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1 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

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2 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

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3 Basic concepts

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4 Production 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 machineBuffer

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5 Production lines or tranfer lines

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6 Production 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)

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7 Machine reliability model TBF = Time Between Failure (T up ) TTR = Time To Repair (T down ) MTBF = Mean TBF MTTR = Mean TTR Failure rate = 1/MTBF Repair rate = 1/MTTR UP DOWN

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8 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.

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9 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

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10 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.

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11 Mathematical models of buffers Buffering capacity of a moving convey where l = length of the convey v = speed of the convey K = maximum number of carriers in the convey

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12 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: N BBS = N BAS +1 Starvation : an idle up machine is starved if its upstream buffer is empty/

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13 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

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14 Performance measures Work-in-process of the i-th buffer (WIP i ) Average number of parts contained in the i-th buffer. Total work-in-process (WIP): Average number of parts) in the system WIP = WIP 1 + WIP Probability of blocking (BL i ) Probability of starvation (ST i )

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15 Failure-prone single-machine systems

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16 Throughput 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). UP DOWN

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17 A machine operating at a reduced speed U' < U Operation Dependent Failure case Time Dependent Failure case:

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18 Production lines with unlimited buffers

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19 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

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20 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

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21 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

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22 Case 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 B2 B3

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Master GI Case 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 B3 B2 B1

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24 Production lines without buffers

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25 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

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26 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. M1M2 222 M1 wait

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27 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

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28 Failure-prone lines with identical cycle times Productivity: M1 B1 M2 B2 M3 B3 M4

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29 Proof 1)Each time interval can be decomposed as follows: 2)t n : instant when the line produces n parts. The time interval [0, t n ) includes : a total duration of n of all UP periods, for each machine Mi, n /MTBF i failures requiring with total repair time of (n /MTBF i ) MTTR i. 3) UPM4UPM2UPM1 All UP some machine DOWN

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30 Impact of the length of the line Unlimited buffer U Zero-buffers U n lost capacity The longer the line is, the higher the capacity loss is.

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31 Aggregation of parallel machines and consecutive dependent machines

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32 Aggregation M1 B1 M2M4 B4 M5 M3 M1 B1 M234 B4 M56 M6 B6 M7 B6 M7 consecutive dependent machines parallel machines

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33 Aggregation of parallel machines M1M1 M2M2 MSMS M eq Identical parallel machines : i = = 1/U, i =, i = U eq = S×U eq = eq =

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34 Aggregation of parallel machines M1M1 M2M2 MSMS M eq Non Identical parallel machines : i = 1/U i, i, i, e i = 1/(1+ i / i ) av failure frequency availability of M eq

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35 Aggregration of consecutive dependent machines M1M1 MSMS M2M2 Failure rate equivalence... M eq Machines of identical cycle time : i = = 1/U, i, i Flow rate equivalence Average stoppage time equivalence

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36 Aggregration of consecutive dependent machines Machines of nonidentical cycle time : i = 1/U i, i, i All machines slowed down to slowest one : U = min{U i, i= 1,..., S} Reduced failure rate : i = U i /U i Equivalent machine cycle time : U eq = U Failure rate equivalence : eq = i i Flow rate equivalence : TH eq = TH(L) Average stoppage time equivalence

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37 Two-machine production lines with intermediate buffer

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38 Motivation & 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

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39 Motivation & cost of intermediate buffers New phenomena: Blocking Starvation M1M3M2M4 Failure of M3 M1M3M2M4 3 time units after the failure where is the cycle time

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40 CTMC model of reliable 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/p 1 on M1 and 1/p 2 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

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41 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 01 K+1K+2 p1 p2 p1 p2 p1 p2 … CTMC model of reliable line with exponential processing times

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42 CTMC model equivalent to M/M/1/(K+2). with = p 1 /p 2, corresponding to the traffic intensity. M1 B M2 01 KK+2 p1 p2 p1 p2 p1 p2 … CTMC model of reliable line with exponential processing times

