Linear Programming Part 2

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Linear ProgrammingPart 2

• By
• Anita Lee-Post

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LP solution methods

• Graphical solution method
• Corner-point solution method
• Excel Solver solution method

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Graphical solution method

• Limited to 2 decision variables problems
• Assign X1 and X2 to the vertical and horizontal
  axes
• Plot constraint equations
• Determine the area of feasibility
• Plot the objective function
• Find the optimum point

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LP example 1

• Minimize Cost 5X1 3X2
• Subject to
• 100X1 100X2 gt 4000
• 200X1 400X2 gt 10000
• 200X1 100X2 gt 5000
• X1, X2 gt 0

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Graphical solution method

• Step 1. Assign decision variables to axes

X1
X2
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Graphical solution method continued

• Step 2. Plot constraint 100X1100X2 gt 4000
• Plot the line 100X1 100X2 4000 ? (0, 40),
  (40,0)
• Determine the constraint region ? (0, 0) lt 4000,
  thus, the constraint region is to the right of
  the line

X1
40
X2
(0,0)
40
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Graphical solution method continued

• Step 2. Plot constraint 200X1400X2 gt 10000
• Plot the line 200X1 400X2 10000 ? (0, 25),
  (50,0)
• Determine the constraint region ? (0, 0) lt 10000,
  thus, the constraint region is to the right of
  the line

X1
50
40
X2
(0,0)
40
25
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Graphical solution method continued

• Step 2. Plot constraint 200X1100X2 gt 5000
• Plot the line 200X1 100X2 5000 ? (0, 50),
  (25,0)
• Determine the constraint region ? (0, 0) lt 5000,
  thus, the constraint region is to the right of
  the line

X1
50
40
25
X2
(0,0)
40
25
50
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Graphical solution method continued

• Step 3. Determine the area of feasibility
• The area of feasibility is found by intersecting
  the three constraint regions

X1
50
Area of feasibility
40
25
X2
(0,0)
40
25
50
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Graphical solution method continued

• Step 4. Plot the objective function cost 5X1
  3X2
• Assume an arbitrary cost, e.g., 150
• Plot the line 5X1 3X2 150 ? (0, 50), (30,0)

X1
Area of feasibility
30
X2
(0,0)
50
Cost 150
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Graphical solution method continued

• Step 5. Find the optimal point
• The region to the left of the iso-cost line
  (Cost150) is represented by Cost lt 150 ? Move
  the iso-cost line leftwards until it reaches the
  most extreme point of the area of feasibility.
• The extreme point is the optimal point ? (10, 30)

X1
Area of feasibility
30
(10,30)
X2
Cost 150
50
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LP example 2

• Maximize Profit 40X1 30X2
• Subject to
• X1 lt 400
• X2 lt 700
• X1 X2 lt 800
• X1 2X2 lt 1000
• X1, X2 gt 0

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Graphical solution method
X1

• Following the 5-step process, the optimal point
  is found to be X1400, X2300

1000
800
X2700
X12X21000
X1X2800
(400,300)
X1400
400
300
Profit 12000
Area of feasibility
X2
700
800
500
400
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Corner-point solution method

• Use in conjunction with the graphical solution
  method to pinpoint the optimal point
  algebraically
• Find the co-ordinates of each corner point of the
  feasible region by simultaneously solving the
  equations of a pair of intersecting lines
• Substitute the values of the co-ordinates in the
  objective function

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LP example 1
Corner Point Intersecting lines Coordinates Cost5X1 3X2
A X20 200X1 400X210000 X150, X20 Cost250
B 100X1 100X2 4000 200X1 400X210000 X130, X210 Cost180
C 100X1 100X2 4000 200X1 100X2 5000 X110, X230 Cost140
D X10 200X1 100X2 5000 X10, X250 Cost150
X1
Optimal solution
A
B
C
D
X2
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LP example 2
Corner Point Intersecting lines Coordinates Profit40X1 30X2
A X20 X1400 X1400, X20 Profit16000
B X1400 X1 2X2 1000 X1400, X2300 Profit25000
C X10 X1 2X2 1000 X10, X2500 Profit15000
D X10 X20 X10, X20 Profit0
X1
Optimal solution
A
B
X1400
X1 2X2 1000
C
D
X2