Introduction to Operations Research Answers to selected problems

Transcription

1 Exercise session 1 Introduction to Operations Research Answers to selected problems 1. Decision variables: Prof. dr. E-H. Aghezzaf, assistant D. Verleye Department of Industrial Management (IR18) x 1 = number of barrels of beer produced x 2 = number of barrels of ale produced LP formulation: Maximize Z = 5x 1 + 2x 2, 5x 1 + 2x x 1 + x 2 25 and x 1, x 2 0. (Availability of corn) (Availability of hops) 2. Decision variables: x 1 = ton of alloy 1 produced x 2 = ton of alloy 2 produced Since x 1 + x 2 = 1, these are also the percentages. LP formulation: Minimize Z = 190x x 2, x 1 + x 2 = 1 (Producing 1 ton) 3.2 3x 1 + 4x (Carbon requirement) 1.8 2x x (Silicon requirement) 0.9 x x (Nickel requirement) 42000x x (Tensile strength requirement) and x 1, x

2 3. Decision variables: x 11 = amount of steel 1 produced at mill 1 (ton) x 12 = amount of steel 1 produced at mill 2 (ton) x 13 = amount of steel 1 produced at mill 3 (ton) x 21 = amount of steel 2 produced at mill 1 (ton) x 22 = amount of steel 2 produced at mill 2 (ton) x 23 = amount of steel 2 produced at mill 3 (ton) LP formulation: Minimize Z = 10x x x x x x 23, x 11 + x 12 + x (Minimum production steel 1) x 21 + x 22 + x (Minimum production steel 2) 20x x (Available furnace time at mill 1) 24x x (Available furnace time at mill 2) 28x x (Available furnace time at mill 3) and x ij 0, i = 1, 2; j = 1, 2, A graphical representation of the problem: Decision variables: x LA H = amount of barrels shipped from LA to Houston (million barrels/year) x LA NY = amount of barrels shipped from LA to NY (million barrels/year) x C H = amount of barrels shipped from Chicago to Houston (million barrels/year) x C NY = amount of barrels shipped from Chicago to NY (million barrels/year) y LA = added refining capacity for LA (million barrels) 2

3 y C = added refining capacity for Chicago (million barrels) LP formulation: Maximize Z = 10 (20x LA H + 15x LA NY + 18x C H + 17x C NY ) 120y LA 150y C x LA H + x LA NY 2 + y LA (Don t ship more from LA than available) x C H + x C NY 3 + y C (Don t ship more from Chicago than available) x LA H + x C H 5 (Amount shipped to Houston Houston Demand) x LA NY + x C NY 5 (Amount shipped to NY NY Demand) and all variables Decision variables: x i = number of hours workers spend on machine i each week, i = 1, 2, 3 p i = units of product i produced each week, i = 1, 2, 3 LP formulation: Maximize Z = 6p 1 + 8p p 3, 2p 1 + 3p 2 + 4p 3 x 1 (process time machine 1 assigned time machine 1) 3p 1 + 5p 2 + 6p 3 x 2 (process time machine 2 assigned time machine 2) 4p 1 + 7p 2 + 9p 3 x 3 (process time machine 3 assigned time machine 3) x (assigned time machine 1 available time) x (assigned time machine 2 available time) x (assigned time machine 3 available time) x 1 + x 2 + x (total workers time) and x i, p i 0, i = 1, 2, Decision variables: S ij = type i trucks sold during year j P ij = type i trucks produced during year j I ij = number of type i trucks in inventory at end of year j LP formulation: 3

4 Maximize Z = 20S S S S S S 23 15P 11 15P 12 15P 13 14P 21 14P 22 14P 23 2I 11 2I 12 2I 13 2I 21 2I 22 2I 23, S 11 P 11 + I 11 = 0 (Year 1 production type 1) S 12 P 12 I 11 + I 12 = 0 (Year 2 production type 1) S 13 P 13 I 12 + I 13 = 0 (Year 3 production type 1) S 21 P 21 + I 21 = 0 (Year 1 production type 2) S 22 P 22 I 21 + I 22 = 0 (Year 2 production type 2) S 23 P 23 I 22 + I 23 = 0 (Year 3 production type 2) P 11 + P (Year 1 capacity limit) P 12 + P (Year 2 capacity limit) P 13 + P (Year 3 capacity limit) S (Year 1 maximum type 1 sold) S (Year 2 maximum type 1 sold) S (Year 3 maximum type 1 sold) S (Year 1 maximum type 2 sold) S (Year 2 maximum type 2 sold) S (Year 3 maximum type 2 sold) 5P P P 13 5P 21 5P 22 5P 23 0 (pollution emission limit) 8. Decision variables: x 11 = units of product 1 produced on machine 1 x 12 = units of product 1 produced on machine 2 x 21 = units of product 2 produced on machine 1 x 22 = units of product 2 produced on machine 2 4

