MIT Manufacturing Systems Analysis Lecture 10 12

Transcription

1 2.852 Manufacturing Systems Analysis 1/91 Copyright 2010 c Stanley B. Gershwin. MIT Manufacturing Systems Analysis Lecture Transfer Lines Long Lines Stanley B. Gershwin Massachusetts Institute of Technology Spring, 2010

2 2.852 Manufacturing Systems Analysis 2/91 Copyright c 2010 Stanley B. Gershwin. Long Lines M 1 B1 M 2 B2 M 3 B3 M 4 B4 M B Difficulty: 5 5 M 6 No simple formula for calculating production rate or inventory levels. State space is too large for exact numerical solution. If all buffer sizes are N and the length of the line is k, the number of states is S = 2 k (N + 1) k 1. if N = 10 and k = 20, S = Decomposition seems to work successfully.

3 2.852 Manufacturing Systems Analysis 3/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Concept Decomposition works for many kinds of systems, and extending it is an active research area. We start with deterministic processing time lines. Then we extend decomposition to other lines. Then we extend it to assembly/disassembly systems without loops. Then we look at systems with loops. Etc., etc. if there is time.

4 2.852 Manufacturing Systems Analysis 4/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Concept Conceptually: put an observer in a buffer, and tell him that he is in the buffer of a two-machine line. Question: What would the observer see, and how can he be convinced he is in a two-machine line?

5 2.852 Manufacturing Systems Analysis 5/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Concept Decomposition breaks up systems and then reunites them. Construct all the two-machine lines.

6 2.852 Manufacturing Systems Analysis 6/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Concept Evaluate the performance measures (production rate, average buffer level) of each two-machine line, and use them for the real line. This is an approximation; the behavior of the flow in the buffer of a two-machine line is not exactly the same as the behavior of the flow in a buffer of a long line. The two-machine lines are sometimes called building blocks.

7 2.852 Manufacturing Systems Analysis 7/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Concept Consider an observer in Buffer B i. Imagine the material flow process that the observer sees entering and the material flow process that the observer sees leaving the buffer. We construct a two-machine line L(i) (ie, we find machines M u (i) and M d (i) with parameters r u (i), p u (i), r d (i), p d (i), and N(i) = N i ) such that an observer in its buffer will see almost the same processes. The parameters are chosen as functions of the behaviors of the other two-machine lines.

8 2.852 Manufacturing Systems Analysis 8/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Concept M i 2 B i 2 M i 1 B i 1 B i M i+1 B i+1 M M i i+2 B i+2 M i+3 Line L(i) M (i) u M (i) d

9 2.852 Manufacturing Systems Analysis 9/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Concept There are 4(k 1) unknowns for the deterministic processing time line: r u (1),p u (1),r d (1),p d (1), r u (2),p u (2),r d (2),p d (2),..., r u (k 1),p u (k 1),r d (k 1),p d (k 1) Therefore, we need 4(k 1) equations, and an algorithm for solving those equations.

10 2.852 Manufacturing Systems Analysis 10/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Overview The decomposition equations relate r u (i), p u (i), r d (i), and p d (i) to behavior in the real line and in other two-machine lines. Conservation of flow, equating all production rates. Flow rate/idle time, relating production rate to probabilities of starvation and blockage. Resumption of flow, relating r u (i) to upstream events and r d (i) to downstream events. Boundary conditions, for parameters of M u (1) and M d (k 1).

11 2.852 Manufacturing Systems Analysis 11/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Overview All the quantities in all these equations are specified parameters, or unknowns, or functions of parameters or unknowns derived from the two-machine line analysis. This is a set of 4(k 1) equations.

12 2.852 Manufacturing Systems Analysis 12/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Overview Notation convention: Items that pertain to two-machine line L(i) will have i in parentheses. Example: r u (i). Items that pertain to the real line L will have i in the subscript. Example: r i.

13 2.852 Manufacturing Systems Analysis 13/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Conservation of Flow E(i) = E(1),i = 2,..., k 1. Recall that E(i) is a function of the unknowns r u (i), p u (i), r d (i), and p d (i). (It is also a function of N(i), but N(i) is known.) We know how to evaluate it easily, but we don t have a simple expression for it. This is a set of k 2 equations.

14 2.852 Manufacturing Systems Analysis 14/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Flow Rate-Idle Time E i = e i prob [n i 1 > 0 and n i < N i ] where e i = r i r i + p i Problem: This expression involves a joint probability of two buffers taking certain values at the same time. But we only know how to evaluate two-machine, one-buffer lines, so we only know how to calculate the probability of one buffer taking on a certain value at a time.

15 2.852 Manufacturing Systems Analysis 15/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Flow Rate-Idle Time Observation: Reason: prob (n i 1 = 0 and n i = N i ) 0. M i 1 B i 1 M i B i M i+1 0 N i The only way to have n i 1 = 0 and n i = N i is if M i 1 is down or starved for a long time and M i is up and M i+1 is down or blocked for a long time and to have exactly N i parts in the two buffers.