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43 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 = p 2 (1- 0 ) = p 1 (1- K+2 ) = p 1 (1- K+2 )/(1- K+3 ) Mean WIP CTMC model of reliable line with exponential processing times

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44 Blocking prob. starving prob Zero-buffer Unlimited buffer Example : p1 = 10, p2 = 9, = 10/9 CTMC model of reliable line with exponential processing times

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45 Exponential model of Failure-prone lines Assumptions: Machines can break down. Exponentially distributed times to failures and time to repair with TBF i = EXP( i ) and TTR i = EXP ( i ) Exponential processing times with T 1 = EXP(p 1 ) and T 2 = EXP(p 2 ) Buffer of capacity K Each machines holds the part in process. M1 never starved and M2 never blocked M1 B M2

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46 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 M i is UP and 0 if M i is DOWN The state vector ( 1, 2, x) is a continuous time Markov chain Exponential model of Failure-prone lines

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47 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 p1 p2 2 22 1 11 Analytical expressions of steady-state probabilities available in the book of SB Gershwin Can be used to evaluate the performance measures Exponential model of Failure-prone lines

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48 Slotted time model of a failure-prone line Assumptions 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: M 1 blocked if B is full, M 2 starved if B is empty A machine M i in state W in t breaks down in period t+1 with proba p i and, with proba 1 - p i, moves to state W or B or I. A machine M i in state R in t moves to state W in t+1 with proba r i and, with proba 1 - r i, 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

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49 Slotted time model of a failure-prone line Discrete 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) 0} 0≤x(t) ≤ N M1 B M2

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50 Slotted time model of a failure-prone line Discrete Time Markov chain 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-p 1 )(1-p 2 ) (1,1,x) + r 1 (1-p 2 ) (0,1,x) + (1-p 1 )r 2 (1,0,x) + r 1 r 2 (0,0,x) (0,0,x) = (1-r 1 )(1-r 2 ) (0,0,x) + p 1 (1-r 2 ) (1,0,x) + (1-r 1 )p 2 (0,1,x) + p 1 p 2 (0,0,x) (1,0,x) = (1-p 1 )p 2 (1,1,x-1) + (1- p 1 )(1-r 2 ) (1,0,x-1) + r 1 p 2 (0,1,x-1) + r 1 (1-r 2 ) (0,0,x-1) (0,1,x) = p 1 (1-p 2 ) (1,1,x+1) + (1-r 1 )(1-p 2 ) (0,1,x+1) + p 1 r 2 (1,0,x+1) + (1-r 1 )r 2 (0,0,x+1) Other boundary equations can be derived similarly. M1 B M2

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51 Slotted time model of a failure-prone line Discrete Time Markov chain Performance measures : Efficiency of M 1 : E 1 = 1 = 1, x < N 1, 2, x) Efficiency of M 2 : E 2 = 2 = 1, x > 0 1, 2, x) Throughput rate : TH = E 1 ×U = E 2 ×U WIP : x 1, 2, x) Probability of starvation : (0, 1, 0) Probability of blocking : (1, 0, N) M1 B M2

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52 Efficiency of the line E(N) = probability a machine is producing Slotted time model of a failure-prone line Analytical results WIP to be determined with expressions of steady-state probabilities available in the book of S.B. Gerswhin

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53 Continuous flow models of failure-prone lines Assumptions M1 B M2 Only synchronuous lines are considered, i.e. U i = 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

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54 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): TH Continuous (h) < TH Discret (h) < TH Continuous (h+2) where TH Discret (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 lines How good is continuous flow approximation

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55 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: u i (t) : production rate of Mi at t Continuous flow models of failure-prone lines Dynamic behavior

