56:171 Homework Exercises -- Fall 1993 Dennis Bricker Dept. of Industrial Engineering The University of Iowa

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1 56:7 Homework Exercises -- Fall 99 Dennis Bricker Dept. of Industrial Engineering The University of Iowa Homework # Matrix Algebra Review: The following problems are to be found in Chapter 2 of the text, Operations Research (2nd edition) by W. Winston: (.) Exercise # 4&6, p. 2 Use the Gauss-Jordan method to determine whether each of the following linear systems has no solution, a unique solution, or an infinite number of solutions. Indicate the solutions (if any exist). 2x - x 2 + x + x 4 = 6 x + x 2 + x = 4 2x 2 +2x = 4 x +2x 2 + x = 4 x 2 - x = (2.) Exercise #2, p. 6 Determine if the following set of vectors is independent or linearly dependent: V = 2, 2, (.) Exercise # 2, p. 42: Find A - (if it exists) for the following matrix: Linear Programming Model Formulation: Formulate a Linear Programming model for each problem below, and solve it using LINDO (available both on the Apollo workstations and Macintoshes of ICAEN.) Be sure to state precisely the definitions of your decision variables, and briefly explain the purpose of each type of constraint. State verbally the optimal solution. (All exercises are from Chapter of the text. For instructions on LINDO, see 4.7 and the appendix of chapter 4 of the text.) (4.) Exercise #2, page 98 (Furniture manufacturing) "Furnco manufactures tables and chairs. A table requires 4 board feet of wood, and a chair requires board feet of wood. Wood may be purchased at a cost of $ per board foot, and 4, board feet of wood are available for purchase. It takes 2 hours of skilled labor to manufacture an unfinished table or an unfinished chair. Three more hours of skilled labor will turn an unfinished table into a finished table, and 2 more hours of skilled labor will turn an unfinished chair into a finished chair. A total of 6 hours of skilled labor are available (and have already been paid for). All furniture produced can be sold at the following unit 56:7 HW Solutions '94 9/9/97 page

2 prices: unfinished table, $7; finished table, $4; unfinished chair, $6; finished chair, $. Formulate an LP that will maximize the contribution to profit from manufacturing tables and chairs." Homework # Solutions Matrix Algebra Review: The following problems are to be found in Chapter 2 of the text, Operations Research (2nd edition) by W. Winston: (.) Exercise # 4&6, p. 2 Use the Gauss-Jordan method to determine whether each of the following linear systems has no solution, a unique solution, or an infinite number of solutions. Indicate the solutions (if any exist). 2x - x 2 + x + x 4 = 6 (A) x + x 2 + x = 4 2x 2 +2x = 4 x +2x 2 + x = 4 (B) x 2 - x = Solution: System (A) has infinite number of solutions, since X = a, X 2 = b, X = 4 a b, X 4 = 2 a + 2b, where a R,b R. System (B) has unique solution X = X 2 = X =. (2.) Exercise #2, p. 6 Determine if the following set of vectors is independent or linearly dependent: 2 V =, 2, Solution: Let a,b,c R, we consider the following system 2 2a + b + c = a + b 2 + c =,i.e., a + 2b + c = c = The only solution is a=b=c=. Hence by the definition, we have that the set of vectors is independent. linearly (.) Exercise # 2, p. 42: Find A - (if it exists) for the following matrix: :7 HW Solutions '94 9/9/97 page 2

3 Solution: Hence A = Linear Programming Model Formulation: Formulate a Linear Programming model for each problem below, and solve it using LINDO (available both on the Apollo workstations and Macintoshes of ICAEN.) Be sure to state precisely the definitions of your decision variables, and briefly explain the purpose of each type of constraint. State verbally the optimal solution. (All exercises are from Chapter of the text. For instructions on LINDO, see 4.7 and the appendix of chapter 4 of the text.) (4.) Exercise #2, page 98 (Furniture manufacturing) "Furnco manufactures tables and chairs. A table requires 4 board feet of wood, and a chair requires board feet of wood. Wood may be purchased at a cost of $ per board foot, and 4, board feet of wood are available for purchase. It takes 2 hours of skilled labor to manufacture an unfinished table or an unfinished chair. Three more hours of skilled labor will turn an unfinished table into a finished table, and 2 more hours of skilled labor will turn an unfinished chair into a finished chair. A total of 6 hours of skilled labor are available (and have already been paid for). All furniture produced can be sold at the following unit prices: unfinished table, $7; finished table, $4; unfinished chair, $6; finished chair, $. Formulate an LP that will maximize the contribution to profit from manufacturing tables and chairs." Solution: Suppose we produce T: number of unfinished tables, T2: number of finished tables, C: number of unfinished chairs, C2: number of finished chairs. LP model : Max 7T+4T2 (profit for tables) +6C+C2 (profit for chairs) -4(T+T2) (cost for tables) -(C+C2) (cost for chairs) s.t. 2T+5T2+2C+4T2<=6 (available board feet of wood) 4T+4T2+C+C2<=4 (available working hours) T,T2,C,C2>= 56:7 HW Solutions '94 9/9/97 page

4 LINDO output : MAX T + T2 + C + 8 C2 SUBJECT TO 2) 2 T + 5 T2 + 2 C + 4 C2 <= 6 ) 4 T + 4 T2 + C + C2 <= 4 END : go LP OPTIMUM FOUND AT STEP 2 OBJECTIVE FUNCTION VALUE ) VARIABLE VALUE REDUCED COST T T C. 5. C2.74. ROW SLACK OR SURPLUS DUAL PRICES 2) ) NO. ITERATIONS= 2 DO RANGE(SENSITIVITY) ANALYSIS?? y RANGES IN WHICH THE BASIS IS UNCHANGED OBJ COEFFICIENT RANGES VARIABLE CURRENT ALLOWABLE ALLOWABLE COEF INCREASE DECREASE T INFINITY T INFINITY C. 5. INFINITY C2 8. INFINITY 5.4 RIGHTHAND SIDE RANGES ROW CURRENT ALLOWABLE ALLOWABLE RHS INCREASE DECREASE 2 6. INFINITY (.) Consider the following LP: Homework # 2 56:7 HW Solutions '94 9/9/97 page 4

