1. QMB 4701 MANAGERIAL OPERATIONS ANALYSIS
Chapter 3
Linear Programming: Graphical Solution Procedure & Spreadsheet Solution Procedure
Jay Marshall Teets Decision and Information Sciences
University of Florida

2. Agenda Graphical linear programming Additional considerations
Example: Par, Inc. Problem
Graphical solution procedure
Example: Dorian problem
Additional considerations
Degeneracy
Inconsistent problem
Multiple optimal solutions
Unbounded problem
Spreadsheet solution basics
Ask students: how much they know about spreadsheets, input formula, absolute
reference, copy and paste.

3. Example: Par, Inc. Problem
Par, Inc: Manufacturer of golf supplies
New products: Standard golf bags and Deluxe golf bags
Manufacturing operations:
Objective: Maximize total profit
Cutting & Dyeing
Sewing
Finishing
Inspection & Packaging

4. Par Inc. Model Formulation
1. What is the objective in words?
2. What are the constraints in words?
Maximize the total profit
Cutting & Dyeing Cutting & Dyeing
Hours Allocated  Hours Available
Sewing Hours Allocated  Sewing Hours Available
Inspection & Packaging Inspection & Packaging
Hours Allocated  Hours Available
Finishing Hours Allocated  Finishing Hours Available

5. Par Inc. Model Formulation
3. What are the decision variables?
4. Formulate the objective function:
5. Formulate the constraints:
6. Do we need non-negative constraints?
x1 = Number of Standards Produced
x2 = Number of Deluxes Produced
Profit Contribution Maximize 10 x1 + 9 x2
Subject to:
Cutting & Dyeing: /10 x x2  630
Sewing /2 x1 + 5/6 x2  600
Finishing x1 + 2/3 x2  708
Inspection & Packaging /10 x1 + 1/4 x2  135
Logic x1  0 , x2  0

6. How to Solve It? Trial and Error? x1 = 600, x2 = 100? Profit = 6900
520
383.33
666.67 85
x1 = 700,
x2 = 100?
Profit =7900
590
433.33
766.67 95
Profit Contribution Maximize 10 x1 + 9 x2
Subject to:
Cutting & Dyeing: /10 x x2  630
Sewing /2 x1 + 5/6 x2  600
Finishing x1 + 2/3 x2  708
Inspection & Packaging /10 x1 + 1/4 x2  135
Logic x1  0 , x2  0
Feasible
Solution
Infeasible
Solution

7. Graphical Approach Solution: any production plan Solutions
Feasible solution:
solution that satisfies all constraints
Feasible solutions
A. Determine feasible region
(draw constraint lines)
B. Construct isoprofit lines
C. Locate the optimal solution
Optimal solutions
Optimal solution:
feasible solution that achieves
the maximal objective value

8. Constraints Determine Feasible Region
Approach:
1. Draw lines representing the constraints solved as equalities
2. Show the direction of feasibility
3. The feasible region is the set of values which simultaneously satisfies all the constraints
Cutting & Dyeing /10 x x2  630
Sewing /2 x /6 x2  600
Finishing x /3 x2  708
Inspection & Packaging /10 x /4 x2  135
x1 0, x2 0,

9. A. Constraints Determine Feasible Region

10. A. Constraints Determine Feasible Region
Redundant
constraint

11. Construct Isoprofit Lines
Approach:
1. Select a constant C
2. Draw the “profit line” 10 x1 + 9 x2 = C
3. Construct parallel profit lines
Slope of the objective function:
x2= -10/9 x1+C/9

12. Locate The Optimal Solution - Observations
1. The optimal solution cannot fall in the interior of the feasible region
2. The optimal solution must occur at a “corner point” of the feasible region
3. In searching for the optimal solution we only have to examine the corner points
Optimal Solution

13. Locate The Optimal Solution
Find extreme points (corner points) and calculate objective values
Extreme Points: the vertices or “corner points” of the feasible region.
With two variables, extreme points are determined by the intersection
of the constraint lines. 5
I &P 4
C&D 3 F 1 2

14. Locate The Optimal Solution - Find Extreme Points
Example: Find extreme point #3:
Intersection of the cutting & dyeing constraint and the finishing constraint:
Cutting & Dyeing: 7/10 x x2 =
Finishing x /3 x2 =
Solve the two equations for x1 and x2:
7/10 x1 + x2 =  7/10 x1 = 630  x2  x1 =  10/7 x2
x1 + 2/3 x2 =  (900  10/7 x2) +2/3 x2 = 708
 /21 x /21 x2 = 708   16/21 x2 =  192  x2 = 21/16  192 = 252
x1 = /7 x2  x1 =  10/7 (252)  x1 =  = 540

15. Slack and Binding ConstraintsF
C&D
I &P
Binding constraints
Slack constraint
At optimal solution (540, 252)
Hrs used Hrs available
C&D: =
S: <
F: =
I&P: <
Binding
Slack = 600  480 = 120
Slack = 135  117 = 18

16. Summary of Graphical Approach
A. Graph Feasible Region
Draw each constraint line
For each constraint, plot points on the x1 axis (x2=0) and x2 axis (x2=0) and connect the points with a line.
Indicate the feasible area with arrow
Check if (0,0) is include, if so, draw arrow towards (0,0). If not, draw arrow away from (0,0). If (0,0) is on the constraint line, you must choose another point to use as a check
Hatch the area in common for all constraints and their respective arrows
B. Find Extreme Points
Set the intersecting constraints equal to one another - basic algebra

17. Summary of Graphical Approach
C. Find optimal solution
Option #1
List all the extreme points and calculate objective function values for each extreme point.
Choose the point with the highest (profit) or lowest (cost) value
Option #2
Find the slope of the objective function
Choose which extreme point is last touched by the isoprofit/isocost line as it is increased/decreased. This point is the optimal point
Calculate the objective value for the optimal point.

