1. Linear Programming

2. Objectives Requirements for a linear programming model.
Graphical representation of linear models.
Linear programming results:
Unique optimal solution
Alternate optimal solutions
Unbounded models
Infeasible models
Extreme point principle.

3. Objectives - continued
Sensitivity analysis concepts:
Reduced costs
Range of optimality--LIGHTLY
Shadow prices
Range of feasibility--LIGHTLY
Complementary slackness
Added constraints / variables
Computer solution of linear programming models
WINQSB
EXCEL
LINDO

4. 3.1 Introduction to 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.
SHOW FORMAT

5. The Importance of Linear Programming
Many real static problems lend themselves to linear
programming formulations.
Many real problems can be approximated by linear models.
The output generated by linear programs provides
useful “what’s best” and “what-if” information.

6. Assumptions of Linear Programming (p. 48)
The decision variables are continuous or divisible, meaning that eggs or airplanes is an acceptable solution
The parameters are known with certainty
The objective function and constraints exhibit constant returns to scale (i.e., linearity)
There are no interactions between decision variables

7. Methodology of Linear Programming
Determine and define the decision variables
Formulate an objective function
verbal characterization
Mathematical characterization
Formulate each constraint

8. 3.2 THE GALAXY INDUSTRY PRODUCTION PROBLEM - A Prototype Example
Galaxy manufactures two toy models:
Space Ray.
Zapper.
Purpose: to maximize profits
How: By choice of product mix
How many Space Rays?
How many Zappers?
A RESOURCE ALLOCATION PROBLEM

9. Galaxy Resource Allocation
Resources are limited to
1200 pounds of special plastic available per week
40 hours of production time per week.
All LP Models have to be formulated in the context of a production period
In this case, a week

10. Marketing requirement
Total production cannot exceed 800 dozens.
Number of dozens of Space Rays cannot exceed number of dozens of Zappers by more than 450.
Technological input
Space Rays require 2 pounds of plastic and
3 minutes of labor per dozen.
Zappers require 1 pound of plastic and
4 minutes of labor per dozen.

11. Current production plan calls for:
Producing as much as possible of the more profitable product, Space Ray ($8 profit per dozen).
Use resources left over to produce Zappers ($5 profit
per dozen).
The current production plan consists of:
Space Rays = 550 dozens
Zapper = 100 dozens
Profit = 4900 dollars per week

12. Management is seeking a production schedule that will increase the company’s profit.

13. MODEL FORMULATION Decisions variables: Objective Function:
X1 = Production level of Space Rays (in dozens per week).
X2 = Production level of Zappers (in dozens per week).
Objective Function:
Weekly profit, to be maximized

14. The Objective Function
Each dozen Space Rays realizes $8 in profit.
Total profit from Space Rays is 8X1.
Each dozen Zappers realizes $5 in profit.
Total profit from Zappers is 5X2.
The total profit contributions of both is
8X1 + 5X2
(The profit contributions are additive because of the linearity assumption)

15. we have a plastics resource constraint, a production time constraint, and two marketing constraints.
PLASTIC: each dozen units of Space Rays requires 2 lbs of plastic; each dozen units of Zapper requires 1 lb of plastic and within any given week, our plastic supplier can provide 1200 lbs.

16. The Linear Programming Model
Max 8X1 + 5X2 (Weekly profit)
subject to
2X1 + 1X2 < = (Plastic)
3X1 + 4X2 < = (Production Time)
X1 + X2 < = (Total production)
X X2 < = (Mix)
Xj> = 0, j = 1,2 (Nonnegativity)

17. 3.4 The Set of Feasible Solutions for Linear Programs
The set of all points that satisfy all the constraints of the model is called a
FEASIBLE REGION

18. Using a graphical presentation
we can represent all the constraints,
the objective function, and the three
types of feasible points.

19. There are three types of feasible pointsX2
The plastic constraint:
2X1+X2<=1200
1200
The Plastic constraint
Total production constraint:
X1+X2<=800
Infeasible
600
Production mix constraint:
X1-X2<=450
Production Time
3X1+4X2<=2400
Feasible X1
600
800
Interior points.
There are three types of feasible points
Boundary points.
Extreme points.

