1. Linear Programming Models: Graphical and Computer Methods
Chapter 7
Linear Programming Models: Graphical and Computer Methods

2. Chapter Outline 7.1 Introduction
7.2 Requirements of a Linear Programming Problem
7.3 Formulating LP Problems
7.4 Graphical Solution to an LP Problem
7.5 Solving LP Problem (QM for Window and Excel)
7.6 Solving Minimization Problems
7.7 Four Special Cases in LP
7.8 Sensitivity Analysis

3. Linear Programming: An Overview
Objectives of business decisions frequently involve maximizing profit or minimizing costs.
Linear programming is an analytical technique in which linear algebraic relationships represent a firm’s decisions, given a business objective, and resource constraints.
Steps in application:
Identify problem as solvable by linear programming.
Formulate a mathematical model of the unstructured problem.
Solve the model.
Implementation

4. Requirements of LP Problems and LP Basic Assumptions
Decision variables - mathematical symbols representing levels of activity of a firm.
Objective function - a linear mathematical relationship describing an objective of the firm, in terms of decision variables - this function is to be maximized or minimized.
Constraints – requirements or restrictions placed on the firm by the operating environment, stated in linear relationships of the decision variables.
Parameters - numerical coefficients and constants used in the objective function and constraints.
Assumptions
Proportionality - The rate of change (slope) of the objective function and constraint equations is constant.
Additivity - Terms in the objective function and constraint equations must be additive.
Divisibility -Decision variables can take on any fractional value and are therefore continuous as opposed to integer in nature.
Certainty - Values of all the model parameters are assumed to be known with certainty (non-probabilistic).
Non-negativity - all answers or variables are greater than or equal to (≥) zero
Decision variables: The unknowns to be determined by the solution to the model.
Objective function: The goal desired to be achieved by the system.
Constraints: Limitation on variables to satisfy the restriction of the modeled system
Parameters - numerical coefficients and constants used in the objective function and constraints.
Linearity requires the following assumptions:
Proportionality - a change in a variable results in a proportionate change in that variable's contribution to the value of the function.
Additivity - the function value is the sum of the contributions of each term.
Divisibility - the decision variables can be divided into non-integer values, taking on fractional values. Integer programming techniques can be used if the divisibility assumption does not hold.
In addition to these linearity assumptions, linear programming assumes certainty; that is, that the coefficients are known and constant.

5. Minime Restaurant Pho (40K VND/bowl) Hu tieu (50K VND/bowl) Chef
(works 40 min./day)
Meat
(120 grs meat/day)
I can prepare 1 bowl of:
- Pho in 1 min.
- Hu tieu in 2 min.
1 bowl of:
- Pho needs 4 gr meat
- Hu tieu neads 3 gr. meat

6. A Maximization Example (1 of 3) Resource Requirements
LP Model Formulation
A Maximization Example (1 of 3)
Product mix problem – Minime Noodle Restaurant
How many bowls of Pho and Hu tieu should be prepared to maximize profits given labor and materials constraints?
Product resource requirements and unit profit:
Products
Resource Requirements
Profits
Labor (min./bowl)
Meat (gr./bowl)
(K. VND/bowl)
Pho 1 4 40
Hu tieu 2 3 50
Resource
Availability
40 min.
120 grs.

7. A Maximization Example (2 of 3) Resource Requirements
LP Model Formulation
A Maximization Example (2 of 3)
Products
Resource Requirements
Profits
Labor (min./bowl)
Meat (gr./bowl)
(K. VND/bowl)
Pho 1 4 40
Hu tieu 2 3 50
Avail. Resource
120
Trial and error
All for Pho:
40 bowls bowls 30 bowls x 40= 1200 (chef’s time surplus)
All for Hu tieu:
20 bowls bowls 20 bowls x 50= 1000 (meat surplus)
Half for Pho and Half for Hu tieu
PHO bowls bowls 15 bowls x 40= 600 (chef’s time surplus)
HU TIEU bowls bowls 10 bowls x 50= 500 (meat surplus)
Time
Meat
Profit

8. A Maximization Example (3 of 3)
LP Model Formulation
A Maximization Example (3 of 3)
Decision x1 = # of bowls of Pho to prepare per day
Variables: x2 = # of bowls of Hu tieu to prepare per day
Objective Maximize Z = 40x1 + 50x2
Function: Where Z = profit per day
subject to (s.t.):
Resource x1 + 2x2  40 minutes of labor
Constraints: x1 + 3x2  120 grs. of meat
Non-Negativity x1  0; x2  0
Constraints:

9. Feasible and Infeasible Solutions
A feasible solution does not violate any of the constraints:
Example x1 = 5 bowls of Pho
x2 = 10 bowls of Hu tieu
Z = 40x1 + 50x2 = 700 (K.VND)
Labor constraint check:
1(5) + 2(10) = 25 < 40 minutes, within constraint
Meat constraint check:
4(5) + 3(10) = 70 < 120 grams, within constraint
An infeasible solution violates at least one of the constraints:
Example x1 = 10 bowls of Pho
x2 = 20 bowls of Hu tieu
Z = (K.VND)
1(10) + 2(20) = 50 > 40 minutes, violates constraint

10. Graphical Solution of LP Models
Graphical solution is limited to linear programming models containing only two decision variables (can be used with three variables but only with great difficulty).
Graphical methods provide visualization of how a solution for a linear programming problem is obtained.

11. Graphical Solution of Maximization Model (1 of 11)
Coordinate Axes
Graphical Solution of Maximization Model (1 of 11)
Maximize Z = 40x1 + 50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Figure 2.2 Coordinates for Graphical Analysis

12. Graphical Solution of Maximization Model (2 of 11)
Labor Constraint
Graphical Solution of Maximization Model (2 of 11)
Maximize Z = 40x1 + 50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Figure 2.3 Graph of Labor Constraint

13. Graphical Solution of Maximization Model (3 of 11)
Labor Constraint Area
Graphical Solution of Maximization Model (3 of 11)
Maximize Z = 40x1 + 50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Figure 2.4 Labor Constraint Area

14. Graphical Solution of Maximization Model (4 of 11)
Meat Constraint Area
Graphical Solution of Maximization Model (4 of 11)
Maximize Z = 40x1 + 50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Figure 2.5 Clay Constraint Area

15. Graphical Solution of Maximization Model (5 of 11)
Both Constraints
Graphical Solution of Maximization Model (5 of 11)
Maximize Z = 40x1 + 50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Figure 2.6 Graph of Both Model Constraints

16. Feasible Solution Area
Graphical Solution of Maximization Model (6 of 11)
Maximize Z = 40x1 + 50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Figure Feasible Solution Area

17. Graphical Solution Methods
ISOPROFIT METHOD
1. Graph all constraints and find the feasible region.
2. Select a specific profit (or cost) line and graph it to find the slope.
3. Move the objective function line in the direction of increasing profit (or decreasing cost) while maintaining the slope. The last point it touches in the feasible region is the optimal solution.
4. Find the values of the decision variables at this last point and compute the profit (or cost).
CORNER POINT METHOD
2. Find the corner points of the feasible reason.
3. Compute the profit (or cost) at each of the feasible corner points.
4. select the corner point with the best value of the objective function found in Step 3. This is the optimal solution.

18. Graphical Solution of Maximization Model (7 of 11)
Isoprofit Line Method
Graphical Solution of Maximization Model (7 of 11)
Maximize Z = 40x1 + 50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Construct the isoprofit line:
Z = 40x1 + 50x2
x2 = Z/ /50x1
= Z/50 -4/5x1
Figure 2.8 Objection Function Line for Z = $800

19. Graphical Solution of Maximization Model (8 of 11)
Isoprofit Line Method
Graphical Solution of Maximization Model (8 of 11)
Maximize Z = 40x1 + 50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Figure 2.9 Alternative Objective Function Lines

20. Graphical Solution of Maximization Model (9 of 11)
Isoprofit Line Method
Graphical Solution of Maximization Model (9 of 11)
Maximize Z = 40x1 + 50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Figure Identification of Optimal Solution

21. Optimal Solution Coordinates
Graphical Solution of Maximization Model (10 of 11)
Maximize Z = 40x1 + 50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Figure 2.11 Optimal Solution Coordinates

22. Extreme (Corner) Point Method
Graphical Solution of Maximization Model (11 of 11)
Maximize Z = 40x1 + 50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Figure Solutions at All Corner Points

23. Using Excel’s Solver to Solve LP Problems
The Solver tool in Excel can be used to find solutions to
LP problems
Integer programming problems
Noninteger programming problems
Solver may be sensitive to the initial values it uses
Solver is limited to 200 variables and 100 constraints
It can be used for small real world problems
Add-ins like What’s Best! Can be used to expand Solver’s capabilities