5 Minimize z = x + x 2 s.t. x - 2x 2 2 x + x 2 x, x 2 a. Plot the feasible region of the LP, and identify the extreme points. b. Convert the constraints (excluding the nonnegativity constraints) into equations, by the use of "surplus" variables. c. How many basic variables will the set of two equality constraints have? d. How many basic solutions (both feasible and infeasible) will the two equality constraint have, i.e., how many ways may these basic variables be chosen from the set of four variables in the equations? e. Use the Big-M method to find the optimal solution. Plot the path which is taken on your graph in (a), and indicate which variables are basic at each basic solution along the path. f. Use the two-phase method to find the optimal solution. If the path followed to the optimum is different from that in (e), plot it, and again indicate which variables are basic at each basic solution along this path. (2.) Exercise #9, p. 6 Linear Programming Model Formulation: Formulate a Linear Programming model for each problem below, and solve it using LINDO (available both on the Apollo workstations and Macintoshes of ICAEN.) Be sure to state precisely the definitions of your decision variables, and briefly explain the purpose of each type of constraint. State verbally the optimal solution. (For instructions on LINDO, see 4.7 and the appendix of chapter 4 of the text.) Brady Corporation produces cabinets. Each week, they require 9, cu ft of processed lumber. They may obtain processed lumber in two ways. First, they may purchase lumber from an outside supplier and then dry it at their kiln. Second, they may chop down logs on their land, cut them into lumber at their sawmill, and finally dry the lumber at their kiln. Brady can purchase grade or grade 2 lumber. Grade lumber costs $ per cu ft and when dried yields.7 cu ft of useful lumber. Grade 2 lumber costs $7 per cu ft and when dried yields.9 cu ft of useful lumber. It costs the company $ per cu ft to chop down a log. After being cut and dried, one cubic foot of log yields.8 cu ft of lumber. Brady incurs costs of $4 per cu ft of lumber dried. It costs $2.5 per cu ft of logs sent through their sawmill. Each week, the sawmill can process up to 5, cu ft of lumber. Each week, up to 4, cu ft of grade lumber and up to 6, cu ft of grade 2 lumber can be purchased. Each week, 4 hours of time are available for drying lumber. the time it takes to dry cu ft of grade lumber, grade 2 lumber, or logs is as follows: grade : 2 seconds grade 2:.8 seconds log:. seconds Formulate an LP to help Brady minimize the weekly cost of meeting the demand for processed lumber. Homework #2 Solutions (.) Consider the following LP: Minimize z = x + x 2 s.t. x - 2x 2 2 x + x 2 x, x 2 a. Plot the feasible region of the LP, and identify the extreme points. b. Convert the constraints (excluding the nonnegativity constraints) into equations, by the use of "surplus" variables. c. How many basic variables will the set of two equality constraints have? 56:7 HW Solutions '94 9/9/97 page 5

6 d. How many basic solutions (both feasible and infeasible) will the two equality constraint have, i.e., how many ways may these basic variables be chosen from the set of four variables in the equations? e. Use the Big-M method to find the optimal solution. Plot the path which is taken on your graph in (a), and indicate which variables are basic at each basic solution along the path. f. Use the two-phase method to find the optimal solution. If the path followed to the optimum is different from that in (e), plot it, and again indicate which variables are basic at each basic solution along this path. Solution: (a). x2 (,) X-2X2 2 (,) (2,) (8/,/) (,) feasible region x (,-) X+X2 (b) Min s.t. X+X2 X-2X2-s=2 X+X2-s2= X,X2,s,s2 Fig. Feasible region for the problem. Here s and s2 are slack variables. (c). Two basic variables.(since we have two constraints) 4 (d). Because we have 4 variables and 2 constraints, hence we have 2 = 4! 2!(4 2)! = 6 basic solutions.(the extreme points in Figure ). Two of them are feasible and four of them are infeasible. (e) Big-M Method. Min (-Z)+X+X2+Ma+Ma2 s.t. X-2X2-s+a=2 X+X2-s2+a2= X,X2,s,s2,a,a2, where aand a2 are artificial varibles, and M is a very large scalar. The system can be rewritten as Min (-Z)+X+X2+M(2-X+2X2+s)+M(-X-X2+s2) s.t. X-2X2-s+a=2 X+X2-s2+a2= X,X2,s,s2,a,a2 Simplex tableau: 56:7 HW Solutions '94 9/9/97 page 6

7 -Z X -2M X2 +M -2 s M - s2 M - a a2 RHS -5M 2 basic vars -Z=-5M a=2 a2= -Z X X2 -M+7-2 s -M+ - s2 M - a 2M- - a2 RHS -M-6 2 basic vars -Z=-M-6 X=2 a2= -Z X X2 s s2 a a2 RHS 2/ 7/ M-(2/) -25/ -/ / -2/ -/ * * * * 8/ / Because all reduced costs are nonnegative, the solution is optimal. The path is (,) ---> (2,) ---> (8/,/) (see Fig ). basic vars -Z=-25/ X=8/ X2=/ (f). Two Phase Method. Min w=a+a2 s.t. (-Z)+X+X2= X-2X2-s+a=2 X+X2-s2+a2= X,X2,s,s2,a,a2 -W -Z X Phase (I) we want to minimize W. -W -Z X -2 X2-2 X2-2 s - s - s2 - s2 - a a a2 a2 RHS 2 RHS :7 HW Solutions '94 9/9/97 page 7

8 -W -Z X -2 X2-2 s - s2 - a a2 RHS W -Z X X s - - s2 - a a2 RHS W -Z X X2 s 2/ -/ / s2 7/ -2/ -/ a * * * a2 * * * RHS -25/ 8/ / All artifical variables are now non-basic, we can drop the phase one objective row and the variables from the tableau. -Z X X2 s s2 RHS 2/ 7/ -25/ -/ -2/ 8/ / -/ / artifical Note that this is optimal. The path is (,) -->(2,) -->(8/,/).(see Fig.). (2.) Exercise #9, p. 6 Linear Programming Model Formulation: Formulate a Linear Programming model for each problem below, and solve it using LINDO (available both on the Apollo workstations and Macintoshes of ICAEN.) Be sure to state precisely the definitions of your decision variables, and briefly explain the purpose of each type of constraint. State verbally the optimal solution. (For instructions on LINDO, see 4.7 and the appendix of chapter 4 of the text.) Brady Corporation produces cabinets. Each week, they require 9, cu ft of processed lumber. They may obtain processed lumber in two ways. First, they may purchase lumber from an outside supplier and then dry it at their kiln. Second, they may chop down logs on their land, cut them into lumber at their sawmill, and finally dry the lumber at their kiln. Brady can purchase grade or grade 2 lumber. Grade lumber costs $ per cu ft and when dried yields.7 cu ft of useful lumber. Grade 2 lumber costs $7 per cu ft and when dried yields.9 cu ft of useful lumber. It costs the company $ per cu ft to chop down a log. After being cut and dried, one cubic foot of log yields.8 cu ft of lumber. Brady incurs costs of $4 per cu ft of lumber dried. 56:7 HW Solutions '94 9/9/97 page 8

