LP Formulation - PowerPoint Presentation

Slide1

LP Formulation

Practice Set 1Slide2

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.

Problem 1. Optimal Product Mix

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 excelSlide3

Decision Variables

x1 : volume of product 1 x

2

: volume of product 2x3 : volume of product 3Objective Function Max Z = 50 x

1 +20 x2 +25 x

3

ConstraintsResources9 x1 +3 x2 +5 x3  500 5 x1 +4 x2 +  350 3 x1 + +2 x3  150 Market x3  20Nonnegativityx1  0, x2  0 , x3 0

Problem 1. FormulationSlide4

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?

Problem 2

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 500x +300y  Slide5

Problem

2. FormulationWrite the constraints.   x+y

≤ 10 (max acreage) x+y

≥ 7 (min acreage) 200x + 100y ≤ 1200 (cost) x + 2y ≤ 12 (time) x

≥ 0, y ≥ 0 (non-negativity)Slide6

Problem

3. Marketing : narrative

A department store want to maximize exposure.

There are 3 media; TV, Radio, Newspapereach ad will have the following impactMedia Exposure (people / ad) CostTV 20000 15000Radio 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. Slide7

Problem

3. Marketing : formulation

Decision variables

x1 = Number of ads in TVx2 = Number of ads in R x

3 = Number of ads in NMax

Z = 20

x1 + 12x2 +9x315x1 + 6x2 + 4x3  100 x1  4 x2  10 x3  7 x1 + x2 + x3  15x1, x2, x

3  0Slide8

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 8Women 1.5 0.75 10Children 1 0.67 6Total 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/hrPart time workers work 20 hrs/w and are paid $10/hrWe 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. Slide9

Problem

4. Decision variables

X

1 : Volume of production of Men’s glovesX2 : Volume of production of Women’s glovesX3

: Volume of production of Children’s glovesY1 : Number of full time employees

Y

2 : Number of part time employeesSlide10

Problem

4. Constraints

Row material constraint

2X1 + 1.5X2 + X3  5000Full time employeesY1  20

Relationship between the number of Full and Part time employeesY1  2

Y2

Labor Required.5X1 + .75X2 + .67X3  40 Y1 + 20Y2Objective FunctionMax Z = 8X1 + 10X2 + 6X3 - 520 Y1 - 200 Y2Non-negativityX1 , X2 , X

3 , Y1 ,

Y2  0Slide11

Problem

4. Excel Solution Slide12

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) (mg/tbsp) ( c/tbsp)Strawberry flavoring 1 50 20 3 10Cream 75 100 0 8 8Vitamin supplement 0 0 50 1 25Artificial sweetener 0 120 0 2 15Thickening agent

30 80 2 2.5 6This 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.Slide13

Decision variables

Decision Variables

X

1 : tbsp of strawberryX2 : tbsp

of creamX3 : tbsp

of vitamin

X4 : tbsp of Artificial sweetenerX5 : tbsp of thickening Slide14

Constraints

Objective Function

Min Z =

10X1 + 8X2 + 25 X

3 + 15

X

4 + 6 X5 Calories 50X1 + 100 X2 + 120 X4 + 80 X5  38050X1 + 100 X2 + 120 X4 + 80 X5  420Calories from fat X1 + 75 X2 + 30 X5  0.2(50X1 + 100 X2 + 120 X4 + 80 X5) Vitamin20X1 + 50 X3 + 2 X5  50Strawberry and sweetenerX1  2 X4 Thickeners3X1 + 8X2 + X3 + 2 X4

+ 2.5 X5 = 15

Non-negativityX

1

, X

2

, X

3

,

X

4

,

X

5



0Slide15

Constraints

Objective Function

Min Z =

10X1 + 8X2 + 25 X

3 + 15

X

4 + 6 X5 Calories 50X1 + 100 X2 + 120 X4 + 80 X5  38050X1 + 100 X2 + 120 X4 + 80 X5  420Calories from fat -9X1 + 55 X2 -24X4 +14X5  0Vitamin20X1 + 50 X3 + 2 X5  50Strawberry and sweetenerX1 -2 X4  0Thickeners3X1 + 8X2 + X3 +

2 X4 + 2.5

X5

= 15

Non-negativity

X

1

, X

2

, X

3

,

X

4

,

X

5



0Slide16

ConstraintsSlide17

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 900Hours of wiring/unit 2 1.5 3Hours of harnessing/unit 1 2 1Cost 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.