5 LP formulation: Minimize Z = 1.5x x x x 22 x 11 + x (Minimum production product 1) x 21 + x (Minimum production product 2) x 12 ) x 11 1(x 2 12 x 11 ) 0 (Half of product 1 on machine 1) x 22 ) 0 (Half of product 2 on machine 2) 0.7x x (Machine 1 availability) 0.8x x (Machine 2 availability) 0.75x x x 12 + x (Labor availability) and x ij 0, i = 1, 2; j = 1, Decision variables: x ij = number of officers who get off days i and j of the week (1 = Saturday,..., 7 = Friday). Note that for example on Saturday, = 2 officers have to take a day off. This reasoning leads us to the following LP formulation: LP formulation: Maximize Z = x 12 + x 23 + x 34 + x 45 + x 56 + x 67 + x 17 x 12 + x 13 + x 14 + x 15 + x 16 + x 17 = 2 x 12 + x 23 + x 24 + x 25 + x 26 + x 27 = 12 x 13 + x 23 + x 34 + x 35 + x 36 + x 37 = 12 x 14 + x 24 + x 34 + x 45 + x 46 + x 47 = 6 x 15 + x 25 + x 35 + x 45 + x 56 + x 57 = 5 x 16 + x 26 + x 36 + x 46 + x 56 + x 67 = 14 x 17 + x 27 + x 37 + x 47 + x 57 + x 67 = 9 and x ij 0, i = 1..6, j = 2..7, i < j. 10. Decision variables: x ij = money invested at beginning of month i for a period of j months. LP formulation: 5

6 Maximize Z = 1.08x x x x 41 x 11 + x 12 + x 13 + x 14 = 200 (month 1) x 21 + x 22 + x 23 = x 11 (month 2) x 31 + x 32 = x x 21 (month 3) x 41 = x x x 31 (month 4) and x ij 0, i = 1..4, j = 1..5 i. 11. Decision variables: x 1 = amount of germanium melted with method 1 x 2 = amount of germanium melted with method 2 g 1R = amount of grade 1 germanium refired g 2R = amount of grade 2 germanium refired g 3R = amount of grade 3 germanium refired d R = amount of detective germanium refired LP formulation: Minimize Z = 50x x (g 1R + g 2R + g 3R + d R ) x 1 + x 2 + g 1R + g 2R + g 3R + d R (capacity constraint) 0.3x x 2 (1 0.3)g 1R d R 3000 (monthly demand grade 1 transistors) 0.2x x g 1R 0.6g 2R d R 3000 (monthly demand grade 2 transistors) 0.15x x g 1R + 0.3g 2R 0.5g 3R + 0.2d R 2000 (monthly demand grade 3 transistors) 0.05x x g 1R + 0.3g 2R + 0.5g 3R + 0.1d R 1000 (monthly demand grade 4 transistors) 0.3x x 2 d R (amount of detective germanium available for refiring) 0.3x x 2 g 1R (amount of grade 1 germanium available for refiring) 0.2x x 2 g 2R (amount of grade 2 germanium available for refiring) 0.15x x 2 g 3R (amount of grade 3 germanium available for refiring) and x 1, x 2, g 1R, g 2R, g 3R, g 4R, d R Decision variables: Let shift i = 1: Midnight - 6 AM 2: 6 AM - Noon 6

7 3: Noon 6 - PM 4: 6 PM - Midnight x 1 = Workers who work shifts 1 and 2 x 2 = Workers who work shifts 2 and 3 x 3 = Workers who work shifts 3 and 4 x 4 = Workers who work shifts 4 and 1 i t = customers left after end of shift t s t = number of people served during shift t LP formulation: Minimize Z = 120(x 1 + x 2 + x 3 + x 4 ) + 5(i 1 + i 2 + i 3 + i 4 ) i 1 = 100 s 1 i 2 = i s 2 i 3 = i s 3 i 4 = i s 4 i 4 = 0 s 1 50x x 4 s 2 50x x 2 s 3 50x x 2 s 4 50x x 3 and all variables Decision variables: L = gallons bought at Los Angeles H = gallons bought at Houston N = gallons bought at New York M = gallons bought at Miami IL = gallons on hand when landing in LA IH = gallons on hand when landing in Houston IN = gallons on hand when landing in NY IM = gallons on hand when landing in Miami LP formulation: 7

8 Minimize Z = 88L + 15H + 105N + 95M, IL = IM + M 2700 ( 1 + (IL + IM + M)/2000 ) IH = IL + L 1500 ( 1 + (IL + L + IH)/2000 ) IN = IH + H 1700 ( 1 + (IH + H + IN)/2000 ) IM = IN + N 1300 ( 1 + (IM + IN + N)/2000 ) IL + L 12, 000 IH + H 12, 000 IN + N 12, 000 IM + M 12, 000 IL, IH, IN, IM 600 and L, H, N, M Decision variables: Let XY represent the amount of currency X converted to currency Y, where each letter represents the first letter of the corresponding currency (P = pound,...). Furthermore, F X represents the final amount of the currency, while X is the unconverted amount. LP formulation: Maximize F D F P F M F Y F D D 1.697P D MD = 0 F P P DP MP = 0 F M M 1.743DM P M = 0 F Y 138.3DY 234.7P Y MY = 0 F D >= 6 F P >= 3 F M >= 1 F Y >= 10 D + DP + DM + DY = 8 P D + P + P M + P Y = 1 MD + MP + M + MY = 8 and all variables 0 8

