TRIBHUVAN UNIVERSITYInstitute of Engineering Pulchowk Campus An Assignment On Operational Research Submitted by: Submitted to: Ajeya Acharya Dr. Jagat Shrestha 073MSTR252 Class Teacher Submission Date: 2017/12/20 Assignment 1 3.1-1. for each of the following constraints, draw a separate graph to show the nonnegative solutions that satisfy this constraint. (a) x1 + 3x2 <= 6 (b) 4x1 + 3x2<= 12 (c) 4x1 + x2 <= 8 (d)Now combine these constraints into a single graph to show the feasible region for the entire set of functional constraints plus no negativity constraints. Solution: For x1+3x2<=6, Taking boundary condition x1+3x2=6 (i) For 4x1+x2<=8, Boundary condition: 4x1+x2=8 (ii) Solving (i) and (ii) We get x1=16/11 X2=18/11 ⇒(x1,x2)= (16/11,18/11) Ajeya Acharya 1 073MSTR252 5 3 3.5 0 0 0.75 EQN-3 2 1. 3.5 3 1.25 eqn-2 2 1.25 eqn-3 4 1 2.5 1 1.1.5 1 1.5 4 4.5 x y 0 3 EQN-2 1 2.75 4 0 3 2.5 4x1+3x2=12 1 0.1 Solution x y 0 6 EQN-1 1 3 eqn-1 2 0 7 6 5 4 3 x1= 6-3x2 2 1 0 0 0.75 6 0.5 2 2.5 x y 0 2 1 1.5 3.5 2 2.5 2 7 0.5 3 0.Assignment 1 Question No.25 1.5 8 0 1 4x1+x2=8 0.5 5 0.5 0 0 1 2 3 4 5 6 7 8 9 Ajeya Acharya 2 073MSTR252 .5 2 1. Assignment 1 COMBINED GRAPH eqn-3 eqn-1 eqn-2 7 6 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 9 feasible region Ajeya Acharya 3 073MSTR252 . identifying both the activities and the resources. (c) Use the graphical model to solve this model. This may force the company to lower the price they charge and so lower the profit made for each wood framed window. Doug makes the wood frames. How would the optimal solution change if he makes only 5 wood frames per day? (a) The question is analogous to Prototype example 3.1 for this problem. and can make 6 per day.Assignment 1 3.1. (b) Formulate a linear programming model for this problem.) Employer Product1 Product 2 Available glass per day Wood frame Aluminium (sq. w≤6. The capacity of each person is limited similar as the capacity of machines is fixed in the example. Linda makes the aluminum frames. a≤4 6w+8a≤48 w≥0 a≥0 Subjected to: Z =60w+30a Ajeya Acharya 4 073MSTR252 . Each wood- framed window uses 6 square feet of glass and each aluminum-framed window uses 8 square feet of Glass. (a) Describe the analogy between this problem and the Wyndor Glass Co. ft. 3. W=no of wood frame window produced per day A= no of aluminum frame window produced per day Then.1. this question has three employees analogous to three plants in example. which makes two different kinds of handcrafted windows: a wood-framed and an aluminum-framed window. Bob forms and cuts the glass.1-6.) Doug 6 0 36 Linda 0 8 32 Bob 6 8 48 Profit per piece $60 $30 (b) Let. As in example. The company wishes to determine how many windows of each type to produce per day to maximize total profit. Then construct and fill in a table like Table 3. and can make 4 per day. (d) A new competitor in town has started making wood-framed windows as well. The production of wood framed window is dependent on capacity of Doug and bob analogous to Plant 1 and Plant 3 in example. The Whitt Window Company is a company with only three employees. The production of aluminum frame window depend upon capacity of Linda and bob analogous to Plant 2 and Plant 3 in example. ft. problem discussed in Sec. How would the optimal solution change (if at all) if the profit per wood-framed window decreases from $60 to $40? From $60 to $20? (e) Doug is considering lowering his working hours. and can make 48 square feet of glass per day. Data for Whitt window Company Glass required per window (sq. which would decrease the number of wood frames he makes per day. They earn $60 profit for each wood-framed window and $30 profit for each aluminum-framed window. 