16 2.852 Manufacturing Systems Analysis 16/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Flow Rate-Idle Time Then prob [n i 1 > 0 and n i < N i ] = prob [NOT {n i 1 = 0 or n i = N i }] = 1 prob [n i 1 = 0 or n i = N i ] = 1 { prob (n i 1 = 0) + prob (n i = N i ) prob (n i 1 = 0 and n i = N i )} 1 { prob (n i 1 = 0) + prob (n i = N i )}

17 2.852 Manufacturing Systems Analysis 17/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Flow Rate-Idle Time Therefore Note that E i e i [1 prob (n i 1 = 0) prob (n i = N i )] prob (n i 1 = 0) = p s (i 1); prob (n i = N i ) = p b (i) Two of the FRIT relationships in lines L(i 1) and L(i) are E(i) = e u (i)[1 p b (i)] ; E(i 1) = e d (i 1)[1 p s (i 1)]

18 2.852 Manufacturing Systems Analysis 18/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Flow Rate-Idle Time or, so (replacing with =), p s (i 1) = 1 E i = e i [1 { 1 E(i 1) e d (i 1) ; p b(i) = 1 } { E(i 1) 1 e d (i 1) The goal is to have E = E i = E(i 1) = E(i), so { } E(i) = e i [1 1 E(i) e d (i 1) { 1 E(i) e u (i) }] E(i) e u (i) }] E(i) e u (i)

19 2.852 Manufacturing Systems Analysis 19/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Flow Rate-Idle Time Since we can write e d (i 1) = r d (i 1) p d (i 1) + r d (i 1) ; e r u (i) u(i) = p u (i) + r u (i), p d (i 1) p u (i) + = r d (i 1) r u (i) 1 + E(i) 1 e i 2, i = 2,..., k 1 This is a set of k 2 equations.

20 2.852 Manufacturing Systems Analysis 20/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Resumption of Flow M i 4 B i 4 M i 3 B i 3 M i 2 B i 2 M i 1 B i 1 M i B i M i+1 B i+1 M i+2 B i+2 M i+3 M (i) u M (i) d When the observer sees M u (i) down, M i may actually be down...

21 2.852 Manufacturing Systems Analysis 21/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Resumption of Flow M i 4 B i 4 M i 3 B i 3 M i 2 B i 2 M i 1 B i 1 M i B i M i+1 B i+1 M i+2 B i+2 M i+3 0 M (i) u M (i) d... or, M i 1 may be down and B i 1 may be empty,...

22 2.852 Manufacturing Systems Analysis 22/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Resumption of Flow M i 4 B i 4 M i 3 B i 3 M i 2 B i 2 M i 1 B i 1 M i B i M i+1 B i+1 M i+2 B i+2 M i M (i) u M (i) d... or M i 2 may be down and B i 1 and B i 2 may be empty,...

23 2.852 Manufacturing Systems Analysis 23/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Resumption of Flow M i 4 B i 4 M i 3 B 0 i 3 M i 2 B i 2 M i 1 B i 1 M i B i M i+1 B i+1 M i+2 B i+2 M i M (i) u M (i) d... or M i 3 may be down and B i 1 and B i 2 and B i 3 may be empty,...

24 2.852 Manufacturing Systems Analysis 24/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Resumption of Flow M i 4 B i 4 0 M i 3 B 0 i 3 M i 2 B i 2 M i 1 B i 1 M i B i M i+1 B i+1 M i+2 B i+2 M i M (i) u M (i) d... etc.

25 2.852 Manufacturing Systems Analysis 25/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Resumption of Flow M i 4 B i 4 M i 3 B i 3 M i 2 B i 2 M i 1 B i 1 M i B i M i+1 B i+1 M i+2 B i+2 M i+3 M (i 1) u M (i 1) d Similarly for the observer in B i 1.

26 2.852 Manufacturing Systems Analysis 26/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Resumption of Flow M i 4 B i 4 M i 3 B i 3 M i 2 B i 2 M i 1 B i 1 M i B i M i+1 B i+1 M i+2 B i+2 M i+3 M (i 1) u M (i 1) d M i 4 B i 4 M i 3 B i 3 M i 2 B i 2 M i 1 B i 1 M i B i M i+1 B i+1 M i+2 B i+2 M i+3 0 M (i) u M (i) d Comparison

27 2.852 Manufacturing Systems Analysis 27/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Resumption of Flow M i 4 B i 4 M i 3 B i 3 M i 2 B i 2 M i 1 B i 1 M i B i M i+1 B i+1 M i+2 B i+2 M i+3 0 M (i 1) u M (i 1) d M i 4 B i 4 M i 3 B i 3 M i 2 B i 2 M i 1 B i 1 M i B i M i+1 B i+1 M i+2 B i+2 M i M (i) u M (i) d