9 Exercise session 2 1. Graphical solution: The solution space is convex, so an optimal solution exists. We note that the line representing the objective function (drawn here for z = 5) is parallel to the corn constraint. All solutions in the line segment (x 1 = 10, x 2 = 5) (x 1 = 12, x 2 = 0) are optimal solutions, with z = Second tableau: Basic var Eq. z x 1 x 2 x 3 s 1 s 2 RHS Ratio z (0) s 1 (1) 0 0 2/5 9/5 1-1/5 3 15/2 x 1 (2) 0 1 3/5 6/5 0 1/5 3 5 All coefficients 0, so this solution is optimal. The NBV x 2 has coefficient 0, so alternative optima exist. We enter x 2 into the basis (instead of x 1 ). Third tableau: Basic var Eq. z x 1 x 2 x 3 s 1 s 2 RHS z (0) s 1 (1) 0-2/ /3 1 x 2 (2) 0 5/ /3 5 9

10 The two optimal solutions are (x 1 = 3, s 1 = 3),(x 2 = 5, s 1 = 1) with objective value z = Final tableau: 6. Final tableau: Basic var Eq. z x 1 x 2 s 1 s 2 RHS z (0) 1 0-7/3-4/3 0-8 x 1 (1) 0 1 1/3 1/3 0 2 s 2 (2) 0 0 7/3 1/3 1 2 Basic var Eq. z x 1 x 2 a 1 s 2 s 3 RHS z (0) 1 0 7/2 M+5/ x 1 (1) 0 1 1/2 1/ s 2 (2) 0 0 1/2-1/ s 2 (3) 0 0 3/2-1/ Any solution on the line segment with endpoints ( 1 2, 0) and (1 3, 2 ) is an optimal 3 solution for the LP. The middle point of this segment has coordinates (x 1, x 2 )= ( 5 12, 1 ), which is a third optimal solution Final tableau: Basic var Eq. z x 1 x 2 e 1 a 1 e 2 a 2 RHS z (0) 1 0 -M+5 -M M 0 5M -6 x 1 (1) a 2 (2) Since all coefficients are negative, this is an optimal tableau. In fact,a 2 is still in the basis, which means that the LP is infeasible. 9. (a) The bfs are the corner points of the feasible region, shown in the graph: 10

11 (b) Initial tableau: Basic var Eq. z x 1 x 2 s 1 s 2 RHS Ratio z (0) s 1 (1) s 2 (2) Second tableau: Basic var Eq. z x 1 x 2 s 1 s 2 RHS Ratio z (0) x 1 (1) s 2 (2) Third tableau: Basic var Eq. z x 1 x 2 s 1 s 2 RHS Ratio z (0) x 1 (1) x 2 (2) Final tableau: 11

12 10. (a) b 0 and c 0. Basic var Eq. z x 1 x 2 s 1 s 2 RHS z (0) s 1 (1) x 2 (2) (b) b 0 and c = 0 - If a 2 > 0, a 3 0 we can pivot in x 4 to obtain an alternative optimum. - If a 3 > 0, a 2 0 we can pivot in x 5 and obtain an alternative optimum. - If a 2, a 3 > 0 Ratio Test has to be used to determine leaving variable (c) c, a 2, a 3 < 0 ensures that x 1 can be made arbitrarily large causing z to become arbitrarily large. 11. (a) b 0 is required since we assume that the LP is written in standard form. - If c 1 = 0 and c 2 0 we can pivot in x 1 to obtain an alternative optimum. - If c 1 0, c 2 0 and a 2 > 0 we can pivot in x 5 and obtain an alternative optimum. - If c 2 = 0, a 1 > 0 and c 1 0 we can pivot in x 2 and obtain an alternative optimum. (b) b < 0 (c) b = 0 (d) b 0 makes the solution feasible. If c 2 < 0 and a 1 0 we can make x 2 as large as desired and obtain an unbounded solution. (e) b 0 makes the current basic solution feasible. For x 6 to replace x 1 we need c 1 < 0 (this ensures that increasing x 1 will increase z and we need Row 3 to win the ratio test for x 1. This requires 12 a 3 b. 12

13 Exercise session 3 1. A graphical representation of the problem: (a) (b) 20 c c 1 /50 1/2 25 c (c) Current basis remains optimal for 20 b The shadow price S.P.1 = 20. (d) The current basis remains optimal for b 2 80/3. S.P.2 = 0 since Hops is not a binding constraint. (e) 20 b 3 50, S.P.3 = 10. (f) On pound = 16 ounces, so each S.P. would be divided by 16. (g) If we look at the graph we can see: 0 c 1 25 : (0, 20) is optimal z = c 1 (0) + 50(20) = c : ( 40 3, 40 3 ) is optimal z = c 1( 40 3 ) + 50(40 3 ) = c : (20, 0) is optimal z = 20c 1 Plot: c 1 13

14 (h) Graph: (i) Graph: 14

15 (j) Graph: 2. (a) S.P. = 50 cents (b) S.P. + $ 1 = $ 1.75 (c) $ 6 + R.C. for x 1 = $ 6.25 (d) $ $0.75 = $ 105 (e) $ $ 2 = $ (a) The dual of the LP is: Minimize Y = 6y 1 + 8y 2 + 2y 3, y 1 + 6y 2 5 y 1 + y 3 1 y 1 + y 2 + y 3 2 and y 1, y 2, y 3 0. The optimal solution is given by the coefficients of the slack variables: y 2 = 0, y 2 = 5 6, y 3 = 7, giving the optimal solution Y = Z = 9. 6 (b) 0 c 1 6. (c) The basis remains optimal as long as: c 2 7/6 4. (a) $ $10x 88 = $