Problem 6. Make / buy decision : Narrative representationSlide18

x

1 = 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

buyThe Objective FunctionMinimize the total cost of filling the order.MIN: 50x1 + 83x2 + 130x3 + 61y1 + 97y2 + 145y3 Problem 6. Make / buy decision : decision variablesSlide19

Demand Constraints

x1 + y1 = 3,000 } model 1x2 + y2 = 2,000 } model 2x3 + y3

= 900 } model 3Resource Constraints2x1

+ 1.5x2 + 3x3 <= 10,000 } wiring1x1 + 2.0x2 + 1x3 <= 5,000 } harnessingNonnegativity Conditionsx1

, x2, x3, y1, y2, y3 >= 0

Problem

6. Make / buy decision : ConstraintsSlide20

Problem

6.

Make / buy decision : Excel Slide21

Do we really need 6 variables

? x1 + y1 = 3,000 ===> y1 = 3,000 - x1 x2 + y

2 = 2,000 ===> y2 = 2,000 - x2

x3 + y3 = 900 ===> y3 = 900 - x3 The objective function was MIN: 50x1 + 83x2 + 130x

3 + 61y1 + 97y2 + 145y3Just replace the valuesMIN: 50x

1

+ 83x2 + 130x3 + 61 (3,000 - x1 ) + 97 ( 2,000 - x2) + 145 (900 - x3 )MIN: 507500 - 11x1 -14x2 -15x3We can even forget 507500, and change the the O.F. into MIN - 11x1 -14x2 -15x3 or MAX + 11x1 +14x2 +15x3 Problem 6. Make / buy decision : ConstraintsSlide22

Resource Constraints

2x1 + 1.5x2 + 3x3 <= 10,000 } wiring1x1 + 2.0x2 + 1x3 <= 5,000 } harnessing

Demand Constraintsx

1 <= 3,000 } model 1x2 <= 2,000 } model 2x3 <= 900 } model 3Nonnegativity Conditions

x1, x2, x3 >= 0

Problem

6. Make / buy decision : ConstraintsMAX + 11x1 +14x2 +15x3Slide23

MIN: 50x1 + 83x2 + 130x3

+ 61y1 + 97y2 + 145y3Demand Constraintsx1

+ y1 = 3,000 } model 1

x2 + y2 = 2,000 } model 2x3 + y3 = 900 } model 3

Resource Constraints2x1 + 1.5x2 + 3x

3

<= 10,000 } wiring1x1 + 2.0x2 + 1x3 <= 5,000 } harnessingNonnegativity Conditionsx1, x2, x3, y1, y2, y3 >= 0Problem 6. Make / buy decision : Constraints y1 = 3,000- x1 y2 = 2,000-x2 y3 = 900-x3

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 <= 900Slide24

Problem

6.

Make / buy decision : ConstraintsSlide25

You are given the following linear programming model in algebraic form, where, X

1 and X2 are the decision variables and Z is the value of the overall measure of performance.Maximize Z = X

1 +2 X

2Subject to Constraints on resource 1: X1 + X2 ≤ 5 (amount available) Constraints on resource 2: X1 + 3X2 ≤ 9 (amount available)AndX1 , X2 ≥ 0

Problem 7Slide26

Identify the objective function, the functional constraints, and the non-negativity constraints in this model.

Objective Function  Maximize Z = X1 +2 X

2Functional constraints

 X1 + X2 ≤ 5, X1 + 3X2 ≤ 9Is (X1 ,X2) = (3,1) a feasible solution?

3 + 1

≤

5, 3 + 3(1) ≤ 9  yes; it satisfies both constraints. Is (X1 ,X2) = (1,3) a feasible solution?1 + 3 ≤ 5, 1 + 3(9) > 9  no; it violates the second constraint. Problem 7Slide27

You are given the following linear programming model in algebraic form, where, X

1 and X2 are the decision variables and Z is the value of the overall measure of performance.Maximize Z = 3X

1 +2 X

2Subject to Constraints on resource 1: 3X1 + X2 ≤ 9 (amount available) Constraints on resource 2: X1 + 2X2 ≤ 8 (amount available)AndX1 , X2 ≥ 0

Problem 8Slide28

Identify the objective function,

Maximize Z = 3X1 +2 X2

the functional constraints, 3X1 + X2 ≤ 9 and X1 + 2X2 ≤ 8

the non-negativity constraintsX1 , X2 ≥ 0 Is (X1 ,X2

) = (2,1) a feasible solution? 3(2

) + 1

≤ 9 and 2 + 2(1) ≤ 8 yes; it satisfies both constraintsIs (X1 ,X2) = (2,3) a feasible solution? 3(2) + 3 ≤ 9 and 2 + 2(3) ≤ 8 yes; it satisfies both constraintsIs (X1 ,X2) = (0,5) a feasible solution?3(0) + 5 ≤ 9 and 0 + 2(5) >

8 no; it violates the second constraintProblem 8