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lecture26 - MIT 2.852 Manufacturing Systems Analysis Lectu...

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Unformatted text preview: MIT 2.852 Manufacturing Systems Analysis Lectu re 26 Miscellany Stanley B. Gershwin Spnng,2004 Copyright @2004 Stanley B. Gershwin. All rights reserved. Subcontracting/ Capacity ExpanSIon c We have assumed that capacity and demand are stationary. o In reality, * demand sometimes spikes, and manufacturers can often obtain additional temporary capacity by paying a premium andlor reserving it in advance. * Goods obtained from temporary capacity cost more than from regular capacity. 0 Mechanisms: * subcontracting, overtime, temporary workers, reserve equipment. Copyright @2004 Stanley B. Gershwin. All rights reserved. Subcontracting/ Events Capacity ExpanSIon 1. Manufacturer, in making long terms plans and forecasts, sees the need for infrequent, unpredictable, temporary expansions of capacity. 2. Manufacturer makes arrangements and pays up front. 0 Writes a contract with another manufacturer to guarantee a price for goods, or o Buys reserve equipment, or 3. Manufacturer operates normally until demand spikes. 4. Manufacturer obtains goods from extra capacity until demand returns to normal. Copyright @2004 Stanley B. Gershwin. All rights reserved. 3 Subcontracting/ Questions Capacity ExpanSIon o How should manufacturer decide when to use high-cost capacity? 0 How much should manufacturer be willing to pay for the availability of high—cost capacity? Copyright @2004 Stanley B. Gershwin. All rights reserved. Subcontracting/ Model Capacity 0 Extension of Bielecki-Kumar problem. a Two production rate variables: 1:. for regular capacity and v for high-priced capacity. a Simplifying assumption: no failures in high-priced capacity. 0 < app; 0 S ’v S V o Demand randomness: d = DLOW + (Ida) HIGH — DLow) Copyright @2004 Stanley B. Gershwin. All rights reserved. Subcontracting/ Model Capacity 0 Dynamics of cap, ad: like unreliable machines with parameters T'p, pp and 1rd, pd. 0 Dynamics of surplus: . Cost to be minimized is T JV 2 E/ (g(:I:) —|—Mu +Kv)dt o where M and K are cost per item for the two kinds of production (K > M) and 9(32) is the same as before. Copyright @2004 Stanley B. Gershwin. All rights reserved. Subcontracting/ Model Capacity ExpanSIon 0 Let JV be the expected amount of demand satisfied if the maximum extra capacity is V. o Let Jo and J0 be the expected amount of demand satisfied and the optimal cost if no extra capacity is ever used. 0 Let the sales price per item be 71'. 0 Then the expected value of the extra capacity availability is [the profit with extra capacity] — [the profit without extra capacity], or . Writing JV and Jo for a finite time interval [0, T] is a tricky It is easy for an infinite intervai, but then we have to discount cost. Copyright @2004 Stanley B. Gershwin. All rights reserved. 7 Subcontracting/ Solution Capacity ExpanSIon o The Bellman equation is an extension of the one in Section 9.3 with: *four discrete states, due to machine failures and demand changes; and *two control variables 15 and v, and *linear terms in u and v in the cost function. Copyright @2004 Stanley B. Gershwin. All rights reserved. Subcontracting/ Solution Capacity ExpanSIon o The part of the Bellman equations that involve u and v is: dW min {Mu + K’v + —(:c, 1, ad)(u + ’v — (1)} “’3’” d3: Copyright @2004 Stanley B. Gershwin. All rights reserved. 9 Subcontracting/ Solution Capacity ExpanSIon . To obtain an: dW ifM —|— —(:13, 1,0501) < 0, then u = (4 d3: dW if M —|— —(:13, 1, and) = 0, then ???? d3: , dW IfM -|— d—(m, 1,056,) < 0, then ’u = 0 a: Let Zu(1, ad) 2 a: satisfy M —|— %($$ 1, ad) 2 0. Copyright @2004 Stanley B. Gershwin. All rights reserved. 