16 (b) The allowable increase for Type 1 machines is less than one, so we cannot answer this question. (c) At the moment we have 260 tons available, of which 1.2 is slack. As a consequence, the dual price is 0 and Carco should not be willing to pay anything for an extra ton of steel. (d) Now, the requirement is on 88 cars. The allowable decrease is 3 so we now that the current basis will not change. The dual price is -20, so if we only need 86 cars, our profit will increase to $ $20 ( 2) = $ (e) $600 1, 2$400 = $120 > 0, so Carco should consider producing jeeps. 5. (a) The dual of this LP is: Minimize Y = 50y y y 3, y 1 + 2y 2 + y 3 3 y 1 y 2 + y 3 4 y 1 + y 2 1 and y 1 0, y 2 0, y 3 u.r.s. The optimal solution can be read from the optimal primal tableau: y 1 = 1, y 2 = M M = 0, y 3 = 3 + M M = 3, which gives Y = 80. (b) c 1 4. (c) To avoid that entering x 1 into the basis improves the objective function, we need c (a) $2, 50 + $10, 00 = $12, 50 (b) $75 (c) $4250 5($75) = $3875 (d) Objective function coefficient for wheat is now 26(5) = $130. The allowable decrease for wheat is $30 so the current basis remains optimal. The decision variables remain unchanged, but the new z value is given by 130(25)+200(20) 10(350) = $3750. (e) $120 1$75 3$12, 5 = $7, 5 > 0, so Leary should consider producing barley. 7. (a) The dual of this LP is: Minimize Y = 4y 1 + 6y 2 + 7y 3, 2y 1 + 3y 2 + 4y 3 3 y 1 + 2y 2 + 2y 3 1 and y 1 0, y 2 0, y 3 u.r.s. 16

17 The optimal dual solution is w = 9/2, y 1 = 0, y 2 = 1, y 3 = 3/2. (b) If we increase the RHS of the third constraint by b 3, we find that: 1 b 3 1 If b 3 = 1/2, the new optimal solution is: 9/2 + 1/2(3/2) = 21/4. 9. (a) The dual of this LP is: Minimize Y = 6y 1 + 3y 2 + 3y 3, 4y 1 + y 2 + 3y 3 4 3y 1 + 2y 2 + y 3 1 and y 1 0, y 2 0, y 3 u.r.s. The optimal dual solution is w = 18/5, y 1 = 0, y 2 = 1/5, y 3 = 7/5. (b) Our dual would have an additional constraint: y 1 + y 2 + y 3 1 which is not satisfied for the current solution, so the basis would change. 10. (a) The dual of this LP is: Minimize Y = 18000y y y y 4, 15y y 2 + 3y 3 + y y y 2 + 4y 3 5 8y 1 + 4y 2 + 2y 3 4 and y 1, y 2, y 3 0,y 4 0 The optimal dual solution is w = 4500, y 1 = 0.5, y 2 = y 3 = 0, y 4 = 4.5 (b) By pivoting in the non-basic variable s 2 (with coefficient 0 in the optimal tableau!) we can obtain an alternative optimal solution z = 4, 500, HB = 375, s 2 = 1, 500, s 3 = 5, 250, RS = 1, 000. This solution uses only 16,500 sewing minutes. (c) The shadow price of sewing time is 0. Thus even though all 18,000 minutes of available sewing time is used, additional sewing time will not increase revenue. This is because additional sewing time is worthless in the alternative optimal solution which uses only 16,500 minutes of sewing time. (d) Revenue will decrease by 100( $4.5) = $

18 Exercise session 4 1. (a) Graphical representation: LP formulation: Minimize 400x x x x x x x x x 33, x 11 + x 12 + x 13 = 35 x 21 + x 22 + x 23 = 30 x 31 + x 32 + x 33 = 35 x 11 + x 21 + x 31 = 30 x 12 + x 22 + x 32 = 30 x 13 + x 23 + x 33 = 20 x 1D + x 2D + x 3D = 20 x ij 0, i = 1, 2, 3, j = 1, 2, 3, D (b) Vogel s Method: 18

19 19

20 (c) Stepping stone: All indices positive, so solution is optimal. Transportation simplex: u 1 = 0 u 1 + v 1 = c 11 = 400 v 1 = 400 u 1 + v 2 = c 12 = 420 v 2 = 420 u 2 + v 2 = c 22 = 420 u 2 = 0 u 2 + v 4 = c 24 = 0 v 4 = 0 u 3 + v 2 = c 32 = 415 u 3 = 5 u 3 + v 3 = c 33 = 410 v 3 = 415 For each unused cell, we now calculate the reduced cost: c 13 = c 13 u 1 v 3 = = 25 c 14 = c 14 u 1 v 4 = = 0 c 21 = c 21 u 2 v 1 = = 25 c 23 = c 23 u 2 v 3 = = 25 c 31 = c 31 u 3 v 1 = 420 ( 5) 400 = 25 c 34 = c 34 u 3 v 4 = 0 ( 5) 0 = word processing files should be stored on the hard disk, 100 word processing files in the computer memory, 100 packaged programs on tape, and 100 data files on tape. 3. Car 2/Car 4 for call 1, Car 5 for call 2, Car 4/Car 3 for call city blocks have to be traveled in total. 7. Let i = flights that leave from NY, j = flights that leave from Chicago. Optimal assignment: x 13 = x 24 = x 35 = x 41 = x 56 = x 67 = x 72 = 1. Total downtime = 25 hours. 20