2.2. (4.5).5). (1.0) Z1=$300 Z2=$360 Z3=$300 Z4=$120 Z5=$0 Hence Optimal Solution is to produce 5 wood frame window and 2 aluminum frame window.25.Assignment 1 (c) The corner points of solutions (a.66). (2.w) are: 0. (4.66). Z=40+30a Z1=$240 Z2=$270 Z3=$220 Z4=$120 Z5=$0 Optimal solution remains same If Profit of wood frame decreases from $60 to $20 Z= 20w+30a Z1=$120 Z2=$150 Z3=$140 Z4=$120 Z5=$0 Still Optimal solution does not change. (4. 6). (4. (d) If the profit of wood decreases from $60 to $40. Ajeya Acharya 5 073MSTR252 .0) To maximize profit: Z1=30*0+60*6=$360 Z2=$390 Z3=$300 Z4=$120 Z5=$0 Hence.5. optimal solution is to produce 1 aluminum frame window and 6 window frame windows. 6). 0) and (0.0) and (0. (e) Change in constraint w≤5 Feasible solution are (0. 25 Assignment 1 073MSTR252 Ajeya Acharya 2 4. No.25 6 1.5 7 0.5 eqn-2 a<=4 3 3. Q.75 4 3 eqn-3 6w+8a<=48 5 2.75 8 0 wood aluminium wood and alminium x axis y axis 9 8 7 6 5 WOOD 4 3 2 1 0 0 1 2 3 4 5 6 7 ALUMINIUM 6 .6 w a 0 6 eqn-1 w<=6 1 5.1. 3. (a) Formulate a linear programming model for this problem. and each unit of product 2. y>=0 Subjected to: Z=x+2y Boundary equations are: x+3y=20 (i) x+y=150 (ii) y=60 (iii) x=0 (iv) y=0 (v) Solve (i) and (ii) x=125 and y=25 Hence the corner points are: (0. the most feasible solution is to produce 125 no of product 1 and 25 no of product 2. For each unit of product 1. (125. The WorldLight Company produces two light fixtures (products 1 and 2) that require both metal frame parts and electrical components. up to 60 units.25) and (150. 3 units of frame parts and 2 units of electrical components are required.0). Any excess over 60 units of product 2 brings no profit. Ajeya Acharya 7 073MSTR252 . gives a profit of $2. 1 unit of frame parts and 2 units of electrical components are required. so such an excess has been ruled out. What is the resulting total profit? Solution: x=no of product 1 y=no of product 2 Constraints: x+3y<=20 2x+2y<=300 y<=60 x>=0.1-8. Z1=$0 Z2=$120 Z3=$140 Z4=$175 Z5=$150 From graph. Each unit of product 1 gives a profit of $1. (0. (b) Use the graphical method to solve this model. The company has 200 units of frame parts and 300 units of electrical components.60). For each unit of product 2. 0). Management wants to determine how many units of each product to produce to maximize profit.Assignment 1 3. 0) 0 20 40 60 80 100 120 140 160 Ajeya Acharya 8 073MSTR252 .8 x y x y 70 20 60 90 60 26 58 95 55 32 56 100 50 (0.60) (20. Assignment 1 Qun 3.60) 60 38 54 105 45 44 52 110 40 50 50 115 35 56 48 120 30 50 62 46 125 25 68 44 130 20 74 42 135 15 80 40 140 10 40 86 38 145 5 92 36 150 0 eqn-1 98 34 eqn-2 104 32 30 110 30 eqn-3 116 28 122 26 (125.0) 0 (0.25) 128 24 20 feasible region 10 (150.1. Thus there is no optimal solution. x2 ≥ 0.2-6. (d) For objective functions where this model has no optimal solution. does this mean that there are no good solutions according to the model? Explain.10) and Z=10 c) Maximize Z=x1-x2 No. the objective function value is maximum by sliding the objective function line to the right. If not. (a) Demonstrate that the feasible region is unbounded. find it. What probably went wrong when formulating the model? Solution: The boundary equation are: -x1 + 3x2 =30 (i) -3x1 + x2 = 30 (ii) x1 = 0 (iii) x2 = 0 (iv) a) From graph it is clear that solution is unbound. explain why not. (c) Repeat part (b) when the objective is to maximize Z =x1-x2. Suppose that the following constraints have been provided for a linear programming model.Assignment 1 3. b) Maximize Z=-x1+x2 yes optimal solution is (0. Ajeya Acharya 9 073MSTR252 . does the model have an optimal solution? If so. (b) If the objective is to maximize Z=-x1 +x2. This can be done forever. -x1 + 3x2 ≤ 30 -3x1 + x2 ≤ 30 And x1 ≥ 0. 