28 2.852 Manufacturing Systems Analysis 28/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Resumption of Flow M i 4 B i 4 M i 3 B 0 i 3 M i 2 B i 2 M i 1 B i 1 M i B i M i+1 B i+1 M i+2 B i+2 M i+3 0 M (i 1) u M (i 1) d M i 4 B i 4 M i 3 B 0 i 3 M i 2 B i 2 M i 1 B i 1 M i B i M i+1 B i+1 M i+2 B i+2 M i M (i) u M (i) d

29 2.852 Manufacturing Systems Analysis 29/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Resumption of Flow M i 4 B i 4 0 M i 3 B 0 i 3 M i 2 B i 2 M i 1 B i 1 M i B i M i+1 B i+1 M i+2 B i+2 M i+3 0 M (i 1) u M (i 1) d M i 4 B i 4 0 M i 3 B 0 i 3 M i 2 B i 2 M i 1 B i 1 M i B i M i+1 B i+1 M i+2 B i+2 M i M (i) u M (i) d

30 2.852 Manufacturing Systems Analysis 30/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Resumption of Flow That is, when the Line L(i) observer sees a failure in M u (i), M (i) u M (i) d either real machine M i is down, M i 4 B i 4 M B M B i 3 i 3 M B M B M i 2 i 2 i 1 i 1 i M i i+1 B i+1 M B i+2 i+2 i+3 or Buffer B i 1 is empty and the Line L(i 1) observer sees a failure in M u (i 1). M B i 4 i 4 M B i 3 i 3 M B i 2 i 2 M B i 1 i 1 M i B i M B i+1 i+1 M B i+2 i+2 M i+3 0 M (i 1) u 0 M (i 1) d Note that these two events are mutually exclusive. Why?

31 2.852 Manufacturing Systems Analysis 31/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Resumption of Flow Also, for the Line L(j) observer to see M u (j) up, M j must be up and B j 1 must be non-empty. Therefore, {α u (j, τ) = 1} {α j (τ) = 1} and {n j 1 (τ 1) > 0} {α u (j, τ) = 0} {α j (τ) = 0} or {n j 1 (τ 1) = 0}

32 2.852 Manufacturing Systems Analysis 32/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Resumption of Flow Then r u (i) = prob [α u (i,t + 1) = 1 α u (i,t) = 0] [ = prob {α i (t + 1) = 1} and {n i 1 (t) > 0} ] {α i (t) = 0} or {n i 1 (t 1) = 0}

33 2.852 Manufacturing Systems Analysis 33/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Resumption of Flow To express r u (i) in terms of quantities we know or can find, we have to simplify prob (U V or W), where U = {α i (t + 1) = 1} and {n i 1 (t) > 0} V = {α i (t) = 0} W = {n i 1 (t 1) = 0} Important: V and W are disjoint. prob (V and W) = 0.

34 2.852 Manufacturing Systems Analysis 34/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Resumption of Flow prob (U V or W) = prob (U and (V or W)) prob (V or W) = prob ((U and V) or (U and W)) prob (V or W) = = prob (U and V) prob (V or W) prob (U V)prob (V) prob (V or W) + + prob (U and W) prob (V or W) prob (U W)prob (W) prob (V or W)

35 2.852 Manufacturing Systems Analysis 35/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Resumption of Flow prob (V) prob (W) = prob (U V) + prob (U W) prob (V or W) prob (V or W) Note that so prob (V V or W) = prob (V and (V or W)) prob (V or W) = prob (V) prob (V or W) prob (U V or W) = prob (U V)prob (V V or W) +prob (U W)prob (W V or W).

36 2.852 Manufacturing Systems Analysis 36/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Resumption of Flow Then, if we plug U, V, and W from Slide 33 into this, we get where r u (i) = A(i 1)X(i) + B(i)X (i), i = 2,..., k 1 A(i 1) = prob (U W) = prob [ n i 1 (t) > 0 and α i (t + 1) = 1 ] n i 1 (t 1) = 0,

37 2.852 Manufacturing Systems Analysis 37/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Resumption of Flow X(i) = prob (W [ V or W) ] = prob n i 1 (t 1) = 0 n i 1 (t 1) = 0 or α i (t) = 0, B(i) = prob (U V) = prob [n i 1 (t) > 0 and α i (t + 1) = 1 α i (t) = 0], X (i) = prob (V V or W) = prob [α i (t) = 0 {n i 1 (t 1) = 0 or α i (t) = 0}].