9 It costs $2.5 per cu ft of logs sent through their sawmill. Each week, the sawmill can process up to 5, cu ft of lumber. Each week, up to 4, cu ft of grade lumber and up to 6, cu ft of grade 2 lumber can be purchased. Each week, 4 hours of time are available for drying lumber. the time it takes to dry cu ft of grade lumber, grade 2 lumber, or logs is as follows: grade : 2 seconds grade 2:.8 seconds log:. seconds Formulate an LP to help Brady minimize the weekly cost of meeting the demand for processed lumber. Solution: Let G be the # of cu ft of grade lumber by purchasing, G2 be the # of cu ft of grade 2 lumber by purchasing, LOG be the # of cu ft of the coperation's own lumber. LP Model : Min G+7G2 (purchase cost) +4(G+G2+LOG) (dry cost for lumber) +LOG (cost for chopping) +2.5LOG (cost for sawmill) s.t. G 4 (available cu ft of grade per week) G2 6 (available cu ft of grade 2 per week) LOG 5 (available cu ft of own lumber per week).7g+.9g2+.8log 9 (constraint for demand) 2G+.8G2+.LOG 44 (available hours for drying) G,G2,LOG LINDO output : MIN 7 G + G LOG SUBJECT TO G <= 4 G2 <= 6 LOG<= 5.7 G +.9 G2 +.8 LOG >= 9 2 G +.8 G2 +. LOG <= 44 END : go LP OPTIMUM FOUND AT STEP 4 OBJECTIVE FUNCTION VALUE ) VARIABLE VALUE REDUCED COST G 4.. G LOG ROW SLACK OR SURPLUS DUAL PRICES 2) ) ) ) :7 HW Solutions '94 9/9/97 page 9

10 6) NO. ITERATIONS= 4 DO RANGE(SENSITIVITY) ANALYSIS?? y RANGES IN WHICH THE BASIS IS UNCHANGED OBJ COEFFICIENT RANGES VARIABLE CURRENT ALLOWABLE ALLOWABLE COEF INCREASE DECREASE G INFINITY G LOG RIGHTHAND SIDE RANGES ROW CURRENT ALLOWABLE ALLOWABLE RHS INCREASE DECREASE INFINITY INFINITY Homework #. For each of the tableaus below, the objective is to be maximized, and the variable in the first column is (- z). Is the tableau optimal? If optimal, is the optimum unique? If not optimal, circle every possible pivot location which might be selected by the simplex method. Indicate if an unbounded objective is detected. (a) (b) (c) (d) :7 HW Solutions '94 9/9/97 page

11 (e) (f) For each of the tableaus below, the objective is to be minimized, and the variable in the first column is (- z). Is the tableau optimal? If optimal, is the optimum unique? If not optimal, circle every possible pivot location which might be selected by the simplex method. Indicate if an unbounded objective is detected. (g) (h) (i) (j) (k) (l) :7 HW Solutions '94 9/9/97 page

12 Consider the following LP in its initial tableau: Without pivoting in the tableau as in the ordinary simplex method, find the missing values in the following tableau which occurs after several iterations, as in the revised simplex metod: 4-2?? -7? 2-4? 5? - 2?? 8? -5 7? (Explain briefly how you determine the missing values.) 4. Consider the following problem: "A manufacturer produces two types of plastic cladding. These have the trade names Ankalor and Beslite. One yard of Ankalor requires 8 lb of polyamine, 2.5 lb of diurethane and 2 lb of monomer. A yard of Beslite needs lb of polyamine, lb of diurethane, and 4 lb of monomer. The company has in stock 8, lb of polyamine, 2, lb of diurethane, and, lb of monomer. Both plastics can be produced by alternate parameter settings of the production plant, which is able to produce sheeting at the rate of 2 yards per hour. A total of 75 production plant hours are available for the next planning period. The contribution to profit on Ankalor is $/yard and $2/yard on Beslite. The company has a contract to deliver at least, yards of Ankalor. What production plan should be implemented in order to maximize the contribution to the firm's profit from this product division." Decision variables: A = Number of yards of Ankalor produced B = Number of yards of Beslite produced LP model: Maximize A + 2 B subject to 8 A + B 8, (Polyamine) 2.5 A + B 2, (Diurethane) 2 A + 4 B, (Monomer) A + B 9, (Plant capacity) A, (Contract) A, B The LINDO solution is:? GO LP OPTIMUM FOUND AT STEP 2 OBJECTIVE FUNCTION VALUE ) 42. VARIABLE VALUE REDUCED COST A.. B 56.. ROW SLACK OR SURPLUS DUAL PRICES 2). 2. ) ) :7 HW Solutions '94 9/9/97 page 2

13 5) 4.. 6). -6. NO. ITERATIONS =2 RANGES IN WHCH THE BASIS IS UNCHANGED OBJ COEFFICIENT RANGES VARIABLE CURRENT ALLOWABLE ALLOWABLE COEF INCREASE DECREASE A. 6. INFINITY B 2. INFINITY 7.5 RIGHTHAND SIDE RANGES ROW CURRENT ALLOWABLE ALLOWABLE RHS INCREASE DECREASE INFINITY INFINITY INFINITY a. How many yards of each product should be manufactured? b. How much of each raw material will be used, and how much will be unused? c. Suppose that the company can purchase 2 pounds of additional polyamine for $.5 per pound. Should they make the purchase? Why or why not? d. If the profit contribution from Ankalor were to decrease to $8/yard, will the optimal solution change? Why or why not? e. If the profit contribution from Beslite were to increase to $25/yard, will the optimal solution change? Why or why not? f. If the company could deliver less than the contracted amount of Ankalor by forfeiting a penalty of $8/yard, should they do so? If so, how much should they deliver? Homework # Solutions. For each of the tableaus below, the objective is to be maximized, and the variable in the first column is (- z). Is the tableau optimal? If optimal, is the optimum unique? If not optimal, circle every possible pivot location which might be selected by the simplex method. Indicate if an unbounded objective is detected. (a) (b) Not optimal Optimal but not unique. Since the variable of column 8 can enter to the basic without changing the objective value. 56:7 HW Solutions '94 9/9/97 page

