MGMT 372. (Updated: February 8, 2000, Homework 5. ² Problem 1, p. 260 (graph and solve). ² Problem 2, p. 260 (graph and solve).

Transcription

1 MGMT 372 (Updated: February 8, 2000, Homework 5 11:39 am) Problem 1, p 260 (graph and solve) Problem 2, p 260 (graph and solve) Problem 3, p 260 (graph and solve) Problem 4, p 260 (graph and solve) Problem 5, p 260 (graph and solve) Cover Page

2 MGMT 372 Solutions HW 5 (Suppl Ch 5) Page 1: Problem 1 Problem 1 (a) max z =4 +3 : ST Material: ; Labor: ; NN: ; 0: To graph the feasible region we plot the equations corresponding to each one of the constraints and nd their respective correct sides For each constraint we obtain two points of the equality Material: Note that (0,0) is on the correct side Labor: Note that (0,0) is on the correct side Material: 6 +4 =48 Labor: 4 +8 =80 Page 1: Problem 1

3 MGMT 372 Solutions HW 5 (Suppl Ch 5) Page 2: Problem 1 FIRST METHOD: \Total Enumeration" The rst method consists in enumerating all the extreme points and evaluating the objective function at each one of them; we then select the best one By solving the corresponding system of equations, we see that the four extreme points in this feasible region are (0; 0); (8; 0); (2; 9), and (0; ); and their corresponding objective function values are 0; 32; 35, and 30, respectively Therefore, the optimum is =2; = 9 with an objective function value of 35 Please note: Total enumeration works when the feasible region is bounded If the feasible region is unbounded, the LP may not have an optimal solution SECOND METHOD: \Objective Function Translation" Let z =12 Then4 +3 = 12, so two points of these equation are: We can now include the objective function value in the graph as follows z =12 By \moving" the objective function in a parallel way, we see that the two candidates for optimality are (0; 0) and the intersection of the equations 6 +4 =48and4 +8 = 80 The solution to this system of equations is =2; =9 Page 2: Problem 1

4 MGMT 372 Solutions HW 5 (Suppl Ch 5) Page 3: Problem z =12 z =0 z =35 We evaluate the objective function at these two points and we have that z =0for(0; 0), and z =35for (2; 9) Since we are maximizing, the optimum is =2; = 9, and the value of the objective function is 35 (b) max z =2 + : ST R: +4 40; S: ; T: ; NN ; 0: Similarly to part (a) above we obtain the feasible region by plotting each constraint Constraint R: Note that (0,0) is not on the correct side Page 3: Problem 1

5 MGMT 372 Solutions HW 5 (Suppl Ch 5) Page 4: Problem 1 Constraint S: Note that (0,0) is not on the correct side Constraint T: Note that (0,0) is not on the correct side FIRST METHOD: The three extreme points of this feasible region are the intersection of R and S; S and T; and R and T Solving the three systems of equations we nd that the points are (2:57; 3:57); (9; 2:5), and (1:5; 6:25), respectively Now we evaluate the objective function at each one of them The values are 4084, 4300, and 6550 respectively Hence, the optimum is =1:5; =6:25 with an objective function value of 6550, because we are maximizing Page 4: Problem 1

6 MGMT 372 Solutions HW 5 (Suppl Ch 5) Page 5: Problem 1 SECOND METHOD: Let z =50 Then2 + = 50, so two points of these equation are: We can now include the objective function in the graph as follows 15 z = By \moving" the objective function in a parallel way, we see that the two candidates for optimality are the intersection of R and S, and R and T 15 z =65:50 z =50 z =40: The two points to evaluate are (2:57; 3:57) and (1:5; 6:25) The values of the objective function are 4084 and 6550, respectively Thus, the optimum is =1:5; =6:25 with an objective function value of 6550, because we are maximizing Page 5: Problem 1

7 MGMT 372 Solutions HW 5 (Suppl Ch 5) Page 6: Problem 1 (c) max z =6A +3B: ST Material: 20A +6B 600; Machinery: 25A +20B 00; Labor: 20A +30B 1200; NN A; B 0: We need two points for each one of the equations corresponding to the constraints Material: A B Note that (0,0) is on the correct side Machinery: A B Note that (0,0) is on the correct side Labor: A B Note that (0,0) is on the correct side Page 6: Problem 1