38 2.852 Manufacturing Systems Analysis 38/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Resumption of Flow To evaluate A(i 1) = prob» n i 1(t) > 0 and α i (t + 1) = 1 ni 1(t 1) = 0 : Note that For Buffer i 1 to be empty at time t 1, Machine M i must be up at time t 1 and also at time t. It must have been up in order to empty the buffer, and it must stay up because it cannot fail. Therefore α i (t) = 1. For Buffer i 1 to be non-empty at time t after being empty at time t 1, it must have gained 1 part. For it to gain a part when α i (t) = 1, M i must not have been working (because it was previously starved). Therefore, M i could not have failed and A(i 1) can therefore be written» A(i 1) = prob n i 1(t) > 0 ni 1(t 1) = 0

39 2.852 Manufacturing Systems Analysis 39/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Resumption of Flow A(i 1) = prob» n i 1(t) > 0 ni 1(t 1) = 0 For Buffer i 1 to be empty, M i 1 must be down or starved. For M i 1 to be starved, M i 2 must be down or starved, etc. Therefore, saying M i 1 is down or starved is equivalent to saying M u (i 1) is down. That is, if n i 1(t 1) = 0 then α u (i 1, t 1) = 0. Conversely, for Buffer i 1 to be non-empty, M i 1 must not be down or starved. That is, if n i 1(t) > 0, then α u (i 1, t) = 1. Therefore, A(i 1) = prob» α u (i 1, t) = 1 αu (i 1, t 1) = 0 = r u (i 1)

40 2.852 Manufacturing Systems Analysis 40/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Resumption of Flow Similarly, B(i) = prob [n i 1(t) > 0 and α i (t + 1) = 1 α i (t) = 0] Note that if α i (t) = 0, we must have n i 1(t) > 0. Therefore or, so B(i) = prob [α i (t + 1) = 1 α i (t) = 0], B(i) = r i r u (i) = r u (i 1)X(i) + r i X (i),

41 2.852 Manufacturing Systems Analysis 41/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Resumption of Flow Interpretation so far: r u (i), the probability that M u (i) goes from down to up, is r i times the probability that M u (i) is down because M i is down plus r u (i 1) times the probability that M u (i) is down because M u (i 1) is down and B i 1 is empty.

42 2.852 Manufacturing Systems Analysis 42/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Resumption of Flow X(i)= the probability that M u (i) is down because M u (i 1) is down and B i 1 is empty; X (i) = the probability that M u (i) is down because M i is down. Since these are the only two ways that M u (i) can be down, X (i) = 1 X(i)

43 2.852 Manufacturing Systems Analysis 43/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Resumption of Flow X(i) = prob [ ] n i 1 (t 1) = 0 n i 1 (t 1) = 0 or α i (t) = 0 = prob [n i 1 (t 1) = 0 and {n i 1 (t 1) = 0 or α i (t) = 0}] prob [n i 1 (t 1) = 0 or α i (t) = 0] = prob [n i 1 (t 1) = 0] prob [n i 1 (t 1) = 0 or α i (t) = 0] = p s (i 1) prob [n i 1 (t 1) = 0 or α i (t) = 0]

44 2.852 Manufacturing Systems Analysis 44/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Resumption of Flow To analyze the denominator, note {n i 1(t 1) = 0 or α i (t) = 0} = {α u (i) = 0} by definition; prob [n i 1(t 1) = 0 or α i (t) = 0] prob [{n i 1(t 1) = 0 or α i (t) = 0} and n i (t 1) < N i ] because prob [n i 1(t 1) = 0 and n i (t 1) = N i ] 0 so the denominator is, approximately, Recall that this is equal to prob [α u (i) = 0 and n i (t 1) < N i ] p u (i) r u (i) prob [αu(i) = 1 and ni (t 1) < Ni ] = p u (i) E(i) r u (i)

45 2.852 Manufacturing Systems Analysis 45/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Resumption of Flow Therefore, and X(i) = p s (i 1)r u (i) p u (i)e(i) r u (i) = r u (i 1)X(i) + r i (1 X(i)), i = 2,..., k 1 This is a set of k 2 equations.

46 2.852 Manufacturing Systems Analysis 46/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Resumption of Flow By the same logic, r d (i 1) = r d (i)y (i) + r i (1 Y(i)), i = 2,..., k 1 where Y (i) = p b (i)r d (i 1). p d (i 1)E(i 1) This is a set of k 2 equations. We now have 4(k 2) = 4k 8 equations.

47 2.852 Manufacturing Systems Analysis 47/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Boundary Conditions M d (1) is the same as M 1 and M d (k 1) is the same as M k. Therefore r u (1) = r 1 p u (1) = p 1 r d (k 1) = r k p d (k 1) = p k This is a set of 4 equations. We now have 4(k 1) equations in 4(k 1) unknowns r u (i), p u (i), r d (i), p d (i), i = 1,..., k 1.