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56 M1 B M2 u1(t)u2(t) 1(t) 2(t) x(t) h u1 = U 1 = 1 u2 = U 2 = 1 u1 = U 1 = 1 u2 = 0 2 = 0 F2 blocking R2F1R1 Starving u1 = 0 1 = 1 u2 = 0 2 = 0 u1 = U 1 = 1 u2 = U 2 = 1 u1 = 0 1 = 0 u2 = U 2 = 1 u1 = 0 1 = 0 u2 = 0 2 = 1 F2 x(t) Continuous flow models of failure-prone lines Dynamic behavior

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57 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 Continuous flow models of failure-prone lines Dynamic behavior

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58 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)/e 2 Probability of blocking of M1: pb(h) = 1 – E(h)/e 1 Isolated efficiency of Mi :e i = 1/(1+I i ) = i /( i + i ) Mean buffer level: Q(h) Continuous flow models of failure-prone lines Performance measures online proof

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59 Case I 1 = 1 / 1 I 2 = 2 / 2 Continuous flow models of failure-prone lines Analytical solution

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60 Case I 1 = 1 / 1 I 2 = 2 / 2 = I Continuous flow models of failure-prone lines Analytical solution

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61 Case U=1, = 0,1 In this case, Q(h) = 0,5h zero-buffer infinite buffer Continuous flow models of failure-prone lines Analytical solution

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62 Numerical results U = 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?

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63 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? Numerical results U = 1, 1 = 0.1, 2 = 0.1, 2 = 0.1 WIP 1 = 0,14 1 = 0,12 1 = 0,10 1 = 0,08 1 = 0,06 Buffer capacity

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64 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? Numerical results U = 1, 1 = 0.1, 2 = 0.1, 2 = 0.1

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65 Improvement to non- bottleneck machine. Same graph for improvement of machine 2 Numerical results U = 1, 1 = 0.1, 1 = 0.1, 2 = 0.1, 2 = 0.1 Throughput Buffer capacity

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66 Inventory increases as the (non-bottleneck upstream machine is improved and as the buffer space is increased. Numerical results U = 1, 1 = 0.1, 1 = 0.1, 2 = 0.1, 2 = 0.1 Average inventory Buffer capacity

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67 Inventory decreases as the (non-bottleneck) downstream machine is improved Inventory increases as the buffer space isincreased. Numerical results U = 1, 1 = 0.1, 1 = 0.1, 2 = 0.1, 2 = 0.1 Average inventory Buffer capacity

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68 1 and 1 vary together and 1 /( 1 + 1 ) = 0.9 Answer: short and frequent failures. Why? Numerical results U = 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?

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69 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 lines Reversiblity M1 B M2 M1 B M2 L L' Proof for the continuous L2 line

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70 A continuous time Markov process with hybrid state space characterized by Internal state distribution (0 < x < h): F 1 2 (x) = P{ 1 (t) = 1, 2 (t) = 2, 0 < x(t) x} f 1 2 (x) = d F 1 2 (x) /dx Boundary distribution: P 1 2 (0), P 1 2 (h) M1 B M2 u1(t)u2(t) 1(t) 2(t) x(t), h Continuous flow models of failure-prone lines Dynamic behavior

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71 Case I 1 = 1 / 1 I 2 = 2 / 2

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72 Case I 1 = 1 / 1 I 2 = 2 / 2 =I

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73 Long failure-prone production lines

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74 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 2 M (N+1) M-1. For M = 10 and N = 100, there are over 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

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75 Notation Given isolated machine performances: Ii = i/ i e i = 1/(1+Ii) : isolated efficiency of Mi e i U : isolated productivity of Mi Unknown system performance measures: E i : Probability that Mi is producing TH i = E i U: throughput rate of Mi ps i : probability of starvation of Mi pb i : proba of blockage of Mi M1 B1 M2 B2 M3 B3 M4

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76 Aggregation method Equivalent 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 Repair time of M12 = Average stoppage time of M2 in L12:

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77 Aggregation method Equivalent machine Repeating the aggregation process leads to an approximated estimation of the throughput of the line. M12 B2 M3 B3 M4M1 B1 M2 B2 M3 B3 M4M123 B3 M4M1234

