MIT Manufacturing Systems Analysis Lectures 6 9: Flow Lines

2.852 Manufacturing Systems Analysis 1/165 Copyright 2010 c Stanley B. Gershwin. MIT 2.852 Manufacturing Systems Analysis Lectures 6 9: Flow Lines Models That Can Be Analyzed Exactly Stanley B. Gershwin http://web.mit.edu/manuf-sys Massachusetts Institute of Technology Spring, 2010

2.852 Manufacturing Systems Analysis 2/165 Copyright c 2010 Stanley B. Gershwin. Models The purpose of an engineering or scientific model is to make predictions. Kinds of models: Mathematical: aggregated behavior is described by equations. Predictions are made by solving the equations. Simulation: detailed behavior is described. Predictions are made by reproducting behavior. Models are simplifications of reality. Models that are too simple make poor predictions because they leave out important features. Models that are too complex make poor predictions because they are difficult to analyze or are time-consuming to use, because they require more data, or because they have errors.

2.852 Manufacturing Systems Analysis 3/165 Copyright c 2010 Stanley B. Gershwin. Flow Line... also known as a Production or Transfer Line. M 1 B 1 M 2 B 2 M 3 B 3 M 4 B 4 M 5 B 5 M 6 Machine Buffer Machines are unreliable. Buffers are finite.

2.852 Manufacturing Systems Analysis 4/165 Copyright c 2010 Stanley B. Gershwin. Flow Line Motivation Economic importance. Relative simplicity for analysis and for intuition.

2.852 Manufacturing Systems Analysis 5/165 Copyright c 2010 Stanley B. Gershwin. Flow Line Buffers and Inventory Buffers are for mitigating asynchronization (ie, they are shock absorbers). Buffer space and inventory are expensive.

2.852 Manufacturing Systems Analysis 6/165 Copyright c 2010 Stanley B. Gershwin. Flow Line Analysis Difficulties Complex behavior. Analytical solution available only for limited systems. Exact numerical solution feasible only for systems with a small number of buffers. Simulation may be too slow for optimization.

2.852 Manufacturing Systems Analysis 7/165 Copyright c 2010 Stanley B. Gershwin. Flow Line Output Variability 4000 Weekly Production 2000 Production output from a simulation of a transfer line. 0 0 20 40 60 80 100 Week

2.852 Manufacturing Systems Analysis 8/165 Copyright c 2010 Stanley B. Gershwin. Flow Line Usual General Assumptions Unlimited repair personnel. Uncorrelated failures. Perfect yield. The first machine is never starved and the last is never blocked. Blocking before service. Operation dependent failures.

2.852 Manufacturing Systems Analysis 9/165 Copyright c 2010 Stanley B. Gershwin. Single Reliable Machine If the machine is perfectly reliable, and its average operation time is τ, then its maximum production rate is 1/τ. Note: Sometimes cycle time is used instead of operation time, but BEWARE: cycle time has two meanings! The other meaning is the time a part spends in a system. If the system is a single, reliable machine, the two meanings are the same.

2.852 Manufacturing Systems Analysis 10/165 Copyright c 2010 Stanley B. Gershwin. Single Unreliable Machine ODFs Operation-Dependent Failures A machine can only fail while it is working. IMPORTANT! MTTF must be measured in working time! This is the usual assumption. Note: MTBF = MTTF + MTTR

2.852 Manufacturing Systems Analysis 11/165 Copyright c 2010 Stanley B. Gershwin. Single Unreliable Machine Production rate If the machine is unreliable, and its average operation time is τ, its mean time to fail is MTTF, its mean time to repair is MTTR, then its maximum production rate is ( ) 1 MTTF τ MTTF + MTTR

2.852 Manufacturing Systems Analysis 12/165 Copyright c 2010 Stanley B. Gershwin. Single Unreliable Machine Production rate Proof Machine UP Machine DOWN Average production rate, while machine is up, is 1/τ. Average duration of an up period is MTTF. Average production during an up period is MTTF/τ. Average duration of up-down period: MTTF + MTTR. Average production during up-down period: MTTF/τ. Therefore, average production rate is (MTTF/τ)/(MTTF + MTTR).

2.852 Manufacturing Systems Analysis 13/165 Copyright c 2010 Stanley B. Gershwin. Single Unreliable Machine Geometric Up- and Down-Times Assumptions: Operation time is constant (τ). Failure and repair times are geometrically distributed. Let p be the probability that a machine fails during any given operation. Then p = τ/mttf.

2.852 Manufacturing Systems Analysis 14/165 Copyright c 2010 Stanley B. Gershwin. Single Unreliable Machine Let r be the probability that M gets repaired in during any operation time when it is down. Then r = τ/mttr. Then the average production rate of M is ( ) 1 r. τ r + p (Sometimes we forget to say average. )

2.852 Manufacturing Systems Analysis 15/165 Copyright c 2010 Stanley B. Gershwin. Single Unreliable Machine Production Rates So far, the machine really has three production rates: 1/τ when it is up (short-term capacity), 0 when it is down (short-term capacity), (1/τ)(r/(r + p)) on the average (long-term capacity).

2.852 Manufacturing Systems Analysis 16/165 Copyright c 2010 Stanley B. Gershwin. Infinite-Buffer Line M 1 B1 M 2 B2 M 3 B3 M 4 B4 M 5 B5 M 6 Assumptions: A machine is not idle if it is not starved. The first machine is never starved.

2.852 Manufacturing Systems Analysis 17/165 Copyright c 2010 Stanley B. Gershwin. Infinite-Buffer Line M 1 B1 M 2 B2 M 3 B3 M 4 B4 M 5 B5 M 6 The production rate of the line is the production rate of the slowest machine in the line called the bottleneck. Slowest means least average production rate, where average production rate is calculated from one of the previous formulas.

2.852 Manufacturing Systems Analysis 18/165 Copyright c 2010 Stanley B. Gershwin. Infinite-Buffer Line M 1 B1 M 2 B2 M 3 B3 M 4 B4 M 5 B5 M 6 Production rate is therefore P = min i and M i is the bottleneck. 1 τ i ( ) MTTF i MTTF i + MTTR i

2.852 Manufacturing Systems Analysis 19/165 Copyright c 2010 Stanley B. Gershwin. Infinite-Buffer Line M 1 B1 M 2 B2 M 3 B3 M 4 B 4 M 5 B5 M 6 The system is not in steady state. An infinite amount of inventory accumulates in the buffer upstream of the bottleneck. A finite amount of inventory appears downstream of the bottleneck.

2.852 Manufacturing Systems Analysis 20/165 Copyright c 2010 Stanley B. Gershwin. Infinite-Buffer Line M 1 B1 M 2 B2 M 3 B3 M 4 B 4 M 5 B5 M 6 B6 M 7 B 7 M 8 B 8 M 9 B9 M 10 14000 12000 10000 8000 Buffer 1 Buffer 2 Buffer 3 Buffer 4 Buffer 5 Buffer 6 Buffer 7 Buffer 8 Buffer 9 6000 4000 2000 0 0 100000 200000 300000 400000 500000 600000 700000 800000 900000 1e+06

2.852 Manufacturing Systems Analysis 21/165 Copyright c 2010 Stanley B. Gershwin. Infinite-Buffer Line M 1 B1 M 2 B2 M 3 B3 M 4 B 4 M 5 B5 M 6 B6 M 7 B 7 M 8 B 8 M 9 B9 M 10 The second bottleneck is the slowest machine upstream of the bottleneck. An infinite amount of inventory accumulates just upstream of it. A finite amount of inventory appears between the second bottleneck and the machine upstream of the first bottleneck. Et cetera.

2.852 Manufacturing Systems Analysis 22/165 Copyright c 2010 Stanley B. Gershwin. Infinite-Buffer Line M 1 B1 M 2 B2 M 3 B3 M 4 B 4 M 5 B5 M 6 B6 M 7 B 7 M 8 B 8 M 9 B9 M 10 12000 10000 8000 Buffer 1 Buffer 2 Buffer 3 Buffer 4 Buffer 5 Buffer 6 Buffer 7 Buffer 8 Buffer 9 6000 4000 2000 0 0 100000 200000 300000 400000 500000 600000 700000 800000 900000 1e+06 A 10-machine line with bottlenecks at Machines 5 and 10.