14 (c) (d) Not optimal. (e) (f) Not optimal. Unbounded. Not optimal. Not optimal. 2. For each of the tableaus below, the objective is to be minimized, and the variable in the first column is (- z). Is the tableau optimal? If optimal, is the optimum unique? If not optimal, circle every possible pivot location which might be selected by the simplex method. Indicate if an unbounded objective is detected. (g) (h) (i) (j) Not optimal. Not optimal. Not optimal. 56:7 HW Solutions '94 9/9/97 page 4

15 (k) (l) Not optimal. Not optimal. Not optimal.. Consider the following LP in its initial tableau: Without pivoting in the tableau as in the ordinary simplex method, find the missing values in the following tableau which occurs after several iterations, as in the revised simplex method: 4-2?? -7? 2-4? 5? - 2?? 8? -5 7? (Explain briefly how you determine the missing values.) Solution: 2 2 B={2,,6}, A B = 2 (A B ) = A = 2 2 b=[4,8 5] T 7 2 Thus, (A B ) A = 5 2, (A B ) b = 2 8 = 2, and :7 HW Solutions '94 9/9/97 page 5

16 [ ] 2 4 C B ( A B ) b = = Therefore the table becomes: Consider the following problem: "A manufacturer produces two types of plastic cladding. These have the trade names Ankalor and Beslite. One yard of Ankalor requires 8 lb of polyamine, 2.5 lb of diurethane and 2 lb of monomer. A yard of Beslite needs lb of polyamine, lb of diurethane, and 4 lb of monomer. The company has in stock 8, lb of polyamine, 2, lb of diurethane, and, lb of monomer. Both plastics can be produced by alternate parameter settings of the production plant, which is able to produce sheeting at the rate of 2 yards per hour. A total of 75 production plant hours are available for the next planning period. The contribution to profit on Ankalor is $/yard and $2/yard on Beslite. The company has a contract to deliver at least, yards of Ankalor. What production plan should be implemented in order to maximize the contribution to the firm's profit from this product division." Decision variables: A = Number of yards of Ankalor produced B = Number of yards of Beslite produced LP model: Maximize A + 2 B subject to 8 A + B 8, (Polyamine) 2.5 A + B 2, (Diurethane) 2 A + 4 B, (Monomer) A + B 9, (Plant capacity) A, (Contract) A, B The LINDO solution is:? GO LP OPTIMUM FOUND AT STEP 2 OBJECTIVE FUNCTION VALUE ) 42. VARIABLE VALUE REDUCED COST A.. B 56.. ROW SLACK OR SURPLUS DUAL PRICES 2). 2. ) ) 6.. 5) 4.. 6). -6. NO. ITERATIONS =2 RANGES IN WHCH THE BASIS IS UNCHANGED OBJ COEFFICIENT RANGES VARIABLE CURRENT ALLOWABLE ALLOWABLE COEF INCREASE DECREASE A. 6. INFINITY 56:7 HW Solutions '94 9/9/97 page 6

17 B 2. INFINITY 7.5 RIGHTHAND SIDE RANGES ROW CURRENT ALLOWABLE ALLOWABLE RHS INCREASE DECREASE INFINITY INFINITY INFINITY a. How many yards of each product should be manufactured? Produce Ankalor yards and Beslite 56 yards. b. How much of each raw material will be used, and how much will be unused? Polyamine will be used 8 lb, diurethane will be used lb. and 69 lb. unused, and monomer will used 284 lb. and 6 lb. unused. c. Suppose that the company can purchase 2 pounds of additional polyamine for $.5 per pound. Should they make the purchase? Yes Why or why not? Since () the allowable increase for polyamine is 4, and (2) $.5<$2(dual price). d. If the profit contribution from Ankalor were to decrease to $8/yard, will the optimal solution change? NO. Why or why not? Since the allowable decrease of objective coefficient for Ankalor is infinity. e. If the profit contribution from Beslite were to increase to $25/yard, will the optimal solution change? NO. Why or why not? Because the allowable increase of objective coefficient for Beslite is infinity. f. If the company could deliver less than the contracted amount of Ankalor by forfeiting a penalty of $8/yard, should they do so? NO If so, how much should they deliver? Not deliver. Because $8(penalty)>$6(see dual price). Consider the following LP: Max X - X 2 subject to 2X + X 2 5 Homework # 4 X - 4X 2 6 X, X 2 a. Plot the feasible region of the LP. b. Identify all of the basic solutions of the problem. Which are feasible and which are infeasible? c. Write the dual of this LP. d. Plot the feasible region of the dual LP. e. Identify all of the basic solutions of the dual LP. Which are feasible and which are infeasible? f. Check that the optimal solutions of the primal and dual LP satisfy the complementary slackness conditions. 2. Consider the following four primal LPs: 56:7 HW Solutions '94 9/9/97 page 7

