HW Sol Due 12-6 DP 1-3

# HW Sol Due 12-6 DP 1-3 - HW Due Problem 1 Notes problem...

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Unformatted text preview: HW Due 12-6-2010 Problem 1: Notes problem statement: A company has known weekly orders for varying numbers of a given product over the next 5 weeks. These requirements are 50, 30, 60, 45, and 25. To make a production run for this product, it costs \$25 to setup the machine and each unit produced costs an additional \$100. The campany. has production capacity and storage Space for as many units of the , product as needed; however, it costs \$1 per unit per week to store units needed in them: time periods. The production capacity is such that all so. future period demands could be produced in a single rum Of course, the company would incur a considerable holding cost. Thus, the probiem is to determine in which periods to make production runs and how many items to produce on each run. This analysis is to be based on a minimum cost basis Homework: Resolve this problem with the setup cost of *{S'S’t‘athcr than \$25. (Answer, (80, 0, 60, 70. 0) for a total cost of \$21,220.) , PROGRAM "Wagner/Whitin Problem" _ DEFINITION ” Requirements“ R: ARRAY[ <1..5>] = ( 50,30,60,45,25); " Holding costs" H : ARRAY{<1..5>]=( 1,1,1,1,1); I ” Fixed costs" K : ARRAY[ <1..5> 1 = (55, 55, 55, 55, 55); " Unit cost" C = 100; “Calculate cum. demand over periods i to j" CD(I,J) = SUM{ R[KK]: K W [LI] }; " ------------------------- FEASIBLE SEQUENCES -------------------- " 13.1 : SEQUENCE = <0,CD(1,1),CD(1,2),CD(1,3),CD(1,4), CD(1,5)>; B2 : SEQUENCE = <O,CD(2,2),CD(2,3),CD(2,4), CD(2,5)>; B.3 : SEQUENCE = <0,CD(3,3),CD(3,4), CD(3,5)>; B.4 : SEQUENCE = <0,CD(4,4), CD(4,5)>; B.5 : SEQUENCE = <0, CD(5,5)>;_ " Formulation: n = period" " i = starting inventory" " z = amount produced" " g(n,i,z) : cost in period 11" " t(n,i,z) = ending inventory in period 11" G(N,I,X) = K[N](X>0) + C(X) + H[N](I+X—R[N]); T(N,I,X) = I+X - R[N]; CONTROL LOGIC F.5(I) = MIN{ G(5,I,Z) : ZIN 35 WITH T(S,I,Z) IN <0>}, I ]N 13.5; FMG) ._. MIN{G(M,I,Z) + F.(M+1)( T(M,I,Z) ): z [N BM WITH T(M,I,Z) IN B.(M+1)}, I IN BM, M IN <4,3,2>; F.1(I) = MIN{G(1,I,Z) + F.2(T(1,I,Z)): 2 IN 13.1 WITH T(1,I,Z) IN 13.2 }, 1 IN <0>; END. *******************************************************\$********* STAGE 1 :for F5 I|Z : Value 0. | Opt 25.:2555.. 25. | Opt' 0.:0. STAGE21for F.4 NZ 1 Value 0. I Opt 70.:7080.. 45. | Opt 0.:2555.. 70. | Opt 0.:25.. STAGE?) : for F3 I|Z : Value 0. | Opt 60.:13135.. 60. | Opt 0.:7080.. 105. | Opt 0.:2600.. 130. | Opt 0.:95.. STAGE4tfor F.2 IIZ : Value 0. ] Opt 30.:16190.. 30. | Opt 0.:13135.. 90. | Opt 0.:7140.. 135. | Opt 0.:2705.. 160. | Opt 0.:225.. STAGES : for F.1 I|Z : Value 0. E Opt 80.:21220.. Optimal Solution Stage State Decision 5 1= 0. Z = 80. 4 I: 30. Z = O 3 1= 0. Z — 60. 2 I = 0. Z = 70. 1 I = 25. Z = 0 Recursion F1 = 21220.. 'F2 = 13135.. F3 = 13135.. F4=7080.. F5 = 0. Problem 2': Min cost through a network—problem from Taha, 7-9. El 7—-9 (Shortest Route Problem) The network given in Figure 7—7 gives different routes for reaching city 13 from city A passing through a. number of other cities. The lengths of the individual routes are shown on the arrows. . It is required to determine the shortest route from A to B. Formulate'the a problem as a dynamic programming model. Explicitly deﬁne the stages, states, and return function; then ﬁnd the optimal solution. Foam Toms Figure 7—7 l l l 1 vyﬁrwﬁg+gi>7+¥z) <ﬁx7+7)mE «a: Was-{*w’gﬁ‘E-f‘fr‘é) (“again-\$1337 £2: m (4+ ¥g}§‘~+£g) , (4%“; iii): 7 \$3": WWW“ 2—:‘-—1C:s) ‘ (3%{ﬁ3ﬁéf 9/ R Problem 3: The demand for a product over the next four time periods is 2, 3. 