Linear Programming (LP)

1
Linear Programming (LP)

• Decision Variables
• Objective (MIN or MAX)
• Constraints
• Graphical Solution

2
History of LP

• World War II shortages
• Limited resources
• Research at RAND in Santa Monica
• Examples limited number of machines
• Limited number of skilled workers
• Budget limits
• Time restrictions (deadlines)
• Raw materials

3
LP Requirements

• Single objective MAX or MIN
• Objective must be linear function
• Linear constraints

4
Linear Programming

• I. Profit maximization example
• II. Cost minimization example

5
I. Profit MAX Example

• Source Render and Stair, Quantitative Analysis
  for Management , Ch 2
• Furniture factory
• Decision variables
• X1 number of tables to make
• X2 number of chairs to make

6
Objective MAX profit

• Each table 7 profit
• Each chair 5 profit
• Total profit 7X1 5X2

7
Carpenter Constraint

• Labor Constraint
• 240 hours available per week
• Each table requires 4 hours from carpenter
• Each chair requires 3 hours from carpenter
• 4X1 3X2 lt 240

8
Painter Constraint

• 100 hours available per week
• Each table requires 2 hours from painter
• Each chair requires 1 hour from painter
• 2X1 X2 lt 100

9
Non-negativity constraints

• X1 gt 0
• X2 gt 0
• Cant have negative production

10
Graphical Solution

• Non-negativity constraints imply positive
  (northeast) quadrant

11
X2
X1
0,0
12
Plot carpenter constraint

• Temporarily convert to equation
• 4X1 3X2 240
• Intercept on X1 axis X2 0
• 4X1 3(0) 240
• 4X1 240
• X1 240/4 60
• Coordinate (60,0)

13
X2
X1
0,0
60,0
14
Plot carpenter constraint

• Temporarily convert to equation
• 4X1 3X2 240
• Intercept on X2 axis X1 0
• 4(0) 3X2 240
• 3X2 240
• X2 240/3 80
• Coordinate (0,80)

15
X2
chairs
(0,80)
.
X1tables
(60,0) .
0,0
16
X2
chairs
(0,80)
.
X1tables
(60,0) .
0,0
17
Convert back to inequality

• 4X1 3X2 lt 240

18
X2
chairs
(0,80)
.
X1tables
(60,0) .
0,0
19
Plot painter constraint

• Equation 2X1 X2 100

20
2X1 X2 100
X1 X2
0 100
100/2 50 0
21
X2
chairs
0,100
(0,80)
.
X1tables
(60,0) .
50,0
0,0
22
Feasible Region

• Decision how many tables and chairs to make
• Feasible allocation satisfies all constraints

23
MAXIMUM PROFIT

• Must be on boundary
• If not on boundary, could increase profit by
  making more tables or chairs
• Must be feasible
• Corner Point

24
X2
chairs
NOT FEASIBLE
0,100
(0,80)
.
3
CORNER POINTS
NOT FEASIBLE
X1tables
(60,0) .
50,0
0,0
25
3RD CORNER POINT

• Intersection of 2 constraints
• Temporarily convert to equations
• (1) 4X1 3X2 240
• (2) 2X1 X2 100
• Solve 2 equations in 2 unknowns
• (2)3implies 6X1 3X2 300
• Subtract (1) - 4X1 - 3X2 -240
• 2X1 60
• X1 30

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Substitute into equation

• (1) 4X1 3X2 240
• 4(30) 3X2 240
• 120 3X2 240
• 3X2 240 120 120
• X2 120/3 40
• 3rd corner point (30,40)

27
X2
chairs
NOT FEASIBLE
0,100
(0,80)
.
(30,40)
NOT FEASIBLE
X1tables
(60,0) .
50,0
0,0
28
MAXIMUM PROFIT
X1tables X2chairs Interpret Profit 7X15X2
50 0 Make tables only 7(50)0 350
0 80 Make chairs only 7(0)5(80) 400
30 40 Mix of tables, chairs 7(30)5(40) 410 MAX
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Exam Format

• Make 30 tables and 40 chairs for 410 profit

30
II. Cost minimization example

• Diet problem
• Decision variables number of pounds of brand 1
  and brand 2 to buy to prepare processed food
• Objective Function MINIMIZE cost
• Each pound of brand 1 costs 2 cents, pound of
  brand 2 costs 3 cents
• Objective MIN 2X1 3X2

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CONSTRAINTS

• Each pound of brand 1 has 5 ounces of ingredient
  A, 4 ounces of ingredient B, and 0.5 ounces of
  ingr C
• Each pound of brand 2 has 10 ounces of ingr A
  and 3 ounces of ingr B
• We need at least 90 ounces of ingr A, 48 ounces
  of ingr B, and 1.5 ounces of ingr C

32
CONSTRAINTS

• (A) 5X1 10X2 gt 90
• (B) 4X1 3X2 gt 48
• (C) 0.5X1 gt 1.5
• (D) X1 gt 0
• (E) X2 gt 0

33
X2
X1
34
(A) 5X110X2gt90
X1 X2 INTERCEPT
0 90/109 (0,9)
90/518 0 (18,0)
35
X2
0,9
A
X1
18,0
36
(B) 4X13X2gt48
X1 X2
0 48/316
48/412 0
37
X2
0,16
0,9
B
A
X1
18,0
12,0
38
(C) 0.5X1 gt 1.5
X1 X2
0 1.5/0 undefined, so no intercept on vertical axis
1.5/.5 3 0
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(C) Must be vertical line

• .5X1 1.5
• X1 3

40
X2
C
0,16
0,9
B
A
X1
18,0
12,0
41
X2
C
FEASIBLE REGION UNBOUNDED
0,16
0,9
B
A
X1
18,0
12,0
42
CORNER POINTS

• Only 1 intercept feasible (18,0)
• Solve 2 equations in 2 unknowns
• B and C
• A and B

43
B and C

• B 4X1 3X2 48
• C X1 3
• Substitute X13 into B
• B 4(3) 3X2 48
• 12 3X2 48
• 3X2 36
• X2 12

44
X2
C
FEASIBLE REGION UNBOUNDED
3,12
B
A
X1
18,0
45
A and B

• A 5X1 10X2 90
• B 4X1 3X2 48
• (A)(4) 20X1 40X2 360
• (B)(5) 20X1 15X2 240
• Subtract 25X2 120
• X2 4.8
• Substitute5X1 10(4.8) 90
• X1 8.4

46
X2
C
FEASIBLE REGION UNBOUNDED
3,12
B
8.4,4.8
A
X1
18,0
47
MINIMIZE COST
X1 BRAND 1 X2 BRAND 2 COST 2X13X2
18 0 BRAND1 ONLY 2(18)3(0) 36 CENTS
3 12 2(3)3(12)42 CENTS
8.4 4.8 2(8.4) 3(4.8)31 MIN
48
EXAM FORMAT

• BUY 8.4 POUNDS OF BRAND 1 AND 4.8 POUNDS OF
  BRAND 2 AT COST OF 31 CENTS

49
COMPUTER OUTPUT

• If computer output says no feasible solution,
  no feasible region
• Reason 1 unrealistic constraints
• Reason 2 computer input error

50
No feasible region
51
MULTIPLE OPTIMA

• If 2 corner points have same objective function
  value, ok to pick midway point

52
X2
NOT FEASIBLE
Any point between 2 100 points is MAX
100
.
100
NOT FEASIBLE
.
80
X1
0,0