18 P A : Max 2X + X 2 subject to X + X 2 4 X - X 2 2 X, X 2 P B : Maximize 2X + X 2 subject to X - X 2 4 X - X 2 2 X, X 2 P C : Max 2X + X 2 subject to X + X 2 4 X - X 2 2 X, X 2 P D : Maximize 2X + X 2 subject to X - X 2 4 X - X 2 2 X, X 2 a. Write the dual LP for each of P A, P B, P C, and P D. (Label them D A, etc.) b. Sketch the feasible region (if it exists) for each LP. c. Classify each primal-dual pair according to the categories: i. Primal & Dual both feasible ii. Primal infeasible, Dual feasible & unbounded iii. Primal feasible and unbounded, Dual infeasible iv. Both Primal & Dual infeasible.. Find the dual LP for the LP: Minimize X + 2X 2 - X + 4X 4 subject to X - 2X 2 + X + 4X 4 X 2 + X + 4X 4-5 2X -X 2-7X -4X 4 = 2 X, X 4 4. The following questions refer to the LP for "Production Planning for a Tire Manufacturer" in your notes: a. A Wheeling machine breaks down in June, and 8 hours of production time are lost before the machine is repaired. How much will this increase the objective function? How should the production plan be modified, according to the "substitution rates"? b. How many Nylon tires will be in storage at the end of July? If the storage cost for Nylon tires in July were to drop from per tire to 5 per tire, how many tires should be put into storage at the end of July? c. In the optimal solution, no fiberglass tires are manufactured on the Regal machine in June. Suppose that the production manager mistakenly has such tires manufactured on the Regal machine. How much will this error increase the cost function? How should the production plan be modified to best compensate for this error? Homework #4 Solutions. Consider the following LP: Max X - X 2 subject to 2X + X 2 5 X - 4X 2 6 X, X 2 a. Plot the feasible region of the LP. b. Identify all of the basic solutions of the problem. Which are feasible and which are infeasible? c. Write the dual of this LP. d. Plot the feasible region of the dual LP. 56:7 HW Solutions '94 9/9/97 page 8

19 e. Identify all of the basic solutions of the dual LP. Which are feasible and which are infeasible? f. Check that the optimal solutions of the primal and dual LP satisfy the complementary slackness conditions. Solution: (a). feasible region x2 X-4X2 6 (,5/) (,) (5/2,) (6/,) x (*,**) 2X+X2 5 (,-4) Fig.Feasible region for the problem. (b). There are six basic variables (see Fig ). (,5/), (5/2,), and (6/,) are feasible, the other three ( (,-4), (,) and (*,**) above) are infeasible. (c). (Dual) Min 5Y+6Y2 s.t.. 2Y+Y2 Y-4Y2 - Y, Y2 56:7 HW Solutions '94 9/9/97 page 9

20 (d). Y2 Y-4Y2 - A B C feasible region D E F Y 2Y+Y2 Fig 2 Feasible region for dual system. There are six basic solutions for dual, points A, B,..., and F in Fig 2. But only three are feasible, i.e. points A, B, and C. (e). The optimal solution for primal problem is (X*,X2*)=(6/,), and the optimal solution for dual is (Y*,Y2*)=(,), i.e., point C in Fig. 2. () X*>. Constraint in dual is tight, i.e., 2Y+Y2=2()+()=. (2) Y2*>. Constraint 2 in primal is tight, i.e., X-4X2-6=(6/)-4()-6=. () Constraint 2 in dual is slack, i.e., Y-4Y2=()-4()=-4>-. We have X2*=. (4) Constraint in primal is slack, i.e., 2X+X2-5=2(6/)+()-5>. We have Y*=. 2. Consider the following four primal LPs: P A : Max 2X + X 2 subject to X + X 2 4 X - X 2 2 X, X 2 P B : Maximize 2X + X 2 subject to X - X 2 4 X - X 2 2 X, X 2 P C : Max 2X + X 2 subject to X + X 2 4 X - X 2 2 X, X 2 P D : Maximize 2X + X 2 subject to X - X 2 4 X - X 2 2 X, X 2 a. Write the dual LP for each of P A, P B, P C, and P D. (Label them D A, etc.) b. Sketch the feasible region (if it exists) for each LP. c. Classify each primal-dual pair according to the categories: i. Primal & Dual both feasible ii. Primal infeasible, Dual feasible & unbounded iii. Primal feasible and unbounded, Dual infeasible iv. Both Primal & Dual infeasible. Solution: (a). (D A ) Min 4Y+2Y2 (D B ) Min 4Y+2Y2 s.t. Y+Y2 2 s.t. Y+Y2 2 Y-Y2 -Y-Y2 Y, Y2. Y, Y2. 56:7 HW Solutions '94 9/9/97 page 2

21 (D C ) Min 4Y+2Y2 (D D ) Min 4Y+2Y2 s.t. Y+Y2 2 s.t. Y+Y2 2 Y-Y2 -Y-Y2 Y, Y2. Y, Y2. (b & c) Feasible regions for primal and dual for each problem. (P A ) Primal is feasble. (D A ) No feasible region. X2 Y2 feasible region Y-Y2 Y X Y+Y2 2 (P B ) No feasible region. (D B ) No feasible region. X2 Y2 X Y Y+Y2 2 -Y-Y2 (P C ) Primal is feasible. (D C ) Dual is feasible. X2 Y2 Y-Y2 X-X2 2 X Y X+X2 4 Y+Y2 2 (P D ) Primal is feasible. (D D ) No feasible region. 56:7 HW Solutions '94 9/9/97 page 2

22 X2 Y2 X-X2 2 X-X2 4 X Y -Y-Y2 Y+Y2 2. Find the dual LP for the LP: Minimize X + 2X 2 - X + 4X 4 subject to X - 2X 2 + X + 4X 4 X 2 + X + 4X 4-5 2X -X 2-7X -4X 4 = 2 X, X 4 Solution: Max Y-5Y2+2Y s.t. Y+2Y -2Y+Y2-Y=2 Y+Y2-7Y=- 4Y+4Y2-4Y 4 Y, Y2, Y no restriction in sign. 4. The following questions refer to the LP for "Production Planning for a Tire Manufacturer" in your notes: a. A Wheeling machine breaks down in June, and 8 hours of production time are lost before the machine is repaired. How much will this increase the objective function? How should the production plan be modified, according to the "substitution rates"? b. How many Nylon tires will be in storage at the end of July? If the storage cost for Nylon tires in July were to drop from per tire to 5 per tire, how many tires should be put into storage at the end of July? c. In the optimal solution, no fiberglass tires are manufactured on the Regal machine in June. Suppose that the production manager mistakenly has such tires manufactured on the Regal machine. How much will this error increase the cost function? How should the production plan be modified to best compensate for this error? Solution: (a).from the tableau, one can find the relation between SLK2 and RHS as : 56:7 HW Solutions '94 9/9/97 page 22

