1. Chapter 2 Linear Programming Models: Graphical and Computer Methods

2. Outline
LP Meaning
LP Models
Model Formulation
Solving LP

3. LP Meaning
LP is a powerful quantitative tool used by the bp to obtain optimal solutions to problem that involve restrictions or limitations.
LP techniques consists of sequence of steps that will lead to an optimal solutions to linear-constrained problems.
LP involves the use of straight line relationships to maximize or minimise some functions usually called objective functions.

4. Properties of LP Models
Seek to minimize or maximize
Include “constraints” or limitations
There must be alternatives available
All equations are linear

5. LP models
LP models are mathematical representations of constrained optimization problems.
Four components provide the structure of a LP models
Objective functions
Decision variables
Constraints
Parameters

6. Objective function: mathematical statement of profit(or cost) for a given solution
Decision variables: Amount of either inputs or outputs
Constraints: Limitations that restrict the available alternatives.3 types of constraints ie <=,>= or=

7. Solving the LP:Outline of Graphical procedure
Set up the objective function and the constraints in mathematical format
Plot the constraints
Identify the feasible solution space
Plot the objective function
Determine the optimum solution

8. Steps in Developing a Linear Programming (LP) Model
Formulation
Solution
Interpretation and Sensitivity Analysis

9. Example LP Model Formulation: The Product Mix Problem
Decision: How much to make of > 2 products?
Objective: Maximize profit
Constraints: Limited resources

10. Example: Flair Furniture Co.
Two products: Chairs and Tables
Decision: How many of each to make this month?
Objective: Maximize profit

11. Flair Furniture Co. Data
Tables
(per table)
Chairs
(per chair)
Hours Available
Profit Contribution $7 $5
Carpentry
3 hrs
4 hrs
2400
Painting
2 hrs
1 hr
1000
Other Limitations:
Make no more than 450 chairs
Make at least 100 tables

12. Decision Variables:
T = Num. of tables to make
C = Num. of chairs to make
Objective Function: Maximize Profit
Maximize $7 T + $5 C

13. Have 2400 hours of carpentry time available
Constraints:
Have 2400 hours of carpentry time available
3 T C < (hours)
Have 1000 hours of painting time available
2 T C < (hours)
Nonnegativity:
Cannot make a negative number of chairs or tables
T > 0
C > 0

14. Model Summary Max 7T + 5C (profit) 3T + 4C < 2400 (carpentry hrs)
Subject to the constraints:
3T + 4C < 2400 (carpentry hrs)
2T + 1C < 1000 (painting hrs)
T, C > 0 (nonnegativity)

15. Graphical Solution
Graphing an LP model helps provide insight into LP models and their solutions.
While this can only be done in two dimensions, the same properties apply to all LP models and solutions.

16. Carpentry Constraint Line 3T + 4C = 2400 Intercepts (T = 0, C = 600)
Carpentry
Constraint Line
3T + 4C = 2400
Intercepts
(T = 0, C = 600)
(T = 800, C = 0)
Infeasible
> 2400 hrs
3T + 4C = 2400
Feasible
< 2400 hrs T

17. Painting Constraint Line 2T + 1C = 1000 Intercepts (T = 0, C = 1000)
Painting
Constraint Line
2T + 1C = 1000
Intercepts
(T = 0, C = 1000)
(T = 500, C = 0)
2T + 1C = 1000 T

18. Corner point solution Value of Z at corner point a Value of Z at d
Z= 7T+5C
Za =7.0+5*600
Za=3000
Value of Z at d
Z=7T+5C
Zd=7*500+5*0=3500

19. Use simultaneous equation
Value of Z at b
Z=7T+5C
Use simultaneous equation
(1) 3T + 4C =2400 (carpentry hrs)
(2) 2T + 1C = 1000 (painting hrs)
Multiply equation 2 by 4
(3) 8T+4C =4000
1-3= -5T= -1600
T= 320 now substitute the value of T in any equation
2*320+ 1C=1000
1C=
C= 360
Z= 7*320+5*360=4040

20. LP Characteristics
Feasible Region: The set of points that satisfies all constraints
Corner Point Property: An optimal solution must lie at one or more corner points
Optimal Solution: The corner point with the best objective function value is optimal

21. Special Situation in LP
Redundant Constraints - do not affect the feasible region
Example: x < 10
x < 12
The second constraint is redundant because it is less restrictive.

22. Problem-1
A frim can produce two types of a certain product, namely basic B and deluxeD.The estimated profit per unit is $10 Basic and $15 Deluxe.The firm has 60 hours of labour available per day and 100 units of capital.Each unit of basic product B requires two hrs of labor and four units of capital.Each unit de-luxe product D requires four hrs of labor and five units of capital. How much of each type of product should the firm produce in order to maximise total profit?

23. Model Summary Maximise Z= 10B+15D …objective function Subject to:
2B+4D<= 60… labour constraint
4B+5D<=100...Capital constraint
B>=0,D>= non-negative constraints

24. Plotting the constraints
The labour constraint line:if we use all our 60 hrs of labor on the basic B we can produce 30 units of B but zero D.At the other extreme if use all our labor to produce D we can produce 15 D but Zero B
The capital Constraint line same way we can get B= 25 and D=20

25. Identify the feasible solution space
Feasible region of production area OVWX
Value of Z at corner point V
Zv=10B+15D=10*25+15*0=250
Value of Z at corner point X
Zx=10*0+15*5=225
Value of Z at corner point W
(1)2B+4D=60
(2)4B+5D=100
Here multiply equation 1 by 2 and subtract the equation
4B+8D=120
4B+5D=100
3D= 20 D= 20/3 then substitute D & get the Value of B=16 2/3 and we can solve for Z at W= 266 2/3

26. Problem2
A firm uses three types of factor input, namely labour, capital and raw material in producing two products X and Y. Each unit of X contributes $20 to profit, each unit of Y contributes $30 to profit. To produce one unit of X requires one unit of labour, one unit of capital and two units of raw material. To produce one unit of Y requires one unit of labour, two units of capital and one unit of raw material. The firm has 50 units of labour,80 units of capital and 80 units of raw material available. How much of each type of product should the firm produce in order to maximise total profit?

27. Minimisation Problem-3
A travel company operates two types of vehicle A and B. Vehicle A can carry 40 passengers and 30 tons of baggage; Vehicle B can carry 60 passengers but only 15 tons of baggage. The travel company is contracted to carry at least 960 passengers and 360 tons of baggage per journey. If vehicle A cost $1000 to operate per journey and vehicle B costs $1200 to operate per journey. What choice of vehicles will minimise the total cost

28. Minimise Z= 1000A+1200B
Subject to
40A+60B>= 960…Passengers constraint
30A+15B>=360…baggage constraint
A>=0,B>=0… non negative constraints

29. Using Excel’s Solver for LP
Recall the Flair Furniture Example:
Max 7T + 5C (profit)
Subject to the constraints:
3T + 4C < 2400 (carpentry hrs)
2T + 1C < 1000 (painting hrs)
C < (max # chairs)
T > (min # tables)
T, C > 0 (nonnegativity)
Go to file 2-1.xls