43621-TahaSMCh01 - Chapter 1 What is Operations Research...

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Unformatted text preview: Chapter 1 What is Operations Research? 1-1 Set 2.1a Buy three roundtn'p tickets for the first three weeks only—«cost = 3X $400 == $1200. Though the cost is cheaper, it is not feasible because it covers only three out of the required five weeks. (l,2)—> (t=2) 2,5,10 Total=2+1+10+2+2=17minutes Given a string of length L: (1) h = .3L, w = .2L, Area = .0st (2) h =.1L, w = .41., Area = .0413 Solution 2 is better because the area is lar er L = 2(w + h) Jim w=L/2—-h Curve Fast Joe Curve .500 .200 z 2 wk = h(L/2 —- h) = 1.1112 — 112 Fast .100 .300 (a) 52/5h = L/2 —— 2h = 0 Alternatives: Joe: Prepare for curve or fast ball. Thus, h = U4 and W = U4. Jim: Throw curve of fast ball. 63) Solution is optimal because 2 is a concave Joe tries to improve his batting score and Jim fianction tries to counter Joe’s action by selecting a less favorable strategy. This means that neither player will be satisfied with a single (pure) (a) strategy. Let T = Total tie to move all four individuals to the other side ofthe river. the The problem is not an optimization objective is to determine the transfer schedule situation in the familiar sense in which the that minimizes T. objective is maximized or minimized. Instead, the conflicting situation requires a compromise (b) solution in which neither layer is tempted to Let t = crossing time fiom one side to the change strategy. Game theory (Chapter 14) other. Use codes 1, 2, 5, and 10 to represent provides such a solution Amy, Jim, John, and Kelly. continued .. . Set 1.1a Let L=ops. 1 and 2=20 sec, C=ops. 3 and 4:25 sec, U=op. 5=20 sec Gant chart: L1+load horse I, L2=load horse 2, etc. one joist: 0—--Ll--20—--C l--—45——-—U1+Ll—~-85--—U2+L2————125---U1+L1--— 165—--U2+L2——-205 20-12-40 45-——C2—-«-70 85-——C1-—~l 10 125--—C2-—-140 165431—190 205—-—C2-—-230--—U2--—250 Total = 250 Loaders utilization=[250—(5+25)]/250=88% Cutter utilization=[ZSO-(20+l5+15+1S+lS)]/250=68% two joists: 0---2Ll-——40---—2C 1«----90-—--2(U1+L1)---1 70—--~2Cl~--—220-~-2U1 — -—260 40--~2L2---80 90---2C2~—-- 140 l 70—--2U2--—2 1 0 Total =260 Loaders utilization=[260—( l 0+10)]/260=92% Cutter utilization=[260-(40+30+40)]/’25%58% three joists: O—-—3L1---60 ———— --3C 1-—-—--135——--‘3C2———-—---—2 10-—a-—3U2————270 60-—~3L2--—120 l35—-—--3U1—---~ 1 95 Total =270 Loaders utilization=f270—( 1 5+1 5)]!270=89% Cutter utilization=[270—(60+60)]/270=56% Recommendation: One joist at time gives the smallest time. The problem has other alternatives that combine 1, 2, and 3 joists. Cutter utilization indicates that cutter represents the bottleneck. ...
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This note was uploaded on 09/17/2009 for the course INDUSTRIAL 0906353 taught by Professor Ahmedjradat during the Three '06 term at ADFA.

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43621-TahaSMCh01 - Chapter 1 What is Operations Research...

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