1. Introduction to Linear Programming
Chapter 2
Introduction to Linear Programming

2. Learning Objectives Formulate a maximization problem
Use graphical solution procedure
Identify feasible solution space
Define extreme points and optimal solution
Formulate a minimization problem.
Set up standard form of an LP problem

3. Introduction
Many decisions focus on the best way to achieve the objectives of business subject to the restrictions
Restrictions take the form of limited resources
One objective of firms is to maximize profit or to minimize cost
Linear Programming is to optimize the usage of limited resources
Derives its name from the functional relationships that are linear and the solution consists of mathematical steps, that is, a program

4. Construction of LP Model
Formulation of model that represents the problem, and then solve it
Three steps in applying LP technique
First, problem must be identified
Second, problem must be formulated
Third, model must be solved

5. Model Formulation Consists of certain components
Components include decision variables, objective function, and constraints
Decision variables represent levels of activity
Objective function describes objective of firm
Consists of either maximizing or minimizing some value
Constraints are linear relationships of decision variable
Represent restrictions placed on firm

6. Example Chairs and table from two resources—labor and wood
Labor is 40 hrs/day and wood is 120/lb
Each chair requires 1 hr of labor and 4 lb wood
A table requires 2 hrs of labor and 3 lb of wood
Profit: chair $40/unit and table $50/unit
Determine number of chairs and tables to produce each day in order to maximize profit
Formulate a linear programming model

7. Decision Variables Two decision variables Represented symbolically
x1 = number of chairs to produce
x2 = number of tables to produce

8. Objective Function Objective is to maximize total profit
Sum of profits gained from each chair and table
Profit derived from chairs is $40x1
Profit derived from tables is $50x2
Total profit, Z, can be expressed as $40x1 +50x2
“maximize” refers to maximize total profit:
maximize Z = $40x1 + 50x2

9. Model Constraints Two resources: labor and wood—both are limited
Each chair needs one hr of labor
Labor used for chairs is 1x1 hrs
Each table requires 2 hrs of labor
Labor used for tables 2x2 hrs
Total labor is the sum of the labor used:1x1 + 2 x2
Labor is limited to 40 hrs per day
Labor constraint: 1x1 +2x2 ≤40 hours
“≤“ is employed because 40 hrs of labor is a maximum
Means that it is possible to have idle, or excess capacity

10. Model Constraints-Cont.
Wood constraint is formulated in same way
Each chair requires 4 lb of wood, 4x1 lb; each table requires 3 lb of wood, 3 x2
Wood constraint: 4x1 + 3 x2 ≤ 120 Ib
Final restriction: produce either zero or a positive value
Referred to as nonnegativity, x1 ≥0, x2 ≥ 0

11. Complete LP Model Complete LP model Maximize Z = 40X1+50X2
subject to: 1X1+2X2 ≤ 40 hrs
4X1+3X2 ≤ 120 lb
X1,X2 ≥ 0
Results in numerical values for x1 and x2

12. Feasible Solution Consider x1 = 5 chairs and x2 = 10 tables
Substitute this solution into each constraint
1(5)+2(10)≤40, or 25<40 and 4(5)+3(10)≤ 20, or 50 <120
Since neither one of the constraints is violated, solution is feasible
Substituting in objective function: Z=40(5) + 50(10)=$700
Not know if $700 is the maximum profit

13. Infeasible Solution x1 =10 chairs and x2 = 20 tables
Results Z= $40(10)+50(20)=$1,400
Infeasible: violates labor constraint
Maximize Z without violating constraints
Optimal solution can be found using graphical solution approach

14. Graphical Solution Provides a picture of how a solution is obtained
Limited to only two decision variables
For three decision variables: cumbersome
Useful: gives a "picture" of how a solution is derived
Also provides how computer and mathematical solution approaches work

15. 1st Step in Graphical Solution
Recall the model
Formulated as
Maximize Z = 40X1+50X2
s.t. 1X1+2X2 ≤ 40 hours
4X1+3X2 ≤ 120 lb
X1,X2 ≥ 0
Only positive quadrant is valid
First step: plot constraints
Treat them as equations
Does not reflect entire constraint
3X1+4X2=120
X1+2X2=40

