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Solution Manual for Operations Research: An Introduction, 10th Edition

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    Solutions manual Operations Research: An Introduction Tenth Edition Hamdy A. Taha University of Arkansas/Fayetteville

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    Downloaded from StudyXY.com ® + StudyXY Sd Ye. o> \ | iF ’ pr E \ 3 S Stu dy Anything This ContentHas been Posted On StudyXY.com as supplementary learning material. StudyXY does not endrose any university, college or publisher. Allmaterials posted are under the liability of the contributors. wv 8) www.studyxy.com

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    2 wv Contents 1 What is Operations Research? 2 Modeling with Linear Programming 3 The Simplex Method and Sensitivity Analysis 4 Duality and Post-Optimal Analysis 5 Transportation Model and its Variants 6 Network Models 7 Advanced Linear Programming 8 Goal Programming 9 Integer Liner Programming 10 Heuristic Programming 11 Traveling Salesperson Problem (TSP) 12 Deterministic Dynamic Programming 13 Inventory Modeling (with Introduction to Supply Chains) 14 Review of Probability 15 Decision Analysis and Games 16 Probabilistic Inventory Models 17 Queuing Systems 18 Simulation Modeling 19 Markov Chains 20 Classical Optimization Theory 21 Nonlinear Programming Algorithms Appendix C ~~ AMPL modeling Language i

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    2 wv Chapter 1 What is Operations Research? 1-1

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    > © 5 a Chapter 1 i Weeks 1-4: 2 weekend trips FYV-DEN-FYV and 2 weekend trips DEN-FYV-DEN. Week | (a) Let T = total time to move all four 5: 1 regular trip. individuals to the other side of the river. The Cost: 4x320+1x400= $2200 objective is to determine the transfer schedule | that minimizes T. 2 (b) Let t = crossing time from one side to the . A | other. Use codes 1, 2, 5, and 10 to represent (a) Given a fence of length L: Cl Ny J i ()h=3L, w=2L, Area = 06L2 Amy, Jim, John, and Kelly. (2)h=IL, w= 4L, Area =_04L _ Solution (2) is better because the area is larger ERI a — (L=2(w+h),w=1/2-h (500 (a=) [2 z=wh=h(L/2-h)= Lh/2 - h? IL [G.10)—(t=10 [25,10 | 52/8h=L172-2h=0 [2 [=20@ [5.10 | Thos, h = L/4 and w= L/4. [none [(1,2)~ ¢=2, [12510 | Solution is optimal because z is a concave function | | Total =2 +1 + 10 +2 + 2 = 17 minutes x = cumulative number of drops of balls #1 and #2 S at any floor (problem vnknown) yi = floor from which ith drop of ball #1 occurs. Jim Step 0: Set yo=0,y;=x,and i=1. Curve Fast General step i: Drop ball#] from floor yi. Ifitis J Curve |.500 .200 dented, use ball#2 to check floors yi + 1 toy, -1, °° Fast |.100 .300 in that order. Else, if #1 is not dented, seti=i+ 1, and repeat step i. — Formula for determining y;: Broa ee of fast ball, y: must include the (cumulative) i #1-drops from Joe: Prepare for curve or fast ball oor i floor Yie uw ot the same number ol (b) Joe tries to improve his batting score and Jim Th ps at any Toor yi, ps cannot exceed X= 1. | rie 10 counter Joe's action by selecting a less — Fxeit]) favorable strategy. This means that neither player Eke SEG SPN will be satisfied with a single (pure) strategy. cx | fo lx 2 Doi (= 3 The problem is not an optimization situation in reas wi) =ix-(i-1) the familiar sense in which the objective is Maximum number of #1-drops is x (else Yi Yi | 7 imized or minimized. Instead, the conflicting fori> fe highest floor from which #1 situation requires a compromise solution in which El _2 2 2 neither player is tempted to change strategy. Game Yo=X = (e102 =X" x12 +x/2 = (x i 02 theory (Chapter 14) provides such a solution. For a 100-storey building, y, > 100, or x* +x — 200 > 0. The associated quadratic equation yields x = 13.64 and ~ 14.64. The rounded positive value x = 14 is the smallest integer that satisfies the inequality. 1-2

