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LP Formulation. LP Applications. Product Mix Diet / Blending Scheduling Transportation / Distribution Assignment Portfolio Selection (Quadratic). Product mix problem : Narrative representation . The Quality Furniture Corporation produces benches and tables .
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LP Applications • Product Mix • Diet / Blending • Scheduling • Transportation / Distribution • Assignment • Portfolio Selection (Quadratic)
Product mix problem : Narrative representation The Quality Furniture Corporation produces benches and tables. The firm has two main resources Resources labor and redwood for use in the furniture. During the next production period 1200 labor hours are available under a union agreement. A stock of 5000 pounds of quality redwood is also available.
Product mix problem : Narrative representation Consumption and profit Each bench that Quality Furniture produces requires 4 labor hours and 10 pounds of redwood Each picnic table takes 7 labor hours and 35 pounds of redwood. Total available 1200, 5000 Completed benches yield a profit of $9 each, and tables a profit of $20 each. Formulate the problem to maximize the total profit.
Product Mix : Formulation x1 = number of benches to produce x2 = number of tables to produce Maximize Profit = ($9) x1 +($20) x2 subject to Labor: 4 x1 + 7 x2 1200hours Wood:10 x1 + 35 x2 5000 pounds and x1 0, x2 0. We will now solve this LP model using the Excel Solver.
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 Number ordered 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 The company has 10,000 hours of wiring capacity and 5,000 hours of harnessing capacity.
Make / buy decision : decision variables x1 = Number of model 1 slip rings to make x2 = Number of model 2 slip rings to make x3 = Number of model 3 slip rings to make y1 = Number of model 1 slip rings to buy y2 = Number of model 2 slip rings to buy y3 = Number of model 3 slip rings to buy The Objective Function Minimize the total cost of filling the order. MIN: 50x1 + 83x2 + 130x3 + 61y1 + 97y2 + 145y3
Make / buy decision : Constraints Demand Constraints x1 + y1 = 3,000 } model 1 x2 + y2 = 2,000 } model 2 x3 + y3 = 900 } model 3 Resource Constraints 2x1 + 1.5x2 + 3x3 <= 10,000 } wiring 1x1 + 2.0x2 + 1x3 <= 5,000 } harnessing Nonnegativity Conditions x1, x2, x3, y1, y2, y3 >= 0
Make / buy decision : Constraints Do we really need 6 variables? x1 + y1 = 3,000 ===> y1 = 3,000 - x1 x2 + y2 = 2,000 ===> y2 = 2,000 - x2 x3 + y3 = 900 ===> y3 = 900 - x3 The objective function was MIN: 50x1 + 83x2 + 130x3 + 61y1 + 97y2 + 145y3 Just replace the values MIN: 50x1 + 83x2 + 130x3 + 61 (3,000 - x1 ) + 97 ( 2,000 - x2) + 145 (900 - x3 ) MIN: 507500 - 11x1 -14x2 -15x3 We can even forget 507500, and change the the O.F. into MIN - 11x1 -14x2 -15x3 or MAX + 11x1 +14x2 +15x3
Make / buy decision : Constraints Resource Constraints 2x1 + 1.5x2 + 3x3 <= 10,000 } wiring 1x1 + 2.0x2 + 1x3 <= 5,000 } harnessing Demand Constraints x1 <= 3,000 } model 1 x2 <= 2,000 } model 2 x3 <= 900 } model 3 Nonnegativity Conditions x1, x2, x3 >= 0 MAX + 11x1 +14x2 +15x3
Make / buy decision : Constraints • y1 = 3,000- x1 • y2 = 2,000-x2 • y3 = 900-x3 MIN: 50x1 + 83x2 + 130x3 + 61y1 + 97y2 + 145y3 Demand Constraints x1 + y1 = 3,000 } model 1 x2 + y2 = 2,000 } model 2 x3 + y3 = 900 } model 3 Resource Constraints 2x1 + 1.5x2 + 3x3 <= 10,000 } wiring 1x1 + 2.0x2 + 1x3 <= 5,000 } harnessing Nonnegativity Conditions x1, x2, x3, y1, y2, y3 >= 0 MIN: 50x1 + 83x2 + 130x3 + 61(3,000- x1) + 97(2,000-x2) + 145(900-x3) • y1 = 3,000- x1>=0 • y2 = 2,000-x2>=0 • y3 = 900-x3>=0 • x1 <= 3,000 • x2 <= 2,000 • x3 <= 900
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.
Marketing : formulation Decision variables x1 = Number of ads in TV x2 = Number of ads in R x3 = Number of ads in N Max Z = 20 x1 +12x2 +9x3 15x1 +6x2 + 4x3 100 x1 4 x2 10 x3 7 x1 +x2 + x3 15 x1,x2, x3 0
Problem ( 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) Profit Men 2 .5 8 Women 1.5 .75 10 Children 1 .67 6 Total available material is 5000sq-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.
Decision variables X1 : Volume of production of Men’s gloves X2 : Volume of production of Women’s gloves X3 : Volume of production of Children’s gloves Y1 : Number of full time employees Y2 : Number of part time employees
Constraints Row material constraint 2X1 + 1.5X2 + X3 5000 Full time employees Y1 20 Relationship between the number of Full and Part time employees Y1 2 Y2 Labor Required .5X1 + .75X2 + .67X3 40 Y1 + 20Y2 Objective Function Max Z = 8X1 + 10X2 + 6X3 - 520 Y1 - 200Y2 Non-negativity X1 , X2 , X3 , Y1 ,Y2 0