Chapter 2Linear Programming Models:Graphical and Computer Methods

Outline

• LP Meaning

• LP Models

• Model Formulation

• Solving LP

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.

Properties of LP Models

1) Seek to minimize or maximize

2) Include “constraints” or limitations

3) There must be alternatives available

4) All equations are linear

LP models

• LP models are mathematical representations of constrained optimization problems.

• Four components provide the structure of a LP modelsi. Objective functions

ii. Decision variables

iii. Constraints

iv. Parameters

• 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=

Solving the LP:Outline of Graphical procedure

i. Set up the objective function and the

constraints in mathematical format

ii. Plot the constraints

iii. Identify the feasible solution space

iv. Plot the objective function

v. Determine the optimum solution

Steps in Developing a Linear Programming (LP) Model

1) Formulation

2) Solution

3) Interpretation and Sensitivity Analysis

Example LP Model Formulation:The Product Mix Problem

Decision: How much to make of > 2 products?

Objective: Maximize profit

Constraints: Limited resources

Example: Flair Furniture Co.

Two products: Chairs and Tables

Decision: How many of each to make this month?

Objective: Maximize profit

Flair Furniture Co. DataTables

(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

Decision Variables:

T = Num. of tables to make

C = Num. of chairs to make

Objective Function: Maximize Profit

Maximize $7 T + $5 C

Constraints:

• Have 2400 hours of carpentry time available

3 T + 4 C < 2400 (hours)

• Have 1000 hours of painting time available

2 T + 1 C < 1000 (hours)Nonnegativity:Cannot make a negative number of chairs or tables

T > 0C > 0

Model Summary

Max 7T + 5C (profit)

Subject to the constraints:

3T + 4C < 2400 (carpentry hrs)

2T + 1C < 1000 (painting hrs)

T, C > 0(nonnegativity)

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.

Carpentry

Constraint Line

3T + 4C = 2400

Intercepts

(T = 0, C = 600)

(T = 800, C = 0)

0 800 T

C

600

0

Feasible

< 2400 hrs

Infeasible

> 2400 hrs

3T + 4C = 2400

Painting

Constraint Line

2T + 1C = 1000

Intercepts

(T = 0, C = 1000)

(T = 500, C = 0)

0 500 800 T

C1000

600 a

0

2T + 1C = 1000

Corner point solution

• Value of Z at corner point a• Z= 7T+5C• Za =7.0+5*600• Za=3000

• Value of Z at d• Z=7T+5C• Zd=7*500+5*0=3500

• 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= 1000-640• C= 360• Z= 7*320+5*360=4040

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

Special Situation in LP

1. Redundant Constraints - do not affect the feasible region

Example: x < 10

x < 12

The second constraint is redundant because it is less restrictive.

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?

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

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

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

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?

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

• Minimise Z= 1000A+1200B

• Subject to

• 40A+60B>= 960…Passengers constraint

• 30A+15B>=360…baggage constraint

• A>=0,B>=0… non negative constraints

Using Excel’s Solver for LPRecall the Flair Furniture Example:

Max 7T + 5C (profit)Subject to the constraints:

3T + 4C < 2400 (carpentry hrs)

2T + 1C < 1000 (painting hrs)

C < 450 (max # chairs)

T > 100 (min # tables)

T, C > 0 (nonnegativity)

Go to file 2-1.xls