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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

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

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3 Definitions and justifications

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4 Bernoulli lines Definition 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,... M1 B1 M2 B2 M3 B3 M4 Justification: appropriate for high volume assembly lines.

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5 Bernoulli lines Definition Machines are subject to Time Dependent Failures (TDF). M1 B1 M2 B2 M3 B3 M4 Justifications: For most practical cases, the difference of performance measures with TDF and ODF models is within 1% - 3% (especially when buffers are not too small). The error resulting from the selection of failure model is small with respect to usual errors in identification of reliability parameters (rarelly known with accuracy better than 5% - 10%. The TDF model is simpler for analysis

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6 Bernoulli lines Definition 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 p i and DOWN with proba 1-p i, independent of its status in all previous time slots and independent of the status of remaining system. M1 B1 M2 B2 M3 B3 M4 Justification: It is practical for describing assembly operations where the downtime is typically very short and comparable with the cycle time of the machine.

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7 Bernoulli lines Operating rules M1 B1 M2 B2 M3 B3 M4 A Bernoulli line can be represented by a vector (p 1,..., p M, N 1,..., N M-1 ) of machine reliability parameters and buffer capacities. The time is slotted with the cycle time 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 p i and DOWN with proba (1-p i ) and it is independent of past history and the status of the remaining system

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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.

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9 Transformation of a failure-prone line into a Bernoulli line M1 B1 M2 B2 M3 B3 M4 A failure-prone line with parameters : i = 1/U i, i, i, h i Bernoulli Line transformation = min{ i i} p i = e i / i, with e i = 1/(1+ i / i ) N i = min{h i i i+1, h i i+1 i } + 1 Justifications: From numerical results with real data, the error between the two models is quite small (less than 4%) for the case N i 2 and is up to 7% - 8% for the case N i < 2. The theory and results work for fractional buffer sizes as well.

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10 Transformation of a failure-prone line into a Bernoulli line M1 B1 M2 B2 M3 B3 M4 Why N i = N i = min{h i i i+1, h i i+1 i } +1: A Bernoulli buffer can prevent starvation of the downstream machine and the blockage of upstream machine for a number of time slots at most equal to N i h i i+1 = largest time during which the downstream machine is protected from failure of upstream machine h i i i+1 = fraction of average downtime of the upstream machine that can be accommodated by the buffer. h i i+1 i = fraction of average downtime of the downstream machine that can be accommodated by the buffer. Fractional buffer sizes are allowed in this chapter

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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}.

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12 Two-machine Bernoulli lines

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13 States of the system: Bernoulli machines are memoryless System state = Buffer state x n at the end of time slot n x n is a discrete time Markov chain State transition diagram M1 B M2 p1p1 p2p2 N > 0 01 N-1N p 01 … p 12 p N-2,N-1 p N-1,N p 10 p 21 p N-1,N-2 p N,N-1 p 00 p 11 p NN p N-1,N-1 DTMC model

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14 Blockage of M1 in period n+1 x n = N M1 is UP M2 is DOWN 01 N-1N p 01 … p 12 p N-2,N-1 p N-1,N p 10 p 21 p N-1,N-2 p N,N-1 p 00 p 11 p NN p N-1,N-1 DTMC model Starvation of M2 in period n+1 x n = 0 M2 is UP M1 B M2 p1p1 p2p2 N

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15 DTMC model Transition probabilities p 00 = 1 - p 1 p 01 = p 1 p 10 = (1 - p 1 )p 2 M1 B M2 p1p1 p2p2 N 01 N-1N p 01 … p 12 p N-2,N-1 p N-1,N p 10 p 21 p N-1,N-2 p N,N-1 p 00 p 11 p NN p N-1,N-1 p ii = p 1 p 2 + (1 - p 1 ) (1 - p 2 ) p i,i+1 = p 1 (1 - p 2 ),i = 1,..., N-1 p i+1,i = (1 - p 1 )p 2 p NN = p 1 p 2 + (1 - p 1 ) (1 - p 2 ) + p 1 (1 - p 2 ) = p 1 p 2 + 1 - p 2

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16 DTMC model Steady state distribution 01 N-1N p 01 … p 12 p N-2,N-1 p N-1,N p 10 p 21 p N-1,N-2 p N,N-1 p 00 p 11 p NN p N-1,N-1 Equilibrium equation states {0,1,..., i} : i+1 p i+1,i = i p i,i+1, i < N Normalization equation 0 + 1 +... + N = 1 M1 B M2 p1p1 p2p2 N

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17 DTMC model Steady state distribution 01 N-1N p 01 … p 12 p N-2,N-1 p N-1,N p 10 p 21 p N-1,N-2 p N,N-1 p 00 p 11 p NN p N-1,N-1 To be shown : M1 B M2 p1p1 p2p2 N

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18 DTMC model Steady state distribution 01 N-1N p 01 … p 12 p N-2,N-1 p N-1,N p 10 p 21 p N-1,N-2 p N,N-1 p 00 p 11 p NN p N-1,N-1 Case of identical machines, p 1 = p 2 = p M1 B M2 p1p1 p2p2 N For practical case with p 1, 0 0 i 1/N, i > 0