2.6 COMBINED GRAPH eqn-1 eqn-2 eqn-3 eqn-4 70 60 50 40 30 20 unbound 10 0 -50 -40 -30 -20 -10 0 10 20 30 40 50 60 -10 Ajeya Acharya 10 073MSTR252 . Assignment 1 Qun 3. he has decided to go on a steady diet of only these two foods (plus some liquids and vitamin supplements) for all his meals. Ajeya Acharya 11 073MSTR252 .27 serving of steak and 2.Assignment 1 3. (0.4-7.8).2.72.27. Ralph realizes that this is not the healthiest diet.9). X= daily serving of steak Y= daily serving of potato 5x+15y>=50 20x+5y>=40 15x+2y<=60 z>=0 y>=0 minimization of 4x+2y points are: (0.06 Hence optimal solution is 1.88 Z4=$19. Therefore.09) Z1=$16 Z2=$60 Z3=$10.30). (3.2. (1. He has obtained the following nutritional and cost information: Ralph wishes to determine the number of daily servings (may be fractional) of steak and potatoes that will meet these requirements at a minimum cost. (b) Use the graphical method to solve this model. so he wants to make sure that he eats the right quantities of the two foods to satisfy some key nutritional requirements. (c) Use a computer to solve this model by the simplex method. Ralph Edmund loves steaks and potatoes. (a) Formulate a linear programming model for this problem.9 serving of potato. 8)10 5 (3.2.27. Assignment 1 Qun 3.09) 0 -15 -10 -5 (1.72.7 COMBINED GRAPH eqn-2 eqn-1 eqn-3 eqn-4 eqn-5 40 35 30 (0.9) 0 5 10 15 20 -5 -10 Ajeya Acharya 12 073MSTR252 .4.2.30) 25 20 feasible region 15 (0. 909091 Resource used Carbo 5 15 50 >= 50 protein 20 5 40 >= 40 fat 15 2 24.00 $ 10.00 $ 2.91 Ajeya Acharya 13 073MSTR252 .6. 3. Assignment 1 Question no.90909 <= 60 total Z $ 4.272727 2.1 Ingrident steak potatoes No to make 1. 0 Answer Report Worksheet: [JKS SIR HOMEWORK1.90909091 $E$9<=$G$9 Not Binding 35.909090909 Contin Constraints Cell Name Cell Value Formula Status Slack $E$7 Carbo used 50 $E$7>=$G$7 Binding 0 $E$8 protein used 40 $E$8>=$G$8 Binding 0 $E$9 fat used 24.272727273 Contin $D$4 No to make potatoes 0 2.09090909 Ajeya Acharya 14 073MSTR252 . Iterations Unlimited.4.7 Report Created: 1/2/2018 12:09:14 AM Result: Solver found a solution. Max Integer Sols Unlimited. All Constraints and optimality conditions are satisfied.91 Variable Cells Cell Name Original Value Final Value Integer $C$4 No to make steak 2 1. Assignment 1 Microsoft Excel 16. Use Automatic Scaling Max Subproblems Unlimited.00 $ 10. Solver Engine Engine: Simplex LP Solution Time: 0. Assume NonNegative Objective Cell (Min) Cell Name Original Value Final Value $E$11Z total $ 8. Integer Tolerance 1%.015 Seconds.xlsx]3.000001. Iterations: 2 Subproblems: 0 Solver Options Max Time Unlimited. Precision 0. 67 Z4=$150 Thus. (3. (3. optimal solution (x. (d) Use the spreadsheet to check the following solutions: (x1. (4. y) =(3. 3). (a) Formulate a linear programming model for this problem. Activity 1: x Activity 2: y Constraints: 2x+y≤10 (i) 3x+3y≤20 (ii) 2x+4y≤20 (iii) x≥0 (iv) y≥0 (v) Objective function: Z=20x+30y Solving (i) and (ii).1 You are given the following data for a linear programming problem where the objective is to maximize the profit from allocating three resources to two nonnegative activities.33.0). (4.5).y)=(3. 