21 8. LP: Table: 10. (a) Supply point ij = engine j of company i and Demand point ij = j th engine to go to fire i. All supplies and demands equal 1. 21

22 This formulation picks up the correct costs because earlier engines arriving at a fire have a larger effect on cost than later engines arriving at the same fire. This implies, for example, that the transportation simplex or assignment method will never assign an engine which takes 6 minutes to get to a fire to arrive earlier than an engine taking 4 minutes to get to the same fire (because interchanging these assignments would reduce cost and leave the same engines available for other assignments). (b) it might be optimal for an assignment to have t 12 < t 11, and this would result in incorrect costs being assigned. 22

23 Exercise session 5 1. Shortest path = 1-2-5, cost = Dijkstra: 1-3-4, cost = 2. Shortest path = , cost =1. Since we cannot update node 3 from node 2 (since its distance is already fixed in the second iteration), we fail to obtain the shortest path with Dijkstra s algorithm. 4. Max flow: The total flow = 45. Minimum cut: so-2 with capacity = Max flow: Optimal flow = 9 units. The minimum cut is so-1-3 with capacity =9. 23

24 6. Network: If max flow= 21 then all packages can be loaded. 7. Network representation: Clearly, 1200 cars can be sent (300 directly in first and second hour, 300 each via C2 and C3). 24

25 8. (a) The shortest path from NY to LA = {NY, St.-L.,Ph.,LA}, requiring 2450 gallons of gas. (b) Balanced transportation problem: 9. (a) 10. Prim: (b) Max flow = 750, min cut ={LA} Length of the MST = =

26 Kruskal: Length of MST = =

27 Exercise session 6 1. Let { yi = 1 if we buy any computers from vendor i. y i = 0 otherwise x i = Number of computers purchased from vendor i. Then our IP becomes: Minimize Z = 500x x x y y y 3, x 1 + x 2 + x 3 = 1100 x 1 500y 1 x 2 900y 2 x 3 400y 3 y i {0, 1}, x i 0, i = Tree: 3. Solving the LP relaxation gives us the integer optimal solution x 1 = x 2 = z = After many iterations, we find out that the problem is infeasible: 27

28 5. Let x i = 1 if surgeon i is used and x i = 0 otherwise. Then we can write: Minimize Z = x 1 + x 2 + x 3 + x 4 + x 5 + x 6, x 1 + x 4 1 x 2 + x 3 1 x 1 + x 5 1 x 1 + x 6 1 x 2 + x 3 + x 6 1 x 1 + x 4 1 x 1 + x 2 1 x i {0, 1}, i =

29 6. Let x ij = Time that job i begins its processing on Machine j. For the objective function, we took the average processing time. Minimize Z = (x x x )/3, x 13 x x 14 x x 22 x x 24 x x 33 x x x 21 My 1 x x 11 M(1 y 1 ) x x 32 My 2 x x 22 M(1 y 2 ) x x 33 My 3 x x 13 M(1 y 3 ) x x 14 My 4 x x 24 M(1 y 4 ) x ij 0, i = 1..3, j = 1..4 y i {0, 1}, i = Let x it = 1 if plant i is built at the beginning of year t, and 0 otherwise. We find: Minimize Z = 5 t=1 (20x 1t + 16x 2t + 18x 3t + 14x 4t ) +7.5x x x x x 15 +4x x x x x x x x x x 35 +3x x x x x 45, 5 t=1 x it 1, i = 1..4 t j=1 (70x 1j + 50x 2j + 60x 3j + 40x 4j ) d t, t = 1..5 and x it {0, 1}, i = 1..4, t = 1..5, where d t is the capacity demand during year t. 29

30 8. The IP: Minimize Z = 5 t=1 (1.5x 1t + 0.8x 2t + 1.3x 3t + 0.6x 4t ) + 4 t=1 (1.7y 1t + 1.2y 2t + 1.3y 3t + 0.8y 4t ) + 5 t=2 (1.9z 1t + 1.5z 2t + 1.3z 3t + 1.1z 4t ), 70x 1t + 50x 2t + 60x 3t + 40x 4t d t, t = 1..5 x it x i,t+1 1 w t, t = y it w t, t = 1..4 x it x i,t 1 1 w t, t = z it w t, t = 2..5 and all variables {0, 1} 10. Let y i = 1 if an auditor is based in city i (1 = NY, 2 = Chic, 3 = LA, 4 = Atl) and y i = 0 otherwise. Also let x ij = the number of trips made by an auditor based in city i to region j(j = 1 is Northeast, etc.) Then an appropriate formulation is Minimize Z = 100, 000(y 1 + y 2 + y 3 + y 4 ) x x x x x x x x x x x x x x x x 44 x 11 + x 21 + x 31 + x x 12 + x 22 + x 32 + x x 13 + x 23 + x 33 + x x 14 + x 24 + x 34 + x x 11 + x 12 + x 13 + x y 1 x 21 + x 22 + x 23 + x y 2 x 31 + x 32 + x 33 + x y 3 x 41 + x 42 + x 43 + x y 4 and y j {0, 1}, j = 1..4, x ij integer, i = 1..4, j =

