Here is the standard LP formulation of the problem: Maximize: 3000X + 2000Y Subject to: X + 2Y ≤ 6 (Constraint on raw material A) 2X + Y ≤ 8 (Constraint on raw material B) Y ≤ X + 1 (Demand for interior cannot exceed exterior by 1 ton) Y ≤ 2 (Maximum demand for interior is 2 tons) X, Y ≥ 0 Where: X = Quantity of exterior paint produced (decision variable 1) Y = Quantity of interior paint produced (decision variable 2) The objective is to maximize total profit by choosing the optimal values of X and Y.

1. 1
Linear Programming
Requirements
 Maximize or minimize an objective
 Constraints act as limiting factor
 There must be alternatives available
 Mathematical relationship is linear

2. 2
Basic Assumptions of LP
 Certainty (R/M per unit, Profit per unit)
 Proportionality ( R/M for 10 U = 1 U x 10)
 Additivity ( Profit From A & B =?)
 Divisibility (non integer solution- no of
chair produced per week = 2.5)

3. An Example LP Problem
Blue Ridge Hot Tubs produces two types of hot
tubs: Type X & Type Y.
There are 200 pumps, 1566 hours of labor,
and 2880 feet of tubing available.
Type X Type Y
Pumps 1 1
Labor 9 hours 6 hours
Tubing 12 feet 16 feet
Unit Profit $350 $300

4. 5 Steps In Formulating LP Models:
1. Understand the problem.
2. Identify the decision variables.
X1=number of Type X to produce
X2=number of Type Y to produce
3. State the objective function as a linear
combination of the decision variables.
MAX: 350X1 + 300X2

5. 5 Steps In Formulating LP
Models
(continued)
4. State the constraints as linear combinations
of the decision variables.
1X1 + 1X2 <= 200 } pumps
9X1 + 6X2 <= 1566 } labor
12X1 + 16X2 <= 2880 } tubing
5. Identify any upper or lower bounds on the
decision variables.
X1 >= 0
X2 >= 0

6. LP Model for Blue Ridge Hot
Tubs
MAX: 350X1 + 300X2
S.T.: 1X1 + 1X2 <= 200
9X1 + 6X2 <= 1566
12X1 + 16X2 <= 2880
X1 >= 0
X2 >= 0

7. Solving LP Problems:
An Intuitive Approach
 Idea: Each Type X (X1) generates the highest unit profit
($350), so let’s make as many of them as possible!
 How many would that be?
 Let X2 = 0
 1st constraint: 1X1 <= 200
 2nd constraint: 9X1 <=1566 or X1 <=174
 3rd constraint: 12X1 <= 2880 or X1 <= 240
 If X2=0, the maximum value of X1 is 174 and the total
profit is $350*174 + $300*0 = $60,900
 This solution is feasible, but is it optimal?
 No!

8. Solving LP Problems:
A Graphical Approach
 The constraints of an LP problem
defines its feasible region.
 The best point in the feasible region is
the optimal solution to the problem.
 For LP problems with 2 variables, it is
easy to plot the feasible region and
find the optimal solution.

9. X2
X1
250
200
150
100
50
0
0 50 100 150 200 250
(0, 200)
(200, 0)
boundary line of pump constraint
X1 + X2 = 200
Plotting the First Constraint

10. X2
X1
250
200
150
100
50
0
0 50 100 150 200 250
(0, 261)
(174, 0)
boundary line of labor constraint
9X1 + 6X2 = 1566
Plotting the Second Constraint

11. X2
X1
250
200
150
100
50
0
0 50 100 150 200 250
(0, 180)
(240, 0)
boundary line of tubing constraint
12X1 + 16X2 = 2880
Feasible Region
Plotting the Third Constraint

12. X2
Plotting A Level Curve of the
Objective Function
X1
250
200
150
100
50
0
0 50 100 150 200 250
(0, 116.67)
(100, 0)
objective function
350X1 + 300X2 = 35000

13. A Second Level Curve of the
Objective Function
X2
X1
250
200
150
100
50
0
0 50 100 150 200 250
(0, 175)
(150, 0)
objective function
350X1 + 300X2 = 35000
objective function
350X1 + 300X2 = 52500

14. Using A Level Curve to Locate
the Optimal Solution
X2
X1
250
200
150
100
50
0
0 50 100 150 200 250
objective function
350X1 + 300X2 = 35000
objective function
350X1 + 300X2 = 52500
optimal solution

15. Calculating the Optimal
Solution
 The optimal solution occurs where the “pumps” and
“labor” constraints intersect.
 This occurs where:
X1 + X2 = 200 (1)
and 9X1 + 6X2 = 1566 (2)
 From (1) we have, X2 = 200 -X1 (3)
 Substituting (3) for X2 in (2) we have,
9X1 + 6 (200 -X1) = 1566
which reduces to X1 = 122
 So the optimal solution is,
X1=122, X2=200-X1=78
Total Profit = $350*122 + $300*78 = $66,100

16. Enumerating The Corner Points
X2
X1
250
200
150
100
50
0
0 50 100 150 200 250
(0, 180)
(174, 0)
(122, 78)
(80, 120)
(0, 0)
obj. value = $54,000
obj. value = $64,000
obj. value = $66,100
obj. value = $60,900
obj. value = $0

17. Summary of Graphical
Solution
to LP Problems
1. Plot the boundary line of each constraint
2. Identify the feasible region
3. Locate the optimal solution by either:
a. Plotting level curves
b. Enumerating the extreme points

18. Special Conditions in LP
Models
 A number of anomalies can occur in LP
problems:
 Alternate Optimal Solutions
 Redundant Constraints
 Unbounded Solutions
 Infeasibility

19. Example of Alternate Optimal
Solutions
X2
X1
250
200
150
100
50
0
0 50 100 150 200 250
450X1 + 300X2 = 78300
objective function level curve
alternate optimal solutions

20. Example of a Redundant Constraint
X2
X1
250
200
150
100
50
0
0 50 100 150 200 250
boundary line of tubing constraint
Feasible Region
boundary line of pump constraint
boundary line of labor constraint

21. Example of an Unbounded Solution
X2
X1
1000
800
600
400
200
0
0 200 400 600 800 1000
X1 + X2 = 400
X1 + X2 = 600
objective function
X1 + X2 = 800
objective function
-X1 + 2X2 = 400

22. 22

 There are no points that satisfy both constraints, hence this problem has
no feasible region, and no optimal solution.
x2
x1
4x1 + 3x2 < 12
2x1 + x2 > 8
3 4
4
8

23. 23
The XYZ Company owns a small paint factory that produces
both exterior and interior house paints. Two basic raw
materials A and B are required to manufacture the paints.
Maximum availability of raw material A and B are 6 tons and 8
tons per day respectively. To produce 1 ton of exterior paint
they need 1 ton of A and 2 tons of B. To produce 1 ton of
interior paint they need 2 tons of A and 1 ton of B. A market
survey has established that the daily demand for interior paint
cannot exceed that of exterior paint by 1 ton. The survey also
shows that the maximum demand for interior paint is limited
to 2 tons per day. Profit for exterior and interior paint is Tk.
3000 and Tk. 2, 000 respectively.
 (A) Represent the problem in standard LP form.