HW8_sol - IE 335 Operations Research Optimization Solutions...

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Unformatted text preview: IE 335 Operations Research - Optimization Solutions to Homework 8 Spring 2011 Problem 33 7-4.(a) x1 = x2 = x3 = Objective function : Constraint 1 : Constraint 2 : Constraint 3 : Number of tickets for media Number of tickets for university Number of tickets for public maximizing the profit limited 10000 seats at least half as many tickets to the university as to the public at least 500 tickets to the media 7-22.(a) $81.667 7-22.(b) For 15000 seats, the additional revenue would be $81.667 × (15000 − 10000) = $408335. For 20000 seats, the additional revenue would be $81.667 × (20000 − 10000) = $816670 7-22.(c) For $50, the revenue lost would be 6333.333 × ($100 − $50) = $316667. For $30, the revenue lost would be at least $6333.333 × ($100 − $45) = $348333 and at most $6333.333 × ($100 − $30) = $443333. 7-22.(d) For 20%, x∗ = 500 ≤ 0.2x∗ = 0.2 × (3166.667) so the optimal solution does not change. However, 1 2 for 10%, x∗ = 500 0.1x∗ = 0.1 × (3166.667) so the optimal solution will change. 1 2 Problem 34 7-23.(a) x1 = x2 = x3 = yj = Number of tons of pulp used Number of tons of recycled paper used Number of tons of recycled newsprint Number of tons produced by process j = {1, 2, 3, 4} The objective function minimizes the total material cost. The first 3 main constraints balance material consumed by the various processes with the amount purchased. The next constraint restricts pulp availability to 80 tons. The last main constraint assures 100 tons of output. 1 7-23.(b) maximize 80v4 + 100v5 v2 ≤ 50 v3 ≤ 20 − 3 v1 + v5 ≤ 0 − v1 − 4v2 + v5 ≤ 0 − v1 − 12v3 + v5 ≤ 0 − 8v2 + v5 ≤ 0 v5 ≥ 0 v4 ≤ 0 v1 , v2 , v3 URS (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) subject to v1 + v4 ≤ 100 ∗ ∗ ∗ ∗ 7-23.(c) x∗ = 80, x∗ = 480, x∗ = 0, y1 = 0, y2 = 80, y3 = 0, y4 = 20 1 2 3 ∗ ∗ ∗ ∗ ∗ 7-23.(d) v1 = 200, v2 = 50, v3 = 20, v4 = −100, v5 = 400 7-23.(e) Substitute dual variable values in the constraints of part (b), and verify that all the constraints are satisfied. Primal Objective Function Value = 100(80) + 50(480) + 20(0) = 32000 Dual Objective Function Value = 80(−100) + 100(400) = 32000 ∗ 7-23.(f ) The marginal cost is the optimal dual on the demand constraint v5 = $400 per ton. ∗ 7-23.(g) Each extra ton of pulp would change cost by v4 = −100 per ton. 7-23.(h) $150 is within the objective function coefficient range (−∞, 200], so the increase would be at primal optimal rate of 80 or 80(150 − 100) = $4000. 7-23.(i) $20 is below the range [25, 60]. With cost reductions helping more and more, the savings would be at least the extension of the primal variable rate of 480(50 − 20) = $14400. 7-23.(j) $75 is above the range [25, 60]. With cost worsening hurting less and less, the cost would be at least the extension of the primal variable rate to the end of the range, or 480(60 − 50) = $4800 and at most its extension over the full change 480(75 − 50) = $12000. 7-23.(k) $60 is below the RHS range [80, ∞). With constraint relaxation helping less and less, the saving would be at least the extension of dual variable rate to the end of the range of 400(100 − 80) = $8000 and at most its extension over the full change 400(100 − 60) = $16000. 7-23.(l) 200 is within the RHS range [80, ∞). With constraint tightening hurting more and more, the extra cost would be at least the extension of the dual variable rate 400(200 − 100) = $40000. 7-23.(m) The price would have to fall below objective function lower range $16.67. 7-23.(o) Yes, the solution would be affected because the current solution 480 400 7-23.