16. Feasible Solution Area
Shading two reflected areas produces an area referred to as feasible solution area
Some points will result in maximum profit
Next is to locate this point
3X1+4X2=120
X1+2X2=40
Feasible region

17. Optimal Solution Point
Select a level of profit
Z=$800=50 x1 +40 x2
Every point will result in a profit of $800
Consider Z=$1,200
Increase Z=$1,600
Outside feasible area
3X1+4X2=120
Z=1200
Z=1600
Z=800
X1+2X2=40
Feasible region

18. Optimal Solution Point-Cont.
Profit increases it moves away from origin
Maximum is at point where objective function line is farthest from the origin and is touching a point in feasible area
B--optimal solution B

19. Extreme Points Third step: solve for x1 and x2
Points on boundary called extreme points
Optimal solution will always occur at an extreme point
Possible solution points A, B, and C
Optimal solution--point B A B C

20. Finding Optimal Point
Point B can be found by solving two equations (x1=24 & x2 =8, Z =$1,360)
Optimal solution can also be found by testing each of the three points
X1=0 chairs
X2=20 tables
Z=$ 1000 A
X1=24 chairs
X2=8 tables
Z=$ 1360 B
X1=30 chairs
X2=0 tables
Z=$ 1200 C

21. Graphical Solution Steps
Steps are as follows
Plot all model constraints as equations
Indicate feasible solution area
Determine corner points; then, find the solution values at each point
Substitute these values into objective function

22. Minimization Problem
Two LP problems, a maximization and a minimization
Minimization is formulated in the same way, except for a few minor differences
Show how to formulate a minimization problem

23. A Minimization Problem
Razi makes fertilizer from two ingredients
Ingredients A and B provide different amounts of potassium and nitrogen
A provides 2 ounces of potassium and 4 ounces of nitrogen, whereas ingredient B provides 3 ounces of nitrogen and 4 ounces of potassium
A and B cost $6 and $3 per pound
How many pounds of each ingredient to put into a bag of fertilizer to meet minimum requirements of 16 ounces of potassium and 24 ounces of nitrogen while minimizing cost
Formulate a LP model

24. Decision Variables and Objective Function
Two decision variables representing the number of lbs of each ingredient
x1=lb of ingredient 1
x2=lb of ingredient 2
Objective: minimize total purchase cost of ingredients
Expressed
Min Z = 6 x1 + 3 x2

25. Model Constraints
Requirements for nitrogen and potassium represent the constraints
Each bag must meet a minimum requirement of nitrogen and potassium
Constraint for potassium: 2x1 +4x2 ≥16
Rather than a “≤” , this requires a “≥”
Because potassium content is a minimum requirement
Amount cannot be less than 16 oz
Constraint for nitrogen: 4x1 +3x2 ≥ 24 oz
Nonnegativity constraints indicate that negative amount of ingredients cannot be purchased

26. Equality Sign Shown two types of LP signs “≤” and “≥”
Third type is an “=“
Specifies a requirement that must be exact
Suppose requirement was exactly
4 x1 + 3 x2 = 24 ounces

27. Complete Model Complete model Min Z = 6 x1 + 3 x2 S.t. 2 x1 + 4 x2 ≥16
x1 and x2 ≥ 0

28. Graphical Solution First, graph the equations
Next, indicate feasible solution area
Third step, locate the optimal point
Optimal solution is located at one of the extreme points A, B, or C A
Feasible area B C

29. Moving Objective Function
Last point that touches feasible area is A
Point A is the closest point (x1 =0 and x2 =8)
Z=$24 A
X1=0
X2=8
Z=24
Feasible area B
Z=6X1-3X2 C

30. Limitation of Graphical Solution
Learned how a LP model is formulated and how a solution can be derived
Graphing provides valuable insight into LP
Limited to only two decision variables
Will show how LP problems can be solved using computer software packages

