recitation01

recitation01 - 2.3 IEOR E4004. SOLUTIONS to Recitation 01...

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Unformatted text preview: 2.3 IEOR E4004. SOLUTIONS to Recitation 01 Consider the linear programming model shown below. a. Graph the feasible region using a 2—di111ensional grid. Show an Minimize z=lrl +.r2 isovalue contom' for the objective SubjeCt to "‘1 + 2x2 3 10 (1) function and indicate the direction ,, v . . .\'1 — .X‘», g 4 (2) of deCIease. Identify the optimal - solution on the graph. XI +1.2 2 3 (3) b. Graphically perform a sensitivity .\'1 + 2x2 2 20 (4) analysis for each of the objective function coefficients and each of the ~"1 2 0, ~"2 2 0 (5) light—hand—side constants. Y unbounded ' 3 Grid size = l feasible region Sensitivity Analysis Variable Anal sis Ob'ective Value: 17.5 Objective Range Lower Range Upper Num. Name Value Status Reduced Cost Coefficient Limit Limit 1 X1 5. Basic 0. 2. 0.5 --- 2 X2 7.5 Basic 0. 1. -4. 4. Constraint Analysis Shadow Constraint Range Lower Range Upper Num. Name Value Status Price Limit Limit Limit 1 Conl 10. Upper -0.75 10. -4. 20. 2 Con2 0. Basic 0. 4. -10. --- 3 Cons 12.5 Basic 0. 8. --- 12.5 4 Con4 20. Lower 1.25 20. 14. - - - a. Maximize z z — 5x1 — 3x2 : 0 3\ + 5x, + x = 15 5x + Zr + .\' = 10 l 2 4 —\ + \‘ + .\ = 2 1 2 > \ + .\' = 2 5 2 6 OPTIMUM 0 1 2 3 4 5 1 d. First we must solve the objective function and constraints in terms of the basic variables 2, x1, x2, x5 and x6. We obtain Maximize : =12.368—O.263x3 —0.842.\'4 subject to 0.263x3 --0.158.\'4 + x2 = 2.468 —0.105x3 + 0.263x4 + x1 = 1.053 —0.368.\'3 + 0.421x4 + x5 = 0.684 —0.263x3 + 0.158x4 + x6 = 0.132 x] 20.j=1._...6 The figure below shows the feasible region in terms of the x3 and x4 variables. The objective function line shown is for z = 10.368. The optimum solution is at x3 =a=0 3.4 Given the following linear program, Minimize z = x1 + 2.5.x‘a+.\'; subject to xl + x, 2 10 .\'j20,j=l,...,3 Algebraically detennine the set of all basic solutions and identify whether 01' not each is feasible. Deteunine the optimum by evaluating the objective function at each feasible point. Set of basic solutions: .\'4 and .\'5 are slacks. Nonbasic Basic Solution Feasibility Objective Vertex Variables Variables (x1 , . . . ,.\'5) (Yes, No) Value #1 (3, 4, 5) (1,2 (0, 10, 0,0, 0) Yes 25 #2 (2, 4, 5) (1, 3) (10, 10, 0, 0,0) Yes 20 #3 2, 3, 5) (l, 4 ) no solution No #4 (2, 3, 4) (1,5) (10,0, 0,0, —10) No 10 #5 (1,4,5) (2, 3) (0,0, 10,0, 0) Yes 25 #6 (1,3,5) (2, 4) (0, 10, 0,0, 0) Yes 25 #7 (1,3,4) (2, 5) (O, 10, 0,0, 0) Yes 25 #8 (1,2,5) (3, 4) (0,0, 10, —10,0) No 10 #9 (l, 2, 4) (3, 5) no solution No #10 (1,2,3) (4, 5) (0,0,0, —10, —10) No 0 The solution is optimum for #2 with objective 20. BI-llVI 1.9 The variables are 2:1: number of Cubs produced per week x2: number of Quickiematic produced per week 13: number of VIPs produced per week max 3:01 + 9:02 + 25173 s.t. 0.1m + 0.22:2 + 0.713 g 250 0.2m] + 0.3512 + 0.11:3 g 350 0.111 + 0.21:2 + 0.3233 g 150 $1 2 250, $2 2 375, 13 Z 150 {total hours of manufacture} {total hours of assemble} {total hours of inspection} 1.11 The variables are: sci]: the amount of distillate j , (j = A, B, C) used to produce product i. 1' = 1 denotes Deluxe 2' = 2 standard and i = 3 economy. Note that we donot have 13A since we are not allowed to use A in economy. 7-9(T1A + 73113 + 9310) + 6-9(I2A + 1‘23 + 9320) + 50(9333 + $30) _0-6(1'1A + 12A) — 0520313 + 17213 + 9333) — 048(110 + 120 + 130) s.t. (2:1,; + 1‘2A + 203,4) 3 4000 (113 + $23 + 133) g 5000 max {available amount of A} {available amount of B } (1'10 + 3:20 + 1‘30) 3 2500 {available amount of C} 331A S 0-6(l'1A + 1‘13 + 7310) {max % of A in deluxe} $221 S 015(12/1 + $23 + I20) {max % of A in standard} I10 2 0-2(f€1A + $13 + $10) {min % of C in deluxe} 1'20 2 0-6(1‘2A + $23 + 1720) {min % of C in standard} 1030 2 0.5(133 + 130) {max % of A in deluxe} Tij Z 0 VUJ) 1.20 Let 12’ and xt— be defined as given in the hint, H is the amount production in month t, I, is the amount of inventory in month t. Also, define the constants dt as the demand in month t. (dt is given) (1521131 11+ +025 2:21 1'; s.t. P1 = 4000 + 23;" » 1'1— Pt = 131—1 + I: — 33¢— 11: 2000+P1— d1 It = 13—1 + P: — d; for t = 2,3, .... 12 It g 10000 for t = 1.2, ..., 12 P¢,I¢,1',+,a:t_ 2 0 Vt min {production in month 1} for t = 2, 3. ..., 12 {production in month t} {Inventory at the end of month 1} {Inventory at the end of month t} {storage constraint} ...
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