31 Exercise session 7 1. Let x be the location of the store. We wish to choose x to minimize f(x) = (x 3) 2 + (x 4) 2 + (x 5) 2 + (x 6) 2 + (x 17) 2 Since we have no side constraints, we can easily find the minimum by: f (x) = 2(x 3 + x 4 + x 5 + x 6 + x 17) = 0, giving x = 7 as a local minimum. However, f (x) = 10 > 0 for all values of x, meaning our function is convex. The local minimum is therefore a global minimum, so the store should be located at x = 7. In general, we get: f (x) = 2(x x 1 + x x 2 + x x x x n ) = 0, giving x = n i=1 x i/n. The store should be located at the arithmetic mean of the location of the n customers. 2. (a) Let R be the number of units of raw materials purchased. We wish to solve: Maximize Z = (49 x 1 )x 1 + (30 2x 2 )x 2 5R, x 1 2R x 2 R x 1, x 2, R 0, i = 1..3 We derive the Hessian of the objective function: H(x 1, x 2, R) = Since all elements in this matrix are 0, so the objective function is concave. Since the constraints are linear, the K-T conditions will yield an optimal solution. The K-T conditions are: 31

32 49 2x 1 λ x 2 λ λ 1 + λ 2 0 λ 1 (x 1 2R) = 0 λ 2 (x 2 R) = 0 x 1 2R 0 x 2 R 0 (49 2x 1 λ 1 )x 1 = 0 (30 4x 2 λ 2 )x 2 = 0 ( 5 + 2λ 1 + λ 2 )R = 0 x 1, x 2, R, λ 1, λ After several iterations, we find that the interval of uncertainty is [1.18, 1.63). This interval has width less than 0.50, so we are finished. (actual maximum occurs for x = 1.5) 4. Iteration 1 f(x 1, x 2 ) = (e (x 1+x 2 ) (1 x 1 x 2 ) 1, e (x 1+x 2 ) (1 x 1 x 2 )). f(0, 1) = ( 1, 0). Thus new point is ( t, 1) where t 0 maximizes f( t, 1) = (1 t)e (t 1) + t. f ( t, 1) = (1 t)e (t 1) e (t 1) + 1 = 0 for 1 = te (t 1) or t = 1. Thus new point is ( 1, 1) Iteration 2 f( 1, 1) = (0, 1) so new point is ( 1, 1 + t) and we choose t 0 to maximize f( 1, 1 + t) = te t + 1. f ( 1, 1 + t) = te t + e t = 0 for t = 1. Thus new point is ( 1, 2). 5. Using the lagrangian multiplier λ, we find x = 6, y = 4, w = 3, λ = 6, z = Using the K-T conditions yields y = 2 and x = 5, with z = 30. Since this is the only point satisfying KT conditions, it must be the optimal solution. 7. We want to maximize Π = (10, a 100p)(p 10) a. Deriving to p and a, we find p = 58 and a = 14,

33 8. Using the hint (with x a), we get the K-T conditions: 1 ln(x) x 2 + λ = 0 λ(a x) = 0 x a λ 0 We find the unique solution to the K- T conditions occurs for x = a, and λ = ln(a) 1. a 2 Thus for b > a e we know that ln(b) < ln(a) or a ln(b) < b ln(a). Taking e to both b a sides we obtain b a < a b. To solve the problem now let a = e and b = π. 10. x = location of the store. We wish to choose x to minimize: f(x) = x 3 + x 4 + x 5 + x 6 + x 17. Problem 4: For any two points on the curve y = x the line joining the two points is never below the curve. Thus x is a convex function of x. From Problem 4 (and the fact that the sum of convex functions is convex) we know that f(x) is a convex function. Thus any local minimum of f(x) will minimize f(x). We claim that choosing x = 5 (the median of the customer s locations) yields a local minimum. To see this suppose that you move the store location λ units to the left of 5 (λ is small). Then customers 3 and 4 travel λ less units to store but customers 5, 6, and 17 travel λ units more to store, so total travel time increases. Suppose you move the store location λ units to the right of 5. Then customers 3,4 and 5 travel λ more units to the store while customers 6 and 17 travel λ less units to the store, and again total travel time increases. Thus x = 5 is a local minimum, and hence a minimum over all possible store locations. In general, if there are n customers, the store should be located at the median value of the customer s locations. If there are an even number (say 2n) of customers, locate the store anywhere between the n th and n + 1 customer. 11. (a) K-T conditions are (ignoring non-negativity) original constraints: x 1 p 1(x 1 ) + p 1 (x 1 ) λ 1 = 0 (1) x 2 p 2(x 2 ) + p 2 (x 2 ) λ 2 = 0 (2) c + λ 1 k 1 + λ 2 k 2 = 0 (3) λ 1 (x 1 k 1 z) = 0 (4) λ 2 (x 2 k 2 z) = 0 (5) λ 1, λ 2 0, x 1 k 1 z, x 2 k 2 z (6) 33