8 MGMT 372 Solutions HW 5 (Suppl Ch 5) Page 7: Problem 1 Thefeasibleregionis: Material Machinery Labor FIRST METHOD: This feasible region has ve extreme points To nd them we solve the corresponding systems of equations and we obtain the points (0; 0); (30; 0); (24; 20); (17:14; 28:57), and (0; 40) Now we evaluate each of them using the objective function 6A +3B and we see that the corresponding values are 0; 180; 204; 188:55, and 120 Our goal is to maximize, so the optimum is A =24;B =20andthe value of the objective function is 204 SECOND METHOD: Let z =120 Then6A +3B = 120, so two points of these equation are: A B Page 7: Problem 1

9 MGMT 372 Solutions HW 5 (Suppl Ch 5) Page 8: Problem 1 The corresponding equation ts in the graph as follows: z =120 Now we can translate the objective function (keeping it always parallel ) and we see that the candidates for optimality are (0; 0) and (24; 20) z =120 z =0 z =204 Their corresponding objective function values are 0 and 204 Then, the optimum is A =24;B =20and the value of the objective function is 204 Page 8: Problem 1

10 MGMT 372 Solutions HW 5 (Suppl Ch 5) Page 9: Problem 2 Problem 2 (a) min z =1:80S +2:20T: ST Potassium: 5S +8T 200; Carbohydrate: 15S +6T 240; Protein: 4S +12T 180; T: T ; NN S; T 0: Once again we obtain two points for each one of the constraints: Potassium: S T Note that (0,0) is not on the correct side Carbohydrate: S T Note that (0,0) is not on the correct side Protein: S T Note that (0,0) is not on the correct side Page 9: Problem 2

11 MGMT 372 Solutions HW 5 (Suppl Ch 5) Page : Problem 2 T: S T 0 20 Note that (0,0) is not on the correct side Thefeasibleregionis: Carbohydrate Potassium Protein T S FIRST METHOD: This feasible region is unbounded and has three extreme points They are the intersection of T = and Potassium; Carbohydrate and Potassium; and Carbohydrate and the vertical axis Note also that the Protein constraint is redundant The points are (24; ); (8; 20), and (0; 40) Since we have an unbounded region we need to evaluate also a point with large coordinates in the directions of unboundedness, say (00; 00) The resulting objective function values are 6520, 5840, 8800, and 4,000, respectively Since the objective is to minimize, the optimum is S =8;T = 20 and the value of the objective function at optimality is 5840 SECOND METHOD: Let z =75 Then1:80S +2:20T = 75, so two points of these equation are: S T Page : Problem 2

12 MGMT 372 Solutions HW 5 (Suppl Ch 5) Page 11: Problem 2 We include the equation of the objective function in the graph z =75 Carbohydrate Potassium Protein T S Now we translate the line and realize that in one direction we nd the point (8; 20) (the intersection of `Carbohydrate' and `Potassium'), and in the other direction there is no limit, so we just take one point far away, say (500; 500) Carbohydrate Potassium Protein T S z =75 z =58:4 The value of the objective function for these two points is 5840 and 2,000 respectively Hence, for this minimization problem the optimum is S = 8;T = 20 with an objective function value of 5840 (b) min z =2 +3 : ST D: ; E: ; F: ; NN ; 0: Page 11: Problem 2

13 MGMT 372 Solutions HW 5 (Suppl Ch 5) Page 12: Problem 2 We need to plot a line for each one of the constraints, so we begin by nding two points for each one of them Constraint D: Note that (0,0) is not on the correct side Constraint E: Note that (0,0) is not on the correct side Constraint F: Note that (0,0) is on the correct side These constraints determine the following feasible region: D E F Page 12: Problem 2

14 MGMT 372 Solutions HW 5 (Suppl Ch 5) Page 13: Problem 2 FIRST METHOD: This feasible region has four extreme points We solve the corresponding systems of equations and see that they are (9; 0); (12; 0); (2:67; 4:67), and (4:2; 1:6) The value of the objective function for each one of these points is 1800, 2400, 1935, and 1320, respectively Since we are minimizing, the optimum is =4:2; =1:6 withanobjectivefunctionvalueof 1320 SECOND METHOD: Let z =18 Then2 +3 = We add this equation to the graph as follows D E F z =18 Page 13: Problem 2

