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final_semi-sample

# final_semi-sample - GE330 Operations Research Methods for...

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Unformatted text preview: GE330: Operations Research Methods for Proﬁt and Value Engineering Final Exam May 7, 2008 NAME: UIN : Please Show all the steps. Good luck and have a nice summer. Sbialiﬁvxﬁ /20 /15 /10 /18 /10 /10 /17 Total /100 NOTE: Formulas are provided on Page 2 and 3. Problem 1. [18 points] Consider the following LP. max 2: 3m + 2222 + 533 5.13. \$1 + 2562 -i- 333 S 430 (1) 3331 -£- 23:3 5 460 (2) :81 + 4332 S 420 (3) \$11 3327 503 Z 0 The optimal tableau is: Basic \$1 x2 1173 m4 595 336 Solution 2 4 0 0 1 2 0 1350 332 -E 1 0 E -Z 0 100 mg g 0 1 0 % 0 230 3:6 2 0 0 -2 1 1 20 where 224, 2:5, and as are slack variables for constraints (1), (2), and (3), respectively. (a) (3 points) Write down the dual LP. (b) (3 points) What is the optimal dual solution? (c) (6 points) If the right-hand side of constraint (1) changes to 370, what is the new optimal value? what is the new optimal solution? ((1) (6 points) In the objective function, if the coefﬁcients of \$1 and :33 are ﬁxed, give the range for the coefﬁcient of 2:2 in which the current solution is still optimal. SDi/wi'imAS: (a) min w = 4750311' 44>de t 42033 3‘ + 331+ 3'5 :5 13‘ + 435 22 31+ it i5 3‘ 351, 55 z o 2-0] F1 1““: (intentionally left blank) C3 3‘ 2 570-480 3—60 DtZDB :0 . \ 3 \. ' , 1 EM Wﬂtfl’djj we. male ([90 +ﬁplﬂgr-‘Dz 2.1) .1720 + Jib} 20 .20 ”lbs + Dz‘FDg 20 5W9“ D5931? We. home, waves. Rs ~20 v-EE) 25 ha ‘wa15 "(0&ij " Tim ﬁnder/o 1)ch fer 11253:,me ! C: I“ 9) W Wmi Wm: £330 - 5w : 2270 law oFng/l Gehvﬁovx: XL: (6:314:13, 2 70 X3: 2&0 M *— zow 23¢ (“60) til-H) a) T0 {WJ UN: @EBDWNW roij, we MK CL 0‘ ‘5 O b D W i Z 4 O O ' l 0 (35‘s &z X: ”2? ‘ O 42" H); 0 (90 0Q} *5 ‘§: 0 r 0 if 0 A30 0 "£6 2 O 0 «Q ( 3- 29 M ‘g '— 90 @ An—ioﬁw—ioﬁs («Lao dlrdle so t‘l‘iﬁil 2’0 Problem 2. [17 points] A company has three plants and three distribution centers. The unit cost for shipping products from each plant to each distribution center, the supply of each plant, and the demand of each distribution center are given in the following table. Notice that the problem is unbalanced, if a unit of demand of a distribution center is not satisﬁed (by any of the plants), a penalty cost is incurred at the rate of \$3, \$5 and \$2 per unit for distribution centers 1, 2, and 3, respectively. Additionally, all the demand at distribution center 3 must be satisﬁed. We want to minimize the total cost. Distribution Center (a) (4 points) Formulate the problem as a balanced standard transportation problem, provide the transportation tableau. (b) {4 points) Find an initial basic feasible solution. ((1) (5 points) Calculate reduced costs for all nonbasic variables in the initial basic feasible solution. (d) (4 points) Choose an entering variable and ﬁnd the leaving variable. Update the current solution. 