2. 1 Additional examples LP

3. Let : X 1, X 2, X 3, ………, X n = decision variables Z = Objective function or linear function Requirement: Maximization of the linear function Z. Z = c 1 X 1 + c 2 X 2 + c 3 X 3 + ………+ c n X n …..Eq (1) subject to the following constraints: …..Eq (2)

4. Formulating LP Problems X-podsBlueBerrysAvailable Hours Department(X 1 )(X 2 )This Week Hours Required to Produce 1 Unit Electronic43240 Assembly21100 Profit per unit$7$5 Decision Variables: X 1 = number of X-pods to be produced X 2 = number of BlueBerrys to be produced Table B.1

5. Formulating LP Problems Objective Function: Maximize Profit = $7X 1 + $5X 2 There are three types of constraints  Upper limits where the amount used is ≤ the amount of a resource  Lower limits where the amount used is ≥ the amount of the resource  Equalities where the amount used is = the amount of the resource

6. Formulating LP Problems Second Constraint: 2X 1 + 1X 2 ≤ 100 (hours of assembly time) Assembly time available Assembly time used is ≤ First Constraint: 4X 1 + 3X 2 ≤ 240 (hours of electronic time) Electronic time available Electronic time used is ≤

7. Graphical Solution  Can be used when there are two decision variables 1.Plot the constraint equations at their limits by converting each equation to an equality 2.Identify the feasible solution space 3.Create an iso-profit line based on the objective function 4.Move this line outwards until the optimal point is identified

8. Graphical Solution 100– – 80 80 – – 60 60 – – 40 40 – – 20 20 – – ||||||||||| 020406080100 Number of BlueBerrys Number of X-pods X1X1X1X1 X2X2X2X2 Assembly (constraint B) Electronics (constraint A) Feasible region Figure B.3

9. Graphical Solution 100– – 80 80 – – 60 60 – – 40 40 – – 20 20 – – ||||||||||| 020406080100 Number of Watch TVs Number of X-pods X1X1X1X1 X2X2X2X2 Assembly (constraint B) Electronics (constraint A) Feasible region Figure B.3 Iso-Profit Line Solution Method Choose a possible value for the objective function $210 = 7X 1 + 5X 2 Solve for the axis intercepts of the function and plot the line X 2 = 42 X 1 = 30

10. Graphical Solution 100– – 80 80 – – 60 60 – – 40 40 – – 20 20 – – ||||||||||| 020406080100 Number of BlueBerrys Number of X-pods X1X1X1X1 X2X2X2X2 Figure B.4 (0, 42) (30, 0) $210 = $7X 1 + $5X 2

11. Graphical Solution 100– – 80 80 – – 60 60 – – 40 40 – – 20 20 – – ||||||||||| 020406080100 Number of BlueBeryys Number of X-pods X1X1X1X1 X2X2X2X2 Figure B.5 $210 = $7X 1 + $5X 2 $350 = $7X 1 + $5X 2 $420 = $7X 1 + $5X 2 $280 = $7X 1 + $5X 2

12. Graphical Solution 100– – 80 80 – – 60 60 – – 40 40 – – 20 20 – – ||||||||||| 020406080100 Number of BlueBerrys Number of X-pods X1X1X1X1 X2X2X2X2 Figure B.6 $410 = $7X 1 + $5X 2 Maximum profit line Optimal solution point (X 1 = 30, X 2 = 40)

13. Corner-Point Method Figure B.7 123 100– – 80 80 – – 60 60 – – 40 40 – – 20 20 – – ||||||||||| 020406080100 Number of BlueBerrys Number of X-pods X1X1X1X1 X2X2X2X2 4

14. Corner-Point Method  The optimal value will always be at a corner point  Find the objective function value at each corner point and choose the one with the highest profit Point 1 :(X 1 = 0, X 2 = 0)Profit $7(0) + $5(0) = $0 Point 2 :(X 1 = 0, X 2 = 80)Profit $7(0) + $5(80) = $400 Point 4 :(X 1 = 50, X 2 = 0)Profit $7(50) + $5(0) = $350

15. Corner-Point Method  The optimal value will always be at a corner point  Find the objective function value at each corner point and choose the one with the highest profit Point 1 :(X 1 = 0, X 2 = 0)Profit $7(0) + $5(0) = $0 Point 2 :(X 1 = 0, X 2 = 80)Profit $7(0) + $5(80) = $400 Point 4 :(X 1 = 50, X 2 = 0)Profit $7(50) + $5(0) = $350 Solve for the intersection of two constraints 2X 1 + 1X 2 ≤ 100 (assembly time) 4X 1 + 3X 2 ≤ 240 (electronics time) 4X 1 +3X 2 =240 - 4X 1 -2X 2 =-200 +1X 2 =40 4X 1 +3(40)=240 4X 1 +120=240 X 1 =30

16. Corner-Point Method  The optimal value will always be at a corner point  Find the objective function value at each corner point and choose the one with the highest profit Point 1 :(X 1 = 0, X 2 = 0)Profit $7(0) + $5(0) = $0 Point 2 :(X 1 = 0, X 2 = 80)Profit $7(0) + $5(80) = $400 Point 4 :(X 1 = 50, X 2 = 0)Profit $7(50) + $5(0) = $350 Point 3 :(X 1 = 30, X 2 = 40)Profit $7(30) + $5(40) = $410

17. 16 The Galaxy Industries Production Problem – A Prototype Example Galaxy manufactures two toy doll models: –Space Ray. –Zapper. Resources are limited to –1000 pounds of special plastic. –40 hours of production time per week.