34 (b) Now firm wishes to maximize f(x 1, x 2 )) = x 1 p 1 (x 1 ) + x 2 p 2 (x 2 ) λ 1 x 1 λ 2 x 2. The optimum is found when: x 1 p 1(x 1 ) + p 1 (x 1 ) λ 1 = 0 x 2 p 2(x 2 ) + p 2 (x 2 ) λ 2 = 0 Clearly any solution to the original problem will satisfy these equations. Also the revenues in (a) are identical to the revenues in (b). Can we show that the costs in (b) are identical to the cost cz incurred in the original problem? Case 1: λ1 > 0, λ 2 > 0 Then x 1 = k 1 z and x 2 = k 2 z and from (3) total cost is c z = λ 1 k1 z+ λ 2 k 2 z = λ 1 x 1 + λ 2 x 2, and both problems incur the same production cost. Case 2: λ1 = 0, λ 2 = 0: From (3) this cannot occur unless c = 0. Case 3: λ 1 > 0, λ 2 = 0: Then (3) and (4) yield x 1 = k 1 z and k 1 λ1 = c. Now total cost is c z = k 1 λ1 z = x 1 λ1 + x 2 λ2, as desired. Case 4: λ1 = 0, λ 2 > 0: The reasoning here is identical to Case 3. (c) Shadow prices. 34

35 Exercise session 8 1. The maximin decision is to mount a small advertising campaign, maximax + minimax regret decision is to mount a large advertising campaign. 2. The reward matrix is: Noble Greek Price Pizza King Price $6 $8 $10 $5 $125 $175 $225 $6 $200 $300 $400 $7 $225 $375 $525 $8 $200 $400 $600 $9 $125 $375 $625 The Maximin action is to charge $7, the Maximax action is to charge $9 and the Minimax regret action is to charge $8. Pizza King maximizes their expected reward by charging $8. 3. Let p = Alden s bid. Then the reward table is as follows: Forbes bid Alden s bid = p $6000 $8000 $11000 p = $ p = $ $2000 $2000 p = $ $5000 Observe that p = $6,000 is dominated by p =$8,000. Either $8,000 or $11,000 is the Maximin action, the Maximax and Minimax regret bid is $11,000. Bidding $11,000 also maximizes expected profit. 4. The optimal solution is to buy the forecast and then act optimally (don t buy the pass if sunny forecast and buy the pass if rainy forecast). If forecast were free it would be worth EVSI = = $1.44 > $1.00. The EVPI = (-15) = $ The doctor should operate now. 35

36 6. The bank should run the survey. If the recommendation is favorable, grant the loan. If not, invest in bonds. EVSI = ,760 = $1020 > $500. EVPI = 0.04* *6000-3,760 = $2, We should hire geologist and if geologist predicts an earthquake, build at Roy Rogers. If the geologist predicts no quake build at Diablo. EVSI = (-14) = $1.1 million > $1 million, so it is worth it to pay $1 million to hire the geologist! EVPI = (-14) = $2 million. 8. Buy toss and if 1 comes up choose die in right hand and if 6 comes up choose die in left hand. EVSI = ( ) = $17. EVPI = = $ (a) From the tree we see that university should not drug test (lower expected cost). Note: we assume that we must act on positive drug test and must not act on negative drug test. (b) From the decision tree we find that Expected Cost for Drug Testing > 0.14*(0.68)*c1 > 0.14*(0.68)*c2 > Expected Cost for not Drug Testing = 0.05*c2, so it is optimal not to test for drugs. 36

37 Exercise session 9 1. Shortest path from node 1 to node 10 is , shortest path from node 2 to node 10 is Retracing the optimal path we find that the salesman should speak in Bloomington on Monday, Tuesday, and Wednesday and speak in Indianapolis on Thursday. 3. Define x t (i) to be a production level during period t with inventory level i at the beginning of the month. One optimal plan is to produce 0 units during month 1, x 2 (0) = 1 unit, and x 3 (0) = 4 units during month 3. Other optimal production plans include: Plan three units produced during month 2 and two units produced during month 4. Plan one unit produced during month 2, two units produced during month 3, and two units produced during month 4. All optimal plans incur a cost of $ It is optimal to produce x 1 (1) = 0 units during month 1; produce x 2 (0) = 0 units during month 2; produce x 3 ( 1) = 0 units during month 3; produce x 4 ( 3) = 5 units during month 4. Production is postponed because the backlogging cost is smaller than the holding cost. 5. Define f t (i) be the minimum cost incurred during months t,t + 1,..3 if the inventory at the beginning of month t is i.let x t (i) be the quantity that should be produced during month t in order to attain f t (i). We find that f 1 (0) = $8,950 and x 1 (0) = 200 radios should be produced during month 1. This yields a month 2 inventory of = 0. Thus during month 2 we produce x 2 (0) = 600 radios. At the beginning of month 3 the inventory will now be = 300. Hence during month 3 x 3 (300) = 0 radios should be produced. Note that the total production cost of this plan is (10) (10) = 8,500 and Total Holding Cost = 1.5(300) = 450 (assessed on inventory at end of month 2). Thus total cost is = $8, 950 = f 1 (0). 6. Define g t to be minimum net cost incurred from time t to end of problem (time t = beginning of year t + 1) given that a new machine has just been purchased at time t. g t includes the cost of buying the machine purchased at time t. Then we find that the machine should always be kept for one year and then traded in. 7. (a) Let f t (w) = minimum cost incurred in meeting demands for years t,... 5 given that (before hiring and firing for year t) w workers are available. Further, let: 37