15 MGMT 372 Solutions HW 5 (Suppl Ch 5) Page 14: Problem 2 And now we translate it in a parallel way in both directions D E F z =18 z =13:2 z =24 The two candidates for optimality are (12; 0) and the intersection of \D" and \E", (4:2; 1:6) Their respective objective function values are 2400 and 1320 Then, the optimum is =4:2; =1:6 andits objective function value 1320 Page 14: Problem 2

16 MGMT 372 Solutions HW 5 (Suppl Ch 5) Page 15: Problem 3 Problem 3 Formulation of the Linear Program - De nition of the variables: H = Number of model H microwave ovens to produce W = Number of model W microwave ovens to produce -Constraints Fabrication: 4H +2W 600; Assembly: 2H +6W 480; NN: H; W 0: - Objective function max z =40H +30W: Linear Program Solution - We begin by obtaining the feasible region Fabrication: H W Note that (0,0) is on the correct side Assembly: H W Page 15: Problem 3

17 MGMT 372 Solutions HW 5 (Suppl Ch 5) Page 16: Problem H W Fabrication Assembly Now we may start looking for the optimum FIRST METHOD: The four points obtained when solving the corresponding systems of equations are (0; 0); (150; 0); (132; 36), and (0; 80) We evaluate the objective function and nd the following pro ts: 0; 6,000; 6,360; and 2,400 Therefore the optimum is H =132andW = 36 and the value of the objective function is 6,360 SECOND METHOD: Let z =2; 400 Then 40H +30W =2; 400 Similarly to problems 1 and 2, we include this equation in the graph and translate it until it touches the last point in the feasible region in both directions The two candidates for optimality are (0; 0) and the intersection of fabrication and assembly, which is (132; 36) The value of the objective function for these two points is $0 and $6; 360 Hence, the optimum is H =132andW =36 Page 16: Problem 3

18 MGMT 372 Solutions HW 5 (Suppl Ch 5) Page 17: Problem H W Fabrication Assembly z =2; 400 z =0 z =6; 360 Page 17: Problem 3

19 MGMT 372 Solutions HW 5 (Suppl Ch 5) Page 18: Problem 4 Problem 4 We take the formulation from homework 4, which is: max z =1:70 +1:50 : ST Peanuts: ; Raisins: ; Maximum number of Deluxe: 1; Maximum number of Standard: 1; NN: ; 0: Now we obtain the feasible region like in the previous problems Raisins Peanuts Standard Deluxe To nd the optimum we can use either of the following two approaches FIRST METHOD: The six points obtained when solving the corresponding systems of equations are (0,0), (1,0), (1,3333), (90,60),(15,1), and (0,1) Its respective objective function values are 000, 18700, 23700, 24300, 19050, and Thus, the optimum is =90; = 60 The value of the objective function at optimality is $24300 Page 18: Problem 4

20 MGMT 372 Solutions HW 5 (Suppl Ch 5) Page 19: Problem 4 SECOND METHOD: Let z =170 Then1:7 +1:5 =170 Similarly to problems 1 and 2, we include this equation in the graph and translate it until it touches the last point in the feasible region in both directions The graph below shows that the two candidates for optimality are (0; 0) and the intersection of raisins and peanuts, which is (90; 60) The value of the objective function for these two points is $0 and $243 Hence, the optimum is =90and = z =0 z =170 z =243 Page 19: Problem 4

21 MGMT 372 Solutions HW 5 (Suppl Ch 5) Page 20: Problem 5 Problem 5 We take the formulation from homework 4, which is: max z =1:50 +1:20 : ST Sugar: ; 200; Flour: ; 0; Time: ; 600; NN: ; 0: The picture below shows the feasible region Time Flour Sugar Nowwe ndtheoptimum FIRST METHOD: The ve points obtained when solving the corresponding systems of equations are (0,0), (600,0), (500,200), (400,300), and (0,600) Their respective objective function values are 0, 900, 990, 960, and 720 Then, the optimum is = 500; = 200 The objective function value at optimality is $990 Page 20: Problem 5

22 MGMT 372 Solutions HW 5 (Suppl Ch 5) Page 21: Problem 5 SECOND METHOD: Let z =600 Then1:5 +1:2 =600 Again, we include this equation in the graph and translate it until it touches the last point in the feasible region in both directions z =600 z =0 z =990 The two candidates for optimality are (0; 0)andtheintersectionof ourandsugar,whichis(500; 200) The value of the objective function for these two points is $0 and $990 Thus the optimum is =500and =200 Page 21: Problem 5