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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
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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
X2 X1 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)
X2 X1 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)
X2 =chairs (0,80) . X1=tables (60,0) . 0,0
X2 =chairs (0,80) . X1=tables (60,0) . 0,0
Convert back to inequality 4X1 + 3X2 < 240
X2 =chairs (0,80) . X1=tables (60,0) . 0,0
Plot painter constraint • Equation: 2X1 + X2 = 100
X2 =chairs 0,100 (0,80) . X1=tables (60,0) . 50,0 0,0
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”
X2 =chairs NOT FEASIBLE 0,100 (0,80) . 3 CORNER POINTS NOT FEASIBLE X1=tables (60,0) . 50,0 0,0
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 = 40 3rd corner point: (30,40)
X2 =chairs NOT FEASIBLE 0,100 (0,80) . (30,40) NOT FEASIBLE X1=tables (60,0) . 50,0 0,0
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
X2 X1
X2 0,9 A X1 18,0
X2 0,16 0,9 B A X1 18,0 12,0
(C) Must be vertical line .5X1 = 1.5 X1= 3
X2 C 0,16 0,9 B A X1 18,0 12,0
X2 C FEASIBLE REGION UNBOUNDED 0,16 0,9 B A X1 18,0 12,0
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
X2 C FEASIBLE REGION UNBOUNDED 3,12 B A X1 18,0
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
X2 C FEASIBLE REGION UNBOUNDED 3,12 B 8.4,4.8 A X1 18,0
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