48 2.852 Manufacturing Systems Analysis 48/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Algorithm FRIT: Upstream equations: p d (i 1) p u (i) = + 2 r d (i 1) r u (i) E(i) e i r u (i) = r u (i 1)X(i) + r i (1 X(i)); X(i) = p u (i) = r u (i) Downstream equations: 1 + E(i) 1 e i 2 «p d (i 1) r d (i 1) r d (i) = r d (i + 1)Y (i + 1) + r i+1(1 Y(i + 1)); Y (i + 1) = p d (i) = r d (i) 1 + E(i + 1) 1 e i+1 2 «p u (i + 1) r u (i + 1) p s (i 1)r u (i) p u (i)e(i) p b (i + 1)r d (i). p d (i)e(i)

49 2.852 Manufacturing Systems Analysis 49/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Algorithm We use the conservation of flow conditions by modifying these equations. Modified upstream equations: r u (i) = r u (i 1)X(i) + r i (1 X(i)); X(i) = p u (i) = r u (i) Modified downstream equations: 1 + E(i 1) 1 e i 2 «p d (i 1) r d (i 1) r d (i) = r d (i + 1)Y (i + 1) + r i+1(1 Y(i + 1)); Y (i + 1) = p d (i) = r d (i) 1 + E(i + 1) 1 e i+1 2 «p u (i + 1) r u (i + 1) p s (i 1)r u (i) p u (i)e(i 1) p b (i + 1)r d (i). p d (i)e(i + 1)

50 2.852 Manufacturing Systems Analysis 50/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Algorithm Possible Termination Conditions: E(i) E(1) < ǫ for i = 2,..., k 1, or The change in each r u (i), p u (i), r d (i), p d (i) parameter, i = 1,..., k 1 is less than ǫ, or etc.

51 2.852 Manufacturing Systems Analysis 51/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Equations Algorithm DDX algorithm : due to Dallery, David, and Xie (1988). 1. Guess the downstream parameters of L(1) (r d (1), p d (1)). Set i = Use the modified upstream equations to obtain the upstream parameters of L(i) (r u (i), p u (i)). Increment i. 3. Continue in this way until L(k 1). Set i = k Use the modified downstream equations to obtain the downstream parameters of L(i). Decrement i. 5. Continue in this way until L(1). 6. Go to Step 2 or terminate.

52 2.852 Manufacturing Systems Analysis 52/91 Copyright c 2010 Stanley B. Gershwin. Decomposition Approximations Is the decomposition exact? NO, because 1. The behavior of the flow in the buffer of a two-machine line is not exactly the same as the behavior of the flow in a buffer of a long line. 2. prob [n i 1 (t 1) = 0 and n i (t 1) = N i ] 0 Question: When will this work well, and when will it work badly?

53 2.852 Manufacturing Systems Analysis 53/91 Copyright c 2010 Stanley B. Gershwin. Examples Three-machine line Three-machine line production rate..8 E p 2 = r 1 = r 2 = r 3 =.2 p 1 =.05 N 1 = N 2 = p 3

54 2.852 Manufacturing Systems Analysis 54/91 Copyright c 2010 Stanley B. Gershwin. Examples Three-machine line Three-machine line total average inventory p 2 = I p 3 r 1 = r 2 = r 3 =.2 p 1 =.05 N 1 = N 2 = 5

55 2.852 Manufacturing Systems Analysis 55/91 Copyright c 2010 Stanley B. Gershwin. Examples Long lines Machines; r=0.1; p=0.01; mu=1.0; N= Average Buffer Level Buffer Number Distribution of material in a line with identical machines and buffers. Explain the shape.

56 2.852 Manufacturing Systems Analysis 56/91 Copyright c 2010 Stanley B. Gershwin. Examples Long lines Analytical vs simulation Time steps Decomp 10,000 50, ,000 Production rate Analytic 10,000 steps 50,000 steps 200,000 steps 12 Average buffer leve Buffer number (Not the same line as in Slide 55.)

57 Examples Long lines Machines; r=0.1; p=0.01; mu=1.0; N=20.0 EXCEPT N(25)= Average Buffer Level 10 5 Same as Slide 55 except that Buffer 25 is now huge. Explain the shape Buffer Number Manufacturing Systems Analysis 57/91 Copyright c 2010 Stanley B. Gershwin.

58 2.852 Manufacturing Systems Analysis 58/91 Copyright c 2010 Stanley B. Gershwin. Examples Long lines Machines; r=0.1; p=0.01; mu=1.0; N= Average Buffer Level 10 5 Upstream half of Slide 57. Explain the shape Buffer Number

59 2.852 Manufacturing Systems Analysis 59/91 Copyright c 2010 Stanley B. Gershwin. Examples Long lines 50 Machines; upstream r=0.1; p=0.01; mu=1.0; N=20.0; N(25)= downstream r=0.15; p=0.01; mu=1.0, N= Average Buffer Level 10 5 Upstream same as Slide 58; downstream faster. Explain the shape Buffer Number

60 2.852 Manufacturing Systems Analysis 60/91 Copyright c 2010 Stanley B. Gershwin. Examples Long lines 50 Machines; upstream r=0.1; p=0.01; mu=1.0; N=20.0; N(25)= downstream r=0.09; p=0.01; mu=1.0, N= Average Buffer Level 10 5 Upstream same as Slide 58; downstream faster. Explain the shape Buffer Number

61 2.852 Manufacturing Systems Analysis 61/91 Copyright c 2010 Stanley B. Gershwin. Examples Long lines 50 Machines; upstream r=0.1; p=0.01; mu=1.0; N=20.0; N(25)= downstream r=0.09; p=0.01; mu=1.0, N= Average Buffer Level Buffer Number Downstream same as downstream half of Slide 57; upstream faster. Explain the shape.