18. Note: Simplex Method 1. Start with a feasible corner point solution
2. Check to see if a feasible neighboring corner
point is better
3. If not, stop; otherwise move to that better neighbor
and return to step 2.

19. Graphing Example: Minimization Problem
Dorian Problem
MIN 50,000 x ,000 x2 ST
HIW: 7x x2  28
HIM: 2x x2  24
x1 0, x2  0 2 4 6 8 10 12 14 16 x 1 ,
3.6
1.4
HIW
HIM

20. Additional Consideration: Degeneracy
Degeneracy: Assume Inspection & Packaging time is 117 hours (instead of 135)
Profit Contribution Maximize 10 x1 + 9 x2 ST
Cutting & Dyeing: /10 x x2  630
Sewing /2 x1 + 5/6 x2  600
Finishing x1 + 2/3 x2  708
Inspection & Packaging /10 x1 + 1/4 x2  117
Logic x1  0 , x2  0

21. Additional Consideration: Inconsistent Problem
Assume there is an additional constraint: need to produce at least 725 standard bags,
i.e., x1  725
Profit Contribution Maximize 10 x1 + 9 x2 ST
Cutting & Dyeing: /10 x x2  630
Sewing /2 x1 + 5/6 x2  600
Finishing x1 + 2/3 x2  708
Inspection & Packaging /10 x1 + 1/4 x2  135
x  725
Logic x1  0 , x2  0

22. Additional Consideration: Multiple Optimal Solutions
Assume the objective function is 12 x1 + 8 x2 (instead of 10 x1 + 9 x2)
Profit Contribution Maximize 12 x1 + 8 x2 ST
Cutting & Dyeing: /10 x x2  630
Sewing /2 x1 + 5/6 x2  600
Finishing x1 + 2/3 x2  708
Inspection & Packaging /10 x1 + 1/4 x2  135
Logic x1  0 , x2  0
Slope of the new objective function:
x2 =-12/8 x1 +C/8
= =-3/2 x1 +C/8

23. Additional Consideration: Unbounded Problem
Assume we have the following problem:
Profit Contribution Maximize 10 x1 + 9 x2 ST
Demand: x x2  1000
x  725
Logic x1  0 , x2  0

24. Excel Solver Solution Procedure
Excel can do the following:
Solve linear programs
Note: Activate the “Assume Linear Model” option if your problem is a LINEAR program, otherwise the Solver will treat it as a NONLINEAR program.
Solve nonlinear programs
Solve integer programs
Perform sensitivity analysis

25. Implementing an LP Model in a Spreadsheet
Organize the data for the model on the spreadsheet including:
Objective Coefficients
Constraint Coefficients
Right-Hand-Side (RHS) values
clearly label data and information
visualize a logical layout
row and column structures of the data are a good start

26. Implementing an LP Model in a Spreadsheet
Reserve separate cells to represent each decision variable
arrange them such that the structure parallels the input data
keep them together in the same place
Right-Hand-Side (RHS) values
Create a formula in a cell that corresponds to the Objective Function
For each constraint, create a formula that corresponds to the left-hand-side (LHS) of the constraint
NOTE: Once you’ve formulated a spreadsheet model for a certain type of application, it’s useful to save it as a template for future use!

27. Solver Terminology Target cell Changing cells Constraint cells
Represents the objective function
Must indicate max or min
Changing cells
Represents the decision variables
Constraint cells
the cells in the spreadsheet that represent the LHS formulas of the constraints in the model
the cells in the spreadsheet that represent the RHS values of the constraints in the model
TIP: If you scale the constraints such that they are all "" or all "", then it's easier to input the constraints as a block. Then, you avoid having to enter each constraint as a separate line.

28. Solver Example: Par, Inc. Model

29. Par, Inc. Model - Answer Report
Solver Parameters: Objective: MAX F9
Variables: B11:C11
Constraints: D3:D6 <= F3:F6
Options: Assume Linear Model
Assume Non-Negative
Optimal objective value
Optimal solution
Constraints
Binding
& Slack
Information

30. Summary of Excel Solver Procedure

31. Solver Example: Minimization Example
Dorian Problem
MIN 50,000 x ,000 x2 ST
7x x2  28
2x x2  24
x1 0, x2  0

32. Solver Example: Minimization Example

33. Summary of Additional Considerations
GRAPHICAL
EXCEL SOLVER
MANAGERIAL IMPLICATION
Degeneracy
More than two constraint lines intersect at one point
Inconsistent Problem
Feasible region does not exist
Could not find a feasible solution
Too many restrictions
Multiple Optimal Solutions
Slope of objective function is same as that of one constraint
Output shows only a single optimal solution
Unbounded Problem
Optimal value is infinite
The set of cell values do not converge
Problem is improperly formulated

34. Summary Graphical solution procedure Additional considerations
Feasible region
Extreme points
Optimal solution
Binding and Slack Constraints
Additional considerations
Degeneracy
Inconsistent problem
Multiple optimal solutions
Unbounded problem
Spreadsheet solution basics