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19 DTMC model Steady state distribution 01 N-1N p 01 … p 12 p N-2,N-1 p N-1,N p 10 p 21 p N-1,N-2 p N,N-1 p 00 p 11 p NN p N-1,N-1 Case of nonidentical machines, i.e. p 1 p 2 M1 B M2 p1p1 p2p2 N

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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

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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

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22 DTMC model Steady state distribution Theorem: 0 = Q(p 1, p 2, N) N = Q(p 2, p 1, N)/(1-p 2 ) (y, x) = 1/ (x, y) Meaning of Q(p 1, p 2, N) : The intermediate buffer is empty Implication : M2 is starved if it is UP Meaning of Q(p 2, p 1, N) : The intermediate buffer is full & its downstream machine does not produce Implication : M1 is blocked if it is UP

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23 DTMC model Performance measures Production rate (PR) PR = p 2 (1 - 0 ) PR = p 1 (1 - N (1-p 2 )) PR = p 2 (1 - Q(p 1, p 2, N)) PR = p 1 (1 - Q(p 2, p 1, N))

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24 DTMC model Performance measures Work In Process (WIP)

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25 DTMC model Blockage and Starvation Blocking probability of M 1 (BL 1 ) BL 1 = p 1 N (1-p 2 ) = p 1 Q(p 2, p 1, N) Starvation probability of M 2 (ST 2 ) ST 2 = p 2 0 = p 2 Q(p 1, p 2, N) Relation with PR PR = p 1 - BL 1 PR = p 2 - ST 2

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

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27 DTMC model Theorem:

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28 Long Bernoulli Lines

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29 DTMC model The vector of buffer states (x 1 (n), x 2 (n),..., x M-1 (n)) is a discrete time Markov chain. Unfortunately, the state space is large with (N 1 +1) (N 2 +1)... (N M-1 +1) states. Analytical formula are not available for performance measures of long Bernoulli lines. Focus on an aggregation approach. M1 B1 M2 B2 M3 B3 M4 p2p2 p3p3 p4p4 N2N2 N3N3 p1p1 N1N1

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30 Idea of the aggregation Backward aggregration p b 3 = production rate of the 2-machine line (M3, B3, M4) Repeating the aggregation process p b i = production rate of the 2-machine line (M i, B i, M b i+1 ) Drawback : is quite different from the production rate of the M-machine line M1 B1 M2 B2 M3 B3 M4 M1 B1 M2 B2 Mb3Mb3 M1 B1 Mb2Mb2 Mb1Mb1

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31 Idea of the aggregation Forward aggreation Forward aggreation is introduced to improve the aggregration. p f i is determine to take into account the starvation of B i-1 in the 2- machine line (M f i-1, B i-1, M b i ) The whole process repeats to futher improved the aggregation Mf2Mf2 B2 Mb3Mb3 M1 B1 Mb2Mb2 Mf3Mf3 B3 Mb4Mb4 Mf4Mf4

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

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

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

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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?

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

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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

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38 Aggregation procedure Home work examples Eight 5-machines with with identical buffer capacity N i = 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

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39 Aggregation procedure Properties Static law of production systems Monotonicity : The production rate PR(p 1,..., p M, N 1,..., N M-1 ) is strictly increasing in N i strictly increasing in p i

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40 Aggregation procedure Properties Reversibility : Consider a line L and its reverse Lr with opposite flow direction. Then, Implications: 1.More capacity at the end of line is not appropriate for buffer capacity assignment 2.If only one buffer is possible and all machines are identical, then it should be in the middle of the line 3.If all machines are identical and a total buffering capacity N* must be allocated, reversibility implies "symmetric assignment ". 4.For 3/, the optimal buffer assignment is of the "inverted bowl" pattern. However, the difference with respect to "equal capacity" assignment is not significant.

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41 Continuous Improvement of Bernoulli Lines

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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?

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43 Constrained Improvability

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

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

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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

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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.

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

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49 WF unimprovable allocation Theorem. The sequence p* i such that i p* 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* = 0.9 5. Compare with equal capacity.

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50 WF continuous improvement WF continuous improvement procedure: Determine WIPi, for all i Determine the buffer with the largest |WIP i - N i /2|. Assume this is buffer k If WIP k - N k /2 > 0, re-allocate a sufficient small amount of work, p k, from M k to M k+1 ; If WIP k - N k /2 <0, re-allocate p k+1 from M k+1 to M k. Return to step 1) Example (home work): Continuous improvement of a 4 machine line with Ni = 5, p* = 0.9 4 and = 0.01. Initially, p = (0.9675, 0.9225, 0.8780, 0.8372)

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51 Improvability wrt WF and BC simultaneously Theorem: A Bernoulli line is unimprovable wrt WF and BC simultaneously iff Corollary. Under condition (WF&BC1), and, moreover where N is the capacity of each buffer, i.e. equal capacity buffers.