62 2.852 Manufacturing Systems Analysis 62/91 Copyright c 2010 Stanley B. Gershwin. Examples Long lines Machines; r=0.1; p=0.01; mu=1.0, N=20.0 EXCEPT N(25)=2000.0, r(26)=.09, p(26)= Average Buffer Level Buffer Number Same as upstream half of Slide 61 except for Machine 26. Explain the shape. How was Machine 26 chosen?

63 2.852 Manufacturing Systems Analysis 63/91 Copyright c 2010 Stanley B. Gershwin. Examples Long lines Bottlenecks Machines; r=0.1; p=0.01; mu=1.0; N=20.0 EXCEPT mu(10)= Average Buffer Level Buffer Number Operation time bottleneck. Identical machines and buffers, except for M 10. Explain the shape.

64 Examples Long lines Bottlenecks Machines; r=0.1; p=0.01; mu=1.0; N=20.0 EXCEPT p(10)= Average Buffer Level 10 5 Failure time bottleneck. Explain the shape Buffer Number Manufacturing Systems Analysis 64/91 Copyright c 2010 Stanley B. Gershwin.

65 2.852 Manufacturing Systems Analysis 65/91 Copyright c 2010 Stanley B. Gershwin. Examples Long lines Bottlenecks Machines; r=0.1; p=0.01; mu=1.0; N=20.0 EXCEPT r(10)= Average Buffer Level 10 5 Repair time bottleneck. Explain the shape Buffer Number

66 2.852 Manufacturing Systems Analysis 66/91 Copyright c 2010 Stanley B. Gershwin. Examples Infinitely long lines Infinitely long lines with identical machines and buffers r i = r p i = p for each i, < i <. N i = N The observer in each buffer sees exactly the same behavior. Consequently, the decomposed pseudo-machines are all identical and symmetric. For each i, r u (i) = r u (i 1) = r d (i) = r d (i 1) p u (i) = p u (i 1) = p d (i) = p d (i 1).

67 2.852 Manufacturing Systems Analysis 67/91 Copyright c 2010 Stanley B. Gershwin. Examples Infinitely long lines Resumption of flow says so r u (i) = r d (i) = r. FRIT says r u (i) = r u (i 1)X(i) + r i (1 X(i)) r u = r u X + r(1 X) p d (i 1) r d (i 1) + p u (i) r u (i) = 1 E (i) + 1 e i 2 2p u r = 1 E + 1 e 2

68 2.852 Manufacturing Systems Analysis 68/91 Copyright c 2010 Stanley B. Gershwin. Examples Infinitely long lines In the last equation, p u is unknown and E is a function of p u. This is one equation in one unknown. Production Rate E, Parts/Cycle r i =.1, p 1 =.1, i= l,..., k N i = 10, i = 1..., k - 1 N i = 5, i = 1,..., k Number of Machines k

69 2.852 Manufacturing Systems Analysis 69/91 Copyright c 2010 Stanley B. Gershwin. Examples Effect of one buffer size on all buffer levels Average Buffer Level n1 n2 n3 n4 n5 n6 n7 Continuous material model. Eight-machine, seven-buffer line. For each machine, r =.075, p =.009, µ = 1.2. For each buffer (except Buffer 6), N = N 6 M 1 B1 B 2 M 3 B3 M 4 M 2 B 4 M 5 B5 M 6 B 6 M 7 B7 M 8

70 2.852 Manufacturing Systems Analysis 70/91 Copyright c 2010 Stanley B. Gershwin. Examples Effect of one buffer size on all buffer levels n1 n2 n3 n4 n5 n6 n7 Average Buffer Level Which n i are decreasing and which are increasing? Why? N 6 M 1 B1 B 2 M 3 B3 M 4 M 2 B 4 M 5 B5 M 6 B 6 M 7 B7 M 8

71 2.852 Manufacturing Systems Analysis 71/91 Copyright c 2010 Stanley B. Gershwin. Examples Buffer allocation Which has a higher production rate? 9-Machine line with two buffering options: 8 buffers equally sized; and M 1 B 1 M 2 B2 M 3 B3 M 4 B 4 M 5 B5 M 6 B 6 M 7 B 7 M 8 B8 M 9 2 buffers equally sized. M 1 M 2 M 3 B 3 M 4 M 5 M 6 B 6 M 7 M 8 M 9

72 2.852 Manufacturing Systems Analysis 72/91 Copyright c 2010 Stanley B. Gershwin. Examples Buffer allocation P buffers 2 buffers Continuous model; all machines have r =.019, p =.001, µ = 1. What are the asymptotes? Is 8 buffers always faster? Total Buffer Space

73 2.852 Manufacturing Systems Analysis 73/91 Copyright c 2010 Stanley B. Gershwin. Examples Buffer allocation buffers 2 buffers P Total Buffer Space Is 8 buffers always faster? Perhaps not, but difference is not significant in systems with very small buffers.

