1. Operations Management
Lecture 7
Linear Programming
PowerPoint presentation to accompany
Heizer/Render
Principles of Operations Management, 7e
Operations Management, 9e

2. Recap Implications of Quality tools and Problem solving Check Sheets
Scatter Diagrams
Cause-and-Effect Diagrams
Pareto Charts
Flowcharts
Histograms
Statistical Process Control (SPC)

3. Outline Linear Programming
Requirements of a Linear Programming Problem
Maximization Problems
Minimization Problems
Linear Programming Methods
Graphical Method
Simplex Method (will do later)

4. Outline – Continued Formulating Linear Programming Problems
Shader Electronics Example
Graphical Solution to a Linear Programming Problem
Graphical Representation of Constraints
Iso-Profit Line Solution Method
Corner-Point Solution Method

5. Outline – Continued Linear Programming Applications
Production-Mix Example
Diet Problem Example
Labor Scheduling Example

6. Learning Objectives
When you complete this module you should be able to:
Formulate linear programming models, including an objective function and constraints
Graphically solve an LP problem with the iso-profit line method
Graphically solve an LP problem with the corner-point method

7. Linear Programming
A mathematical technique to allocate limited resources to achieve an objective
linear programming (LP) is a technique for optimization of a linear objective function, subject to linear equality and linear inequality constraints.
This slide provides some reasons that capacity is an issue. The following slides guide a discussion of capacity.

8. Linear Programming
Linear programming determines the way to achieve the best outcome (such as maximum profit or minimum cost) in a given mathematical model and given some list of requirements represented as linear equations.
It is one of more powerful technique for managerial decisions
This slide provides some reasons that capacity is an issue. The following slides guide a discussion of capacity.

9. Linear Programming
A Linear Programming model seeks to maximize or minimize a linear function, subject to a set of linear constraints.
The linear model consists of the following components:
A set of decision variables.
An objective function.
A set of constraints
This slide provides some reasons that capacity is an issue. The following slides guide a discussion of capacity.

10. Linear Programming
Will find the minimum or maximum value of the objective
Guarantees the optimal solution to the model formulated
This slide provides some reasons that capacity is an issue. The following slides guide a discussion of capacity.

11. LP Applications
Scheduling school buses to minimize total distance traveled
Allocating police patrol units to high crime areas in order to minimize response time
Scheduling tellers at banks so that needs are met during each hour of the day while minimizing the total cost of labor
This slide can be used to frame a discussion of capacity.
Points to be made might include:
- capacity definition and measurement is necessary if we are to develop a production schedule
- while a process may have “maximum” capacity, many factors prevent us from achieving that capacity on a continuous basis.
Students should be asked to suggest factors which might prevent one from achieving maximum capacity.

12. LP Applications
Selecting the product mix in a factory to make best use of machine- and labor-hours available while maximizing the firm’s profit
Picking blends of raw materials in feed mills to produce finished feed combinations at minimum costs
Determining the distribution system that will minimize total shipping cost
This slide can be used to frame a discussion of capacity.
Points to be made might include:
- capacity definition and measurement is necessary if we are to develop a production schedule
- while a process may have “maximum” capacity, many factors prevent us from achieving that capacity on a continuous basis.
Students should be asked to suggest factors which might prevent one from achieving maximum capacity.

13. LP Applications
Developing a production schedule that will satisfy future demands for a firm’s product and at the same time minimize total production and inventory costs
This slide can be used to frame a discussion of capacity.
Points to be made might include:
- capacity definition and measurement is necessary if we are to develop a production schedule
- while a process may have “maximum” capacity, many factors prevent us from achieving that capacity on a continuous basis.
Students should be asked to suggest factors which might prevent one from achieving maximum capacity.

14. Requirements of an LP Problem
There should be an objective function to maximize or minimize some quantity (usually profit or cost)
The restrictions or constraints are present which limit the ability to achieve objective
This slide can be used to frame a discussion of capacity.
Points to be made might include:
- capacity definition and measurement is necessary if we are to develop a production schedule
- while a process may have “maximum” capacity, many factors prevent us from achieving that capacity on a continuous basis.
Students should be asked to suggest factors which might prevent one from achieving maximum capacity.

15. Requirements of an LP Problem
There must be alternative courses of action from which to choose
The objective and constraints in linear programming problems must be expressible in terms of linear equations or inequalities
This slide can be used to frame a discussion of capacity.
Points to be made might include:
- capacity definition and measurement is necessary if we are to develop a production schedule
- while a process may have “maximum” capacity, many factors prevent us from achieving that capacity on a continuous basis.
Students should be asked to suggest factors which might prevent one from achieving maximum capacity.

16. Mathematical formulation of Linear Programming model
Step 1
- Study the given situation
- Find the key decision to be made
Identify the decision variables of the problem
Step 2
Formulate the objective function to be optimized
Step 3
Formulate the constraints of the problem
Step 4
- Add non-negativity restrictions or constraints
The objective function , the set of constraints and the non-negativity restrictions together form an LP model.