2.852 Manufacturing Systems Analysis 23/165 Copyright c 2010 Stanley B. Gershwin. Infinite-Buffer Line M 1 B1 M 2 B2 M 3 B3 M 4 B 4 M 5 B5 M 6 B6 M 7 B 7 M 8 B 8 M 9 B9 M 10 12000 10000 8000 Buffer 4 Buffer 9 6000 4000 2000 0 0 100000 200000 300000 400000 500000 600000 700000 800000 900000 1e+06 Question: What are the slopes (roughly!) of the two indicated graphs?

2.852 Manufacturing Systems Analysis 24/165 Copyright c 2010 Stanley B. Gershwin. Infinite-Buffer Line Questions: If we want to increase production rate, which machine should we improve? What would happen to production rate if we improved any other machine?

2.852 Manufacturing Systems Analysis 25/165 Copyright c 2010 Stanley B. Gershwin. Zero-Buffer Line M 1 M 2 M 3 M 4 M 5 M 6 If any one machine fails, or takes a very long time to do an operation, all the other machines must wait. Therefore the production rate is usually less possibly much less than the slowest machine.

2.852 Manufacturing Systems Analysis 26/165 Copyright c 2010 Stanley B. Gershwin. Zero-Buffer Line M 1 M 2 M 3 M 4 M 5 M 6 Special case: Constant, unequal operation times, perfectly reliable machines. The operation time of the line is equal to the operation time of the slowest machine, so the production rate of the line is equal to that of the slowest machine.

2.852 Manufacturing Systems Analysis 27/165 Copyright c 2010 Stanley B. Gershwin. Zero-Buffer Line Constant, equal operation times, unreliable machines M 1 M 2 M 3 M 4 M 5 M 6 Assumption: Failure and repair times are geometrically distributed. Define p i = τ/mttf i = probability of failure during an operation. Define r i = τ/mttr i probability of repair during an interval of length τ when the machine is down. Operation-Dependent Failures (ODFs): Machines can only fail while they are working.

2.852 Manufacturing Systems Analysis 28/165 Copyright c 2010 Stanley B. Gershwin. Zero-Buffer Line M 1 M 2 M 3 M 4 M 5 M 6 Buzacott s Zero-Buffer Line Formula: Let k be the number of machines in the line. Then P = 1 τ 1 k p i 1 + r i i=1

2.852 Manufacturing Systems Analysis 29/165 Copyright c 2010 Stanley B. Gershwin. Zero-Buffer Line M 1 M 2 M 3 M 4 M 5 M 6 Same as the earlier formula (page 11, page 14) when k = 1. The isolated production rate of a single machine M i is ( ) ( ) 1 1 1 r i p τ 1 + i =. r i τ r i + p i

2.852 Manufacturing Systems Analysis 30/165 Copyright c 2010 Stanley B. Gershwin. Zero-Buffer Line Proof of formula Let τ (the operation time) be the time unit. Assumption: At most, one machine can be down. Consider a long time interval of length Tτ during which Machine M i fails m i times (i = 1,... k). M 3 M 5 M 2 M 3 M 1 M 4 All up Some machine down Without failures, the line would produce T parts.

2.852 Manufacturing Systems Analysis 31/165 Copyright c 2010 Stanley B. Gershwin. Zero-Buffer Line The average repair time of M i is τ/r i each time it fails, so the total system down time is close to Dτ = k i=1 m i τ r i where D is the number of operation times in which a machine is down.

2.852 Manufacturing Systems Analysis 32/165 Copyright c 2010 Stanley B. Gershwin. Zero-Buffer Line The total up time is approximately k Uτ = Tτ i=1 m i τ r i. where U is the number of operation times in which all machines are up.

2.852 Manufacturing Systems Analysis 33/165 Copyright c 2010 Stanley B. Gershwin. Zero-Buffer Line Since the system produces one part per time unit while it is working, it produces U parts during the interval of length Tτ. Note that, approximately, m i = p i U because M i can only fail while it is operational.

2.852 Manufacturing Systems Analysis 34/165 Copyright c 2010 Stanley B. Gershwin. Zero-Buffer Line Thus, or, Uτ = Tτ Uτ k i=1 p i r i, U T = E ODF = 1 k 1 + i=1 p i r i

2.852 Manufacturing Systems Analysis 35/165 Copyright c 2010 Stanley B. Gershwin. Zero-Buffer Line and P = 1 τ 1 k 1 + i=1 p i r i Note that P is a function of the ratio p i /r i and not p i or r i separately. The same statement is true for the infinite-buffer line. However, the same statement is not true for a line with finite, non-zero buffers.

2.852 Manufacturing Systems Analysis 36/165 Copyright c 2010 Stanley B. Gershwin. Zero-Buffer Line Questions: If we want to increase production rate, which machine should we improve? What would happen to production rate if we improved any other machine?

2.852 Manufacturing Systems Analysis 37/165 Copyright c 2010 Stanley B. Gershwin. Zero-Buffer Line ODF and TDF TDF= Time-Dependent Failure. Machines fail independently of one another when they are idle. P TDF = 1 τ k i=1 ( ) r i > P ODF r i + p i

2.852 Manufacturing Systems Analysis 38/165 Copyright c 2010 Stanley B. Gershwin. Zero-Buffer Line P as a function of p i All machines are the same except M i. As p i increases, the production rate decreases. P 0.5 0.45 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 0.2 0.4 0.6 0.8 1 p i

2.852 Manufacturing Systems Analysis 39/165 Copyright c 2010 Stanley B. Gershwin. Zero-Buffer Line P as a function of k All machines are the same. As the line gets longer, the production rate decreases. 1 0.9 0.8 0.7 0.6 0.5 P 0.4 0.3 0.2 0.1 0 0 10 20 30 40 50 60 70 80 90 100 k Infinite buffer production rate Capacity loss

2.852 Manufacturing Systems Analysis 40/165 Copyright c 2010 Stanley B. Gershwin. Finite-Buffer Lines M 1 B1 M 2 B2 M 3 B3 M 4 B4 M 5 B5 M 6 Motivation for buffers: recapture some of the lost production rate. Cost in-process inventory/lead time floor space material handling mechanism

2.852 Manufacturing Systems Analysis 41/165 Copyright c 2010 Stanley B. Gershwin. Finite-Buffer Lines M 1 B1 M 2 B2 M 3 B3 M 4 B4 M 5 B5 M 6 Infinite buffers: no propagation of disruptions. Zero buffers: instantaneous propagation. Finite buffers: delayed propagation. New phenomena: blockage and starvation.

2.852 Manufacturing Systems Analysis 42/165 Copyright c 2010 Stanley B. Gershwin. Finite-Buffer Lines Difficulty: M 1 B1 M 2 B2 M 3 B3 M 4 B4 M 5 B5 M 6 No simple formula for calculating production rate or inventory levels. Solution: Simulation Analytical approximation

2.852 Manufacturing Systems Analysis 43/165 Copyright c 2010 Stanley B. Gershwin. Exact solution is available to model of two-machine line. Discrete time-discrete state Markov process: prob{x(t + 1) = x(t + 1) X(t) = x(t), X(t 1) = x(t 1), X(t 2) = x(t 2),...} = prob{x(t + 1) = x(t + 1) X(t) = x(t)} In the following, we construct prob{x(t + 1) = x(t + 1) X(t) = x(t)} and solve the steady-state transition equations.

2.852 Manufacturing Systems Analysis 44/165 Copyright c 2010 Stanley B. Gershwin. Here, X(t) = (n(t),α 1 (t),α 2 (t)), where n is the number of parts in the buffer; n = 0,1,..., N. α i is the repair state of M i ; i = 1,2. α i = 1 means the machine is up or operational; α i = 0 means the machine is down or under repair.

2.852 Manufacturing Systems Analysis 45/165 Copyright c 2010 Stanley B. Gershwin. Motivation: We can develop intuition from these systems that is useful for understanding more complex systems. Two-machine lines are used as building blocks in decomposition approximations of realistic-sized systems.

2.852 Manufacturing Systems Analysis 46/165 Copyright c 2010 Stanley B. Gershwin. Several models available: Deterministic processing time, or Buzacott model: deterministic processing time, geometric failure and repair times; discrete state, discrete time. Exponential processing time: exponential processing, failure, and repair time; discrete state, continuous time. Continuous material, or fluid: deterministic processing, exponential failure and repair time; mixed state, continuous time. Extensions Models with multiple up and down states.

