Chapter 2
Linear 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 models
i.
ii.
iii.
iv.
Objective functions
Decision variables
Constraints
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. Data
Tables
Chairs
(per table)
(per chair)
Profit
Contribution
$7
$5
Hours
Available
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>0
C>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
C
3T + 4C = 2400
Infeasible
> 2400 hrs
600
Intercepts
(T = 0, C = 600)
(T = 800, C = 0)
Feasible
< 2400 hrs
0
0
800 T
C
1000
Painting
Constraint Line
2T + 1C = 1000
600 a
Intercepts
(T = 0, C = 1000)
(T = 500, C = 0)
0
0
500
800 T
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 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 < 450 (max # chairs)
T
> 100 (min # tables)
T, C > 0
(nonnegativity)
Go to file 2-1.xls