23 ART WN IN SLK WG RN WN SLK7 = SLK2 RN WG IG 25.. RN.. WG 5.. SLK2 will increase 8 hours if we lose 8 hours capacity in Wheeling machine in June. Hence, the objective value will increase 8(.)=2.667, WN will decrease 8(6.667), RN will increase 8(6.667), and SLK will decrease 8(.67). (b). (). From the tableau we can see that there will be no tires stored at the end of July. (2). Since allowable decrease of the objective cofficient range for IN2 is $.2, the optimal solution will not change, i.e., no tire will be stored at the end of July when cost drops from cents per tire to 5 cents per tire. (c). ART WN IN SLK WG RN WN SLK7 = RG RN WG IG 25.. RN.. WG 5.. It is obvious that increasing units of RG will increase (.6)=.6 in objective value. At this time, WN will increase (.8)=8 units, RN will decrease (.8)=8 units, and WG will decrease ()= units. Homework # 5. Consider the transportation problem having the following cost and requirements table: Destination 56:7 HW Solutions '94 9/9/97 page 2

24 2 4 Supply Source Demand 4 a. Write the regular LP formulation of this problem. b. How many basic variables will the LP have (in addition to -z, the objective value)? c. Use each of the following methods to obtain an initial basic feasible solution: i) Northwest Corner Method ii) Least-Cost Method iii) Vogel's Approximation Method d. For each of the basic feasible solutions you found in (c), apply the transportation simplex method to find the optimal solution. e. Compare the number of iterations required starting at each solution obtained in (c). f. What are the values of the dual variables (simplex multipliers) for the optimal solution? 2. A manufacturer of electronic calculators is working on setting up his production plans for the next six months. One product is particularly puzzling to him. The orders on hand for the coming season are: Month Orders January February 5 March 2 April May 2 June 5 The product will be discontinued after satisfying the June demand. Therefore, there is no need to keep any inventory after June. The production cost, using regular manpower, is $ per unit. Producing the calculator on overtime costs an additional $2 per unit. The inventory-carrying cost is $.5 per unit per month. The regular shift production is limited to units per month, and overtime production is limited to an additional 75 units per month. In every month except June, backorders are allowed, i.e. demand can be shipped one month late, but there is an estimated $ per unit loss, including the loss of the customer's "good will". (Orders cannot be satisfied more than month late.) Set up, but do not solve, the transportation tableau which could be used to find the optimal production schedule. Homework #5 Solutions. Consider the transportation problem having the following cost and requirements table: Destination 2 4 Supply Supply Demand 4 a. Write the regular LP formulation of this problem. b. How many basic variables will the LP have (in addition to -z, the objective value)? c. Use each of the following methods to obtain an initial basic feasible solution: i) Northwest Corner Method ii) Least-Cost Method iii) Vogel's Approximation Method d. For each of the basic feasible solutions you found in (c), apply the transportation simplex method to find the optimal solution. 56:7 HW Solutions '94 9/9/97 page 24

25 e. Compare the number of iterations required starting at each solution obtained in (c). f. What are the values of the dual variables (simplex multipliers) for the optimal solution? Solutions: (a) Since the total supply is greater than total demand, we add a dummy column with demand (5+2+)-(+4++)=. Destination 2 4 dummy 5 Supply Let Xij be the number of units transported from supply i to destination j. Min X+7X2+6X+4X4+2X2+4X22+X2+2X24+4X+X2+8X+5X4 s.t. X+X2+X+X4+X5= X2+X22+X2+X24+X25= X+X2+X+X4+X5= X+X2+X= X2+X22+X2=4 X+X2+X= X4+X24+X4= X5+X25+X5= All Xij. (b). m+n-=+5-=7, i.e., seven basic variables. (c). i) Northwest Corner Method: Supply 2 Destination 2 4 dummy X=, X2=2, X22=2, X=, X4=, X5=, X25=. Total cost=44. ii) Least-Cost Method: :7 HW Solutions '94 9/9/97 page 25

26 Supply 2 Destination 2 4 dummy X=X2=X=X4=X5=, X2=2, X2=. Total cost=. iii) Vogel's Approximation Method: Supply 2 Destination X=, X2=2, X2=X24=, X2=2, X5=, X25=. Total cost=4. (d).i) Northwest Corner Method: X25 out of basic. u\v u\v dummy θ θ θ +θ θ +θ θ θ X out of basic. reduced cost: c= c4= c5=2 c2=2 c24=2 c25=5 c=- c2=-6<< reduced cost: c=-5 c4=-4 c5=- c2=2 c2=-5 <<---- c24=- c=4 c2=- 56:7 HW Solutions '94 9/9/97 page 26

27 u\v 7 2 θ θ θ θ X4 out of basis. u\v - -4 X5 out of basis. u\v θ θ θ θ reduced cost: c= c4=-5<<---- c5=-4 c2=2 c24=-4 c25=- c=5 c=6 reduced cost: c= c5=-4<<---- c2=2 c24= c25=- c=5 c=6 c4=5 reduced cost: c= c2> c24> c25> - c> c> -4 c4> c5> Since all reduced costs for non-basic variables are nonnegative, it is an optimal solution. Total cost=29. ii) Least-Cost Method: 4 X out of basis. 4 X2 out of basis θ θ θ θ θ θ θ θ reduced cost: c22=-2 c2=-2<<---- c24=- c25= c=5 c=6 c4=5 c5= reduced cost: c=2 c22=-2<<---- c24=- c25= c=5 c=8 c4=5 c5=4 56:7 HW Solutions '94 9/9/97 page 27

28 2 X2 out of basis θ θ θ θ reduced cost: c2=2 c=2 c24=-<<---- c25= c= c=6 c4= c5=2 reduced cost: c2> c> c2> c25> c> c> c4> c5> Since all reduced costs for non-basic variables are nonnegative, it is an optimal solution. Total cost=29. iii). Vogel's Approximation Method 4 4 X5 out of basis. 4 X25 out of basis. 4 X24 out of basis θ θ θ θ θ +θ θ θ θ θ θ θ reduced cost: c=- c4=-2 c5=-4<<---- c2= c22= c=5 c=5 c4= reduced cost: c= c4=2 c2=- c22=-<<---- c=5 c=9 c4=7 c5=4 reduced cost: c= c4=-<<---- c2=2 c25= c=5 c=6 c4=4 c5=4 56:7 HW Solutions '94 9/9/97 page 28