2.852 Manufacturing Systems Analysis 47/165 Copyright c 2010 Stanley B. Gershwin. Outline: Details of two-machine, deterministic processing time line. Assumptions Performance measures Transient states Transition equations Identities Analytical solution Limits Behavior

2.852 Manufacturing Systems Analysis 48/165 Copyright c 2010 Stanley B. Gershwin. Assumptions, etc. Assumptions, etc. for deterministic processing time systems (including long lines) All operation times are deterministic and equal to 1. The amount of material in Buffer i at time t is n i (t), 0 n i (t) N i. A buffer gains or loses at most one piece during a time unit. The state of the system is s = (n 1,..., n k 1,α 1,...,α k ).

2.852 Manufacturing Systems Analysis 49/165 Copyright c 2010 Stanley B. Gershwin. Assumptions, etc. Operation dependent failures: prob [α i (t + 1) = 0 n i 1 (t) = 0,α i (t) = 1,n i (t) < N i ] = 0, prob [α i (t + 1) = 1 n i 1 (t) = 0,α i (t) = 1,n i (t) < N i ] = 1, prob [α i (t + 1) = 0 n i 1 (t) > 0,α i (t) = 1,n i (t) = N i ] = 0, prob [α i (t + 1) = 1 n i 1 (t) > 0,α i (t) = 1,n i (t) = N i ] = 1, prob [α i (t + 1) = 0 n i 1 (t) > 0,α i (t) = 1,n i (t) < N i ] = p i, prob [α i (t + 1) = 1 n i 1 (t) > 0,α i (t) = 1,n i (t) < N i ] = 1 p i.

2.852 Manufacturing Systems Analysis 50/165 Copyright c 2010 Stanley B. Gershwin. Assumptions, etc. Repairs: prob [α i (t + 1) = 1 α i (t) = 0] = r i, prob [α i (t + 1) = 0 α i (t) = 0] = 1 r i.

Assumptions, etc. Timing convention: In the absence of blocking or starvation: More generally, where n i (t + 1) = n i (t) + α i (t + 1) α i+1 (t + 1). n i (t + 1) = n i (t) + I ui (t + 1) I di (t + 1), I ui (t + 1) = { 1 if αi (t + 1) = 1 and n i 1 (t) > 0 and n i (t) < N i, 0 otherwise. { 1 if αi+1 (t + 1) = 1 and n i (t) > 0 and n i+1 (t) < N i+1 I di (t + 1) = 0 otherwise. 2.852 Manufacturing Systems Analysis 51/165 Copyright c 2010 Stanley B. Gershwin.

2.852 Manufacturing Systems Analysis 52/165 Copyright c 2010 Stanley B. Gershwin. Assumptions, etc. In the Markov chain model, there is a set of transient states, and a single final class. Thus, a unique steady state distribution exists. The model is studied in steady state. That is, we calculate the stationary probability distribution. We calculate performance measures (production rate and average inventory) from the steady state distribution.

2.852 Manufacturing Systems Analysis 53/165 Copyright c 2010 Stanley B. Gershwin. Performance measures The steady state production rate (throughput, flow rate, or efficiency ) of Machine M i is the probability that Machine M i produces a part in a time step. Units: parts per operation time. It is the probability that Machine M i is operational and neither starved nor blocked in time step t. It is equivalent, and more convenient, to express it as the probability that Machine M i is operational and neither starved nor blocked in time step t + 1: E i = prob (α i (t + 1) = 1, n i 1 (t) > 0, n i (t) < N i ) For a useful analytical expression, we must rewrite this so that all states are evaluated at the same time.

2.852 Manufacturing Systems Analysis 54/165 Copyright c 2010 Stanley B. Gershwin. Performance measures E i = prob (α i (t + 1) = 1, n i 1 (t) > 0, n i (t) < N i ) = prob (α i (t + 1) = 1 n i 1 (t) > 0, α i (t) = 1, n i (t) < N i ) prob (n i 1 (t) > 0, α i (t) = 1, n i (t) < N i ) + prob (α i (t + 1) = 1 n i 1 (t) > 0, α i (t) = 0, n i (t) < N i ) prob (n i 1 (t) > 0, α i (t) = 0, n i (t) < N i ). = (1 p i ) prob (n i 1 (t) > 0, α i (t) = 1, n i (t) < N i ) +r i prob (n i 1 (t) > 0, α i (t) = 0, n i (t) < N i ).

2.852 Manufacturing Systems Analysis 55/165 Copyright c 2010 Stanley B. Gershwin. Performance measures In steady state, there is a repair for every failure of Machine i, or Therefore, r i prob (n i 1 (t) > 0,α i (t) = 0,n i (t) < N i ) = p i prob (n i 1 (t) > 0,α i (t) = 1,n i (t) < N i ) E i = prob (α i = 1,n i 1 > 0,n i < N i ).

2.852 Manufacturing Systems Analysis 56/165 Copyright c 2010 Stanley B. Gershwin. Performance measures The steady state average level of Buffer i is n i = n i prob (s). s

2.852 Manufacturing Systems Analysis 57/165 Copyright c 2010 Stanley B. Gershwin. State Space s = (n,α 1,α 2 ) where n = 0,1,..., N α i = 0,1

2.852 Manufacturing Systems Analysis 58/165 Copyright c 2010 Stanley B. Gershwin. Transient states (0,1,0) is transient because it cannot be reached from any state. If α 1 (t + 1) = 1 and α 2 (t + 1) = 0, then n(t + 1) = n(t) + 1. (0,1,1) is transient because it cannot be reached from any state. If n(t) = 0 and α 1 (t + 1) = 1 and α 2 (t + 1) = 1, then n(t + 1) = 1 since M 2 is starved and thus not able to operate. If n(t) > 0 and α 1 (t + 1) = 1 and α 2 (t + 1) = 1, then n(t + 1) = n(t).

2.852 Manufacturing Systems Analysis 59/165 Copyright c 2010 Stanley B. Gershwin. Transient states (0,0,0) is transient because it can be reached only from itself or (0,1,0). It can be reached from itself if neither machine is repaired; it can be reached from (0,1,0) if the first machine fails while attempting to make a part. It cannot be reached from (0,0,1) or (0,1,1) since the second machine cannot fail. Otherwise, if α 1(t + 1) = 0 and α 2(t + 1) = 0, then n(t + 1) = n(t). (1,1,0) is transient because it can be reached only from (0,0,0) or (0,1,0). If α 1(t + 1) = 1 and α 2(t + 1) = 0, then n(t + 1) = n(t) + 1. Therefore, n(t) = 0. However, (1,1,0) cannot be reached from (0,0,1) since Machine 2 cannot fail. (For the same reason, it cannot be reached from (0,1,1), but since the latter is transient, that is irrelevant.) Similarly, (N, 0, 0), (N, 0, 1), (N, 1, 1), and (N 1, 0, 1) are transient.

2.852 Manufacturing Systems Analysis 60/165 Copyright c 2010 Stanley B. Gershwin. State space (α,α ) (0,0) 1 2 n=0 n=1 n=2 n=3 n=4 n=5 n=6 n=7 n=8 n=9 n=10 n=11 n=12 n=13 =N 1 =N (0,1) (1,0) (1,1) key states transient non transient boundary internal transitions out of transient states out of non transient states to increasing buffer level to decreasing buffer level unchanging buffer level

2.852 Manufacturing Systems Analysis 61/165 Copyright c 2010 Stanley B. Gershwin. Transition equations Internal equations 2 n N 2 p(n, 0, 0) = (1 r 1)(1 r 2)p(n, 0, 0) + (1 r 1)p 2p(n, 0, 1) +p 1(1 r 2)p(n,1, 0) + p 1p 2p(n, 1, 1) p(n, 0, 1) = (1 r 1)r 2p(n + 1, 0, 0) + (1 r 1)(1 p 2)p(n + 1, 0, 1) +p 1r 2p(n + 1, 1, 0) + p 1(1 p 2)p(n + 1, 1, 1) p(n, 1, 0) = r 1(1 r 2)p(n 1, 0, 0) + r 1p 2p(n 1, 0, 1) +(1 p 1)(1 r 2)p(n 1, 1, 0) + (1 p 1)p 2p(n 1, 1, 1) p(n, 1, 1) = r 1r 2p(n, 0, 0) + r 1(1 p 2)p(n, 0, 1) + (1 p 1)r 2p(n, 1, 0) +(1 p 1)(1 p 2)p(n, 1, 1)