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    > Chapter 1 3 a \L L 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 1, L2=load horse 2, etc. one joist: 0---L1-—20-—C1-—45-—U1+L1-—85——U2+L2-—125—~-U1+L1—165— U2+H.2---205 20-12-40 45-—-C2--—-70 85—-C1--110 125—-C2---140 165-C1-190 205---C2--230--U2--250 Total = 250 Loaders utilization=[250-(5+25))/250=88% Cutter utilization=[250-(20+15+15+15+15)]/250=68% two joists: 0-—2L1---40—2C1—90—-2(U1+L1)—170——2C1---220—2U1-- -260 40—-2L2—80 90—2C2-—140 170--2U2--210 Total =260 Loaders utilization=[260-(10+10)]/260=92% Cutter utilization=[260-(40+30+40)]/250=58% three joists: 0-—3L1--60—-3C1——-135—---—-3C2-—-—-~210—-3U2--—270 60---3L.2-—120 135-—-3U1-—-195 Total =270 Loaders utilization=[270-(15+15))/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. Note that all ‘dots are indistinguishable even if they are designated as 1, 2, 3, ..., 10. (a) Alternative 1: Move dots 5, 6, and 7 below bottom row, move dots 8 and 9 below new 5, 6, and 7. Move 10 to the bottom. Number of moves = 6. Alternative 2: See part (b). (b) Three moves: Move dot 1 up to the left of dot 8, dot 4 to the right of dot 9, and dot 10 below dots 2 and 3. 10 8 9 5 6 7 123 4 1-3

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    > Chapter 1 = 8 wn @ Alternative 1: Break one end link of each chain and connect to another chain. Four breaks and re- solders, cost =4 x (2 + 3) =20 cents. Alternative 2: See Part (b) (b) Break three links in one chain and use them to connect the remaining three chains: Three breaks and re-solder, cost = 3 x (2 +3) = 15 cents. 9 Represent the selected 2-digit number as 10x+y. The corresponding square number is 10x+y-(x+y)=9x . This means that the selected square will always be 9, 18, 27, ..., or 81. By assigning zero dollars to these squares, the reward is always zero regardless of the rewards assigned to the remaining squares or the number of times the game is repeated. Assign a sequential number x to each cartons, xeX = {1, 2, 3,4, 5,6, 7,8, 9, 10}. (a) Let Y be the sct of cartons so far weighed (initially Y = @). General Step: Randomly select a carton yeX-Y. If y weighs 90 oz, stop. Else, augment y to Y and repeat the General Step. 1 < number of times scale is used < 10. (b) Exactly once! Take x bottles from carton xeX to end up with (142+... +10) = (10+11)/2=55 bottles. Weigh the 55 bottles. If the weight = 550 ~ x, carton x is the defective onc. 1-4

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    TT] a CHAPTER 2 Modeling with Linear Programming | 21

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    Chapter 2 = 3 a @ XX, 21 or =, +x, 21 [1] Quanly diderionts reounlls he 14] (b) X +2x, 23 and x,+3x,%€ Fllrerg ronbnten ech sfsnchin: (Y xz 2X or =X, so @ Fh 23 SK +4 Ke, os% ET Tre =F er SX 5% 20 z= @) (5,40) = (14%) X72 xX, %) =e 2 Ys x tax, Exl+yxy =22 <2¢ (xi +2x4 = 9 gh & infeasible The oda alirr cammt- Co Lira lia] ®) (x, 0) =(2 2) ao a Anis fergpiam. Nondondondy x Ie) z N Can be accoondid gh +4 Hi caer bxzt xy 277 pd 4 aig mixed anlig poipummin] 1x2 +2 = = y xr +1 x2 =o <I feasible (cloptes - | x2 =z =z z= sSx2+9x2 = $18 (0) C5, %,) = (3,15) XX, 20 Ex3+4x1.8=24 =2¢ Vx3+2x)is = 6 26} finerkls “1X3 +1 XLS ==1.85 ixeg = 1.5 2 Zz Sx3+ 4x15 = F221 WB ELxd= 2,1) X), X, Z © ex +axi 216 <RYL po cole (x2+2x1 = 4 <6 ~ixz +lxl ==) < xy = <z Z=Sx2+4x1 = $14 © (ox) =(22-1) x, zo, x, <0, wferutle Conclenni (c) goveo th fet fesaihle Seldon (Xs %)= (2,2) ; de 5, and 5, be He unetdd docly [3 | Emandy & Mian M2. Formis 5, = 24 = (6x45) = 4 fms [day FrMA1 Sy = =f, + 2x) = €-(araxz) = 0 tons Joy 2-2