17. Formulating LP Problems
The product-mix problem at Shader Electronics
Two products
Shader X-pod, a portable music player
Shader BlueBerry, an internet-connected color telephone
Determine the mix of products that will produce the maximum profit

18. Formulating LP Problems
X-pods BlueBerrys Available Hours
Department (X1) (X2) This Week
Hours Required
to Produce 1 Unit
Electronic
Assembly
Profit per unit $7 $5
Table B.1
Decision Variables:
X1 = number of X-pods to be produced
X2 = number of BlueBerrys to be produced

19. Formulating LP Problems
Objective Function:
Maximize Profit = $7X1 + $5X2
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

20. Formulating LP Problems
First Constraint:
Electronic
time available
time used
is ≤
4X1 + 3X2 ≤ 240 (hours of electronic time)
Second Constraint:
Assembly
time available
time used
is ≤
2X1 + 1X2 ≤ 100 (hours of assembly time)

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

22. 4X1 + 3X2 = (I)
2X1 + 1X2 = (II)
If X2 = 0, X1 = 60
Point (60, 0)
If X2 = 0, X1 = 50
Point (50, 0)
If X1 = 0, X2 = 80
Point (0, 80)
If X1 = 0, X2 = 100
Point (0, 100)

23. Graphical Solution X2 II(0, 100) I(0, 80) Assembly (constraint B)–
80 –
60 –
40 –
20 –
| | | | | | | | | | |
Number of BlueBerrys
Number of X-pods X1 X2
II(0, 100)
I(0, 80)
Assembly (constraint B)
Electronics (constraint A)
Feasible region
I (60, 0)
II(50, 0)
Figure B.3

24. Graphical Solution Iso-Profit Line Solution Method $210 = 7X1 + 5X2–
80 –
60 –
40 –
20 –
| | | | | | | | | | |
Number of Watch TVs
Number of X-pods X1 X2
Assembly (constraint B)
Electronics (constraint A)
Feasible region
Figure B.3
Choose a possible value for the objective function
$210 = 7X1 + 5X2
Solve for the axis intercepts of the function and plot the line
X2 = X1 = 30

25. Graphical Solution $210 = $7X1 + $5X2 X2 Number of BlueBerrys (0, 42)–
80 –
60 –
40 –
20 –
| | | | | | | | | | |
Number of BlueBerrys
Number of X-pods X1 X2
Figure B.4
$210 = $7X1 + $5X2
(0, 42)
(30, 0)

26. Graphical Solution $350 = $7X1 + $5X2 $280 = $7X1 + $5X2–
80 –
60 –
40 –
20 –
| | | | | | | | | | |
Number of BlueBeryys
Number of X-pods X1 X2
$350 = $7X1 + $5X2
$280 = $7X1 + $5X2
$210 = $7X1 + $5X2
$420 = $7X1 + $5X2
Figure B.5

27. Optimal solution point
Graphical Solution –
80 –
60 –
40 –
20 –
| | | | | | | | | | |
Number of BlueBerrys
Number of X-pods X1 X2
Maximum profit line
Optimal solution point
(X1 = 30, X2 = 40)
$410 = $7X1 + $5X2
Figure B.6

28. Corner-Point Method 2 3 1 4 X2 Number of BlueBerrys X1–
80 –
60 –
40 –
20 –
| | | | | | | | | | |
Number of BlueBerrys
Number of X-pods X1 X2 2 3 1
Figure B.7 4

29. 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 1 : (X1 = 0, X2 = 0) Profit $7(0) + $5(0) = $0
Point : (X1 = 0, X2 = 80) Profit $7(0) + $5(80) = $400
Point : (X1 = 50, X2 = 0) Profit $7(50) + $5(0) = $350

30. 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
Solve for the intersection of two constraints
2X1 + 1X2 ≤ 100 (assembly time)
4X1 + 3X2 ≤ 240 (electronics time)
4X1 + 3X2 = 240
- 4X1 - 2X2 = -200
+ 1X2 = 40
4X1 + 3(40) = 240
4X = 240
X1 = 30
Point 1 : (X1 = 0, X2 = 0) Profit $7(0) + $5(0) = $0
Point 2 : (X1 = 0, X2 = 80) Profit $7(0) + $5(80) = $400
Point 4 : (X1 = 50, X2 = 0) Profit $7(50) + $5(0) = $350

31. 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 : (X1 = 0, X2 = 0) Profit $7(0) + $5(0) = $0
Point : (X1 = 0, X2 = 80) Profit $7(0) + $5(80) = $400
Point : (X1 = 50, X2 = 0) Profit $7(50) + $5(0) = $350
Point : (X1 = 30, X2 = 40) Profit $7(30) + $5(40) = $410