2.852 Manufacturing Systems Analysis 62/165 Copyright c 2010 Stanley B. Gershwin. Transition equations Lower boundary equations n 1 p(0, 0, 1) = (1 r 1)p(0, 0, 1) + (1 r 1)r 2p(1, 0, 0) +(1 r 1)(1 p 2)p(1, 0, 1) + p 1(1 p 2)p(1, 1, 1). p(1, 0, 0) = (1 r 1)(1 r 2)p(1, 0, 0) + (1 r 1)p 2p(1, 0, 1) + p 1p 2p(1, 1, 1) p(1, 0, 1) = (1 r 1)r 2p(2, 0, 0) + (1 r 1)(1 p 2)p(2, 0, 1)+ p 1r 2p(2, 1, 0) + p 1(1 p 2)p(2, 1, 1) p(1, 1, 1) = r 1p(0, 0, 1) + r 1r 2p(1, 0, 0) + r 1(1 p 2)p(1, 0, 1) +(1 p 1)(1 p 2)p(1, 1, 1) p(2, 1, 0) = r 1(1 r 2)p(1, 0, 0) + r 1p 2p(1, 0, 1) + (1 p 1)p 2p(1, 1, 1)

2.852 Manufacturing Systems Analysis 63/165 Copyright c 2010 Stanley B. Gershwin. Transition equations Upper boundary equations n N 1 p(n 2, 0, 1) = (1 r 1)r 2p(N 1, 0, 0) + p 1r 2p(N 1, 1, 0) +p 1(1 p 2)p(N 1, 1, 1) p(n 1, 0, 0) = (1 r 1)(1 r 2)p(N 1, 0, 0) + p 1(1 r 2)p(N 1, 1, 0) +p 1p 2p(N 1, 1, 1) p(n 1, 1, 0) = r 1(1 r 2)p(N 2, 0, 0) + r 1p 2p(N 2, 0, 1) + (1 p 1)(1 r 2)p(N 2, 1, 0) + (1 p 1)p 2p(N 2, 1, 1) p(n 1, 1, 1) = r 1r 2p(N 1, 0, 0) + (1 p 1)r 2p(N 1, 1, 0) +(1 p 1)(1 p 2)p(N 1, 1, 1) + r 2p(N, 1, 0) p(n, 1, 0) = r 1(1 r 2)p(N 1, 0, 0) + (1 p 1)(1 r 2)p(N 1, 1, 0) +(1 p 1)p 2p(N 1, 1, 1) + (1 r 2)p(N, 1, 0)

2.852 Manufacturing Systems Analysis 64/165 Copyright c 2010 Stanley B. Gershwin. Performance measures E 1 is the probability that M 1 is operational and not blocked: E 1 = p(n, α 1, α 2 ). n < N α 1 = 1 (α,α ) (0,0) 1 2 n=0 n=1 n=2 n=3 n=4 n=5 n=6 n=7 n=8 n=9 n=10 n=11 n=12 n=13 =N 1 =N (0,1) (1,0) (1,1)

2.852 Manufacturing Systems Analysis 65/165 Copyright c 2010 Stanley B. Gershwin. Performance measures E 2 is the probability that M 2 is operational and not starved: E 2 = p(n, α 1, α 2 ). n > 0 α 2 = 1 (α,α ) (0,0) 1 2 n=0 n=1 n=2 n=3 n=4 n=5 n=6 n=7 n=8 n=9 n=10 n=11 n=12 n=13 =N 1 =N (0,1) (1,0) (1,1)

2.852 Manufacturing Systems Analysis 66/165 Copyright c 2010 Stanley B. Gershwin. Performance measures The probabilities of starvation and blockage are: p s = p(0, 0, 1), the probability of starvation, p b = p(n, 1, 0), the probability of blockage. The average buffer level is: n = np(n, α 1, α 2 ). all s

2.852 Manufacturing Systems Analysis 67/165 Copyright c 2010 Stanley B. Gershwin. Identities Repair frequency equals failure frequency For every repair, there is a failure (in steady state). When the system is in steady state, Let then r 1 prob [{α 1 = 0} and {n < N}] = p 1 prob [{α 1 = 1} and {n < N}]. D 1 = prob [{α 1 = 0} and {n < N}], r 1 D 1 = p 1 E 1.

2.852 Manufacturing Systems Analysis 68/165 Copyright c 2010 Stanley B. Gershwin. Identities Proof: The left side is the probability that the state leaves the set of states S 0 = {{α 1 = 0} and {n < N}}. since the only way the system can leave S 0 is for M 1 to get repaired. (M 1 is down, so the buffer cannot become full.) (α,α ) (0,0) 1 2 n=0 n=1 n=2 n=3 n=4 n=5 n=6 n=7 n=8 n=9 n=10 n=11 n=12 n=13 =N 1 =N (0,1) (1,0) (1,1)

2.852 Manufacturing Systems Analysis 69/165 Copyright c 2010 Stanley B. Gershwin. Identities The right side is the probability that the state enters S 0. When the system is in steady state, the only way for the state to enter S 0 is for it to be in set in the previous time unit. S 1 = {{α 1 = 1} and {n < N}} (α,α ) (0,0) 1 2 n=0 n=1 n=2 n=3 n=4 n=5 n=6 n=7 n=8 n=9 n=10 n=11 n=12 n=13 =N 1 =N (0,1) (1,0) (1,1)

2.852 Manufacturing Systems Analysis 70/165 Copyright c 2010 Stanley B. Gershwin. Identities Conservation of Flow E 1 = E 2 = E N 1 N 1 Proof: E 1 = p(n, 1, 0) + p(n, 1, 1), n=0 n=0 N N E 2 = p(n, 0, 1) + p(n, 1, 1). n=1 n=1 N 1 N Then E 1 E 2 = p(n, 1, 0) p(n, 0, 1) n=1 n=0 n=1 N 2 N 2 = p(n + 1, 1, 0) p(n, 0, 1) n=1

2.852 Manufacturing Systems Analysis 71/165 Copyright c 2010 Stanley B. Gershwin. Identities Or, E 1 E 2 = N 2 n=1 ( ) p(n + 1,1,0) p(n,0,1) Define δ(n) = p(n + 1,1,0) p(n,0,1). Then E 1 E 2 = N 2 n=1 δ(n)

2.852 Manufacturing Systems Analysis 72/165 Copyright c 2010 Stanley B. Gershwin. Identities Add lots of lower boundary equations: p(0, 0, 1) + p(1, 0, 0) + p(1, 1, 1) + p(2, 1, 0) = (1 r 1)p(0, 0, 1) + (1 r 1)r 2p(1, 0, 0) +(1 r 1)(1 p 2)p(1, 0, 1) + p 1(1 p 2)p(1, 1, 1) +(1 r 1)(1 r 2)p(1, 0, 0) + (1 r 1)p 2p(1, 0, 1) + p 1p 2p(1, 1, 1) +r 1p(0, 0, 1) + r 1r 2p(1, 0, 0) + r 1(1 p 2)p(1, 0, 1) +(1 p 1)(1 p 2)p(1, 1, 1) +r 1(1 r 2)p(1, 0, 0) + r 1p 2p(1, 0, 1) + (1 p 1)p 2p(1, 1, 1)

2.852 Manufacturing Systems Analysis 73/165 Copyright c 2010 Stanley B. Gershwin. Identities Or, p(0, 0, 1) + p(1, 0, 0) + p(1, 1, 1) + p(2, 1, 0) = Or, p(0, 0, 1) + p(1, 0, 0) + p(1, 0, 1) + p(1, 1, 1) p(2, 1, 0) = p(1, 0, 1) Then δ(1) = 0.