4). (b) Use the graphical method to solve this model. Feasible solutions are (x. 4). 2) Non Feasible solutions are (x.3).3. Z1=$0 Z2=$100 Z3=$166.33. 2). 4).6.Assignment 1 3. (3. 2). x2) = (2. (5. (c) Display the model on an Excel spreadsheet.33) and Z= $166. 4).67 d) solution: Here.3. (4. (2.33. Which of these solutions are feasible? Which of these feasible solutions has the best value of the objective function? (e) Use the Excel Solver to solve the model by the simplex method.3) Ajeya Acharya 15 073MSTR252 .33).0).3. (3.y)=(3. 3). (2. (i)and (iii) & (ii) and (iii) (x. (0.33) Boundary points: (0. (4. 2). y) = (2. 67 8 1 5 0 5 1.Question:3.67 6 2 4 2 4 2.6.67 2 4 2 6 2 4.67 4 3 3 4 3 3.67 EQN-1 EQN-2 EQN-3 12 10 8 6 4 2 0 0 2 4 6 8 10 12 Ajeya Acharya 16 073MSTR252 .67 0 5 1 8 1 5.67 10 0 6 0.1 Assignment 1 EQN-1 EQN-2 EQN-3 X Y X Y X Y 0 10 0 6. 67 Ajeya Acharya 17 073MSTR252 .6. 3.333333 3.333333 Resource used 1 2 1 10 <= 10 2 3 3 20 <= 20 3 2 4 20 <= 20 total Z 20 30 $ 166. Assignment 1 Question no.1 Activity 1 2 No to make 3. Precision 0. Integer Tolerance 1%.000001. $ 166. All Constraints and optimality conditions are satisfied.xlsx]Sheet1 Report Created: 1/1/2018 11:29:54 PM Result: Solver found a solution. Max Integer Sols Unlimited. Iterations Unlimited. Use Automatic Scaling Max Subproblems Unlimited.67 Variable Cells Cell Name Original Value Final Value Integer $C$4 No to make 0 3.016 Seconds. Solver Engine Engine: Simplex LP Solution Time: 0.333333333 Contin Constraints Cell Name Cell Value Formula Status Slack $E$7 used 10 $E$7<=$G$7 Binding 0 $E$8 used 20 $E$8<=$G$8 Binding 0 $E$9 used 20 $E$9<=$G$9 Binding 0 Ajeya Acharya 18 073MSTR252 . Assume NonNegative.333333333 Contin $D$4 No to make 0 3.0 Answer Report Worksheet: [JKS SIR HOMEWORK1. Objective Cell (Max) Cell Name Original Value Final Value $E$11Z total $ . Iterations: 2 Subproblems: 0 Solver Options Max Time Unlimited.Assignment 1 Microsoft Excel 16. 666666667 0 Ajeya Acharya 19 073MSTR252 . Assignment 1 Microsoft Excel 16.xlsx]Sheet1 Report Created: 1/1/2018 11:29:55 PM Variable Cells Final Reduced Objective Allowable Allowable Cell Name Value Cost Coefficient Increase Decrease $C$4 No to make 3.0 Sensitivity Report Worksheet: [JKS SIR HOMEWORK1. Side Increase Decrease $E$7 used 10 0 10 1E+30 0 $E$8 used 20 3.333333333 0 20 10 5 $D$4 No to make 3.H.333333333 0 30 10 10 Constraints Final Shadow Constraint Allowable Allowable Cell Name Value Price R.333333333 20 0 5 $E$9 used 20 5 20 6. 67 Ajeya Acharya 20 073MSTR252 .67 3. Assignment 1 Microsoft Excel 16.333333333 0 100 3.67 Variable Lower Objective Upper Objective Cell Name Value Limit Result Limit Result $C$4 No to make 3.333333333 0 66.333333333 166.67 $D$4 No to make 3.xlsx]Sheet1 Report Created: 1/1/2018 11:29:55 PM Objective Cell Name Value $E$11 Z total $ 166.333333333 166.0 Limits Report Worksheet: [JKS SIR HOMEWORK1. x3) =(1. x2. He now wishes to determine the quantities of the available types of feed (corn.y. along with the daily nutritional requirements and feed costs: (a) Formulate a linear programming model for this problem.2) is feasible solution with daily cost of $348.z≥0 Minimize Objective Function: 84x+72y+60z b) and e) in spread sheet c) (1. What is the daily cost for your solution? (e) Use the Excel Solver to solve the model by the simplex method. 