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52 Unimprovabe allocation wrt WF and BC Unimprovable allocation Theorem. Let N* be a multiple of M-1. Then the series p*i and N*i, which render the line unimprovable wrt WF and BC, are given by PR** can be determined as PR* with N* i. flat inverted bowl WF dist. uniform BC dist.

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Improvabiblity wrt BC Theorem: A Bernoulli line is unimprovable wrt BC iff the quantity is maximized over all sequences N' i such that i N' i = N*. Condition of little practical importance.

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Improvabiblity wrt BC Numerical Fact. The production rate ensured by the buffer capacity allocation defined by (BC1) is almost always the same as the production rate defined by the allocation that minimizes over all sequences N' i such that i N' i = N*. Implication: A line is practically unimprovable wrt BC if the occupancy of each buffer B i-1 is as close to the availability of buffer Bi as possible. MiMi B i-1 BiBi

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Improvabiblity wrt BC BC continuous improvement procedure: Determine WIPi, for all i Determine the buffer with the largest |WIP i - (N i+1 - WIP i+1 )|. Assume this is buffer k If WIP k - (N k+1 - WIP k+1 ) > 0, transfer a unit of capacity from B k to B k+1 ; If WIP k - (N k+1 - WIP k+1 ) < 0, re-allocate a unit from B k+1 to B k. Return to step 1) Example (home work): Continuous improvement of a 11 machine line with pi = 0.8, i = 6, and p6 = 0.6. N* = 24. Determine the unimprovable buffer allocation (PR = 0.5843).

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Unconstrained Improvability

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Bottleneck machine Definition: A machine Mi is the bottleneck machine (BN-m) of a Bernoulli line if Problems with this definition: 1/ Gradient information cannot be measured on shopfloor 2/ No analytical methods for evaluation of the gradients Remark: gradient estimation is possible with sample path approaches (to be addressed).

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Bottleneck machine The best machine is the bottleneck The worst machine is not the bottleneck Machine with the smalllest pi is not always the BN-m Machine with the largest WIP in front is not always the BN-m 0.9 6 0.7 6 0.8 1 0.7 1 0.75 4 0.6 6 0.7 2 0.85 PR/ p0.050.060.280.380.310.170.060.05 WIPi 5.595.390.870.691.681.180.66 0.8 2 0.83 2 0.77 3 0.8 PR/ p0.3690.4520.4430.022

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Bottleneck buffer Definition: A buffer Bi is the bottleneck buffer (BN-b) of a Bernoulli line if Buffer with the smallest Ni is not necessarily the BN-b 0.8 3 0.85 3 2 0.9 PR(Ni+1)0.7690.7660.763

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Bottlenecks in 2-machine lines Theorem: For a 2-machine Bernoulli line, if and only if BL 1 ST 2 ). Remarks : The theorem reformulates partial derivatives in terms of "measurable" and "calculable" probabilities. It offers the possibility to identify BN-m without knowing the parameters of the system. It offers a simple graphic way of representing the BN-m.

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Bottlenecks in 2-machine lines STi00.0215 BLi0.12150 0.9 2 0.8 Arrow in the direction of the inequality of the two probability Arrow pointing to the BN-m

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Bottlenecks in long lines Arrow Assignment Rule: If BL i > ST i+1, assign the arrow pointing from M i to M i+1. If BL i < ST i+1, assign the arrow pointing from M i+1 to M i.

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Bottlenecks in long lines Bottleneck indicator: If there is a single machine with no outgoing arrows, it is the BN-m If there are multiple machines with no outgoing arrows, the one with the largest severity is the Primary BN-m (PBN-m), where the severity of each BN-m is defined by S i = |ST i+1 - BL i | + |BL i-1 -ST i |, i = 2,..., M-1 S 1 = |ST 2 - BL 1 | S M = |BL M-1 -ST M | The BN-b is the buffer immediately upstream the BN-m (or PBN-m) if it is more often starved than blocked, or immediately downstream the BN-m ( or PBN-m) if it is more often blocked than starved. Remark : It was shown numerically that Bottleneck Indicator correctly identifies the BN in most cases.

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BN-m Bottlenecks in long lines Single Bottleneck Multiple Bottlenecks 0.9 6 0.7 6 0.8 1 0.7 1 0.75 4 0.6 6 0.7 2 0.85 STi0000.090.230.10.20.36 BLi 0.40.20.30.140.0300.010 BN-b BN-m 0.9 2 0.5 2 0.9 2 2 2 2 0.6 2 0.9 STi000.390.370.330270.110.41 BLi 0.410.010.030.050.10.1700 BN-b PBN-m

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Potency of buffering Motivation: When the worst machine is not the BN of the system, the buffer capacity is often incorrectly set. Need to assess the buffering quality Definition : The buffering of a production system is weakly potent if the BN-m is the worst machine; otherwise it is not potent potent if it is weakly potent and its production rate is sufficiently close to the BN-m efficiency (i.e. within 5%) strongly potent if it is potent and the system has the smallest possible total buffer capacity.

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