2.852 Manufacturing Systems Analysis 74/165 Copyright c 2010 Stanley B. Gershwin. Identities Now add all the internal equations, after changing the index of two of them: p(n, 0, 0) + p(n 1, 0, 1) + p(n + 1, 1, 0) + p(n, 1, 1) = (1 r 1)(1 r 2)p(n,0, 0) + (1 r 1)p 2p(n, 0, 1) +p 1(1 r 2)p(n, 1, 0) + p 1p 2p(n, 1, 1) (1 r 1)r 2p(n, 0, 0) + (1 r 1)(1 p 2)p(n, 0, 1) +p 1r 2p(n, 1, 0) + p 1(1 p 2)p(n,1, 1) r 1(1 r 2)p(n,0, 0) + r 1p 2p(n, 0, 1) +(1 p 1)(1 r 2)p(n, 1, 0) + (1 p 1)p 2p(n, 1, 1) r 1r 2p(n, 0, 0) + r 1(1 p 2)p(n, 0, 1) + (1 p 1)r 2p(n, 1, 0) +(1 p 1)(1 p 2)p(n, 1, 1)

2.852 Manufacturing Systems Analysis 75/165 Copyright c 2010 Stanley B. Gershwin. Identities Or, for n = 2,..., N 2, p(n,0,0) + p(n 1,0,1) + p(n + 1,1,0) + p(n,1,1) = p(n,0,0) + p(n,0,1) + p(n,1,0) + p(n,1,1), or, or, p(n + 1,1,0) p(n,0,1) = p(n,1,0) p(n 1,0,1) δ(n) = δ(n 1)

2.852 Manufacturing Systems Analysis 76/165 Copyright c 2010 Stanley B. Gershwin. Identities Since we have Therefore QED δ(1) = 0 and δ(n) = δ(n 1), n = 2,..., N 2 δ(n) = 0, n = 1,..., N 2 N 2 E 1 E 2 = δ(n) = 0 n=1

2.852 Manufacturing Systems Analysis 77/165 Copyright c 2010 Stanley B. Gershwin. Identities Alternative interpretation of p(n + 1,1,0) p(n,0,1) = 0: (α,α ) 1 2 (0,0) (0,1) (1,0) (1,1) n n+1 The only way the buffer can go from n + 1 to n is for the state to go to (n,0,1). The only way the buffer can go from n to n + 1 is for the state to go to (n + 1,1,0).

2.852 Manufacturing Systems Analysis 78/165 Copyright c 2010 Stanley B. Gershwin. Identities Flow rate/idle time E = e 1 (1 p b ). Proof: From the definitions of E 1 and D 1, we have prob [n < N] = E + D 1, p 1 or, 1 p b = E + E = r 1 Similarly, E = e 2 (1 p s ) E e 1.

2.852 Manufacturing Systems Analysis 79/165 Copyright c 2010 Stanley B. Gershwin. Analytical Solution 1. Guess a solution for the internal states of the form p(n,α 1,α 2 ) = ξ j (n,α 1,α 2 ) = X n Y α 1 1 Y α 2 2. 2. Determine sets of X j,y 1j,Y 2j that satisfy the internal equations. 3. Extend ξ j (n,α 1,α 2 ) to all of the boundary states using some of the boundary equations. 4. Find coefficients C j so that p(n,α 1,α 2 ) = j C j ξ j (n,α 1,α 2 ) satisfies the remaining boundary equationss and normalization.

2.852 Manufacturing Systems Analysis 80/165 Copyright c 2010 Stanley B. Gershwin. Analytical Solution Internal equations: X n = (1 r 1)(1 r 2)X n + (1 r 1)p 2X n Y 2 + p 1(1 r 2)X n Y 1 + p 1p 2X n Y 1Y 2 X n 1 Y 2 = (1 r 1)r 2X n + (1 r 1)(1 p 2)X n Y 2 + p 1r 2X n Y 1 +p 1(1 p 2)X n Y 1Y 2 X n+1 Y 1 = r 1(1 r 2)X n + r 1p 2X n Y 2 + (1 p 1)(1 r 2)X n Y 1 + (1 p 1)p 2X n Y 1Y 2 X n Y 1Y 2 = r 1r 2X n + r 1(1 p 2)X n Y 2 + (1 p 1)r 2X n Y 1 + (1 p 1)(1 p 2)X n Y 1Y 2

2.852 Manufacturing Systems Analysis 81/165 Copyright c 2010 Stanley B. Gershwin. Analytical Solution Or, 1 = (1 r 1)(1 r 2) + (1 r 1)p 2Y 2 + p 1(1 r 2)Y 1 + p 1p 2Y 1Y 2 X 1 Y 2 = (1 r 1)r 2 + (1 r 1)(1 p 2)Y 2 + p 1r 2Y 1 +p 1(1 p 2)Y 1Y 2 XY 1 = r 1(1 r 2) + r 1p 2Y 2 + (1 p 1)(1 r 2)Y 1 + (1 p 1)p 2Y 1Y 2 Y 1Y 2 = r 1r 2 + r 1(1 p 2)Y 2 + (1 p 1)r 2Y 1 + (1 p 1)(1 p 2)Y 1Y 2

2.852 Manufacturing Systems Analysis 82/165 Copyright c 2010 Stanley B. Gershwin. Analytical Solution Or, 1 = (1 r 1 + Y 1 p 1 )(1 r 2 + Y 2 p 2 ) X 1 Y 2 = (1 r 1 + Y 1 p 1 )(r 2 + Y 2 (1 p 2 )) XY 1 = (r 1 + Y 1 (1 p 1 ))(1 r 2 + Y 2 p 2 ) Y 1 Y 2 = (r 1 + Y 1 (1 p 1 )) (r 2 + Y 2 (1 p 2 ))

2.852 Manufacturing Systems Analysis 83/165 Copyright c 2010 Stanley B. Gershwin. Analytical Solution Since the last equation is a product of the other three, there are only three independent equations in three unknowns here. They may be simplified further: 1 = (1 r 1 + Y 1 p 1 )(1 r 2 + Y 2 p 2 ) XY 1 = r 1 + Y 1 (1 p 1 ) 1 r 1 + Y 1 p 1 X 1 Y 2 = r 2 + Y 2 (1 p 2 ) 1 r 2 + Y 2 p 2

2.852 Manufacturing Systems Analysis 84/165 Copyright c 2010 Stanley B. Gershwin. Analytical Solution Eliminating X and Y 2, this becomes 0 = Y 2 1 (p 1 + p 2 p 1 p 2 p 1 r 2 ) Y 1 (r 1 (p 1 + p 2 p 1 p 2 p 1 r 2 ) + p 1 (r 1 + r 2 r 1 r 2 r 1 p 2 )) which has two solutions: +r 1 (r 1 + r 2 r 1 r 2 r 1 p 2 ), Y 11 = r 1 p 1, Y 12 = r 1 + r 2 r 1 r 2 r 1 p 2 p 1 + p 2 p 1 p 2 p 1 r 2.

2.852 Manufacturing Systems Analysis 85/165 Copyright c 2010 Stanley B. Gershwin. Analytical Solution The complete solutions are: Y 11 = Y 21 = X 1 = 1 r 1 p 1 r 2 p 2 Y 12 = Y 22 = X 2 = r 1 + r 2 r 1 r 2 r 1 p 2 p 1 + p 2 p 1 p 2 p 1 r 2 r 1 + r 2 r 1 r 2 p 1 r 2 p 1 + p 2 p 1 p 2 p 2 r 1 Y 22 Y 12

2.852 Manufacturing Systems Analysis 86/165 Copyright c 2010 Stanley B. Gershwin. Analytical Solution Recall that ξ(n,α 1,α 2 ) = X n Y α 1 1 Y α 2 2. We now have the complete internal solution: p(n,α 1,α 2 ) = C 1 ξ 1 (n,α 1,α 2 ) + C 2 ξ 2 (n,α 1,α 2 ) = C 1 X n 1 Y α 1 11 Y α 2 21 + C 2 X n 2 Y α 1 12 Y α 2 22.

2.852 Manufacturing Systems Analysis 87/165 Copyright c 2010 Stanley B. Gershwin. Analytical Solution Boundary conditions: If we plug the internal expression for ξ(n,α 1,α 2 ) = X n Y α 1 right side of we find 1 Y α 2 2 into the ξ(1,0,1) = (1 r 1 )r 2 ξ(2,0,0) + (1 r 1 )(1 p 2 )ξ(2,0,1)+ p 1 r 2 ξ(2,1,0) + p 1 (1 p 2 )ξ(2,1,1), which implies that ξ(1,0,1) = XY 2 p(1,0,1) = C 1 Y 21 + C 2 X 2 Y 22.