20. 3.5 Solving Graphically for an Optimal Solution

21. We now demonstrate the search for an optimal solution
Start at some arbitrary profit, say profit = $2,000...
Then increase the profit, if possible... X2
1200
...and continue until it becomes infeasible
Profit =$5040
Profit = $ 2, 3, 4,
800
Recall the feasible Region
600 X1
400
600
800

22. Let’s take a closer look at the optimal pointX2
1200
Let’s take a closer look at
the optimal point
800
Infeasible
600
Feasible
region
Feasible
region X1
400
600
800

23. Summary of the optimal solution
Space Rays = 480 dozens
Zappers = 240 dozens
Profit = $5040
This solution utilizes all the plastic and all the production hours.
Total production is only 720 (not 800).
Space Rays production exceeds Zapper by only 240 dozens (not 450).

24. Extreme points and optimal solutions
If a linear programming problem has an optimal solution, it will occur at an extreme point.
Multiple optimal solutions
For multiple optimal solutions to exist, the objective function must be parallel to a constraint that defines the boundary of the feasible region.
Any weighted average of optimal solutions is also an optimal solution.

25. 3.6 The Role of Sensitivity Analysis of the Optimal Solution
Is the optimal solution sensitive to changes in input parameters?
Possible reasons for asking this question:
Parameter values used were only best estimates.
Dynamic environment may cause changes.
“What-if” analysis may provide economical and operational information.

26. 3.7 Sensitivity Analysis of Objective Function Coefficients.
Range of Optimality
The optimal solution will remain unchanged as long as
An objective function coefficient lies within its range of optimality
There are no changes in any other input parameters.
The value of the objective function will change if the coefficient multiplies a variable whose value is nonzero.

27. The effects of changes in an objective function coefficient on the optimal solutionX2
1200
800
600
Max 8x1 + 5x2
Max 4x1 + 5x2
Max 3.75x1 + 5x2
Max 2x1 + 5x2 X1
400
600
800

28. The effects of changes in an objective function coefficient on the optimal solutionX2
1200
Range of optimality
Max8x1 + 5x2 10
Max 10 x1 + 5x2
Max 3.75 x1 + 5x2
3.75
800
600
Max8x1 + 5x2
Max 3.75x1 + 5x2 X1
400
600
800

29. Multiple changes
The range of optimality is valid only when a single objective function coefficient changes.
When more than one variable changes we turn to the
100% rule.
This is beyond the scope of this course

30. Complementary slackness
Reduced costs
The reduced cost for a variable at its lower bound (usually zero) yields:
The amount the profit coefficient must change before
the variable can take on a value above its lower bound.
Complementary slackness
At the optimal solution, either a variable is at its lower bound or the reduced cost is 0.

32. 3.8 Sensitivity Analysis of Right-Hand Side Values
Any change in a right hand side of a binding constraint will change the optimal solution.
Any change in a right-hand side of a non-binding constraint that is less than its slack or surplus, will cause no change in the optimal solution.

33. In sensitivity analysis of right-hand sides of constraints we are interested in the following questions:
Keeping all other factors the same, how much would the optimal value of the objective function (for example, the profit) change if the right-hand side of a constraint changed by one unit?
For how many additional UNITS is this per unit change valid?
For how many fewer UNITS is this per unit change valid?

34. Feasible The Plastic constraint The new Plastic constraintX2
1200
The Plastic constraint
The new Plastic constraint
2x1 + 1x2 <=1200
Maximum profit = 5040
2x1 + 1x2 <=1350
600
Production mix constraint
Production time
constraint
Feasible
Infeasible extreme points X1
600
800

35. Skip this detail Correct Interpretation of shadow prices
Sunk costs: The shadow price is the value of an extra unit of the resource, since the cost of the resource is not included in the calculation of the objective function coefficient.
Included costs: The shadow price is the premium value above the existing unit value for the resource, since the cost of the resource is included in the calculation of the objective function coefficient.