24. Using Excel’s Solver to Solve LP Problems
Maximize Z = 40 x x2
Subject to
x1 + 2x2 40 mins. (labor constraint)
4x1 + 3x2 120 grs. (meat constraint)
x1 , x2 0
Click on “Tools”
to invoke “Solver.”
Objective function
=E6-F6
=E7-F7
=C6*B10+D6*B11
=C7*B10+D7*B11
Decision variables – Pho
(x1)=B10; Hu tieu (x2)=B11

25. Using Excel’s Solver to Solve LP Problems
After all parameters and constraints have been input, click on “Solve.”
Objective function
Decision variables
C6*B10+D6*B11≤40
C7*B10+D7*B11≤120
Click on “Add” to insert constraints

26. Using Excel’s Solver to Solve LP Problems

27. Using QM for Windows and Excel Solving Flair Furniture’s LP Problem
Maximize profit = $70T + $50C
subject to
4T + 3C ≤ 240 (carpentry constraint)
2T + 1C ≤ 100 (painting and varnishing constraint)
T, C ≥ 0 (nonnegativity constraint)
Most organizations have access to software to solve big LP problems
While there are differences between software implementations, the approach each takes towards handling LP is basically the same
Once you are experienced in dealing with computerized LP algorithms, you can easily adjust to minor changes

28. Using QM for Windows Computer screen for input of data
First select the Linear Programming module
Specify the number of constraints (non-negativity is assumed)
Specify the number of decision variables
Specify whether the objective is to be maximized or minimized
For the Flair Furniture problem there are two constraints, two decision variables, and the objective is to maximize profit
Computer screen for input of data

29. Using QM for Windows Computer screen for input of data
Computer screen for output of solution

30. Using QM for Windows
Graphical output of solution

31. A Minimization Example (1 of 6)
LP Model Formulation
A Minimization Example (1 of 6)
Two brands of fertilizer available - Super-Gro, Crop-Quick.
Field requires at least 16 pounds of nitrogen and 24 pounds of phosphate.
Super-Gro costs $6 per bag, Crop-Quick $3 per bag.
Problem: How much of each brand to purchase to minimize total cost of fertilizer given following data ?

32. Fertilizing Farmer’s Field

33. A Minimization Example (2 of 6)
LP Model Formulation
A Minimization Example (2 of 6)
Decision Variables: x1 = bags of Super-Gro
x2 = bags of Crop-Quick
The Objective Function:
Minimize Z = 6x1 + 3x2
Model Constraints:
2x1 + 4x2  16 lb (nitrogen constraint)
4x1 + 3x2  24 lb (phosphate constraint)
x1, x2  0 (non-negativity constraint)

34. LP Model Formulation and Constraint Graph
A Minimization Example (3 of 6)
Minimize Z = $6x1 + $3x2
subject to: 2x1 + 4x2  16
4x1 + 3x2  24
x1, x2  0
Figure Graph of Both Model Constraints

35. Feasible Solution Area A Minimization Example (4 of 6)
Minimize Z = $6x1 + $3x2
subject to: 2x1 + 4x2  16
4x1 + 3x2  24
x1, x2  0
Figure 2.17 Feasible Solution Area

36. Optimal Solution Point A Minimization Example (5 of 6)
Minimize Z = $6x1 + $3x2
subject to: 2x1 + 4x2  16
4x1 + 3x2  24
x1, x2  0
Construct the objective line:
Z = 6x1 + 3x2
x2 = Z/3 – 6/3x1
x2 = Z/3 -2x1
Figure Optimum Solution Point

37. A Minimization Example (6 of 6)
Graphical Solutions
A Minimization Example (6 of 6)
Minimize Z = $6x1 + $3x2
subject to: 2x1 + 4x2  16
4x1 + 3x2  24
x1, x2  0
Figure 2.19 Graph of Fertilizer Example

38. Four Special Cases in LP
Four special cases and difficulties arise at times when using the graphical approach to solving LP problems
Infeasibility
Unboundedness
Redundancy
Alternate Optimal Solutions

39. Four Special Cases in LP
Infeasibility-A problem with no feasible solution
Exists when there is no solution to the problem that satisfies all the constraint equations
No feasible solution region exists
8 – –
6 –
4 –
2 –
0 – X2
| | | | | | | | | | X1
Region Satisfying Third Constraint
Region Satisfying First Two Constraints

40. Four Special Cases in LP
Unboundedness- A solution region unbounded to the right
In a graphical solution, the feasible region will be open ended
This usually means the problem has been formulated improperly
15 –
10 –
5 –
0 – X2
| | | | | X1
X1 ≥ 5
X2 ≤ 10
Feasible Region
X1 + 2X2 ≥ 15

