1. OPS 301 Module B and Additional Topics in Linear Programming
Dr. Steven Harrod

2. Topics Definition of Linear Programming Steps to Formulate a Problem
Applications
Mathematical Requirements
Linear functions
Convex, non-negative feasible region
Single objective
Steps to Formulate a Problem
Solution Techniques
Graphical
Simplex

3. Linear Programming
A mathematical tool to make multiple decisions affecting a single outcome or objective
Guaranteed to find the optimal solution (best possible set of decisions) relative to the stated objective and linear assumptions
This slide provides some reasons that capacity is an issue. The following slides guide a discussion of capacity.

4. 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
Determining the best diet for persons or animals based on any combination of nutrition, cost, taste, etc.
Determining production, routes, and storage locations 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.

5. Requirements of an LP Problem
Linear
Acceptable: X1, 2X2, 9
Not acceptable: X2, X3, |X|, √X
Convex A line drawn between any two points on the region will not travel outside the region
Single 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.

6. Convex and Linear X2 Feasible region X1 – 80 – 60 – 40 – 20 –
80 –
60 –
40 –
20 –
Feasible region
| | | | | | | | | | | X1

7. Linear, But Not Convex X2 Feasible region X1 – 80 – 60 – 40 – 20 –
80 –
60 –
40 –
20 –
Feasible region
| | | | | | | | | | | X1

8. Convex, But Not Linear X2 Feasible region X1 – 80 – 60 – 40 – 20 –
80 –
60 –
40 –
20 –
Feasible region
| | | | | | | | | | | X1

9. Formulating an LP What is the goal? What do you control?
“Objective Function”
Maximize – make as large as possible
Minimize – make as small as possible
What do you control?
Called “Decision Variables”
Must be non-negative
What are the limits?
Resources not to exceed (<=)
Minimum requirements to meet (>=)
Items that must be used/served exactly (=)
Constraints are labeled “s.t.” for “subject to”

10. Example
Young MBA Erica Cudahy may invest up to $1,000. She can invest her money in stocks and loans. Each dollar invested in stocks yields 10¢ profit, and each dollar invested in a loan yields 15¢ profit. At least 30% of all money invested must be in stocks, and at least $400 must be in loans.
source: Winston, Operations Research, 2004

11. Example What is the goal? Maximize profit.
What do we control? Dollar amount invested each in stocks and loans.
What are the limits?
At least $400 invested in loans
30% invested in stocks
$1000 maximum available to invest

12. Mathematical Formulation
Label our decision variables
X1 dollar amount invested in stocks
X2 dollar amount invested in loans
Write objective function
the value resulting from our decisions
0.10 X X2

13. Mathematical Formulation
Write Limits
$1000 available to invest X1 + X2 <= 1000
at least 30% of all money in stocks 0.3 (X1 + X2) <= X1
Arrange all variables on left side
– 0.7 X X2 <=0
at least $400 in loans X2 >=400

14. Complete Formulation max 0.10 X1 + 0.15 X2 s.t. X1 + X2 <= 1000

15. 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

16. Graphical Solution

17. Graphical Example max 7 X1 + 5 X2 s.t. 4X1 + 3X2 ≤ 240 2X1 + 1X2 ≤ 100

18. Graphical Solution 2X1 + 1X2 ≤ 100 4X1 + 3X2 ≤ 240 X2 Feasible region–
80 –
60 –
40 –
20 –
2X1 + 1X2 ≤ 100
4X1 + 3X2 ≤ 240
Feasible region
| | | | | | | | | | | X1
Figure B.3

19. Graphical Solution Iso-Profit Line Solution Method $210 = 7X1 + 5X2–
80 –
60 –
40 –
20 –
| | | | | | | | | | |
Number of Watch TVs
Number of Walkmans 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

20. Graphical Solution $210 = $7X1 + $5X2 X2 (0, 42) (30, 0) X1 – 80 –
80 –
60 –
40 –
20 –
$210 = $7X1 + $5X2
(0, 42)
(30, 0)
| | | | | | | | | | | X1
Figure B.4