36. Range of feasibility
The set of right - hand side values for which the same set of constraints determines the optimal extreme point.
The range over-which the same variables remain in solution (which is another way of saying that the same extreme point is the optimal extreme point)
Within the range of feasibility, shadow prices remain constant; however, the optimal objective function value and decision variable values will change if the corresponding constraint is binding

37. 3.9 Other Post Optimality Changes SKIP THIS
Addition of a constraint.
Deletion of a constraint.
Addition of a variable.
Deletion of a variable.
Changes in the left - hand side technology coefficients.

38. 3.10 Models Without Optimal Solutions
Infeasibility: Occurs when a model has no feasible point.
Unboundedness: Occurs when the objective can become infinitely large.

39. Infeasibility No point, simultaneously,
lies both above line and below lines and 1 2 3 2 1 3

40. Unbounded solution
Maximize
the Objective Function
The feasible region

41. A cost minimization diet problem
Navy Sea Ration
A cost minimization diet problem
Mix two sea ration products: Texfoods, Calration.
Minimize the total cost of the mix.
Meet the minimum requirements of
Vitamin A, Vitamin D, and Iron.

42. Decision variables The Model
X1 (X2) -- The number of two-ounce portions of Texfoods (Calration) product used in a serving.
The Model
Minimize 0.60X X2
Subject to
20X X Vitamin A
25X X Vitamin D
50X X Iron
X1, X
Cost per 2 oz.
% Vitamin A
provided per 2 oz.
% required

43. The Graphical solution5
The Iron constraint
Feasible Region 4
Vitamin “D” constraint 2
Vitamin “A” constraint 2 4 5

44. Summary of the optimal solution
Texfood product = 1.5 portions (= 3 ounces)
Calration product = 2.5 portions (= 5 ounces)
Cost =$ 2.15 per serving.
The minimum requirements for Vitamin D and iron are met with no surplus.
The mixture provides 155% of the requirement for Vitamin A.

45. 3. 12 Computer Solution of Linear
3.12 Computer Solution of Linear Programs With Any Number of Decision Variables
Linear programming software packages solve large linear models.
Most of the software packages use the algebraic technique called the Simplex algorithm.
The input to any package includes:
The objective function criterion (Max or Min).
The type of each constraint:
The actual coefficients for the problem.

46. The typical output generated from linear programming software includes:
Optimal value of the objective function.
Optimal values of the decision variables.
Reduced cost for each objective function coefficient.
Ranges of optimality for objective function coefficients.
The amount of slack or surplus in each constraint.
Shadow (or dual) prices for the constraints.
Ranges of feasibility for right-hand side values.

47. WINQSB Input Data for the Galaxy Industries Problem
Variable and
constraint name can be
changed here
WINQSB Input Data for the Galaxy Industries Problem
Click to solve
Variables are
restricted to >= 0
No upper bound

48. Simplex Algorithm Basics
Starting at a feasible extreme point, the algorithm proceeds from extreme point to extreme point until one is found that is better than all neighboring extreme points
The transition from one extreme point to the next is called an iteration.
The algorithm chooses which extreme point to go to next based on the fastest rate of improvement

49. An iteration
One non-basic variable enters the basis and becomes a basic variable
One basis variable exits the basis and becomes non-basic
The basis variables re-solved in terms of the non-basis variables
The non-basis variables are fixed, usually at their lower bounds, which is zero, usually

50. Basis and non-basis variables
The basis variable values are free to take on values other than their lower bounds
The non-basis variables are fixed at their lower bounds (0)
THERE ARE ALWAYS AS MANY BASIS VARIABLES AS THERE ARE CONSTRAINTS, ALWAYS

51. Another problem--10 products
max 10x x x3 + 5 x4 + 8 x5 + 17x6 + 3 x7 + 9x8 + 5x9 + 11x10
s.t.
2x1 + x2 + 3x3 + x4 + 2x5 + 3x6 + x7 + 3x8 + 2x9 + x10 <= 100
all xi >= 0
How many basis variables?
How many products should we be making?