1. Linear Programming
Chapter 13 Supplement

2. Pottery Example
Beaver Creek Pottery Company is located on a Native American reservation. Each day, the company has available 40 hours of labor and 120 pounds of clay. The firm makes two products, bowls and mugs. A bowl requires 1 hour of labor and 4 pounds of clay. A mug requires 2 hours of labor and 3 pounds of clay. The firm's profit is $40 per bowl and $50 per mug. The company wants to maximize profit.

3. Business Objective: Determine
Pottery Example
Business Objective: Determine
Number of bowls to make
Number of mugs to make
to maximize profit.
Decision variables:
x = number of bowls to make
y = number of mugs to make

4. Objective Function Profit Maximize Z = 40x + 50y $40 per bowl (x)
$50 per mug (y)
Maximize Z = 40x + 50y

5. Constraint Table Constraints: x + 2y < 40 4x + 3y < 120
Constrained
Quantity
Max. or
Minimum
Bowls x
Mugs y
Type
Labor 1 4 2 3
< 40
120
Clay
Constraints: x + 2y < 40
4x + 3y < 120

6. Nonnegativity Constraints
We cannot make a negative amount of either product
Add nonnegativity constraints
x > 0
y > 0

7. Problem Statement Maximize Z = 40x + 50y subject to x + 2y < 40

8. Linear Programming Terminology
Feasible solution: Any solution which satisfies all constraints.
Feasible region: Set of points which satisfy all constraints.
Optimal solution(s): A point which
Satisfies all constraints
Maximizes or minimizes the value of the objective function.

9. Solving a Linear Programming Problem by the Graphical Method
(1) Set up the problem:
Decision variables and their definitions.
Write objective function and state whether it should be maximized or minimized.
Write the constraints as mathematical
inequalities or equalities.
(2) Draw the feasible region.

10. Solving a Linear Programming Problem by the Graphical Method(2)
(3) Determine which point(s) in the feasible
region give an optimal solution
Point(s) which maximize or minimize the objective function.
Note: To use the graphical method, the problem must have only 2 variables.

11. To Draw the Feasible Region
Convert each constraint into an equation.
Draw the corresponding line.
The set of points bounded by these lines is the feasible region.

12. Problem Statement Maximize Z = 40x + 50y subject to x + 2y < 40

13. Feasible Region for This Problem
This region is bounded by the lines
x + 2y = 40 (Labor)
4x + 3y = 120 (Clay)
x = 0
y = 0

14. Plot x + 2y = 40 x-intercept: Set y = 0. x + 0 =40 ==> x = 40.
Point is (40,0)
y-intercept: Set x=0.
(0)+2y=40 ==> y = 20.
Point is (0,20)

15. Points in feasible
region satisfy
all constraints. 40
Clay
Mugs (y) 30 20 10
Feasible
Region
Labor 10 20 30 40
Bowls (x)

16. Iso-Profit Lines
A set of points on which the objective function is constant
Example: The set of points satisfying
40x + 50y = 800
x-intercept: (20,0)
y-intercept: (0,16)

17. Iso-profit Line 40 Clay 30 Mugs 20 10 Labor 10 20 30 40 Bowls

18. 40 Clay 30 Mugs 20 10 Labor 10 20 30 40 Bowls Iso-profit Lines
for profits of
$800 and $1,000 40
Clay 30
Mugs 20 10
Labor 10 20 30 40
Bowls

19. Iso-profit Lines 40 Clay 30 Mugs 20 10 Labor 10 20 30 40 Bowls
Optimal Point=(24,8) 10
Labor 10 20 30 40
Bowls

20. Optimal Solution Satisfies x + 2y = 40 and 4x + 3y = 120
Solution is (24,8): 24 bowls and 8 mugs
Verification:
24 + 2(8) = 40
4(24) + 3(8) = 120

21. Maximum Profit Optimal solution is (24,8) Objective function is
40x + 50y
$40(24) + $50(8) = $1,360
Maximum profit is $1,360

22. Fundamental Theorem of Linear Programming
In a linear programming problem, the optimal solution occurs at an extreme point (vertex or corner point) of the feasible region.

23. Objective of Linear Programming
Maximize the use of resources (or minimize costs) to achieve competitive priorities.
Optimum solution is a base case which must be adjusted to reflect business realities.

24. Linear Programming Model
Decision variables are mathematical symbols representing activity levels.
Objective function is a mathematical, linear function which represents the organization’s objectives.
Used to compare alternative courses of action.
© 1998 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 2/e
Ch 11 Supp - 2

25. Linear Programming Model (2)
Constraints are mathematical, linear relationships representing restrictions on decision making
Resource constraints.
Policy or legal constraints.
Sales constraints.
Constraints may be <, =, or >.
© 1998 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 2/e
Ch 11 Supp - 2

26. Linear programming maximizes or minimizes the objective function subject to constraints.

27. Uses of Linear Programming
Production Scheduling
Maximize profit
Minimize cost
Different objectives yield different schedules.
Determine product or service mix
Maximize revenue
Minimize costs

28. Uses of Linear Programming (2)
Scheduling labor in services
Minimize cost
Production-location problem
Allocate products and customers to plants to minimize total cost of production and distribution

29. Uses of Linear Programming (3)
Distribution
Minimize cost
Facility location
Minimize transportation cost.
Emergency response systems
Minimize average response time.

30. Conditions to Use Linear Programming
Objective function must be linear
Constraints must be linear inequalities or linear equations

31. Methods for Solving LP Problems
Graphical method: limited to 2 variables
Any linear programming problem
Simplex method
Karmarkar’s algorithm
Transportation method:
Facility location problems
Production-location problem: minimize total cost of production and shipment to customers

32. Slack Variables
Slack = unused amount of any resource or constrained quantity
S1 = Amount of unused labor
= 40 - (x + 2y) = 40 - x - 2y = (2)
= 0. Optimum solution uses all labor.
S2 = Amount of unused clay
= (4x + 3y) = x - 3y
= (24) - 3(8) = 0.
Optimum solution uses all clay.

33. Sensitivity Analysis (Ranging)
Dual value or shadow price
Incremental increase or decrease in the objective function if one more unit of a resource is added.
The amount we would pay to get one more unit of a resource.
Valid only for resources which have no slack.

34. Shadow Price for Labor Dual value = shadow price = $16.
If labor hours increase from 40 to 41, profit increases from $1360 to $1376.
Amount to buy
= upper bound - original value
= 80 hours - 40 hours = 40 hours

35. Mugs
Optimal point = (0,40). Profit = $2,000 40 30
Clay 20
New
Labor 10
Labor 10 20 30 40 80
Bowls