21. Graphical Solution $350 = $7X1 + $5X2 $280 = $7X1 + $5X2–
80 –
60 –
40 –
20 –
$350 = $7X1 + $5X2
$280 = $7X1 + $5X2
$210 = $7X1 + $5X2
$420 = $7X1 + $5X2
| | | | | | | | | | | X1
Figure B.5

22. Optimal solution point
Graphical Solution X2 –
80 –
60 –
40 –
20 –
Maximum profit line
Optimal solution point
(X1 = 30, X2 = 40)
$410 = $7X1 + $5X2
| | | | | | | | | | | X1
Figure B.6

23. Corner-Point Method 2 3 1 4 X2 X1 – 80 – 60 – 40 – 20 –
80 –
60 –
40 –
20 – 2 3
| | | | | | | | | | | 1 X1
Figure B.7 4

24. 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 : (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

25. 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

26. 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 : (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
Point 3 : (X1 = 30, X2 = 40) Profit $7(30) + $5(40) = $410

27. Beyond 2 Variables, Matrix Algebra
To solve problems of more than two variables with more complex feasible regions (decimal and fractional values) requires computers and matrix algebra
Matrix algebra allows easy calculation of corner points
Simplex method finds optimal corner point

28. Matrix Algebra max 7 X1 + 5 X2 s.t. 4X1 + 3X2 ≤ 240 2X1 + 1X2 ≤ 100
Define 4 matrices:

29. Matrix Formulation
max cx
s.t.
Ax <=b
x>=0

30. Linear System Solution
IF Ax=b, and A is “Square” (number of rows equals number of columns)
DEFINE the inverse of A: A-1
THEN x=A-1b
Use spreadsheet to calculate inverse
How to convert Ax<=b to Ax=b?

31. Slack Variables “S” max 7 X1 + 5 X2 + 0 S1 + 0 S2 s.t.
{one of X1, X2, S1, S2} = 0
X1, X2, S1, S2 >=0

32. Instructions for Slack
Add one new slack variable to each constraint
Slack variables added to objective with zero coefficient
Square matrix by choosing some variables equal to zero

33. Slack Variables “S”
New matrices:

34. Solution Method Type all of these matrices into Excel In each row of ?
Enter a single 1
Fill the rest of the row with zeros
Do not put 1 in same column on multiple rows
Do not leave cells blank, zeros are required
Label matrix “x” your corner point

35. Solution Method Select the cells of the corner point
Enter formula =mmult(minverse(A matrix cells), b matrix cells)
Keep all cells selected
Hold keys ctrl & shift, then press Enter.
If successful, number values should display in corner point
Pick a new cell
Label “Objective Value”
Enter formula =mmult(c matrix cells, corner point cells)
Again hold keys ctrl & shift, then press enter

36. Finding Corner Points
Select different combinations of variables to set to zero by moving 1 value within row
Round corner points to 3 decimals (#.###)
Ignore corner points found with negative values
Objective value calculated is value given displayed choices or decisions for variables
Given all combinations, all possible corner points, the optimal solution will be one of these

37. Simplex Method
Realistic problems quickly grow to have more corner points than humanly possible to examine
Simplex method starts from a simple point and jumps from point to point until reaching optimal, then stops.
Does not usually need to cover a large number of points
Solution can be proven optimal without listing all corner points
Developed by George Dantzig in the late 1940s
Most computer-based LP packages use the simplex method

38. Simplex Spreadsheet Simplex spreadsheet provided for you
Follow instructions on spreadsheet
Simplex method uses additional matrix algebra to calculate a directional indicator to next best corner point
Simplex also uses matrix algebra to identify stopping point

39. Conclusion Understand Solution Techniques
Use of LP
Mathematical Requirements
Steps to Formulate a Problem
Solution Techniques
Graphical
Simplex
Matrix Algebra Foundation of Advanced Work