29 reduced cost: c= c2> c24> c25> c> c> c4> c5> Since all reduced costs for non-basic variables are nonnegative, it is an optimal solution. Total cost=29. (e). Northwest Corner Method iterations. Least-Cost Method iterations. Vogel's Approximation Method iterations. (f) The values of dual variables are shown on the above optimal matrix (u and v). 2. A manufacturer of electronic calculators is working on setting up his production plans for the next six months. One product is particularly puzzling to him. The orders on hand for the coming season are: Month Orders January February 5 March 2 April May 2 June 5 The product will be discontinued after satisfying the June demand. Therefore, there is no need to keep any inventory after June. The production cost, using regular manpower, is $ per unit. Producing the calculator on overtime costs an additional $2 per unit. The inventory-carrying cost is $.5 per unit per month. The regular shift production is limited to units per month, and overtime production is limited to an additional 75 units per month. Backorders are allowed, in that demand can be shipped one month late, but there is an estimated $ per unit loss, including the loss of the customer's "good will". Set up but do not solve the transportation tableau which could be used to find the optimal production schedule. Solution: Since the total supply is greater than the total demand, we add a dummy column with capacity=5 (total supply-total demand=5). The cost for the dummy column is zero. 56:7 HW Solutions '94 9/9/97 page 29

30 JAN FEB MAR APRIL MAY JUNE Dummy Supply JAN RT OT FEB RT OT MAR RT.5.5 OT APRIL RT.5 OT MAY RT.5 OT JUNE RT OT Demand Total 5 Note: RT means regular time, OT means overtime. Homework # 6. Consider the assignment problem having the following cost matrix: \ 2 4 A: 4 B: 4 C: 2 D: 2 2 a. Consider this as a transportation problem. What should be the supplies and demands? b. Use the "Least Cost Rule" or Vogel's Approximation Method to get an initial basic feasible solution. Be sure to indicate which variables are basic. (This may require some arbitrary choices.) c. Starting with the solution indicated in (b), apply the transportation simplex method to find the optimal solution. How many iterations, i.e., changes of basis, were required? In how many iterations was the objective function improved? d. Apply the Hungarian method to the same problem. How many iterations were required? 2. Three new automatic feed devices have been made available for existing punch presses. Six presses in the plant can be fitted with this equipment. The plant superintendent estimates that the increased output, together with the labor saved, will result in the following dollar increase in profits per day: Punch Press \ A B C D E F Device : : : a. Convert this assignment problem with an equal number of rows and columns. b. Convert the objective from maximization to minimization by minimizing the negative of the increased profits. 56:7 HW Solutions '94 9/9/97 page

31 c. How can the cost matrix be modified so that the costs are nonnegative as required in the Hungarian method? (What happens if you perform row reduction on a row?) d. Find the assignment of the feed devices to the punch presses which will yield the greatest increase in profits. Homework #6 Solutions. Consider the assignment problem having the following cost matrix: \ 2 4 A: 4 B: 4 C: 2 D: 2 2 a. Consider this as a transportation problem. What should be the supplies and demands? b. Use the "Least Cost Rule" or Vogel's Approximation Method to get an initial basic feasible solution. Be sure to indicate which variables are basic. (This may require some arbitrary choices.) c. Starting with the solution indicated in (b), apply the transportation simplex method to find the optimal solution. How many iterations, i.e., changes of basis, were required? In how many iterations was the objective function improved? d. Apply the Hungarian method to the same problem. How many iterations were required? Solution : a. The supplies and demands should all be one. b. Using the Least Cost Rule,we obtain the following initial basic feasible solution. 2 4 A 4 B 4 C D Note that this is one of several feasible solutions which might be found by Least Cost Rule. The number of basic variables is 7 (=m+n-), and so three zero shipments must be basic. Suppose that we select X B2, X C, and X D (indicated by the in the boxes above). Next we calculate the reduced costs for all nonbasic variables. 56:7 HW Solutions '94 9/9/97 page

32 u\v A B 2 C D reduced cost for non-basic vars: CA=5 CA2= CA4=4 CB=2 CB4= CC= CC4=5 CD2=-2 CD= Since reduced cost for D2(row D, column 2) is negative, it will be chosen to be basic variable. 2 4 u\v - - A B 2 4 +θ θ 4 C D 2 θ θ 2 2 It is easy to find that θ=. We arbitrarily choose B2 to be removed from the basic set and then have the following. (Note we could instead choose D instead of B2 to be removed.) A B C D u\v reduced cost for non-basic vars: CA= CA2= CA4=2 CB2=2 CB=4 CB4= CC= CC4= CD=2 Because reduced costs of all non-basics are nonnegative, so it is optimal. We used one iteration, but obtained no improvement in objective. Total cost=. d. First we use row reduction and obtain the following cost matrix. 56:7 HW Solutions '94 9/9/97 page 2

33 2 4 A 4 B 4 C 2 2 D 2 2 Then by using column reduction in above cost matrix, we obtain the following cost matrix. 2 4 A B 2 4 C 2 D It is obvious that four lines are needed to cover all zeros in above cost matrix. Hence we stop and can assign as A->2,B->,C->,D->4 or A->,B->,C->2,D->4. Total cost=. 2. Three new automatic feed devices have been made available for existing punch presses. Six presses in the plant can be fitted with this equipment. The plant superintendent estimates that the increased output, together with the labor saved, will result in the following dollar increase in profits per day: Punch Press \ A B C D E F Device : : : a. Convert this assignment problem with an equal number of rows and columns. b. Convert the objective from maximization to minimization by minimizing the negative of the increased profits. c. How can the cost matrix be modified so that the costs are nonnegative as required in the Hungarian method? (What happens if you perform row reduction on a row?) d. Find the assignment of the feed devices to the punch presses which will yield the greatest increase in profits. Solution :a. 56:7 HW Solutions '94 9/9/97 page

34 A B C D E F dummy dummy dummy 56:7 HW Solutions '94 9/9/97 page 4

35 b. A B C D E F dummy dummy dummy c. Subtract min{-22,-7,-22,...,-7}=-2 from all the elements of above matrix, we obtain the following matrix, which has nonnegative costs: A B C D E F 2 dummy dummy dummy d. Applying the row reduction of Hungarian method, we can obtain the following table. A B C D E F dummy dummy dummy Since we need four lines to cover all zeros in matrix, so we stop and can assign either --->A, 2 --->D, --->E, or --->C, 2 --->D, --->E. 56:7 HW Solutions '94 9/9/97 page 5