38 h t = Workers hired at beginning of year t d t = Workers fired (discharged) at beginning of year t. w t = Workers required during year t. We assume that newly hired workers do not quit during their first year, and that salaries are paid to workers working any portion of a year. Then f t (w) = min {10, 000h t + 20, 000d t + 30, 000(w + h t d t ) + f t+1 (h t +.9(w d t ))}, where h t and d t must satisfy 0 h t, 0 d t w, and h t + w d t w t. We work back to compute f 1 (20). (b) Add the condition that if w < w t, f t (w) =. This will ensure that each month s requirement is met. Also note that d t must satisfy w d t w t. 8. Add 1 spare unit to each of the systems. The probability that all three systems will work is We choose to begin by putting a Type 2 item in the knapsack. This leaves us with an 8-3 = 5 pound knapsack so we put another Type 2 item in the knapsack. This leaves us with a 5-3 = 2 pound knapsack so we next put in a Type 3 item to fill the knapsack. Thus the knapsack should be filled with two Type 2 items and one Type 3 item. There are, of course, other optimal solutions such as using two Type 1 items. 38

39 Exercise session Transition matrix: P = /3 1/3 1/ /3 1/3 1/3 2 1/3 1/3 1/ /3 1/3 1/ /3 1/3 1/3 2. Transition matrix: P = /3 2/3 2 1/9 4/9 4/9 3. (a) State 4 is transient. (b) States 1, 2, 3, 5, 6 (c) {1,3,5} and{2,6} are closed sets. (d) No, because state 4 is transient. 4. (a) π 1 = 0.6, π 2 = 0.4. (b) π 1 = 16/25, π 2 = 1/5, π 3 = 4/25. (c) m 11 = 1.56, m 12 = 5, m 13 = 6.25, m 23 = 1.25, m 21 = 2.81, m 31 = 1.56, m 22 = 5, m 32 = 5, m 33 = If we do not replace a fair car we face the following transition matrix (note: we observe state at beginning of year before a car is replaced): P = G F BD G F BD The Expected Cost per Year is given by 1000(.64) (.25) (.11) = $1785. If we do replace a fair car, the Expected Net Cost Per Year is $1700, so the car should be replaced as soon as it becomes fair. 39

40 6. Let the state be the age of the machine at the beginning of the month. For example, if state is 3, the machine is 3 months old at the beginning of the month and starts its fourth month of operation. For Policy 1, the expected cost per month = $431.5, for Policy 2 the expected cost per month = $410.5, for Policy 3 the Expected Cost Per Month= $600 Thus Policy 2(replace after 2 months) is the cheapest policy. 7. Expected Cost/Month for policy 1 = $30, for policy 2 this is $35. Thus Policy 1 has a lower expected monthly cost. 8. (a) Element 1-1 of (I Q) 1 R =.705. (b) Element 2-2 of (I Q) 1 R =.30. (c) Sum of the elements in row 1 of (I Q) 1 = = (a) Sum of the elements in row 1 of (I Q) 1 = = 11.0 days. (b) Good Prediction = (.65) + 300(.50) + 200(.51) = 627 Fair Prediction = (.20) + 300(.30) + 200(.25) = 280 Critical Prediction = (.05) + 300(.12) + 200(.20) = 131 (c) Let G = steady state number in good condition, F = steady state in fair condition, C = steady state number in critical condition: F +.51C =.35G G +.25C =.70F G +.12F =.80C Solving we find G = F = 114 C = 47.7 (d) Entry 1-1 of (I Q) 1 R = (a) Let A = steady state number of gallons of A sold per week, B = steady state number of gallons of B sold per week, C = steady state number of gallons of C sold per week. Then A = 1, 000, 000 p B + p C p A p A + p B + p C B = 1, 000, 000 p A + p C p B p A + p B + p C 40

41 C = 1, 000, 000 p A + p B p C p A + p B + p C (b) Currently A s market share is ( )/( ) = 55.6%. If we cut percentage of defective juice cans in half p A =.05 and we find A s market share is now ( )/( )=.75. Annually improved quality increases profit by $10, 088, 000 = 52($10, 000)( ) > $1, 000, 000, so we should go for the improved quality. 41

42 Exercise session /7 4. (a) 5/6 (b) 25/6 (c) 1/2 minute 5. (a) 1.33 customers (b) 3 minutes (c) 16/81 6. (a) 2.45 (b) 14.8 cars/hour (c) 0.23 h 7. (a) 8/13-1/13-4/13 (b) 6/13 lines (c) 30/13 calls per hour will be lost 8. Let CNC = College ambulance is answering non-college call CC = College ambulance is answering college call CI = College ambulance is idle DNC = Downtown ambulance is answering non-college call DC = Downtown ambulance is answering college call DI = Downtown ambulance is idle Possible states (numbered 1-9) are State 1: (CNC, DNC) State 2: (CNC, DC) State 3: (CNC, DI) State 4: (CC, DNC) State 5: (CC, DC) State 6: (CC, DI) State 7: (CI, DNC) State 8: (CI, DC) State 9: (CI,DI) (a) 0.24 (b) (c) 8.1 % of all calls are considered lost to the system (d) We assume that half of the service time is the time required for an ambulance to go from its base to a patient. A total of ( )3 = 2.76 college calls per hour are handled. 3(π 3 + π 6 ) calls wait an average of 5/2 minutes and 3(π 7 +π 8 +π 9 ) calls wait an average of 4/2 minutes. Thus a College call waits an average of 3(π 3 + π 6 )2, 5 + 3(π 7 + π 8 + π 9 )2 =

43 2.08. In the same way we find that the average waiting time for a non-college call is 2.25 min. Thus, on average, a non-college call waits longer than a college call. 43