41. Four Special Cases in LP
Redundancy- A problem with a redundant constraint
A redundant constraint is one that does not affect the feasible solution region
30 –
25 –
20 –
15 –
10 –
5 –
0 – X2
| | | | | | X1
2X1 + X2 ≤ 30
Redundant Constraint
X1 ≤ 25
X1 + X2 ≤ 20
Feasible Region

42. Four Special Cases in LP
Alternate optimal solutions
Occasionally two or more optimal solutions may exist
Graphically this occurs when the objective function’s isoprofit or isocost line runs perfectly parallel to one of the constraints
8 –
7 –
6 –
5 –
4 –
3 –
2 –
1 –
0 – X2
| | | | | | | | X1 A
Optimal Solution Consists of All Combinations of X1 and X2 Along the AB Segment
Isoprofit Line for $8 B
Isoprofit Line for $12 Overlays Line Segment AB
Feasible Region

43. Sensitivity Analysis
Sensitivity analysis is used to determine effects on the optimal solution within specified ranges for the objective function coefficients, constraint coefficients, and right hand side (RHS) values.
Sensitivity analysis provides answers to certain what-if questions.

44. Range of Optimality
A range of optimality of an objective function coefficient is found by determining an interval for the objective function coefficient in which the original optimal solution remains optimal while keeping all other data of the problem constant. The value of the objective function may change in this range.
Graphically, the limits of a range of optimality are found by changing the slope of the objective function line within the limits of the slopes of the binding constraint lines. (This would also apply to simultaneous changes in the objective coefficients.)
The slope of an objective function line, Max c1x1 + c2x2, is
-c1/c2
The slope of a constraint, a1x1 + a2x2 = b, is -a1/a2

45. Shadow Price
A shadow price for a RHS value (or resource limit) is the amount the objective function will change per unit increase in the right hand side value of a constraint.
Graphically, a shadow price is determined by adding +1 to the right hand side value in question and then resolving for the optimal solution in terms of the same two binding constraints.
The shadow price is equal to the difference in the values of the objective functions between the new and original problems.
The shadow price for a nonbinding constraint is 0.

46. Example: Sensitivity Analysis
Solve graphically for the optimal solution:
Max z = 5x1 + 7x2
s.t x < 6
2x1 + 3x2 < 19
x1 + x2 < 8
x1, x2 > 0

47. Example: Sensitivity Analysis
Graphical Solution x2 8 7 6 5 4 3 2 1
x1 + x2 < 8
Max 5x1 + 7x2
x1 < 6
Optimal x1 = 5, x2 = 3
z = 46
2x1 + 3x2 < 19 x1

48. Example: Sensitivity Analysis
Range of Optimality for c1
The slope of the objective function line is -c1/c2.
The slope of the 1st binding constraint, x1 + x2 = 8, is -1,
the slope of the 2nd binding constraint, 2x1 + 3x2 = 19, is -2/3.
Find the range of values for c1 (with c2 staying 7) such that
the objective function line slope lies between that of the two
binding constraints:
-1 < -c1/7 < -2/3
Multiplying through by -7 (and reversing the inequalities):
14/3 < c1 < 7

49. Example: Sensitivity Analysis
Range of Optimality for c2
Find the range of values for c2 ( with c1 staying 5) such that the objective function line slope lies between that of the two binding constraints:
-1 < -5/c2 < -2/3
Multiplying by -1: > 5/c2 > 2/3
Inverting, < c2/5 < 3/2
Multiplying by 5: < c2 < 15/2

50. Example: Sensitivity Analysis
Shadow Prices
Constraint 1: Since x1 < 6 is not a binding constraint, its shadow price is 0.
Constraint 2: Change the RHS value of the 2nd constraint to 20 and resolve for the optimal point determined by the last two constraints: 2x1 + 3x2 = 20 and x1 + x2 = 8.
The solution is x1 = 4, x2 = 4, z = 48.
Hence, the shadow price = znew - zold = = 2.
Constraint 3: Change the RHS value of the 3rd constraint to 9 and resolve for the optimal point determined by the last two constraints: 2x1 + 3x2 = 19 and x1 + x2 = 9.
The solution is: x1 = 8, x2 = 1, z = 47.
Hence, the shadow price = znew - zold = = 1.

51. Homework 06
7.15, 7.16, 7.20, 7.22, 7.24, 7.26, 7.28