74 2.852 Manufacturing Systems Analysis 74/91 Copyright c 2010 Stanley B. Gershwin. Long Lines More Models Discrete Material Exponential Processing Time and Continuous Material Models New issue: machines may operate at different speeds. Blockage and starvation may be caused by differences in machine speeds, not only failures. Decomposition of these classes of systems is similar to that of discrete-material, deterministic-processing time lines except The two-machine lines have machines with 3 parameters (r u (i), p u (i), µ u (i); r d (i), p d (i), µ d (i)). More equations 6(k 1) are therefore needed. Exponential decomposition is described in the book in detail; continuous material decomposition was not developed until after book was written.

75 2.852 Manufacturing Systems Analysis 75/91 Copyright c 2010 Stanley B. Gershwin. Long Lines Exponential Processing Time Model The observer thinks he is in a two-machine exponential processing time line with parameters r u (i)δt = p u (i)δt = probability that M u (i) goes from down to up in (t, t + δt), for small δt; probability that M u (i) goes from up to down in (t, t + δt) if it is not blocked, for small δt; µ u (i)δt = probability that a piece flows into B i in (t, t + δt) when M u (i) is up and not blocked, for small δt; r d (i)δt = p d (i)δt = probability that M d (i) goes from down to up in (t, t + δt), for small δt; probability that M d (i) goes from up to down in (t, t + δt) if it is not starved, for small δt; µ d (i)δt = probability that a piece flows out of B i in (t, t + δt) when M d (i) is up and not starved, for small δt.

76 2.852 Manufacturing Systems Analysis 76/91 Copyright c 2010 Stanley B. Gershwin. Long Lines Exponential Processing Time Model Equations We have 6(k 1) unknowns, so we need 6(k 1) equations. They are Interruption of flow, relating p u (i) to upstream events and p d (i) to downstream events, Resumption of flow, Conservation of flow, Flow rate/idle time, Boundary conditions. All of these, except for the Interruption of Flow equations, are similar to those of the deterministic processing time case.

77 2.852 Manufacturing Systems Analysis 77/91 Copyright c 2010 Stanley B. Gershwin. Long Lines Exponential Processing Time Model Interruption of Flow The first two sets of equations describe the interruptions of flow caused by machine failures. By definition, [ ] p u (i)δt = prob α u (i; t + δt) = 0 α u (i; t) = 1 and n i (t) < N i, or, p u (i)δt = prob [ ] M u (i) down at t + δt M u (i) up and n i < N i at t.

78 2.852 Manufacturing Systems Analysis 78/91 Copyright c 2010 Stanley B. Gershwin. Long Lines Exponential Processing Time Model Interruption of Flow We define the events that a pseudo-machine is up or down as follows: M u (i) is down if 1. M i is down, or 2. n i 1 = 0 and M u (i 1) is down. M u (i) is up for all other states of the transfer line upstream of Buffer B i. Therefore, M u (i) is up if 1. M i is operational and n i 1 > 0, or 2. M i is operational, n i 1 = 0 and M u (i 1) is up.

79 2.852 Manufacturing Systems Analysis 79/91 Copyright c 2010 Stanley B. Gershwin. Long Lines Exponential Processing Time Model Interruption of Flow After a lot of equation manipulation, we get: and similarly, p u (i) = p i + r u (i 1)p(i 1; 001). E u (i) p d (i) = p i+1 + r d (i + 1)p(i + 1; N10). E d (i) in which p(i 1; 001) is the steady state probability that line L(i 1) is in state (0, 0, 1) and p(i + 1; N10) is the steady state probability that line L(i + 1) is in state (N i+1, 1, 0).

80 2.852 Manufacturing Systems Analysis 80/91 Copyright c 2010 Stanley B. Gershwin. Long Lines Exponential Processing Time Model Resumption of Flow pi 1(0, 0, 1)ru (i)µu(i) r u (i) = r u (i 1) p u (i)p(i) +r i 1 i = 2,, k 1 p i 1(0, 0, 1)r u (i)µ u (i) p u (i)p(i) «, r d (i) = r d (i + 1) pi+1(ni+1, 1, 0)r d (i)µ d (i) p d (i)p(i) +r i+1 1 i = 1,, k 2 p i+1(n i+1, 1, 0)r d (i)µ d (i) p d (i)p(i) «,

81 2.852 Manufacturing Systems Analysis 81/91 Copyright c 2010 Stanley B. Gershwin. Long Lines Exponential Processing Time Model Conservation of Flow P(i) = P(1),i = 2,..., k 1.