310 kg proteins and 170 kg vitamins daily. (b) Display the model on an Excel spreadsheet. To supplement several food products grown on the farm.4 Fred Jonasson manages a family-owned farm. d) The solution vary. How many units of each nutritional ingredient would this diet provide daily? (d) Take a few minutes to use a trial-and-error approach with the spreadsheet to develop your best guess for the optimal solution. 2) is a feasible solution and. the objective is to determine which mix will meet certain nutritional requirements at a minimum cost. Fred also raises pigs for market. and alfalfa) that should be given to each pig.2. Since pigs will eat any mix of these feed types.Assignment 1 3. tankage.6. (a) x: kg requirement of corn y: kg requirement of tankage z: kg requirement of corn alfalfa 90x+80y+40z≥200 30x+80y+60z≥180 10x+20y+60z≥150 x. 2. what the daily cost would be for this diet. if so. The number of units of each type of basic nutritional ingredient contained within a kilogram of each feed type is given in the following table. (c) Use the spreadsheet to check if (x1. This diet will provide 210 kg carbohydrate. Ajeya Acharya 21 073MSTR252 . 7143 Ajeya Acharya 22 073MSTR252 .4 corn tankage alfalfa kilogram 1.1429 >= 150 total Z 84 72 60 241.142857 0 2.6. Assignment 1 qun no 3.428571 Nutrition used Carbohy 90 20 40 200 >= 200 proteins 30 80 60 180 >= 180 vitamins 10 20 60 157. Integer Tolerance 1%. Assignment 1 Microsoft Excel 16.142857143 Contin $D$4 kilogram tankage 2 0 Contin $E$4 kilogram alfalfa 2 2. Max Integer Sols Unlimited.7142857 Variable Cells Cell Name Original Value Final Value Integer $C$4 kilogram corn 1 1. Iterations Unlimited. Precision 0.xlsx]3. Solver Engine Engine: Simplex LP Solution Time: 0.016 Seconds. All Constraints and optimality conditions are satisfied.4 Report Created: 1/1/2018 11:56:58 PM Result: Solver found a solution.1428571 $F$9>=$H$9 Not Binding 7. Iterations: 4 Subproblems: 0 Solver Options Max Time Unlimited. Use Automatic Scaling Max Subproblems Unlimited. Assume NonNegative Objective Cell (Min) Cell Name Original Value Final Value $F$11Z total 348 241.142857143 Ajeya Acharya 23 073MSTR252 .000001.6.0 Answer Report Worksheet: [JKS SIR HOMEWORK1.428571429 Contin Constraints Cell Name Cell Value Formula Status Slack $F$7 Carbohy used 200 $F$7>=$H$7 Binding 0 $F$8 proteins used 180 $F$8>=$H$8 Binding 0 $F$9 vitamins used 157. H.428571429 0 60 11.6.xlsx]3.27272727 22.142857143 0 84 51 37.771428571 200 25 80 $F$8 proteins used 180 0.71428571 $E$4 kilogram alfalfa 2.0 Sensitivity Report Worksheet: [JKS SIR HOMEWORK1.1428571 0 150 7.485714286 180 120 6 $F$9 vitamins used 157.66666667 Constraints Final Shadow Constraint Allowable Allowable Cell Name Value Price R.71428571 72 1E+30 17.4 Report Created: 1/1/2018 11:56:58 PM Variable Cells Final Reduced Objective Allowable Allowable Cell Name Value Cost Coefficient Increase Decrease $C$4 kilogram corn 1.142857143 1E+30 Ajeya Acharya 24 073MSTR252 . Assignment 1 Microsoft Excel 16. Side Increase Decrease $F$7 Carbohy used 200 0.2 $D$4 kilogram tankage 0 17. 428571429 241.6.7142857 #N/A #N/A $D$4 kilogram tankage 0 0 241.428571429 2.0 Limits Report Worksheet: [JKS SIR HOMEWORK1. Assignment 1 Microsoft Excel 16.4 Report Created: 1/1/2018 11:56:58 PM Objective Cell Name Value $F$11 Z total 241.7142857 #N/A #N/A $E$4 kilogram alfalfa 2.7142857 Variable Lower Objective Upper Objective Cell Name Value Limit Result Limit Result $C$4 kilogram corn 1.142857143 241.xlsx]3.142857143 1.7142857 #N/A #N/A Ajeya Acharya 25 073MSTR252 .
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