31. Slack Variable Graphical does not provide all the information
Need to transform the model into standard form
Necessary for interpretation of LP solutions
Recall the maximization example
Solution occurred at an extreme point
Constraints must all be in the form of equations “=“
Procedure for transforming “≤” constraints into “=“
Achieved by adding a new variable, called a slack variable

32. Addition of Slack Variable
Model constraints
x1 + 2x2 ≤ 40 hr of labor
4x1 + 3x2 ≤120 lb of wood
Addition of s1 and s2 to the constraint results
x1 + 2x2 + s1 = 40 hrs of labor
4 x1 + 3x2 + s2 = 120 Ib of wood
s1 and s2 will make the left-hand side=right-hand side
Represent amount of unused labor and amount of unused wood
Slack variables represent amount of unused resources

33. Slack Variables in Objective Function
Objective function represents the profit gained from the production of chairs and tables
What do slack variables s1 and s2 contribute?
Contribute nothing to profit because they represent unused resources
Write objective function
max Z = $40x1 + $50x2 +0s1 + 0s2
Slack variables can have only nonnegative values
x1, x2, s1, and s2 ≥ 0

34. Complete Model in Standard Form
Standard form with slack variables
Max Z=$40x1+50x2+0s1+0s2
s.t. x1+2x2+s1=40 4x1 +3x2+s2=120
x1, x2, s1, s2 ≥ 0
Shows slack variables at each point
X1=0 chairs
X2=20 tables
Z=$ 1000
S1=0
S2=60 A
X1=24 chairs
X2=8 tables
Z=$ 1360
S1=0
S2=0
X1=30 chairs
X2=0 tables
Z=$ 1200
S1=10
S2=0 B C

35. Converting ≥ Constraints
“≥” cannot be converted to equations by adding slack variables
Subtract a surplus variable
Slack variable reflects unused resources
Surplus variable reflects the excess above a minimum resource requirement level
Surplus variable is represented symbolically by s
Nitrogen constraint, 2x1+4x2 - s1=16
Extra amount of nitrogen above the minimum requirement of 16 ounces
Contributes nothing to the overall cost of the model
Will not affect the purchase cost

36. Standard Form of Minimization Problem
Standard form LP model
min Z = $6x1+3x2+0s1+0s2
s.t. 2x1+4x2 –s1=16
4x1+3x2 – s2=24
x1, x2, s1, s2 ≥ 0
Requires variables on left side and numerical values to the right of inequality or equality sign
Necessary condition to input problems into computer programs

37. A Simple LP Model
A company produces two types of house paints, interior and exterior. Two basic raw materials, A and B, are used to manufacture the paints.The maximum availability of A is 6 tons a day and 8 tons a day for B. The daily requirements of the raw materials per ton of interior and exterior paints are as follows
Tons of Row Material per Ton of Paint
Exterior
Interior
Max availability
Raw Material A 1 2 6
Raw Material B 8

38. Another Simple LP Model
A company produces two types of house paints, interior and exterior
Two basic raw materials, A and B, are used to manufacture the paints
The maximum availability of A is 6 tons a day and 8 tons a day for B
The daily requirements of the raw materials per ton of interior and exterior paints are as follows
Tons of Row Material per Ton of Paint
Exterior
Interior
Max. availability
Raw Material A 1 2 6
Raw Material B 8

39. Example-Cont.
The sales manager has stated that the daily demand for interior paint cannot exceed exterior paint by more than 1 ton
She/he has also stated that the maximum demand for interior paint is limited to 2 tons per day
The selling price is $3000/ton for exterior and $2000/ton for interior paint
How much interior and exterior paints should the company produce daily to maximize gross income?