(p) No, the solution would not change because the current solution with x3 = 0 ≤ 400. 2 Problem 35 Rock Paper Scissors Rock 0 1 -1 v1 Player C Paper -1 0 1 v2 Scissors 1 -1 0 v3 x1 x2 x3 Player R xi = vi = probability of choosing action i for player R probability of choosing action i for player C (a) maximize min {x2 − x3 , −x1 + x3 , x1 − x2 } x1 , x2 , x3 ≥ 0 (12) (13) (14) (15) Linearize to LP: maximize z z + x1 − x3 ≤ 0 z − x1 + x2 ≤ 0 x1 + x2 + x3 = 1 x1 , x2 , x3 ≥ 0 (b) minimize max {−v2 + v3 , v1 − v3 , −v1 + v2 } v1 , v2 , v3 ≥ 0 (22) (23) (24) (25) Linearize to LP: minimize w w − v1 + v3 ≥ 0 w + v1 − v2 ≥ 0 v1 + v2 + v3 = 1 v1 , v2 , v3 ≥ 0 (c) Primal problem: (26) (27) (28) (29) (30) (31) (16) (17) (18) (19) (20) (21) subject to x1 + x2 + x3 = 1 subject to z − x2 + x3 ≤ 0 subject to v1 + v2 + v3 = 1 subject to w + v2 − v3 ≥ 0 3 maximize z →y1 →y2 →y3 →y4 z + x1 − x3 ≤ 0 z − x1 + x2 ≤ 0 x1 + x2 + x3 = 1 x1 , x2 , x3 ≥ 0 (32) (33) (34) (35) (36) (37) subject to z − x2 + x3 ≤ 0 Dual problem: minimize y4 subject to y1 + y2 + y3 = 1 y2 − y3 + y4 ≥ 0 − y1 + 0 + y3 + y4 ≥ 0 y1 − y2 + y4 ≥ 0 y1 , y2 , y3 ≥ 0 y4 is URS Let y1 = v1 , y2 = v2 , y3 = v3 , y4 = w. Then, the dual of LP in part (a) is the LP in part (b). 1 1 1 (d) Set x1 = x2 = x3 = 3 . Therefore, min 1 − 3 , − 1 + 3 , 1 − 1 = 0. 3 3 3 3 1 11 1 1 1 (e) Set v1 = v2 = v3 = 3 . Therefore, max − 3 + 3 , 3 − 3 , − 1 + 3 = 0. 3 1 (f ) By strong duality, “playing each alternative with probability 3 ” is optimal for both primal and dual LP. (38) (39) (40) (41) (42) (43) (44) Problem 36 6-2.(a) Graphical Solution: See the following page. 6-2.(b) The most rapid improving direction is the negation of the gradient of the objective function, since it’s a minimization problem ∆w = (−9, −1) 6-2.(c) No active constraints 6-2.(d) w(0) = (2, 3) strictly satisfies all constraints and none of the constraints are active. 6-2.(e) 3(2) + 3(2) − 6 2(2) + 3(3) − 6 6 λmax = min , = 3(9) + 2(1) 2(9) + 3(1) 29 w(1) = w(0) + λmax ∆w = 4 81 , 29 29 Note: The other alternative is first write out the line function crosing w(0) with direction ∆w (∆w gives the slope of the line), and then calculate the intersection point with the constraint line which gives you the point w(1) . Finally calculate the distance between w(0) and w(1) , which gives you the λmax . 6-2.(f ) See the graph on next page 6-2.(g) At the boundary, we confront the limits on feasible directions. Problem 37 6-3.(b) Yes 6-3.(d) No, because it visits the non-optimal boundary point. 6-3.(f ) No, because the first point is on the boundary. 4 6−2.(a) 4 3.5 3w1 + 2w2 = 6 2w1 + 3w2 = 6 9 w 1 + w 2 = 27 9 w 1 + w 2 = 18 w(0) = (2, 3) 6 29 ) ∗ 3 w = (0, 3) 4 w(1) = ( 29 , 2.5 w2 2 1.5 1 0.5 0 0 0.5 1 1.5 2 w1 2.5 3 3.5 4 : \Users\Timnt\Documents\gamsdir\projdir\hw5_7_23.l:: C:::::::::::::::::::::::::::::::::::::::::::::::::::t s 2 3 :2 :51:11 1age:1 ::::1:SSSS:1e1:235::11111S1:23.5.2:1111SS:1indows:::::::::::::13111111:21:11:31:1age:1 ::::2:S:e:n:e:r:a:l:::S:l:g:e:m:r:a:i:c:::S:o:d:e:l:i:n:g:::S:m:s:t:e:m ::::3:C:o:m:p:i:l:a:t:i:o:n ::::1: ::::5: ::::1::::1::sets ::::7::::2::i:"decision:1ariamle:sumscript" ::::1::::3::11;71; ::::1::::1::: :::1: :::5::parameter:c"i":"unit:o":omjecti1e:"unction" :1 :::1: :::1::11:111;2:51;3:21;1:1;5:1;1:1;7:11; :1 :::1: :::7::: :2 :::1: :::1::positi1e:1ariamle:1"i"; :3 :::1: :::1::"ree:1ariamle:cost; :1 :::1: ::11 :5 ::: : :::1: ::11::ssuations :1 :::1: 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This note was uploaded on 04/27/2011 for the course IE 335 taught by Professor Smith during the Spring '11 term at Southwestern.

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