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    . Chapter 2 ) 3 2 X - © -X+x,= 1 fe t % reference paint: Yo [5] Xa Gplimuin c rd xX (12) 2=$/3 -5 s k ¢ * g x 4 ZX -. © (© X4 % a p . ©) 6x, +94 224 Gotimiim.: 6.2) 2g FOR Gm 2 hd 4 % iP S. [6] ©) No feasible Space + %. . . 8] . So | fed rig pe = day wry Gf pradack ca) Cb) < 2 4 2 Matimuze 2 = 2X, + 3x, KA ¥ s.t. we SY 10%, + Sx; =6oo [0] J P Xe 6X, +2X, s6&o0 (2) &x, +10 X, <= goo © td) ‘ = loo & oN) a] A @ x %a- imum: 9 x pA td a) % rAd 2=#/9-50 So x ey 8 x, 22 f SS x apprmuen: ° c \og ! (2,2) . r-tis Oplirasrn. OCcurs od A: 4 % = 52.99 Xp = (4-12 x’ 2 = % 148.24 continued. — 2-3

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    Chapter 2 oo 3 K = number df umils ff A EIB Y T i Xp= Tuber. of sonal of B X= ¢ weet am B Maximize z= 20x +50% Maximize Z= -05X +-08X2 5.4 I TP rmx, +57, $0 X, 2-25 (44%) p+ Kz X, © -S(XHK) Xx, £ loo Xx, = SX 2x, + 4X, © 240 XX & S000 X, Xp 20 Xi, X 20 vo 2) 2 Optimum 50 \3 Ls X,=180 * x, 2500 ey, X= #2500 B | v z=Fazs i 7 a 7 X => ~ ; ad) : 5 ; Optimal occas af B- ee %= number. A pradiad ernie 2) X= 80 units Ay = rams § Amari conan Xy= Ro Units Maire Z = /S00X, +1000 % z $2,600 s g +. Xx, +X, £30 X, = ruarmber of 51 pects fay. 10] x 2/0 + Pullen at re sve % > 10 aximize 2 = r ‘2 St Xi Keg Kak Zo Foo Goo (Xe ophmum: ox, S50, 9SX E580 KF X=20 . Xo %,+% $30 X= 10 00 %2,£580 # Zz ="Yg, 0 fesse x,2\0 °° Cphimisen ° 3a (by Change x40, $30 Ho 4%, €3/ ° 700 ul Get z = #4500 Optrmmrn bobuch az=P41500- go,000 = F150 ; Fologn: . wo % = €50 cheess Conclusion: Any ized crumre s p © Xoo 18743 bars wilt Le phe prodiiad = 2=$285% vo 2-4

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    > Chapter 2 3 = wv X= undzgt Hlulios A 3] ape X= aril off Arla. s 120 fagno KEP 7774 Kui = x ea F= 8X, +10x, a ARE 6 SX, FSX, = /S0 L EAE I SNE % x, > Bo doo 3oo , < 152 Ophraum : Xm 200, X, = C0 , 72 =$26150 © =z 40 Ares allocation: 673 Grane, $3) when Xa» £200 x, = play 4 for cl 502s siwlrorsy, 18) Maxeruze Z = 2X +x, SE. xX, + x, Sl0 ES ERS PY) 3 N phim dyladers' e Ea %,=4 Rrwes 5 3 Kya € ewes < 4 2=14 "pleasarits” - : N N a «FT 10 x, po \ ' ; hel ON acl x, Diy mip Gp fot 16] pe N Xo.» Duby br. 4 Byps - Maximize Z= 8% +5K2 oe Sb. AX + Xx S900 a X, < /50 8 X, € 200 A ol HE EE NN X,. 5, 30 a " 400 Zt X= Nbr. of Frans Loxeo 14) optimum; d . Xp= nbr. of chest boxes X= f00 Gpa! Magimize Z= X * 135%, 200 £200 X, 220 Ger 2 S54. 2K 19%, Z€0 Kee? z= 51800 x, < 200 XxX, € /20 . X,,%, 20 200 x, continued. continued. 2-5

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    > Chapter 2 3 a X) = Aadip mirada VT] Optimim occurs a A: Xo = TV rumelio X,= 5128 tons por Gur Magimize z= a5 Xp = 70.256 Fons Lom aim g 2a > Z- 53546 eof Seem - y pack £5 Ophmal rats = 5028 _ 5 Xi>2z or -x+2X, $0 10-256 xX; 1 2 (4) RIX, +. 9X, 2(zo+)=21 X, £900, X,,%Z° Ghimum z = 16[538 th of Sham 200 |X AZ =161538-153844 = 7492 1b ty 0 x, =Nbr. of radi commencints 19 . beyond the frst 7 usa X, = Abe. § TV ads byord 7 first wo | Maximize Z = 2000+ 3000X,+ Sort 2000 ky, S-t. Z00(X +1) +2000(,+1) < 20,000 5X r300X; =loveo 300CK+) © Fx 20000 2000 (XK, +1) €-§X 20,000 2 . oC) <Page . Soo 1000 on Oplempnn oceeens at A: Makrmize. Z = R000,+3000X,1 7900 X= 60.61 winulss 770 ® X= 30-3 wrinubis SF 3ogx,+ 200%, = 17700 z = §/8.18 3c0x, & /S700 ® 7 1400 X= tons of CG Consumed ptr 18] 000% = ° ® Xe» Fons f Cs amasmed pin Rewr 1s Xp 20 Maximize 7 = /R000X, + 9000 X, Xs st. 1800 XX, + 2/00), £ 2006(X%+X,) i ~200X, + [0ax, $0 ® 2X, + TX, =20 B85 Xy5 X, 20 2 - ¥2 & EY 20 8 N a. x Optimum blohrm » wt & Radio Commevcesds = 2.3340 = 53.33 ro TV ads = (+1 =2 ZN - a Py P18 14 j=) 5 Z = 107666.67 77000 = /I4E y= ES I= Fo == 3 continued 2-6