0»). Tetelc gab-13?”;- ge+\$o+2o 1:60 60<70 7°me Clﬂwwmclt «10+%9+2.0 :70 539 we ml to add dummy stTPlY with WWW to 1 2 3 WWW i0 (intentionally left blank) E E at ~— 0 Problem 3. [10 points] Consider the following integer program. max z = 3931 + \$62 + 3273 S.t. —331 + 2562 + \$3 5 4 4332 — 3533 S 2 {1:1 — 3332 + 233 S 3 3:1, 372, 3:3 2 0 and integer The following tableau is the optimal tableau of the LP relaxation. Basic 331 (1)2 333 :84 \$5 335 Sol 2 0 0 0 2 3 5 29 \$3 33 '3 3 \$2 \$3 \$1 1_ 5i D D C) O—I O C: C: l—' II—mII—ml Imp-ml ,_| (a) (5 points) Suppose we use the branch—and—bound method, which variables can be the branching variables? Write down the two sub problems for each possible branching variable. (b) (5 points) Derive a valid cutting plane based on the optimal tableau. {00% @Y ”(5 Com iii-L “cl/UL brow/Mug, Uan‘ahle - if 5le . “ll/LU,“ X19. 5 + Of‘lalrwo/L ’FY‘OBlQM “M I 6 Jr swat/ml IWDIDlim if (Y 5 M“ X; g 3 ‘+ erljl’fxwl ?Wbl1\m (X534 "i aﬂjlwuL TWlDlP/e lb)" 543 +%m+ 7‘11”“ 531%,: “s 15+ m ﬁve + [0+J-5’Xg T Mime: g+ % “l ‘1 ‘1 3 . Problem {18 points} Customers arrl - at a one-window drivehin bank according to a Poisson distribution, with a v an of 15 per hour. The servr : time per customer is exponential, with a mean of 3 minutes. Ther- :r. e four spaces in front of the window, 'ciuding the car being served. Other arriving cars line : outside this 4-car spacer (a) (3 points) What is the probabi‘ > that an arriving car can enter a e of the 4-car space? (13) (3 points) How long is an arriving cus . - er expected to » it before starting service? (c) (4 points) How many car spaces should be pro ' - a in front of the window (including the car being served) so that an arriving car can ﬁnd a sp - the : at least 95% of the time? (d) (4 points) If cars are not allowed to ait outside the 4—car ace, what is the probability that an arriving car will leave without 1' ing service? (6) (4 points) Since the arr' . rate increases to 35 customers per hour, the i ;. l k decided to open a new window with the y. e average service time. How long is an arriving custo : expected to spend before ﬁnishi : service. NOTE: For this part, the cars are allowed to wait outsi- - he 4—car space. (intentionaily left blank) [glam-l P _,: [tr-15)?“ h ['FN.“ a, 4 Pat 3 u I”)? z (037 D <93 (MM : (GD/WM “:5; F:%§:JLT , ' h D “*1 Fo:[:{%+~§-{~(TMJS)] 113367 hzo .. 19C?“ 2" w 2 £27: . L1 (cm! (of? F0 7 W; 2 E? “3‘! "£3 how “-13.37%?“ 1/ ‘ : Ci.