18. 17 Marketing requirement –Total production cannot exceed 700 dozens. –Number of dozens of Space Rays cannot exceed number of dozens of Zappers by more than 350. Technological input –Space Rays requires 2 pounds of plastic and 3 minutes of labor per dozen. – Zappers requires 1 pound of plastic and 4 minutes of labor per dozen. The Galaxy Industries Production Problem – A Prototype Example

19. 18 The current production plan calls for: –Producing as much as possible of the more profitable product, Space Ray ($8 profit per dozen). –Use resources left over to produce Zappers ($5 profit per dozen), while remaining within the marketing guidelines. The current production plan consists of: Space Rays = 450 dozen Zapper = 100 dozen Profit = $4100 per week The Galaxy Industries Production Problem – A Prototype Example 8(450) + 5(100)

20. 19 Management is seeking a production schedule that will increase the company’s profit.

21. 20 A linear programming model can provide an insight and an intelligent solution to this problem.

22. 21 :Decisions variables: –X 1 = Weekly production level of Space Rays (in dozens) –X 2 = Weekly production level of Zappers (in dozens). Objective Function: – Weekly profit, to be maximized The Galaxy Linear Programming Model

23. 22 Max 8X 1 + 5X 2 (Weekly profit) subject to 2X 1 + 1X 2  1000 (Plastic) 3X 1 + 4X 2  2400 (Production Time) X 1 + X 2  700 (Total production) X 1 - X 2  350 (Mix) X j > = 0, j = 1,2 (Nonnegativity) The Galaxy Linear Programming Model

24. 23 Using a graphical presentation we can represent all the constraints, the objective function, and the three types of feasible points.

25. 24 The non-negativity constraints X2X2 X1X1 Graphical Analysis – the Feasible Region

26. 25 1000 500 Feasible X2X2 Infeasible Production Time 3X 1 +4X 2  2400 Total production constraint: X 1 +X 2  700 (redundant) 500 700 The Plastic constraint 2X 1 +X 2  1000 X1X1 700 Graphical Analysis – the Feasible Region

27. 26 1000 500 Feasible X2X2 Infeasible Production Time 3X 1 +4X2  2400 Total production constraint: X 1 +X 2  700 (redundant) 500 700 Production mix constraint: X 1 -X2  350 The Plastic constraint 2X 1 +X 2  1000 X1X1 700 Graphical Analysis – the Feasible Region There are three types of feasible points Interior points. Boundary points.Extreme points.

28. 27 Solving Graphically for an Optimal Solution

29. 28 The search for an optimal solution Start at some arbitrary profit, say profit = $2,000... Then increase the profit, if possible......and continue until it becomes infeasible Profit =$4360 500 700 1000 500 X2X2 X1X1

30. 29 Summary of the optimal solution Summary of the optimal solution Space Rays = 320 dozen Zappers = 360 dozen Profit = $4360 –This solution utilizes all the plastic and all the production hours. –Total production is only 680 (not 700). –Space Rays production exceeds Zappers production by only 40 dozens.

31. 30 –If a linear programming problem has an optimal solution, an extreme point is optimal. Extreme points and optimal solutions

32. 31 For multiple optimal solutions to exist, the objective function must be parallel to one of the constraints Multiple optimal solutions Any weighted average of optimal solutions is also an optimal solution.

33. Copyright 2006 John Wiley & Sons, Inc.Supplement 13-32 Minimization Problem CHEMICAL CONTRIBUTION BrandNitrogen (lb/bag)Phosphate (lb/bag) Gro-plus24 Crop-fast43 Minimize Z = $6x 1 + $3x 2 subject to 2x 1 +4x 2  16 lb of nitrogen 4x 1 +3x 2  24 lb of phosphate x 1, x 2  0

34. Copyright 2006 John Wiley & Sons, Inc.Supplement 13-33 14 14 – 12 12 – 10 10 – 8 8 – 6 6 – 4 4 – 2 2 – 0 0 – |22|222 |44|444 |66|666 |88|888 |10 12 14 x1x1x1x1 x2x2x2x2 A B C Graphical Solution x 1 = 0 bags of Gro-plus x 2 = 8 bags of Crop-fast Z = $24 Z = 6x 1 + 3x 2

35. Dual problem (2 vars primal) 34