10 Subcontracting/ Solution Capacity ExpanSIon a To obtain 1:: dW ifK—I— —(a3,1,c1¢d) < 0, then 1) = V div . dW If K —|— d—(m, 1, and) = 0, then ???? a: . dW IfK —|— d—(m,l,ad) < 0, then 1) = 0 {I} Let Z1,(1, ad) 2 a; satisfy K —|— %(SE, 1, ad) 2 0. Copyright @2004 Stanley B. Gershwin. All rights reserved. 11 Subcontracting/ Solution Capacity ExpanSIon a Let us assume that J and therefore W is convex and continuous. 0 Then W is decreasing and then increasing, and dW/da: is increasing. - Therefore, the value of a: such that i—Z’ch, 1, cud) = —M is greater than the value of a: such that Cit—ft”: 1, and) = —K since —M > —K. Copyright @2004 Stanley B. Gershwin. All rights reserved. 12 Subcontracting/ Solution Capacity ExpanSIon if€B<Zu(1,cxd),u=u ifaz<Zv(1,ad),v=V ifa: = Zu(1,ad), ???? ifa: = Zv(l,cv.d), ???? ifm>Zu(1,cxd),u=0 ifat>Zv(1,crd),V=0 Copyright @2004 Stanley B. Gershwin. All rights reserved. 13 Subcontracting/ Solution Capacity ExpanSIon Z1,(1,05d) Zu(1, ad) Copyright @2004 Stanley B. Gershwin. All rights reserved. 14 Subcontracting/ Solution Capacity olfaPZI * If :1: > Zu(1,ad), then u = 0 and v = 0. k If and is a low demand state and a; = Zu(1, ad), then u=d=DLowandv=0. * If Zv(1,acd) < a: < Zn(1,ad), then u = p and v = 0. at If and is a high demand state and a: = Z1,(1, cud), then u = H and’tJZd—HZDWGH—fl. *lfm < Zv(1,c1¢d), thenu = pandv = V Copyright @2004 Stanley B. Gershwin. All rights reserved. 15 Subcontracting/ Solution Capacity olfacp =0,thenu=0. * Let Zv(0, ad) 2 a: satisfy K —|— dfillies, 0, ad) 2 0. 1% Whether v = 0 or v = d or v = V depends on whether :1: is greater than, equal to, or less than Z1,(0, cud). Copyright @2004 Stanley B. Gershwin. All rights reserved. 15 Subcontracting/ Solution Capacity 0 Assume that V > DHIGH- at a: will never be greater than max Zu(ap, 05d) , except possibly initially. * a: will never be less than min Zv(ap, ad) , except possibly initially. Copyright @2004 Stanley B. Gershwin. All rights reserved. 17 Subcontracting/ Extensions Capacity ExpanSIon . Multiple subcontractors at different prices. . Discounted cost. 0 Customers balking — dynamically reducing demand — at long lead times. Copyright @2004 Stanley B. Gershwin. All rights reserved. 18 Sim le Yield M: 31 M2 32 M3 doe a G A. 7-0“ ”1 N1 M2 N2 ”3 r1 ’2 amp I"3 p, F5 FE; . Continuous material model. a At M2 a fraction q is allowed to go to M3; 1 — q is scrapped. o idea: transform this system into one we already know how to analyze. Copyright @2004 Stanley B. Gershwin. All rights reserved. 19 Sim le Yield Mo el 0 Simplifying assumption: Assume that the fraction q of flow out of M2 that is scrapped is constant. 0 During an interval [t, t —|— M] when M2 is operating at rate m2(t), m2(t)6t leaves B1 and qm2(t)6t enters B2. *That is, during [13, t —|— 61:], B1 decreases by m2(t)6t/N2 (as a fraction of its size) and B2 increases by qmg (t)6t/N2. 0 We construct a new system whose behavior is a simple transformation of this, and which we know how to analyze. . In the new system, that the amount that enters 3; during [t, t —|— 6t], is the same as the amount that leaves B1. Copyright @2004 Stanley B. Gershwin. All rights reserved. 20 Sim le Yield Mo el M! 31 M2 32 M3 “1 N1 “2 W2 ”3 ’1 ’2 ’5 p, F5 155 0 Choose Mi 2 M1; Bi 2 Bl; M; = M2. - Choose N; so that the fraction that B; gains is the is the same the fraction B2 gains during [t, t + 6t]. *m2(t)6t/N£ = qm2(t)6t/N2, 50 N5 = Ng/q. Copyright @2004 Stanley B. Gershwin. All rights reserved. 21 Sim le Yield Mo el By similar logic, 0M3 = ua/q and 0p3=p3/q '71:; 2 7'3 Copyright @2004 Stanley B. Gershwin. All rights reserved. 22 Sim le Yield EXtens'ms Mo el . Multiple stages with imperfect yield. 0 Split and merge. Copyright @2004 Stanley B. Gershwin. All rights reserved. 23 Hard Problems Remaining . Variance of production . Variance of lead time . Integration of setups with real-time control Copyright @2004 Stanley B. Gershwin. All rights reserved. 24 ...
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