern

  38. Aggregation procedure Home work examples Eight 5-machines with with identical buffer capacity Ni = N varying from 1 to 20 L1 : p = [0.9; 0.9; 0.9; 0.9; 0.9] :uniform pattern L2 : p = [0.9; 0.85; 0.8; 0.75; 0.7] : decreasing efficiency L3 : p = [0.7; 0.75; 0.8; 0.85; 0.9] : increasing efficiency L4 : p = [0.9; 0.85; 0.7; 0.85; 0.9] : bowl pattern L5 : p = [0.7; 0.85; 0.9; 0.85; 0.7] : inverted bowl pattern L6 : p = [0.7; 0.9; 0.7; 0.9; 0.7] : oscillating L7 : p = [0.9; 0.7; 0.9; 0.7; 0.9] : oscillating L8 : p = [0.75; 0.75; 0.95; 0.75; 0.75] : single bottleneck

  39. Aggregation procedure Properties Static law of production systems • Monotonicity : • The production rate PR(p1, ..., pM, N1, ..., NM-1) is • strictly increasing in Ni • strictly increasing in pi

  40. Aggregation procedure Properties Reversibility : Consider a line L and its reverse Lr with opposite flow direction. Then, Implications: More capacity at the end of line is not appropriate for buffer capacity assignment If only one buffer is possible and all machines are identical, then it should be in the middle of the line If all machines are identical and a total buffering capacity N* must be allocated, reversibility implies "symmetric assignment". For 3/, the optimal buffer assignment is of the "inverted bowl" pattern. However, the difference with respect to "equal capacity" assignment is not significant.

  41. Continuous Improvement of Bernoulli Lines

  42. Two improvability concepts Constrained improvability : Can a production system be improved by redistributing its limited buffer capacity and workforce resources? Unconstrained Improbability : Identify the bottleneck resource (buffer capacity or machine capability) such that its improvement best improves the system?

  43. Constrained Improvability

  44. Resource constraints Buffer capacity constraint (BC): Workforce constraints (WF): Production rates of the machines depend on workforce assignment

  45. Definitions • Definition: A Bernoulli line is • improvable wrt BC if there exists a buffer assignment N'i such that SiN'i = N* and • PR(p1, ..., pM, N'1, ..., N'M-1) > PR(p1, ..., pM, N1, ..., NM-1) • improvable wrt WF if there exists a workforce assignment p'i such that Pi p'i = p* and • PR(p'1, ..., p'M, N1, ..., NM-1) > PR(p1, ..., pM, N1, ..., NM-1) • improvable wrt BC and WF simultaneously if there exist sequences N'i and p'i such that Si N'i = N*, Pi p'i = p* and • PR(p'1, ..., p'M, N'1, ..., N'M-1) > PR(p1, ..., pM, N1, ..., NM-1)

  46. Improvability with respec to WF Theorem: A Bernoulli line is unimprovable wrt WF iff where are the steady states of the recursive aggregation procedure. Corollary. Under condition (WF1), which implies Half buffer capacity usage

  47. Improvability with respec to WF WF-improvability indicator: A Bernoulli line is practically unimprovable wrt workforce if each buffer is, on the average, close to half full.

  48. WF unimprovable allocation Unimprovable allocation Theorem. If Si Ni-1 ≤ M/2, then the series x(n) defined below converges to PR* where

  49. WF unimprovable allocation Theorem. The sequence p*i such that Pip*i = p*, which renders the line unimprovable wrt WF, is given by Corollary. If all buffers are of equal capacity, i.e. Ni = N, then which is a "flat" inverted bowl allocation. Example : M = 5, Ni = 2, p* = 0.95. Compare with equal capacity.

  50. WF continuous improvement • WF continuous improvement procedure: • Determine WIPi, for all i • Determine the buffer with the largest |WIPi - Ni/2|. Assume this is buffer k • If WIPk - Nk/2 > 0, re-allocate a sufficient small amount of work, epk, from Mk to Mk+1; If WIPk - Nk/2 <0, re-allocate epk+1 from Mk+1 to Mk. • Return to step 1) • Example (home work): Continuous improvement of a 4 machine line with Ni = 5, p* = 0.94 and e = 0.01. Initially, p = (0.9675, 0.9225, 0.8780, 0.8372)

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