LP Formulation Practice Set 1 Problem 1 Optimal

LP Formulation Practice Set 1

Problem 1. Optimal Product Mix Management is considering devoting some excess capacity to one or more of three products. The hours required from each resource for each unit of product, the available capacity (hours per week) of the three resources, as well as the profit of each unit of product are given below. Sales department indicates that the sales potentials for products 1 and 2 exceeds maximum production rate, but the sales potential for product 3 is 20 units per week. Formulate the problem and solve it using excel LP-Formulation Ardavan Asef-Vaziri June-2013 2

Problem 1. Formulation Decision Variables x 1 : volume of product 1 x 2 : volume of product 2 x 3 : volume of product 3 Objective Function Max Z = 50 x 1 +20 x 2 +25 x 3 Constraints Resources 9 x 1 +3 x 2 +5 x 3 500 5 x 1 +4 x 2 + 350 3 x 1 + +2 x 3 150 Market x 3 20 Nonnegativity x 1 0, x 2 0 , x 3 0 LP-Formulation Ardavan Asef-Vaziri June-2013 3

Problem 2 A farmer has 10 acres to plant in wheat and rye. He has to plant at least 7 acres. However, he has only $1200 to spend and each acre of wheat costs $200 to plant and each acre of rye costs $100 to plant. Moreover, the farmer has to get the planting done in 12 hours and it takes an hour to plant an acre of wheat and 2 hours to plant an acre of rye. If the profit is $500 per acre of wheat and $300 per acre of rye, how many acres of each should be planted to maximize profits? State the decision variables. x = the number of acres of wheat to plant y = the number of acres of rye to plant Write the objective function. maximize 500 x +300 y LP-Formulation Ardavan Asef-Vaziri June-2013 4

Problem 2. Formulation Write the constraints. x+y ≤ 10 x+y ≥ 7 200 x + 100 y ≤ 1200 x + 2 y ≤ 12 x ≥ 0, y ≥ 0 LP-Formulation (max acreage) (min acreage) (cost) (time) (non-negativity) Ardavan Asef-Vaziri June-2013 5

Problem 3. Marketing : narrative A department store want to maximize exposure. There are 3 media; TV, Radio, Newspaper each ad will have the following impact Media Exposure (people / ad) Cost TV 20000 15000 Radio 12000 6000 News paper 9000 4000 Additional information 1 -Total budget is $100, 000. 2 -The maximum number of ads in T, R, and N are limited to 4, 10, 7 ads respectively. 3 -The total number of ads is limited to 15. LP-Formulation Ardavan Asef-Vaziri June-2013 6

Problem 3. Marketing : formulation Decision variables x 1 = Number of ads in TV x 2 = Number of ads in R x 3 = Number of ads in N Max Z = 20 x 1 + 12 x 2 +9 x 3 15 x 1 + 6 x 2 + 4 x 3 x 1 x 2 x 3 x 1 + x 2 + x 3 100 4 10 7 15 x 1, x 2, x 3 0 LP-Formulation Ardavan Asef-Vaziri June-2013 7

Problem 4. ( From Hillier and Hillier) Men, women, and children gloves. Material and labor requirements for each type and the corresponding profit are given below. Glove Material (sq-feet) Labor (hrs) Contribution Margin Men 2 0. 5 8 Women 1. 5 0. 75 10 Children 1 0. 67 6 Total available material is 5000 sq-feet. We can have full time and part time workers. Full time workers work 40 hrs/w and are paid $13/hr Part time workers work 20 hrs/w and are paid $10/hr We should have at least 20 full time workers. The number of full time workers must be at least twice of that of part times. Labor is considered fixed cost not variable. LP-Formulation Ardavan Asef-Vaziri June-2013 8

Problem 4. Decision variables X 1 : Volume of production of Men’s gloves X 2 : Volume of production of Women’s gloves X 3 : Volume of production of Children’s gloves Y 1 : Number of full time employees Y 2 : Number of part time employees LP-Formulation Ardavan Asef-Vaziri June-2013 9

Problem 4. Constraints Row material constraint 2 X 1 + 1. 5 X 2 + X 3 5000 Full time employees Y 1 20 Relationship between the number of Full and Part time employees Y 1 2 Y 2 Labor Required. 5 X 1 +. 75 X 2 +. 67 X 3 40 Y 1 + 20 Y 2 Objective Function Max Z = 8 X 1 + 10 X 2 + 6 X 3 - 520 Y 1 - 200 Y 2 Non-negativity X 1 , X 2 , X 3 , Y 1 , Y 2 0 LP-Formulation Ardavan Asef-Vaziri June-2013 10

