LP with graphical solution.

Additional examples
LP
1
Let:
X1, X2, X3, ………, Xn = decision variables
Z = Objective function or linear function
Requirement: Maximization of the linear function Z.
Z = c1X1 + c2X2 + c3X3 + ………+ cnXn
…..Eq (1)
subject to the following constraints:
…..Eq (2)
Formulating LP Problems
Hours Required
to Produce 1 Unit
Department
Electronic
Assembly
Profit per unit
X-pods
(X1)
BlueBerrys
(X2)
Available Hours
This Week
4
2
$7
3
1
$5
240
100
Table B.1
Decision Variables:
X1 = number of X-pods to be produced
X2 = number of BlueBerrys to be produced
Formulating LP Problems
Objective Function:
Maximize Profit = $7X1 + $5X2
There are three types of constraints
 Upper limits where the amount used is ≤ the
amount of a resource
 Lower limits where the amount used is ≥ the
amount of the resource
 Equalities where the amount used is = the
amount of the resource
Formulating LP Problems
First Constraint:
Electronic
time used
is ≤
Electronic
time available
4X1 + 3X2 ≤ 240 (hours of electronic time)
Second Constraint:
Assembly
time used
is ≤
Assembly
time available
2X1 + 1X2 ≤ 100 (hours of assembly time)
Graphical Solution
 Can be used when there are two
decision variables
1. Plot the constraint equations at their
limits by converting each equation to an
equality
2. Identify the feasible solution space
3. Create an iso-profit line based on the
objective function
4. Move this line outwards until the optimal
point is identified
Graphical Solution
X2
100 –
–
Number of BlueBerrys
80 –
–
60 –
–
40 –
–
20 –
–
Figure B.3
Assembly (constraint B)
|–
0
Electronics (constraint A)
Feasible
region
|
|
20
|
|
40
|
|
60
|
Number of X-pods
|
80
|
|
100
X1
Graphical Solution
Iso-Profit
Line Solution Method
X
2
Choose a possible
100 – value for the objective function
–
Number of Watch TVs
80 –
$210
–
Assembly (constraint B)
= 7X1 + 5X2
60 –
Solve for the axis– intercepts of the function and plot the line
40 –
–
20 –
–
Figure B.3
|–
0
Electronics (constraint A)
Feasible
X2 = 42
region
|
|
20
|
|
40
X1 = 30
|
|
60
|
Number of X-pods
|
80
|
|
100
X1
Graphical Solution
X2
100 –
–
Number of BlueBerrys
80 –
–
60 –
–
40 –
$210 = $7X1 + $5X2
(0, 42)
–
(30, 0)
20 –
–
Figure B.4
|–
0
|
|
20
|
|
40
|
|
60
|
Number of X-pods
|
80
|
|
100
X1
Graphical Solution
X2
100 –
$350 = $7X1 + $5X2
–
Number of BlueBeryys
80 –
$280 = $7X1 + $5X2
–
60 –
$210 = $7X1 + $5X2
–
40 –
–
$420 = $7X1 + $5X2
20 –
–
Figure B.5
|–
0
|
|
20
|
|
40
|
|
60
|
Number of X-pods
|
80
|
|
100
X1
Graphical Solution
X2
100 –
Maximum profit line
–
Number of BlueBerrys
80 –
–
60 –
Optimal solution point
(X1 = 30, X2 = 40)
–
40 –
–
$410 = $7X1 + $5X2
20 –
–
Figure B.6
|–
0
|
|
20
|
|
40
|
|
60
|
Number of X-pods
|
80
|
|
100
X1
Corner-Point Method
X2
100 –
2
–
Number of BlueBerrys
80 –
–
60 –
–
3
40 –
–
20 –
–
Figure B.7
1
|–
0
|
|
20
|
|
40
|
4
|
60
|
Number of X-pods
|
80
|
|
100
X1
Corner-Point Method
 The optimal value will always be at a corner
point
 Find the objective function value at each
corner point and choose the one with the
highest profit
Point 1 :
(X1 = 0, X2 = 0)
Profit $7(0) + $5(0) = $0
Point 2 :
(X1 = 0, X2 = 80)
Profit $7(0) + $5(80) = $400
Point 4 :
(X1 = 50, X2 = 0)
Profit $7(50) + $5(0) = $350
Corner-Point Method
 The optimal value will always be at a corner
Solve for the intersection of two constraints
point
(electronics
1 + 3X2 ≤ 240
 Find the4X
objective
function
value time)
at each
2X1 + and
1X2 ≤choose
100 (assembly
time)
corner point
the one
with the
highest profit
4X1 + 3X2 = 240
- 4X1 - 2X2 = -200
Point 1 :
(X1 = 0, X2 = 0)
+ 1X2 = 40
Point 2 :
(X1 = 0, X2 = 80)
4X1 + 3(40) = 240
4X1 + 120 = 240
Profit $7(0) + $5(0) = $0
X1 = 30
Point 4 :
(X1 = 50, X2 = 0)
Profit $7(50) + $5(0) = $350
Profit $7(0) + $5(80) = $400
Corner-Point Method
 The optimal value will always be at a corner
point
 Find the objective function value at each
corner point and choose the one with the
highest profit
Point 1 :
(X1 = 0, X2 = 0)
Profit $7(0) + $5(0) = $0
Point 2 :
(X1 = 0, X2 = 80)
Profit $7(0) + $5(80) = $400
Point 4 :
(X1 = 50, X2 = 0)
Profit $7(50) + $5(0) = $350
Point 3 :
(X1 = 30, X2 = 40)
Profit $7(30) + $5(40) = $410
The Galaxy Industries Production Problem –
A Prototype Example
• Galaxy manufactures two toy doll models:
– Space Ray.