40. Decision Variables
Determining the amounts (in tons) of interior and exterior paints to be produced
x1= tons produced daily of exterior paint
x2= tons produced daily of interior paint

41. Objective Function
The gross income from selling x1 tons is 3x1 (in $1000)
The gross income from selling x2 tons is 2x2 (in $1000)
Total gross income = 3x1+2x2, or
Z = 3x1+2x2

42. Constraints
Constraints impose restrictions on the usage of resources. Three types of restrictions in this example:
The usage restriction
The demand restriction
Nonnegativity restriction

43. The Usage Restriction (Usage of raw material by both paints) 
(maximum raw material availability)
For raw material A
x1+2x2  6
For raw material B
2x1+x2  8
Exterior x1
Interior x2
Max availability
Material A 1 2 6
Material B 8

44. Demand and Nonnegativity Restrictions
Demand restriction
1. Interior paint (x2) cannot exceed exterior paint (x1) by more than 1 ton/day
x2- x1  1
2. Maximum demand for interior paint is limited to 2 tons per day; or
x2 2
Nonnegativity restriction
Amount produced of each paint cannot be negative
x1 and x2 0

45. Complete Model Max Z = 3x1+2x2 s.t. x1+2x2  6 (1) 2x1+x2  8 (2)

46. Feasible Solution Feasible Solution
The values of the variables x1 and x2 are said to have a feasible solution if they satisfy all the constraints of the model
Example:
x1=1, x2=2; … solution
x1=0, x2=3; … solution

47. Graphical Solution
Plot the feasible space which satisfies all the constraints simultaneously
The nonnegativity restrictions (x1 and x2 0) confine all the feasible values to the first quadrant
Set all () by (=) for each constraint, yielding a straight-line equation
Plot each line on the (x1 , x2) plane
The feasible space is indicated by the direction of the arrow on the associated straight line

48. Solution Space--ABCDEF5
x2>0
2x1+x2<8
-x1+x2<1 2 3
x1+2x2<6
x2<2 4
x1>0 1 6

49. Optimum Solution Point
The optimum solution point can be determined by observing the direction in which the objective function 3x1+2x2 increases
Move the revenue line “uphill” to the point where any further increases in revenue would result an infeasible solution X2
optimum point
x1=10/3
x2=4/3
z=38/3 E D C F B A X1

50. Determination of Optimum Point
Point C is the intersection of line (1) and (2)
x1+2x2=6
2x1+x2=8
x1=10/3, x2=4/3
Value of the objective function
Z=3 (3 1/3)+ 2 (1 1/3)=38/3 ($1000)

51. Corner (Extreme) Points
Point A,B,C,D,E, and F are called corner points (corner-point feasible)
The optimal solution always occurs at one of the feasible corner (or extreme) points of the solution space E D C F B A

52. Standard Form of the LP Model
Z= value overall measure of performance
xj=level of activity j(for j=1,..n)
cj =increase in Z that would result from each unit increase in level of activity j
bi= amount of resources i that is available for allocation to activities (for i=1,2, .. m)
Max Z= c1x1+c2x2+ … cnxn
a11x1+a12x2+...+a1n xn  b1
a21x1+a22x2+...+a2n xn  b2 .
am1x1+am2x2+...+amn xn  bm
x1,x2,x3 ,… , xn0

53. Class Problem
A meat packing company produces a hot dog mixture in 1000-pound batches containing chicken and beef. The cost per pound of each of these ingredients is as follows:
Ingredient Cost/lb Chicken $ Beef $5
Each batch has the following recipe requirements:
A: At least 500 pounds of chicken
B: At least 200 pounds of beef
the ratio of chicken to beef must be at least 2 to 1. The company wants to the optimal mixture of ingredients that minimize cost. Formulate the problem.

54. Complete Model Identify the decision variables
Formulate the obj fcn as Min z=… +…
Establish model constraints as …

55. Summary Provided you with a good start in LP modeling
Learned how to formulate two simple LP models, a maximization and a minimization problem
Showed how to use the graphical approach to find optimal solution point and how to put a linear programming model in the standard format

56. HW Assignments
Problem set 2.1A: prob. 2. Problem Set 2.2A: Prob. 5, 11 and 12 Problem set 2.2B: prob. 4. Problem set 3.1A: prob. 1 and prob. 2.
Problem set 3.6A: Problem 2. Problem set 3.6B: Problem 3. Problem set 2.3D: Problem 3. These problems can be solved using either by Solver or TORA.