2.852 Manufacturing Systems Analysis 88/165 Copyright c 2010 Stanley B. Gershwin. Analytical Solution Recall that Then p(2, 1, 0) = p(1, 0, 1). C 1 X 2 1 Y 11 + C 2 X 2 2 Y 12 = C 1 X 1 Y 21 + C 2 X 2 Y 22, or, ( ) ( ) C 1 X1 2 Y 11 C 1 X 1 Y 21 + C 2 X2 2 Y 12 C 2 X 2 Y 22 = 0, or, C 1 X 1 ( X 1 Y 11 Y 21 ) + C 2 X 2 ( X 2 Y 12 Y 22 ) = 0,

2.852 Manufacturing Systems Analysis 89/165 Copyright c 2010 Stanley B. Gershwin. Analytical Solution Recall Consequently, or, Therefore, X 2 = Y 22 Y 12 C 1 X 1 ( X 1 Y 11 Y 21 ) = 0, if ( ) r1 r 2 C 1 = 0, p 1 p 2 r 1 r 2, then C 1 = 0. p 1 p 2

2.852 Manufacturing Systems Analysis 90/165 Copyright c 2010 Stanley B. Gershwin. Analytical Solution In the following, we assume But what happens when And what does r 1 p 1 = r 1 p 1 = r 1 p 1 r 2 p 2 mean? r 2 p 2? r 2 p 2 and we drop the j subscript.

2.852 Manufacturing Systems Analysis 91/165 Copyright c 2010 Stanley B. Gershwin. Analytical Solution Combining the following two boundary conditions... r 1 p(0, 0, 1) = (1 r 1 )r 2 p(1, 0, 0) + (1 r 1 )(1 p 2 )p(1, 0, 1) +p 1 (1 p 2 )p(1, 1, 1). gives or, p(1, 1, 1) = r 1 p(0, 0, 1) + r 1 r 2 p(1, 0, 0) + r 1 (1 p 2 )p(1, 0, 1) +(1 p 1 )(1 p 2 )p(1, 1, 1) p(1, 1, 1) = r 2 p(1, 0, 0) + (1 p 2 )CXY 2 + (1 p 2 )p(1, 1, 1) p 2 p(1, 1, 1) = r 2 p(1, 0, 0) + (1 p 2 )CXY 2. There are three unknown quantities: p(1, 0, 0), p(1, 1, 1), and C.

2.852 Manufacturing Systems Analysis 92/165 Copyright c 2010 Stanley B. Gershwin. Analytical Solution Another boundary condition, p(1, 0, 0) = (1 r 1)(1 r 2)p(1, 0, 0) + (1 r 1)p 2p(1, 0, 1) + p 1p 2p(1, 1, 1) can be written (r 1 + r 2 r 1r 2)p(1, 0, 0) = (1 r 1)p 2CXY 2 + p 1p 2p(1, 1, 1). which also has three unknown quantities: p(1, 0, 0), p(1, 1, 1), and C. If we eliminate p(1, 1, 1) and simplify, we get (r 1 + r 2 r 1r 2 p 1r 2)p(1, 0, 0) = (p 1 + p 2 p 1p 2 p 2r 1)CXY 2. From the definition of Y 22 (slide 85), p(1, 0, 0) = CX.

2.852 Manufacturing Systems Analysis 93/165 Copyright c 2010 Stanley B. Gershwin. Analytical Solution If we plug this into the last equation on slide 91, we get or p 2 p(1,1,1) = CX(r 2 + (1 p 2 )Y 2 ) p(1,1,1) = Finally, the first equation on slide 91 gives p(0,0,1) = CX CX p 2 r 1 + r 2 r 1 r 2 r 1 p 2 p 1 + p 2 p 1 p 2 r 1 p 2. r 1 + r 2 r 1 r 2 r 1 p 2 r 1 p 2. The upper boundary conditions are determined in the same way.

2.852 Manufacturing Systems Analysis 94/165 Copyright c 2010 Stanley B. Gershwin. Analytical Solution Summary of Steady-State Probabilities: Boundary values p(0, 0, 0) = 0 r 1 + r 2 r 1 r 2 r 1 p 2 p(0, 0, 1) = CX r 1 p 2 p(0, 1, 0) = 0 p(0, 1, 1) = 0 p(1, 0, 0) = CX p(1, 0, 1) = CXY 2 p(1, 1, 0) = 0 p(1, 1, 1) = CX p 2 r 1 + r 2 r 1 r 2 r 1 p 2 p 1 + p 2 p 1 p 2 r 1 p 2 p(n 1, 0, 0) = CX N 1 p(n 1, 0, 1) = 0 p(n 1, 1, 0) = CX N 1 Y 1 p(n 1, 1, 1) = p(n, 0, 0) = 0 p(n, 0, 1) = 0 CX N 1 p 1 r 1 + r 2 r 1 r 2 p 1 r 2 p 1 + p 2 p 1 p 2 p 1 r 2 p(n, 1, 0) = CX N 1 r 1 + r 2 r 1 r 2 p 1 r 2 p 1 r 2 p(n, 1, 1) = 0

2.852 Manufacturing Systems Analysis 95/165 Copyright c 2010 Stanley B. Gershwin. Analytical Solution Summary of Steady-State Probabilities: Internal states, etc. where p(n, α 1, α 2 ) = CX n Y α1 1 Y α2 2, 2 n N 2; α 1 = 0, 1; α 2 = 0, 1 Y 1 = Y 2 = X = r 1 + r 2 r 1 r 2 r 1 p 2 p 1 + p 2 p 1 p 2 p 1 r 2 r 1 + r 2 r 1 r 2 p 1 r 2 p 1 + p 2 p 1 p 2 r 1 p 2 Y 2 Y 1 and C is a normalizing constant.

2.852 Manufacturing Systems Analysis 96/165 Copyright c 2010 Stanley B. Gershwin. Analytical Solution Observations: Typically, we can expect that r i <.2 since a repair is likely to take at least 5 times as long as an operation. Also, since, typically, efficiency = r i /(r i + p i ) >.7, p i <.4r i, p(0, 0, 1), p(1, 1, 1), p(n 1, 1, 1), p(n, 1, 0) are much larger than internal probabilities. This is because the system tends to spend much more time at those states than at internal states. Refer to transition graph on page 60 to trace out typical scenarios.

2.852 Manufacturing Systems Analysis 97/165 Copyright c 2010 Stanley B. Gershwin. Limits If r 1 0, then E 0, p s 1, p b 0, n 0. If r 2 0, then E 0, p b 1, p s 0, n N. If p 1 0, then p s 0, E 1 p b e 2, n N e 2. If p 2 0, then p b 0, E 1 p s e 1, n e 1. If N and e 1 < e 2, then E e 1, p b 0, p s 1 e 1 e 2.

2.852 Manufacturing Systems Analysis 98/165 Copyright c 2010 Stanley B. Gershwin. Limits Proof: Many of the limits follow from combining conservation of flow and the flow rate-idle time relationship: E = r 1 r 1 + p 1 (1 p b ) = r 2 r 2 + p 2 (1 p s ). The last set comes from the analytic solution and the observation that if e 1 > e 2, X > 1, and if e 1 < e 2, X < 1.

2.852 Manufacturing Systems Analysis 99/165 Copyright c 2010 Stanley B. Gershwin. Behavior 10 8 6 n(t) 4 2 0 0 200 400 600 800 1000 t r 1 =.1, p 1 =.01, r 2 =.1, p 2 =.01, N = 10

2.852 Manufacturing Systems Analysis 100/165 Copyright c 2010 Stanley B. Gershwin. Behavior 100 80 60 n(t) 40 20 0 0 2000 4000 6000 8000 10000 t r 1 =.1, p 1 =.01, r 2 =.1, p 2 =.01, N = 100

2.852 Manufacturing Systems Analysis 101/165 Copyright c 2010 Stanley B. Gershwin. Behavior 100 80 60 n(t) 40 20 0 0 2000 4000 6000 8000 10000 t r i =.1, i = 1, 2, p 1 =.02, p 2 =.01, N = 100

2.852 Manufacturing Systems Analysis 102/165 Copyright c 2010 Stanley B. Gershwin. Behavior 100 80 60 n(t) 40 20 0 0 2000 4000 6000 8000 10000 t r i =.1, i = 1, 2, p 1 =.01, p 2 =.02, N = 100

2.852 Manufacturing Systems Analysis 103/165 Copyright c 2010 Stanley B. Gershwin. Behavior Deterministic Processing Time 0.5 r = 0.14 1 τ = 1. p 1 =.1 r 2 =.1 p 2 =.1 P 0.4 0.3 r = 0.12 1 r = 0.1 1 r = 0.08 1 r = 0.06 1 0 50 100 150 200 N