Problem 4. Excel Solution LP-Formulation Ardavan Asef-Vaziri June-2013 11

Problem 5. From Hillier and Hillier Strawberry shake production Several ingredients can be used in this product. Ingredient calories from fat Total calories Vitamin Thickener Cost ( per tbsp) (per tbsp) (mg/tbsp) ( c/tbsp) Strawberry flavoring 1 50 20 3 10 Cream 75 100 8 Vitamin supplement 0 50 1 25 Artificial sweetener 0 120 2 15 Thickening agent 30 80 2. 5 6 This beverage has the following requirements Total calories between 380 and 420. No more than 20% of total calories from fat. At least 50 mg vitamin. At least 2 tbsp of strawberry flavoring for each 1 tbsp of artificial sweetener. Exactly 15 mg thickeners. Formulate the problem to minimize costs. LP-Formulation Ardavan Asef-Vaziri June-2013 12

Decision variables Decision Variables X 1 : tbsp of strawberry X 2 : tbsp of cream X 3 : tbsp of vitamin X 4 : tbsp of Artificial sweetener X 5 : tbsp of thickening LP-Formulation Ardavan Asef-Vaziri June-2013 13

Constraints Objective Function Min Z = 10 X 1 + 8 X 2 + 25 X 3 + 15 X 4 + 6 X 5 Calories 50 X 1 + 100 X 2 + 120 X 4 + 80 X 5 380 50 X 1 + 100 X 2 + 120 X 4 + 80 X 5 420 Calories from fat X 1 + 75 X 2 + 30 X 5 0. 2(50 X 1 + 100 X 2 + 120 X 4 + 80 X 5) Vitamin 20 X 1 + 50 X 3 + 2 X 5 50 Strawberry and sweetener X 1 2 X 4 Thickeners 3 X 1 + 8 X 2 + X 3 + 2 X 4 + 2. 5 X 5 = 15 Non-negativity X 1 , X 2 , X 3 , X 4 , X 5 0 LP-Formulation Ardavan Asef-Vaziri June-2013 14

Constraints Objective Function Min Z = 10 X 1 + 8 X 2 + 25 X 3 + 15 X 4 + 6 X 5 Calories 50 X 1 + 100 X 2 + 120 X 4 + 80 X 5 380 50 X 1 + 100 X 2 + 120 X 4 + 80 X 5 420 Calories from fat -9 X 1 + 55 X 2 -24 X 4 +14 X 5 0 Vitamin 20 X 1 + 50 X 3 + 2 X 5 50 Strawberry and sweetener X 1 -2 X 4 0 Thickeners 3 X 1 + 8 X 2 + X 3 + 2 X 4 + 2. 5 X 5 = 15 Non-negativity X 1 , X 2 , X 3 , X 4 , X 5 0 LP-Formulation Ardavan Asef-Vaziri June-2013 15

Constraints LP-Formulation Ardavan Asef-Vaziri June-2013 16

Problem 6. Make / buy decision : Narrative representation Electro-Poly is a leading maker of slip-rings. A new order has just been received. Model 1 Model 2 Model 3 3, 000 2, 000 900 Hours of wiring/unit 2 1. 5 3 Hours of harnessing/unit 1 2 1 Cost to Make $50 $83 $130 Cost to Buy $61 $97 $145 Number ordered The company has 10, 000 hours of wiring capacity and 5, 000 hours of harnessing capacity. LP-Formulation Ardavan Asef-Vaziri June-2013 17

Problem 6. Make / buy decision : decision variables x 1 = Number of model 1 slip rings to make x 2 = Number of model 2 slip rings to make x 3 = Number of model 3 slip rings to make y 1 = Number of model 1 slip rings to buy y 2 = Number of model 2 slip rings to buy y 3 = Number of model 3 slip rings to buy The Objective Function Minimize the total cost of filling the order. MIN: 50 x 1 + 83 x 2 + 130 x 3 + 61 y 1 + 97 y 2 + 145 y 3 LP-Formulation Ardavan Asef-Vaziri June-2013 18

Problem 6. Make / buy decision : Constraints Demand Constraints x 1 + y 1 = 3, 000 } model 1 x 2 + y 2 = 2, 000 } model 2 x 3 + y 3 = 900 } model 3 Resource Constraints 2 x 1 + 1. 5 x 2 + 3 x 3 <= 10, 000 } wiring 1 x 1 + 2. 0 x 2 + 1 x 3 <= 5, 000 } harnessing Nonnegativity Conditions x 1, x 2, x 3, y 1, y 2, y 3 >= 0 LP-Formulation Ardavan Asef-Vaziri June-2013 19

Problem 6. Make / buy decision : Excel LP-Formulation Ardavan Asef-Vaziri June-2013 20