4. and 2' units. The cost of placing an order is \$15, independent ofthe number of units ordered. The individual item cost is \$100. and the holding is \$2 per unit per period. Due to the company accounting system. holding costs are charged on the inventory units at the beginning of a period; that is, on entering inventory. A maximum of four units can be held Erom period to period. The company orders from a local supplier, so we can consider that units ordered in a period are available for use in the same period. However. due to the limited production capacity of the supplier, orders are limited to once a period with a maximum of ﬁve units. We want the optimal ordering policy that meets the demands at minimal cos: for the . four time periods. The company currently has no items in inventory and has no :7, 1 0 requirements for a Speciﬁed number in inventory at the end of the planning horizon. Since demands must be met. ordering for future demand is the only viable option for decreasing costs. Problem: Resolve this problem with a setup cost of \$20 and an inventory holding cost of \$3 per unit per period. PROGRAM "prodHW.dpe" "s is entering inventory: storage limit is 0 to 4“ "x is production quantity: limits 0 to 5" "holding on entering inventory: \$35" "production setup cost \$20 (if x > 0)" ”unit cost is \$100K" "d demands must be met" DEFINITION D:ARRAY[ <1..4> ] = (234,2); PC = 20; HO = 3; UC = 100; XDOM : INTERVAL: [0..5]; CONTROL LOGIC F.4(S) = MIN{ PC(X>O) + UC*X + HC*S: X IN XDOM WITH (X+S—D[4]) IN [0..4]},S IN [0..4]; F.3(S) = MIN{ PC(X>0) + UC*X + HC*S + F.4(S+X—D[3]): X 1N XDOM WITH (X+S—D[3]) IN [0..4]},S IN [0..4]; F.2(S) = MIN{ PC(X>0) + UC*X + HC*S + F.3(S+X-D[2]): X IN XDOM WITH (X+S-D[2]) IN [0..4]},S IN [0..4]; F.1(S) = MIN{ PC(X>O) + UC*X + HC*S + F.2(S+X-D[1]): X IN XDOM WITH _(X+S—D[1]) 1N {0..4}},S IN <0>; END. *******=§<*****\$**\$************************************************ STAGE 1 : for F4 SIX : Value 0. I 23220.. 31320.. 4.:420.. 5.520.. Opt 2.:220.. l. ] l.:123.. 2.:223.. 3.:323.. 4.:423.. 5.:523.. Opt 1.1123,. 2. 1 0.:6.. 1.:126.. 2.2226. 3.326.. 45426.. Op .16.. 3. .:9.. 1.2129.. 2.:229.. 3.:329.. O t 0 | 0 pt 0.29.. | 0.:12.. 1.:132.. 2.232.. t O STAGE2 1 for F3 SlX 1 Value 0. I 41640.. 51643.. Opt 4.1640.. 1. | 31543.. 4.1546. 5.1529. Opt 5.1529. 2. | 21446.. 3.1449. 4.1432. 5.1535.. Opt 4.1432. 3. | 1.1349. 2.1352.. 3.1335.. 4.1438. 5.1541.. Opt 3.1335.. 4. | 0.:232.. 1.1255. 2.1238.. 3.1341.. 4.1444. Opt 0.:232.. - STAGE3 1 for F2 SIX 1 Value 0. | 31960.. 4.1949. 5.1952. Opt 4.1949. 1. | 21863.. 3.1852. 4.1855,. 5.1858.. Opt 3.1852. 2. | 1.1766” 2.1755.. 3.1758. 4.1761.. 5.1758. Opt 21755.. 3. [ 0.:649.. 1.1658. 21661.. 3.:664.. 4.1661.. Opt 0.1649. 4. | 0.1541.. 1.1564.. 2.1567.. 3.1564,. Opt 0.1541.. STAGE41f0r F.1 S |X 1 Value 0. | 2.11169. 3.11172. 4.:1175.. 5.11169. Opt 2.11169. Optimal Solution Stage State Decimon Recursion 4 S=0. X=2 F1=1169.. 3 S = 0. X = 4. F2 = 949.. 2 S=1. X=5 F3=529.. 1 S 2 2. X = 0 F4 = 6.. ...
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