2.852 Manufacturing Systems Analysis 104/165 Copyright c 2010 Stanley B. Gershwin. Behavior Discussion: Deterministic Processing Time Why are the curves increasing? Why do they reach an asymptote? What is P when N = 0? P 0.5 0.4 r = 0.14 1 r = 0.12 1 r = 0.1 1 What is the limit of P as N? Why are the curves with smaller r 1 lower? 0.3 0 50 100 150 200 N r = 0.08 1 r = 0.06 1

2.852 Manufacturing Systems Analysis 105/165 Copyright c 2010 Stanley B. Gershwin. Behavior Discussion: Why are the curves increasing? Why different asymptotes? What is n when N = 0? What is the limit of n as N? Why are the curves with smaller r 1 lower? n Deterministic Processing Time 200 150 100 50 0 0 50 100 150 200 N r = 0.14 1 r = 0.12 1 r = 0.1 1 r = 0.08 1 r = 0.06 1

2.852 Manufacturing Systems Analysis 106/165 Copyright c 2010 Stanley B. Gershwin. Behavior Deterministic Processing Time Deterministic Processing Time 200 0.5 r = 0.14 1 150 r = 0.14 1 P 0.4 r = 0.12 1 r = 0.1 1 n 100 r = 0.12 1 r = 0.1 1 0.3 r = 0.08 1 50 r = 0.08 1 0 50 100 150 200 N r = 0.06 1 0 0 50 100 150 200 N r = 0.06 1 What can you say about the optimal buffer size? How should it be related to r i, p i?

2.852 Manufacturing Systems Analysis 107/165 Copyright c 2010 Stanley B. Gershwin. Behavior Questions: If we want to increase production rate, which machine should we improve? What would happen to production rate if we improved any other machine?

2.852 Manufacturing Systems Analysis 108/165 Copyright c 2010 Stanley B. Gershwin. Production rate vs. storage space P 0.8 Improvements to non-bottleneck machine. 0.7 Identical machines Machine 1 more improved Machine 1 improved 0.6 20 40 60 80 100 120 140 160 180 200 Note: Graphs would be the same if we improved Machine 2. N

2.852 Manufacturing Systems Analysis 109/165 Copyright c 2010 Stanley B. Gershwin. Average inventory vs. storage space 200 180 n Machine 1 more improved 160 Inventory increases as the (non-bottleneck) upstream machine is improved and as the buffer space is increased. 140 120 100 80 60 40 Machine 1 improved 20 Identical machines 0 0 20 40 60 80 100 120 140 160 180 200 N

2.852 Manufacturing Systems Analysis 110/165 Copyright c 2010 Stanley B. Gershwin. Average inventory vs. storage space 200 180 n 160 Inventory decreases as the (non-bottleneck) downstream machine is improved. Inventory increases as the buffer space is increased. 140 120 100 80 60 40 20 0 Identical machines Machine 2 improved Machine 2 more improved 20 40 60 80 100 120 140 160 180 200 N

2.852 Manufacturing Systems Analysis 111/165 Copyright c 2010 Stanley B. Gershwin. Frequency and Production Rate Should we prefer short, frequent, disruptions or long, infrequent, disruptions? r 2 = 0.8, p 2 = 0.09, N = 10 0.95 r 1 and p 1 vary together and r 1 r 1+p 1 =.9 Answer: evidently, short, frequent failures. Why? P 0.9 0.85 0.8 0.75 0 0.2 0.4 0.6 0.8 1 r 1

2.852 Manufacturing Systems Analysis 112/165 Copyright c 2010 Stanley B. Gershwin. Frequency and Production Rate P 0.95 0.9 M 2 slow M 1 r 1 =.1, p 1 =.01 0.85 M 2 fast M 2 very slow 0.8 0 2 4 6 8 10 12 14 16 18 20 M 2 r 2 =.1, p 2 =.01 Fast r 2 =.01, p 2 =.001 Slow r 2 =.001, p 2 =.0001 Very slow N

2.852 Manufacturing Systems Analysis 113/165 Copyright c 2010 Stanley B. Gershwin. Frequency and Production Rate 0.95 M 1 r 1 =.1, p 1 =.01 0.9 0.85 M 2 r 2 =.1, p 2 =.01 Fast r 2 =.01, p 2 =.001 Slow r 2 =.001, p 2 =.0001 Very slow 0.8 0 20 40 60 80 100 120 140 160 180 200

2.852 Manufacturing Systems Analysis 114/165 Copyright c 2010 Stanley B. Gershwin. Frequency and Production Rate 0.95 M 1 r 1 =.1, p 1 =.01 0.9 0.85 M 2 r 2 =.1, p 2 =.01 Fast r 2 =.01, p 2 =.001 Slow r 2 =.001, p 2 =.0001 Very slow 0.8 0 200 400 600 800 1000 1200 1400 1600 1800 2000

2.852 Manufacturing Systems Analysis 115/165 Copyright c 2010 Stanley B. Gershwin. Frequency and Production Rate 0.91 0.9 0.89 M 1 r 1 =.1, p 1 =.01 0.88 0.87 0.86 0.85 0.84 M 2 r 2 =.1, p 2 =.01 Fast r 2 =.01, p 2 =.001 Slow r 2 =.001, p 2 =.0001 Very slow 0.83 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 x 10 4

2.852 Manufacturing Systems Analysis 116/165 Copyright c 2010 Stanley B. Gershwin. Frequency and Production Rate 0.95 M 1 r 1 =.1, p 1 =.01 0.9 M 2 0.85 r 2 =.1, p 2 =.01 Fast r 2 =.01, p 2 =.001 Slow r 2 =.001, p 2 =.0001 Very slow 0.8 0 1 2 3 4 5 6 7 8 9 10 x 10 4

2.852 Manufacturing Systems Analysis 117/165 Copyright c 2010 Stanley B. Gershwin. Frequency and Average Inventory 20 18 16 14 M 1 r 1 =.1, p 1 =.01 12 10 8 6 4 2 M 2 r 2 =.1, p 2 =.01 Fast r 2 =.01, p 2 =.001 Slow r 2 =.001, p 2 =.0001 Very slow 0 0 2 4 6 8 10 12 14 16 18 20

2.852 Manufacturing Systems Analysis 118/165 Copyright c 2010 Stanley B. Gershwin. Frequency and Average Inventory 200 180 160 M 1 r 1 =.1, p 1 =.01 140 120 100 80 60 40 20 M 2 r 2 =.1, p 2 =.01 Fast r 2 =.01, p 2 =.001 Slow r 2 =.001, p 2 =.0001 Very slow 0 0 20 40 60 80 100 120 140 160 180 200

2.852 Manufacturing Systems Analysis 119/165 Copyright c 2010 Stanley B. Gershwin. Frequency and Average Inventory 2000 1800 1600 M 1 r 1 =.1, p 1 =.01 1400 1200 1000 800 600 400 200 M 2 r 2 =.1, p 2 =.01 Fast r 2 =.01, p 2 =.001 Slow r 2 =.001, p 2 =.0001 Very slow 0 0 200 400 600 800 1000 1200 1400 1600 1800 2000

2.852 Manufacturing Systems Analysis 120/165 Copyright c 2010 Stanley B. Gershwin. Exponential processing time model Exponential processing time: exponential processing, failure, and repair time; discrete state, continuous time; discrete material. Assumptions are similar to deterministic processing time model, except: µ i δt = the probability that M i completes an operation in (t, t + δt); p i δt = the probability that M i fails during an operation in (t, t + δt); r i δt = the probability that M i is repaired, while it is down, in (t, t + δt); We can assume that only one event occurs during (t, t + δt).

2.852 Manufacturing Systems Analysis 121/165 Copyright c 2010 Stanley B. Gershwin. Exponential processing time model (α,α ) (0,0) 1 2 n=0 n=1 n=2 n=3 n=4 n=5 n=6 n=7 n=8 n=9 n=10 n=11 n=12 n=13 =N 1 =N (0,1) (1,0) (1,1) key transient non transient boundary internal out of transient states out of non transient states to increasing buffer level to decreasing buffer level unchanging buffer level

2.852 Manufacturing Systems Analysis 122/165 Copyright c 2010 Stanley B. Gershwin. Exponential processing time model Performance measures for general exponential lines The probability that Machine M i is processing a workpiece is the efficiency: E i = prob [α i = 1,n i 1 > 0,n i < N i ]. The production rate (throughput rate) of Machine M i, in parts per time unit, is P i = µ i E i.

2.852 Manufacturing Systems Analysis 123/165 Copyright c 2010 Stanley B. Gershwin. Exponential processing time model Conservation of Flow P = P 1 = P 2 =... = P k. This should be proved from the model.