%m:w+ EMMA: I2;%m:n. W§ZW17+ 7&— 10 ' blem 5. [10 points] Joe lov- - eat out in restaurants. His favorite foods are Mexican, Italian, and Chines cm the average, Joe pays \$10 0 : exican meal, \$15 for a Italian meal, and \$12 for a Chines- - -al. Joe’s eating habits are predictable: There » ~ I% chance that today’s meal is a repeat of _ *' erday‘s, and equal probabilities of switching to one of the rem'-'ng two. (NOTE: If you I . ‘- aifﬁculty in solving a system of linear equations, just set it up and then exp ~. the answe - e questions in terms of the variables.) - (a) (4 points) Express the situation -: . arkov in. (b) (4 points) How 111 . noes Joe pay on the average for his I ;. dinner. (c) (2 u ' 5 How Often does Joe eat Chinese food? .’ ‘Tl/m Transition W‘tvii: [1A 1 C Wk ‘ .9, .2 -2 l .2 xi) ~ L C 3.2 ~Z t b b3 .tTT, + .2ii2 tarry) :‘m \ZTi 1+ ~éle. Jr ‘ZTlgT—TYZ “'71 ”l' Tl: fin-Bil |~Tlt3lTL3 FEM-TEE“ ' l .1, * Jinx—:23 ~ 30 EKLH git-lb 4r 5 f 5 .3 ~— % .BUQY7 '5 0107/3 . 11 Problem 6. An item is consumed at te of 45 items per day. The holding cost per un' ay is \$05, and the setup cost is \$200. Suppose that no tage is allowed and that the asng cost per unit is \$10. The lead time is 21 days. (a) {4 points) Determine the optimal inv (b) (6 points) If the man units, What urer now provides a discount of \$3 per un' r orders more than 1000 be the optimal inventory policy. k: ‘0: K3200 .F-w-“ﬁ _ é .. %— ZKD .s ‘f‘ k r, -—-——~—-~ '1 i3 33 .4... .__.— .___ O _?) g ‘ j J l" 60 ' (if 4 L Le: L” LFJ’C: 3 7x67 ‘ ‘D Week? 737?“ ”Lab L: 345 ' Q0 oYelSZY" 500 welt“; Wl/ULMQUEZF inventor] (By/Hfs {30 '54.? . [b3 C; :IO jam: goo 1721090 7 590 £117 Tammi» ‘2 4%0' as (Mm ”iii? L51 “26943 " (oer: <5§4§ so if: DOD .*_-_ {coo 122‘22, LelL‘D: ([45? ' JC " it? go ‘ moniker Woe units WLWQIUEr QQWWV/ divers to \$45 12 Problem 7. [17 points] Consider the following function f(ar:) = f(m1,:cg,\$3) = x? +mg +m§ — :31— 23:3 —w2w3. (a) {3 points) Calculate the stationary point(s) of f(:r). (b) (3 points) For each stationary point, specify whether it is a local minimum or a local maximum. (Hint: Calculate the Hessian matrix) - id be the moving direction? How do . * '-e the step length? (Just give the optimization problem for a - '--' g e s - e - ; , do NOT solve it). (e) (3 points) If we want to minimize f (w) subject to the constraints 321 +302 +323 = 10 and 2331 +3532 + 433 = 20, write down the conditions for 213* being a stationary point of this constraint optimization problem. » .l. ii?“ m l 9 N 1 A“ Z (”‘3 Vim: lime ‘2 0 he; le-le'-2 ' U (ngi ‘ 3 “"1 c O ii73~Hth - 2 a! ‘0 “-i 2/ l ”0 Since oiﬁ‘t [:2] =32 >0 &£i "cil- >0 0 a z E) D (Diet 9 2“} 16%) 0 “l 2, 15 (intentionally left blank) .1 "‘ r . C3 ‘3 ‘4“: 01 A' ==—Vm°)=[ 0 éwﬁ _&w 30 To Tx-MAtl/xm Ste? Mark. we 10W ‘x SD HAM‘ 13mm Mm P5 Mmhwkd (A) * Xk-H T” X3 "’ [HFMQJ4U10VML] _ H; D 0 q A LLDIQJJ)“ [70 2 “I I [D (9 ~r 2' ’3 Z 9) LWu‘M: “(5415415 -' “1‘27‘3'X1Xs‘” (AEC‘Xﬁ-XL‘EX5vgb) w (WWB’Xﬁémg‘zo) ‘ 3L m =2 *1-()\1"'D_ 2":0 am “#1 7f L _ Pgm/ :2N2‘1g‘xvﬁﬁz‘0 (\L a 2 a" . .s \ 3L “Z“Zaﬁz—‘sz'N-“Ak'hzto [QM MW ‘50 B9 SW3? q “(l-1'“! L4? 15 ”5-: [O 2%1+5’\(z+4'}§5:2_0 ‘ 16 ...
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