82 2.852 Manufacturing Systems Analysis 82/91 Copyright c 2010 Stanley B. Gershwin. Long Lines Exponential Processing Time Model Flow Rate/Idle Time The flow rate-idle time relationship is, approximately, P i = e i µ i (1 prob [n i 1 = 0] prob [n i = N i ]). which can be transformed into e i µ i P = ; i = 2,..., k 1. e d (i 1)µ d (i 1) e u (i)µ u (i)

83 2.852 Manufacturing Systems Analysis 83/91 Copyright c 2010 Stanley B. Gershwin. Long Lines Exponential Processing Time Model Flow Rate/Idle Time For the algorithm, we express it as µ u (i) = { } 1 1 e u (i) 1 P(i) + 1 1, e i µ i e d (i 1)µ d (i 1) i = 2,, k 1, µ d (i) = { } 1 1 e d (i) 1 P(i) + 1 1, e i +1µ i +1 e u (i+1)µ u (i+1) i = 1,, k 2.

84 2.852 Manufacturing Systems Analysis 84/91 Copyright c 2010 Stanley B. Gershwin. Long Lines Exponential Processing Time Model Boundary Conditions M d (1) is the same as M 1 and M d (k 1) is the same as M k. Therefore r u (1) = r 1 p u (1) = p 1 µ u (1) = µ 1 r d (k 1) = r k p d (k 1) = p k µ d (k 1) = µ k

85 2.852 Manufacturing Systems Analysis 85/91 Copyright c 2010 Stanley B. Gershwin. Long Lines Exponential Processing Time Example.25 LINE PRODUCTION RATE (UNIT/TIME) Upper Bound Decomposition Simulation i Parameters r i p i µ i N i ρ2 Exponential processing time line 3 machines Upper bound determined by smallest ρ i. Simulation satisfies upper bound; decomposition does not. Why?

86 2.852 Manufacturing Systems Analysis 86/91 Copyright c 2010 Stanley B. Gershwin. Long Lines Continuous Material M 1 B1 M 2 B2 M 3 B3 M 4 B4 M B 5 5 M 6 Conceptually very similar to exponential processing time model. One difference: prob (x i 1 = 0 and x i = N i ) = 0 exactly.

87 2.852 Manufacturing Systems Analysis 87/91 Copyright c 2010 Stanley B. Gershwin. Long Lines Continuous Material Model New approximation New approximation: The observer sees both pseudo-machines operating at multiple rates, but the two-machine lines assume single rates. M (i) u r (i), p (i), µ (i) u u u M (i) d r (i), p (i), µ (i) d d d If this were really a two-machine continuous material line, material would enter the buffer at rate µ u (i) (if M u (i) is up and the buffer is not full) or µ d (i) (if M u (i) and M d (i) are up and the buffer is full and µ d (i) < µ u (i)) or 0; material would exit the buffer at rate µ d (i) (if M d (i) is up and the buffer is not empty) or µ u (i) (if M u (i) and M d (i) are up and the buffer is empty and µ u (i) < µ d (i)) or 0;

88 2.852 Manufacturing Systems Analysis 88/91 Copyright c 2010 Stanley B. Gershwin. Long Lines Continuous Material New approximation M i 2 B i 2 M i 1 B i 1 B i M i+1 B i+1 M M i i+2 B i+2 M i+3 M (i) u M (i) d Assume that... < µ i 2 < µ i 1 < µ i < µ i+1 <... Assume all the machines are up and B i is not full. Then the observer in B i actually sees material entering B i... at rate µ i if B i 1 is not empty; at rate µ i 1 if B i 2 is not empty and B i 1 is empty; at rate µ i 2 if B i 3 is not empty and B i 2 is empty and B i 1 is empty; etc. Therefore, this approximation may break down if the µ i are very different.

89 2.852 Manufacturing Systems Analysis 89/91 Copyright c 2010 Stanley B. Gershwin. Long Lines Continuous Material Equations We have the same 6(k 1) unknowns, so we need 6(k 1) equations. They are, as before, Interruption of flow, Resumption of flow, Conservation of flow, Flow rate/idle time, Boundary conditions. They are the same as in the exponential processing time case except for the Interruption of Flow equations.

90 2.852 Manufacturing Systems Analysis 90/91 Copyright c 2010 Stanley B. Gershwin. Long Lines Continuous Material Interruption of Flow Considerable manipulation leads to and, similarly, p u (i) = p i 1(0, 1, 1)µ u (i) p i 1 + P(i) p i (N i, 1, 1)µ d (i) p i 1(0, 0, 1)µ u (i) P(i) p i (N i, 1, 1)µ d (i) p d (i) = p i+1(n i+1, 1, 1)µ d (i) p i P(i) p i (0, 1, 1)µ u (i) pi+1(n i+1, 1, 0)µ d (i + 1) P(i) p i (0, 1, 1)µ u (i) ««µu (i 1) 1 + µ i «r u (i 1), i = 2,, k 1 ««µd (i + 1) 1) + µ i+1 «r d (i + 1), i = 1,, k 2

91 2.852 Manufacturing Systems Analysis 91/91 Copyright c 2010 Stanley B. Gershwin. To come Assembly/Disassembly Systems Buffer Optimization Effect of Buffers on Quality Loops Real-Time Control????

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