– Zapper.
• Resources are limited to
– 1000 pounds of special plastic.
– 40 hours of production time per week.
16
The Galaxy Industries Production Problem –
A Prototype Example
• Marketing requirement
– Total production cannot exceed 700 dozens.
– Number of dozens of Space Rays cannot exceed
number of dozens of Zappers by more than 350.
• Technological input
– Space Rays requires 2 pounds of plastic and
3 minutes of labor per dozen.
– Zappers requires 1 pound of plastic and
4 minutes of labor per dozen.
17
The Galaxy Industries Production Problem –
A Prototype Example
• The current production plan calls for:
– Producing as much as possible of the more profitable product,
Space Ray ($8 profit per dozen).
– Use resources left over to produce Zappers ($5 profit
per dozen), while remaining within the marketing guidelines.
• The current production plan consists of:
8(450) + 5(100)
Space Rays = 450 dozen
Zapper
= 100 dozen
Profit
= $4100 per week
18
Management is seeking a
production schedule that will
increase the company’s profit.
19
A linear programming model
can provide an insight and an
intelligent solution to this problem.
20
The Galaxy Linear Programming Model
• Decisions variables:
– X1 = Weekly production level of Space Rays (in dozens)
– X2 = Weekly production level of Zappers (in dozens).
• Objective Function:
– Weekly profit, to be maximized
21
The Galaxy Linear Programming Model
Max 8X1 + 5X2
subject to
2X1 + 1X2  1000
3X1 + 4X2  2400
X1 + X2  700
X1 - X2  350
Xj> = 0, j = 1,2
(Weekly profit)
(Plastic)
(Production Time)
(Total production)
(Mix)
(Nonnegativity)
22
Using a graphical presentation
we can represent all the constraints,
the objective function, and the three
types of feasible points.
23
Graphical Analysis – the Feasible Region
X2
The non-negativity constraints
X1
24
Graphical Analysis – the Feasible Region
X2
The Plastic constraint
2X1+X2  1000
1000
Total production constraint:
X1+X2  700 (redundant)
700
500
Infeasible
Production
Time
3X1+4X2  2400
Feasible
500
700
X1
25
Graphical Analysis – the Feasible Region
X2
1000
The Plastic constraint
2X1+X2 1000
Total production constraint:
X1+X2 700 (redundant)
700
500
Production
Time
3X1+4X22400
Infeasible
Production mix
constraint:
X1-X2  350
Feasible
500
700
X1
Interior points. Boundary points. Extreme points.
• There are three types of feasible points
26
Solving Graphically for an
Optimal Solution
27
The search for an optimal solution
X2
1000
Start at some arbitrary profit, say profit = $2,000...
Then increase the profit, if possible...
...and continue until it becomes infeasible
700
Profit =$4360
500
X1
28
500
Summary of the optimal solution
Space Rays = 320 dozen
Zappers
= 360 dozen
Profit = $4360
– This solution utilizes all the plastic and all the production hours.
– Total production is only 680 (not 700).
– Space Rays production exceeds Zappers production by only 40
dozens.
29
Extreme points and optimal solutions
– If a linear programming problem has an optimal
solution, an extreme point is optimal.
30
Multiple optimal solutions
• For multiple optimal solutions to exist, the objective
function must be parallel to one of the constraints
•Any weighted average of
optimal solutions is also an
optimal solution.
31
Minimization Problem
CHEMICAL CONTRIBUTION
Brand
Nitrogen (lb/bag)
Phosphate (lb/bag)
2
4
4
3
Gro-plus
Crop-fast
Minimize Z = $6x1 + $3x2
subject to
2x1 + 4x2  16 lb of nitrogen
4x1 + 3x2  24 lb of phosphate
x 1, x 2  0
Copyright 2006 John Wiley & Sons, Inc.
Supplement 13-32
Graphical Solution
x2
14 –
12 –
x1 = 0 bags of Gro-plus
x2 = 8 bags of Crop-fast
Z = $24
10 –
8–A
Z = 6x1 + 3x2
6–
4–
B
2–
0–
|
2
|
4
Copyright 2006 John Wiley & Sons, Inc.
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6
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8
C
Supplement 13-33
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10
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12
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14
x1
Dual problem (2 vars primal)
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