2.852 Manufacturing Systems Analysis 124/165 Copyright c 2010 Stanley B. Gershwin. Exponential processing time model Flow Rate-Idle Time Relationship The isolated efficiency e i of Machine M i is, as usual, r i e i = r i + p i and it represents the fraction of time that M i is operational. The isolated production rate is ρ i = µ i e i.

2.852 Manufacturing Systems Analysis 125/165 Copyright c 2010 Stanley B. Gershwin. Exponential processing time model The flow rate-idle time relation is E i = e i prob [n i 1 > 0 and n i < N i ]. or P = ρ i prob [n i 1 > 0 and n i < N i ]. This should also be proved from the model.

2.852 Manufacturing Systems Analysis 126/165 Copyright c 2010 Stanley B. Gershwin. Exponential processing time model Balance equations steady state only α 1 = a 2 = 0 : p(n,0,0)(r 1 + r 2 ) = p(n,1,0)p 1 + p(n,0,1)p 2, 1 n N 1, p(0,0,0)(r 1 + r 2 ) = p(0,1,0)p 1, p(n,0,0)(r 1 + r 2 ) = p(n,0,1)p 2.

2.852 Manufacturing Systems Analysis 127/165 Copyright c 2010 Stanley B. Gershwin. Exponential processing time model α 1 = 0, α 2 = 1 : p(n,0,1)(r 1 + µ 2 + p 2 ) = p(n,0,0)r 2 + p(n,1,1)p 1 +p(n + 1,0,1)µ 2,1 n N 1 p(0,0,1)r 1 = p(0,0,0)r 2 + p(0,1,1)p 1 + p(1,0,1)µ 2 p(n,0,1)(r 1 + µ 2 + p 2 ) = p(n,0,0)r 2

2.852 Manufacturing Systems Analysis 128/165 Copyright c 2010 Stanley B. Gershwin. Exponential processing time model α 1 = 1, α 2 = 0 : p(n,1,0)(p 1 + µ 1 + r 2 ) = p(n 1,1,0)µ 1 + p(n,0,0)r 1 +p(n,1,1)p 2,1 n N 1 p(0,1,0)(p 1 + µ 1 + r 2 ) = p(0,0,0)r 1 p(n,1,0)r 2 = p(n 1,1,0µ 1 + p(n,0,0)r 1 + p(n,1,1)p 2

2.852 Manufacturing Systems Analysis 129/165 Copyright c 2010 Stanley B. Gershwin. Exponential processing time model α 1 = 1, α 2 = 1 : p(n, 1, 1)(p 1 + p 2 + µ 1 + µ 2 ) = p(n 1, 1, 1)µ 1 + p(n + 1, 1, 1)µ 2 +p(n, 1, 0)r 2 + p(n, 0, 1)r 1, 1 n N 1 p(0, 1, 1)(p 1 + µ 1 ) = p(1, 1, 1)µ 2 + p(0, 1, 0)r 2 + p(0, 0, 1)r 1 p(n, 1, 1)(p 2 + µ 2 ) = p(n 1, 1, 1)µ 1 + p(n, 1, 0)r 2 + p(n, 0, 1)r 1

2.852 Manufacturing Systems Analysis 130/165 Copyright c 2010 Stanley B. Gershwin. Exponential processing time model Performance measures Efficiencies: E 1 = E 2 = N 1 1 n=0 α 2 =0 N 1 n=1 α 1 =0 p(n,1,α 2 ), p(n,α 1,1).

2.852 Manufacturing Systems Analysis 131/165 Copyright c 2010 Stanley B. Gershwin. Exponential processing time model Production rate: Expected in-process inventory: n = N P = µ 1 E 1 = µ 2 E 2. 1 1 n=0 α 1 =0 α 2 =0 np(n,α 1,α 2 ).

2.852 Manufacturing Systems Analysis 132/165 Copyright c 2010 Stanley B. Gershwin. Solution of balance equations Assume p(n,α 1,α 2 ) = cx n Y α 1 1 Y α 2 2, 1 n N 1 where c,x,y 1,Y 2 are parameters to be determined. Plugging this into the internal equations gives µ 1 ( 1 X p 1 Y 1 + p 2 Y 2 r 1 r 2 = 0 ) r 1 1 p 1 Y 1 + r 1 + p 1 = 0 Y 1 µ 2 (X 1) p 2 Y 2 + r 2 Y 2 + r 2 p 2 = 0

Exponential processing time model These equations can be reduced to one fourth-order polynomial (quartic) equation in one unknown. One solution is Y 11 = r 1 p r 1 2 Y 21 = p 2 X 1 = 1 This solution of the quartic equation has a zero coefficient in the expression for the probabilities of the internal states: 4 p(n, α 1, α 2 ) = c j Xj n Y α1 1j Y α2 2j for n = 1,..., N 1. j=1 The other three solutions satisfy a cubic polynomial equation. Compare with slide 85. In general, there is no simple expression for them. 2.852 Manufacturing Systems Analysis 133/165 Copyright c 2010 Stanley B. Gershwin.

2.852 Manufacturing Systems Analysis 134/165 Copyright c 2010 Stanley B. Gershwin. Exponential processing time model Just as for the deterministic processing time line, we obtain the coefficients c 1, c 2, c 3, c 4 from the boundary conditions and the normalization equation; we find c 1 = 0; (What does this mean? Why is this true?) we construct all the boundary probabilities. Some are 0. we use the probabilities to evaluate production rate, average buffer level, etc; we prove statements about conservation of flow, flow rate-idle time, limiting values of some quantities, etc. we draw graphs, and observe behavior which is qualitatively very similar to deterministic processing time line behavior (e.g., P vs. N, n vs N, etc.). We also draw some new graphs (P vs. µ i, n vs µ i) and observe new behavior. This is discussed below with the discussion of continuous material lines.

2.852 Manufacturing Systems Analysis 135/165 Copyright c 2010 Stanley B. Gershwin. Continuous Material model Continuous material, or fluid: deterministic processing, exponential failure and repair time; mixed state, continuous time.; continuous material. µ i δt = the amount of material that M i processes, while it is up, in (t, t + δt); p i δt = the probability that M i fails, while it is up, in (t, t + δt); r i δt = the probability that M i is repaired, while it is down, in (t, t + δt);

2.852 Manufacturing Systems Analysis 136/165 Copyright c 2010 Stanley B. Gershwin. Model assumptions, notation, terminology, and conventions During time interval (t,t + δt): When 0 < x < N 1. the change in x is (α 1 µ 1 α 2 µ 2 )δt 2. the probability of repair of Machine i, that is, the probability that α i (t + δt) = 1 given that α i (t) = 0, is r i δt 3. the probability of failure of Machine i, that is, the probability that α i (t + δt) = 0 given that α i (t) = 1, is p i δt.

2.852 Manufacturing Systems Analysis 137/165 Copyright c 2010 Stanley B. Gershwin. When x = 0 1. the change in x is (α 1 µ 1 α 2 µ 2 ) + δt (That is, when x = 0, it can only increase.) 2. the probability of repair is r i δt 3. if Machine 1 is down, Machine 2 cannot fail. If Machine 1 is up, the probability of failure of Machine 2 is p2 b δt, where b p 2 µ p 2 =, µ = min(µ 1,µ 2 ) µ 2 The probability of failure of Machine 1 is p 1 δt.

2.852 Manufacturing Systems Analysis 138/165 Copyright c 2010 Stanley B. Gershwin. When x = N 1. the change in x is (α 1 µ 1 α 2 µ 2 ) δt 2. the probability of repair is r i δt 3. if Machine 2 is down, Machine 1 cannot fail. If Machine 2 is up, the probability of failure of Machine 1 is p1 b δt, where b p 1 µ p 1 =. µ 1 The probability of failure of Machine 2 is p 2 δt.

2.852 Manufacturing Systems Analysis 139/165 Copyright c 2010 Stanley B. Gershwin. Transition equations internal f (x,α 1,α 2,t)δx +o(δx) is the probability of the buffer level being between x and x + δx and the machines being in states α 1 and α 2 at time t. (0,0) Transitions into ([x,x+ δ x],1,1) (0,1) δx (1,0) r 1 δt (1,1) r 2 δt x µ δ t 1 x µ δ t + µ δ t 2 1 1 (p +p ) δ t 1 2 x µ δ 2