Linear Programming (LP)

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

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

LP Requirements

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

Linear Programming

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

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

Objective: MAX profit

• Each table: $ 7 profit• Each chair: $ 5 profit• Total profit = 7X1 + 5X2

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 < 240

Painter Constraint

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

Non-negativity constraints

• X1 > 0• X2 > 0• Can’t have negative production

Graphical Solution

Non-negativity constraints imply positive (northeast) quadrant

X1

X2

0,0

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)

X1

X2

0,0 60,0

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)

X1=tables

X2 =chairs

0,0

(0,80).

(60,0) .

X1=tables

X2 =chairs

0,0

(0,80).

(60,0) .

Convert back to inequality

4X1 + 3X2 < 240

X1=tables

X2 =chairs

0,0

(0,80).

(60,0) .

Plot painter constraint

• Equation: 2X1 + X2 = 100

2X1 + X2 = 100

X1 X2

0 100

100/2 = 50 0

X1=tables

X2 =chairs

0,0

(0,80).

(60,0) .

0,100

50,0

Feasible Region

• Decision: how many tables and chairs to make

• Feasible allocation: satisfies all constraints

MAXIMUM PROFIT

• Must be on boundary• If not on boundary, could increase profit by

making more tables or chairs• Must be feasible• “Corner Point”

X1=tables

X2 =chairs

0,0

(0,80).

(60,0) .

0,100

50,0

NOT FEASIBLE

NOT FEASIBLE

CORNER POINTS3

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

Substitute into equation

• (1) 4X1 + 3X2 = 240• 4(30) + 3X2 = 240

120 + 3X2 = 240 3X2 = 240 – 120 = 120 X2 = 120/3 = 403rd corner point: (30,40)

X1=tables

X2 =chairs

0,0

(0,80).

(60,0) .

0,100

50,0

NOT FEASIBLE

NOT FEASIBLE

(30,40)

MAXIMUM PROFIT

X1=tables X2=chairs Interpret Profit=7X1+5X2

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

Exam Format

Make 30 tables and 40 chairs for $410 profit

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

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

CONSTRAINTS

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

X1

X2

(A) 5X1+10X2>90

X1 X2 INTERCEPT

0 90/10=9 (0,9)

90/5=18 0 (18,0)

X1

X2

0,9

18,0A

(B) 4X1+3X2>48

X1 X2

0 48/3=16

48/4=12 0

X1

X2

0,9

18,0A

0,16

12,0

B

(C) 0.5X1 > 1.5

X1 X2

0 1.5/0 undefined, so no intercept on vertical axis

1.5/.5= 3 0

(C) Must be vertical line

.5X1 = 1.5X1= 3

X1

X2

0,9

18,0A

0,16

12,0

B

C

X1

X2

0,9

18,0A

0,16

12,0

B

C FEASIBLE REGION UNBOUNDED

CORNER POINTS

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

B and C

• B: 4X1 + 3X2 = 48• C: X1 = 3• Substitute X1=3 into B• B: 4(3) + 3X2 = 48• 12 + 3X2 = 48• 3X2 = 36• X2 = 12

X1

X2

18,0A

B

C FEASIBLE REGION UNBOUNDED

3,12

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

X1

X2

18,0A

B

C FEASIBLE REGION UNBOUNDED

3,12

8.4,4.8

MINIMIZE COSTX1=BRAND #1

X2=BRAND #2

COST=2X1+3X2

18 0 BRAND#1ONLY

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

EXAM FORMAT

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

COMPUTER OUTPUT

• If computer output says “no feasible solution”, no feasible region

• Reason #1: unrealistic constraints• Reason #2: computer input error

No feasible region

MULTIPLE OPTIMA

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

X2

0,0

$ 100 .

.$ 80

NOT FEASIBLE

NOT FEASIBLE

$ 100

Any point between 2 $100 points is MAX

X1