36 Total profit=68. Homework # 7. Decision Trees The president of a firm in a highly competitive industry believes that an employee of the company is providing confidential information to a competing firm. She is 9% certain that this informer is the treasurer of the firm, whose contacts have been extremely valuable in obtaining financing for the company. If she fires the treasurer and he is indeed the informer, the company gains $K. If he is fired but is not the informer, the company loses his expertise and still has an informer on the staff, for a net loss to the company of $5K. If the president does not fire the treasurer, the company loses $K whether or not he is the informer, since in either case the informer is still with the company. Before deciding the fate of the treasurer, the president could order lie detector tests. To avoid possible lawsuits, such tests would have to be administered to all company employees, at a total cost of $K. Another problem is that lie detector tests are not definitive. If a person is lying, the test will reveal it 9% of the time; but if a person is not lying, the test will incorrectly indicate that he is lying % of the time. a. Write the "pay-off" for each of the terminal nodes (2)-(2) of the decision tree below: b. Write the prior probabilities on the branches from nodes (4) and (5). c. Compute the probability that the treasurer fails the lie detector test, if it is administered: P{T fails} = P{T fails T is culprit}p{t is culprit} + P{T fails T is innocent}p{t is innocent} = x + x = d. Using Bayes' rule, compute the posterior probability that the treasurer is the culprit, if he fails the lie detector test: P{T fails test T is culprit}p{t is culprit} P{T is culprit T fails test} = P{T fails test} = 56:7 HW Solutions '94 9/9/97 page 6

37 = e. Fold back the decision tree, computing the expected payoff at each node. f. What is the optimal strategy? g. What is the maximum expected payoff which is achieved with this strategy? h. What is the EVSI, i.e., the expected value of the lie detector test? i. What is the EVPI, i.e., the expected value of perfect information? 2. Project Scheduling. A building contractor is building a house. He has identified the component activities below, with their predecessors and durations as indicated: ID Description Duration Predecesors A Basement 5 -none- B Erect house frame 5 A C Rough electrical work 2 B D Rough plumbing B E Roofing 2 B F Chimney B G Siding B H Hardwood floors B I Windows & Doors 2 B J Wallboard 4 C, D, F K Trim wood 2 H, I, J L Paint exterior 5 G, I M Paint ceilings 2 J N Paint interior walls 5 K, M O Kitchen cabinets J P Finished plumbing, heat 2 K, O Q Finished electrical O R Finish floors 4 N S Fine grading of site A T Landscape 2 S U Clean up site E, L, P, Q, R, T a. The building contractor has prepared the A-O-A Project Network shown below. There are, however, some "dummy" activities missing, and some "dummy" activities with no direction indicated. Draw the necessary "dummy" activities. 56:7 HW Solutions '94 9/9/97 page 7

38 b. Label each of the nodes so that for each activity from node i to node j, i<j. c. Compute the Early Time (ET) for each node (i.e., event): Node ET LT :7 HW Solutions '94 9/9/97 page 8

39 d. Compute, for each activity, the early start (ES) time, the early finish (EF) time, the late start (LS) time, and the late finish (LF) time, and the (total) float (TF). ID Description Duration ES EF LS LF TF A Basement 5 B Erect house frame 5 C Rough electrical work 2 D Rough plumbing E Roofing 2 F Chimney G Siding H Hardwood floors I Windows & Doors 2 J Wallboard 4 K Trim wood 2 L Paint exterior 5 M Paint ceilings 2 N Paint interior walls 5 O Kitchen cabinets P Finished plumbing, heat 2 Q Finished electrical R Finish floors 4 S Fine grading of site T Landscape 2 U Clean up site e. Circle the ID of each critical activity in the preceding table in (d), and indicate the critical path on the diagram in (a). Homework #7 Solutions. Decision Trees The president of a firm in a highly competitive industry believes that an employee of the company is providing confidential information to a competing firm. She is 9% certain that this informer is the treasurer of the firm, whose contacts have been extremely valuable in obtaining financing for the company. If she fires the treasurer and he is indeed the informer, the company gains $K. If he is fired but is not the informer, the company loses his expertise and still has an informer on the staff, for a net loss to the company of $5K. If the president does not fire the treasurer, the company loses $K whether or not he is the informer, since in either case the informer is still with the company. Before deciding the fate of the treasurer, the president could order lie detector tests. To avoid possible lawsuits, such tests would have to be administered to all company employees, at a total cost of $K. Another problem is that lie detector tests are not definitive. If a person is lying, the test will reveal it 9% of the time; but if a person is not lying, the test will incorrectly indicate that he is lying % of the time. Questions and solutions. a. Write the "pay-off" for each of the terminal nodes (2)-(2) of the decision tree below: 56:7 HW Solutions '94 9/9/97 page 9

40 4K.9 +k 4K..9-5K -K -K. -K 4K -62.5K 9/6 +K -K -62.5K -K 7/6 9/6-5K -K 4K /6 66/84K 8/84 -K +K 66/84K /84 8/84-5K -K -K /84 -K Here we calculate P(T culprit T passes test) and P(T culprit T fails test) by and P(T passes test T culprit)p(t culprit) P(T culprit T passes test) = P(T passes test) = (.)(.9).6 P(T fails test T culprit)p(t culprit) P(T culprit T fails test) = = (.9)(.9) = 8 P( T fails test) b. Write the prior probabilities on the branches from nodes (4) and (5). c. Compute the probability that the treasurer fails the lie detector test, if it is administered: P{T fails} = P{T fails T is culprit}p{t is culprit} + P{T fails T is innocent}p{t is innocent} =.9 x.9 +. x. =.84 d. Using Bayes' rule, compute the posterior probability that the treasurer is the culprit, if he fails the lie detector test: P{T fails test T is culprit}p{t is culprit} P{T is culprit T fails test} = P{T fails test} =.9 _.9 _.84 = 8/84 e. Fold back the decision tree, computing the expected payoff at each node. See decision tree in (a). f. What is the optimal strategy? Fire T without test. g. What is the maximum expected payoff which is achieved with this strategy? 4K. = :7 HW Solutions '94 9/9/97 page 4