Problem 6. Make / buy decision : Constraints Do we really need 6 variables? x 1 + y 1 = 3, 000 ===> y 1 = 3, 000 - x 1 x 2 + y 2 = 2, 000 ===> y 2 = 2, 000 - x 2 x 3 + y 3 = 900 ===> y 3 = 900 - x 3 The objective function was MIN: 50 x 1 + 83 x 2 + 130 x 3 + 61 y 1 + 97 y 2 + 145 y 3 Just replace the values MIN: 50 x 1 + 83 x 2 + 130 x 3 + 61 (3, 000 - x 1 ) + 97 ( 2, 000 - x 2) + 145 (900 - x 3 ) MIN: 507500 - 11 x 1 -14 x 2 -15 x 3 We can even forget 507500, and change the O. F. into MIN - 11 x 1 -14 x 2 -15 x 3 or MAX + 11 x 1 +14 x 2 +15 x 3 LP-Formulation Ardavan Asef-Vaziri June-2013 21

Problem 6. Make / buy decision : Constraints MAX + 11 x 1 +14 x 2 +15 x 3 Resource Constraints 2 x 1 + 1. 5 x 2 + 3 x 3 <= 10, 000 } wiring 1 x 1 + 2. 0 x 2 + 1 x 3 <= 5, 000 } harnessing Demand Constraints x 1 <= 3, 000 } model 1 x 2 <= 2, 000 } model 2 x 3 <= 900 } model 3 Nonnegativity Conditions x 1, x 2, x 3 >= 0 LP-Formulation Ardavan Asef-Vaziri June-2013 22

Problem 6. Make / buy decision : Constraints MIN: 50 x 1 + 83 x 2 + 130 x 3 y 1 = 3, 000 - x 1 + 61 y 1 + 97 y 2 + 145 y 3 y 2 = 2, 000 -x 2 y 3 = 900 -x 3 Demand Constraints x 1 + y 1 = 3, 000 } model 1 x 2 + y 2 = 2, 000 } model 2 MIN: 50 x 1 + 83 x 2 + 130 x 3 + 61(3, 000 - x 1) + 97(2, 000 -x 2) + 145(900 -x 3) x 3 + y 3 = 900 } model 3 Resource Constraints 2 x 1 + 1. 5 x 2 + 3 x 3 <= 10, 000 } wiring 1 x 1 + 2. 0 x 2 + 1 x 3 <= 5, 000 } harnessing Nonnegativity Conditions y 1 = 3, 000 - x 1>=0 y 2 = 2, 000 -x 2>=0 y 3 = 900 -x 3>=0 x 1 <= 3, 000 x 2 <= 2, 000 x 3 <= 900 x 1, x 2, x 3, y 1, y 2, y 3 >= 0 LP-Formulation Ardavan Asef-Vaziri June-2013 23

Problem 6. Make / buy decision : Constraints LP-Formulation Ardavan Asef-Vaziri June-2013 24

Problem 7 You are given the following linear programming model in algebraic form, where, X 1 and X 2 are the decision variables and Z is the value of the overall measure of performance. Maximize Z = X 1 +2 X 2 Subject to Constraints on resource 1: X 1 + X 2 ≤ 5 (amount available) Constraints on resource 2: X 1 + 3 X 2 ≤ 9 (amount available) And X 1 , X 2 ≥ 0 LP-Formulation Ardavan Asef-Vaziri June-2013 25

Problem 7 Identify the objective function, the functional constraints, and the nonnegativity constraints in this model. Objective Function Maximize Z = X 1 +2 X 2 Functional constraints X 1 + X 2 ≤ 5, X 1 + 3 X 2 ≤ 9 Is (X 1 , X 2) = (3, 1) a feasible solution? 3 + 1 ≤ 5, 3 + 3(1) ≤ 9 yes; it satisfies both constraints. Is (X 1 , X 2) = (1, 3) a feasible solution? 1 + 3 ≤ 5, 1 + 3(9) > 9 no; it violates the second constraint. LP-Formulation Ardavan Asef-Vaziri June-2013 26

Problem 8 You are given the following linear programming model in algebraic form, where, X 1 and X 2 are the decision variables and Z is the value of the overall measure of performance. Maximize Z = 3 X 1 +2 X 2 Subject to Constraints on resource 1: 3 X 1 + X 2 ≤ 9 (amount available) Constraints on resource 2: X 1 + 2 X 2 ≤ 8 (amount available) And X 1 , X 2 ≥ 0 LP-Formulation Ardavan Asef-Vaziri June-2013 27

Problem 8 Identify the objective function, Maximize Z = 3 X 1 +2 X 2 the functional constraints, 3 X 1 + X 2 ≤ 9 and X 1 + 2 X 2 ≤ 8 the non-negativity constraints X 1 , X 2 ≥ 0 Is (X 1 , X 2) = (2, 1) a feasible solution? 3(2) + 1 ≤ 9 and 2 + 2(1) ≤ 8 yes; it satisfies both constraints Is (X 1 , X 2) = (2, 3) a feasible solution? 3(2) + 3 ≤ 9 and 2 + 2(3) ≤ 8 yes; it satisfies both constraints Is (X 1 , X 2) = (0, 5) a feasible solution? 3(0) + 5 ≤ 9 and 0 + 2(5) > 8 no; it violates the second constraint LP-Formulation Ardavan Asef-Vaziri June-2013 28