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    > Chapter 2 3 - = 7) X= rusnbon dhl pon Kvn 20] X= numb 4) HiFi wneds 22 ox amb. 9) bliss gain fone x = Poemden of HiFi wrils Maxonize Z= §X + 126 | Conolraunds: |se&. Ex + 4X, © 4F0x-9= F32 Aox, +60 KF 25 x60=/S00 (1) SX, + SX, SAPOX-5E = 412.8 Tox, +60X, ©3xé0s 2100 @) 9%, + 6X2 < EFoX-8Y =422.4 2X, + 4X, €5X =300 (3) ov ’ Xk zo EX, + 4X2 TS, =432 sx, 485%. +5, =HZ.§ Xp ophmumat A: 4x, +6Xy + Sg =422.4 X=1(2 Shiets fo. . | xo = 2 Bousea [Rn Objective function: sl \@ z = $348 [bv Minimize S, +5 +353 =RETI-1SK7Isx) Q A | Thao, mon S,+5,+5; = max ISH T/SX: 7h Maximize Z = ISX, */SX, He A x 6x, +4x, <%32 0 Xo gx +5x% =+z.8 @ X, = Wb. of desks pen day = ax, 6X, =9224 16) X, = Nbr. of Chairs fon dag { © ¥,3X, 20 Lo. 6) . . Maxmwmze Z = 50% +/00X, A (Atternaliveophima x51? Ko, Xe o<y o 7, ¥, = 50.8 unis Err \ Xo 231.63 units X, % ogy ® ING F21238.4 min 750 To xX, £/20, x, € 60 (3,4) TN o ue 4 IN a Ophimum oo xX; Id 60 [~ yA ~N PT rw NE Sg . Wit, x | Cmrem wn a A ©. 0 Ophmum: B “0 2 c | Xx = 90 sk H pop 2 @pTIMUN) Xy = HY Chars E 1.2) 1 7

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    Chapter 2 3 % Xo 3 “ - [28 30 FE A, z * 4 Soo x (+ : , 5 | y “ Soe todo El x Opliamismn X =o, x, =v, Z=o, “ which is monSorsizal (c} 2 = ¥%, = ntumbor of fers foseth in Store 4 Hl M Xg = Alientecef Revd fuseck +n Stured 7] ll Qdditirnal constraint: x < 450 29 Minimize Z = &X 76x 2 s.t. 2 Xx 220 5s Xx </2 1 [rr] “ esx, £/0 20 Optimum " Soe 13 8 xX, aq 500 000 Co 4 fF 2 6 wo NM Gplismarn dwn : . x = 450 Ib Oplomum : X = 350 1 X= 10 Rewrs 2 . $450 Xx; = 70 Krsurs ) 2 = 140 Stes sadex continued. continued... 2-8

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    Chapter 2 5 FT 32] 33 E A= Lif em To X= 15 arid on bl clap hed Ld Xp= 10 bt) dary, form 3 Xa= 104 nicelid “ Leh ted, atk Re fering capacity Z=X+X, [0 bid ’ K Mining 2 = X + X, Miniorize z = X,+ Xp - . Subgect~& Lutpict-& dX, task, 2/0 C4 KX, ! 2 a Xz 4CKTR) 6X, — 4h 20 — 6x, Fo X, <0 x. x RX +l XK, = 1d 12 Ky Zo ASK, + 6X, 230 To . IX, 4 sex, 210 GA aplimim silodirro : ISK) £1, 2F Xs Xe 20 phim Ions Form Toes: LINEAR PROGRAMMING — GRAPHICAL OLUTION FE i vam Spite LINEAR PROGRAMING ~ GRAPHICAL SOLUTION FEEL Deedstoobes ~~ OO OO OO -— i Sam Et ’ Tae mn ? oo / Opeimium w AN : eh TTR Ee 2-9

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