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78 Decomposition method Properties of a continuous line Flow conservation: TH i = TH 1, i = 2, …, K(1) E i = E 1, i = 2, …, K(2) Flow-idle time relation: E i = e i (1- ps i –pb i ), i = 2, …, K (3) Proof: E i = 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} = e i (1- ps i –pb i ) since the proba that Mi is blocked and starved simultaneously is null in continuous flow model. M1 B1 M2 B2 M3 B3 M4

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79 Decomposition method Decomposition Decompose a K-machine line into K-1 lines of two-machines h1 h2h3 M1 B1 M2 B2 M3 B3 M4 L: u u h1 Mu(1) B(1) Md(1) L(1) h2 Mu(2) B(2) Md(2) L(2) h3 Mu(3) B(3) Md(3) L(3) d d u u d d u u d d 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

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80 Notation : I u (i), e u (i), I d (i), e d (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) = E i+1, i = 1, …, K-1(4) ps(i) = ps i+1, i = 1, …, K-1 (5) pb(i) = pb i, i = 1, …, K-1 (6) Decomposition method Decomposition

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81 Apply(3) to L(i): E(i) = e u (i)(1- pb(i) ), i = 1, …, K-1 (7) E(i) = e d (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) = e i (1 - ps(i-1) - pb(i)) Combining with (7) & (8) Id(i-1) + Iu(i) = 1/E(i-1) + Ii –1(II) Decomposition method Decomposition

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82 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) Decomposition method Decomposition

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83 C) 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 idle E(i) u (i) u (i) Decomposition method Decomposition

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84 Combine (9)-(10), u (i) = X. u (i-1) + (1-X) i (III) with X = ps(i-1) / (I u (i).E(i)). Decomposition method Decomposition

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85 Repair time of Md(i) d (i) = Y. u (i+1) + (1-Y) i+1 (IV) with Y = pb(i+1) / (I d (i).E(i)). Boundary equations: u (1) = 1, u (1) = 1, d (K-1) = K, u (K-1) = K, (V) Decomposition method Decomposition

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86 Equation system (I) – (V), (I)E(i) = E(1), i = 1, …, K-1 (II)I d (i-1) + I u (i) = 1/E(i-1) + I i –1 u (i) = X u (i-1) + (1-X) i with (IV) d (i) = Y u (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) Decomposition method Decomposition

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87 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), I u (i) = 1/E(i-1) + I i –1 - I d (i-1) 2.3 From (III)-(I), u (i) = X u (i-1) + (1-X) i with X = ps(i-1) / (I u (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), I d (i) = 1/E(i+1) + I i+1 –1 - I u (i+1) 3.3 From (IV)-(I), d (i) = Y d (i+1) + (1-Y) i+1 with Y = pb(i+1) / (I d (i).E(i+1)) Step 4: Repeat (2) – (3) till convergence, i.e. E(i) = E(1). Decomposition method Decomposition

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88 Distribution of material in a line with average buffer 50 identical machines = 0.01, = 0.1, U = 1, hi = 20

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89 Effect of bottleneck average buffer 50 identical machines : = 0.01, = 0.1, U = 1, hi = 20 except bottleneck at M10 with 10 =0.0375

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90 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

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91 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 M1M2M3 B3 M4M5M6 B6 M7M8M9

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92 Distribution of buffer capacity Throughput Total buffer space All machines have = 0.001, = 0.019, U = 1 What are the asymptotes Is 8 buffers always faster?

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93 Distribution of buffer capacity Throughput Total buffer space Is 8 buffers always faster? Perhaps not, but the difference is not significant in systems with very small buffers.

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94 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.

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95 Design buffer space distribution The common operation time is one operation per minute. The target production rate is 0.88 parts per minute.

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

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97 Design buffer space distribution Are buffers really needed? LineProduction rate with no buffer Case Case Case Yes. How to compute these numbers? (homework)

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98 Design buffer space distribution Optimal buffer space distribution Observation: Buffer space is needed most where buffer level variability is greatest!

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