Linear Programming (LP) Problems MAX (or MIN): c 1 X 1 + c 2 X 2 + … + c n X n Subject to:a 11 X 1 + a 12 X 2 + … + a 1n X n <= b 1 : a j1 X 1 + a j2 X 2 + … + a jn X n >=b j : a m1 X 1 + a m2 X 2 + … + a mn X n = b m

Weekly supply of raw materials: 6 Large Bricks 8 Small Bricks Products: Table Chair Profit = $20/TableProfit = $15/Chair The Lego Production Problem

X1 is the number of Chairs X2 is the number of Tables Large brick constraint X1+2X2 <= 6 Small brick constraint 2X1+2X2 <= 8 Objective function is to Maximize 15X1+20 X2 X1>=0 X2>= 0 Problem Formulation

0 Graphical Solution to the Prototype Problem Chairs Tables X1 + 2 X2 = 6 Large Bricks

0 Graphical Solution to the Prototype Problem Chairs Tables 2 X1 + 2 X2 = 8 Small Bricks

0 Graphical Solution to the Prototype Problem Chairs Tables X1 + 2 X2 = 6 Large Bricks 2 X1 + 2 X2 = 8 Small Bricks

0 Graphical Solution to the Prototype Problem Chairs Tables X1 + 2 X2 = 6 Large Bricks 2 X1 + 2 X2 = 8 Small Bricks

Z = 15 X X2 Lets draw it for 15 X X2 = 30 In this case if # of chair = 0, then # of table = 30/20 = 1.5 if # of table = 0, then # of chair = 30/15 = 2 The Objective Function

0 Graphical Solution to the Prototype Problem Chairs Tables X1 + 2 X2 = 6 Large Bricks 2 X1 + 2 X2 = 8 Small Bricks

We can make Product1 and or Product2. There are 3 resources; Resource1, Resource2, Resource3. Product1 needs one unit of Resource1, nothing of Resource2, and three units of resource3. Product2 needs nothing from Resource1, two units of Resource2, and two units of resource3. Net profit of product 1 and Product2 are 3 and 5, respectively. Formulate the Problem Solve it graphically Solve it using excel. A second example

Objective Function Z = 3 x 1 +5 x 2 Constraints Resource 1 x 1  4 Resource 2 2x 2  12 Resource 3 3 x x 2  18 Nonnegativity x 1  0, x 2  0 Problem 2 : Original version Product 1 needs 1 unit of resource 1, and 3 units of resource 3. Product 2 needs 2 units of resource 2 and 2 units of resource 3 There are 4 units of resource 1, 12 units of resource 2, and 18 units of resource 3

x 2 x 1 Max Z = x 1 + x 2 Subject to x 1  4 2x 2  12 3 x x 2  18 x 1  0, x 2  0 Problem 2 : Original version

x 2 x 1 Max Z = 3x 1 + 5x 2 Subject to x 1  4 2x 2  12 3 x x 2  18 x 1  0, x 2  0 Problem 2

x 2 x 1 Problem 2 : Original version Max Z = 2x 1 + 3x 2 Subject to x1 +3x2  6 4x1 +3x2  12 4x1 +x2  8 x 1  0, x 2  0 x2=0 x1 =6 x1 =0 x2 =2

x 2 x 1 Problem 2 : Original version Max Z = 2x 1 + 3x 2 Subject to x1 +3x2  6 4x1 +3x2  12 4x1 +x2  8 x 1  0, x 2  0 x2=0 x1 =3 x1 =0 x2 =4

x 2 x 1 Problem 2 : Original version Max Z = 2x 1 + 3x 2 Subject to x1 +3x2  6 4x1 +3x2  12 4x1 +x2  8 x 1  0, x 2  0 x2=0 x1 =2 x1 =0 x2 =8 How many corner points: 3 functional constraints plus two non-negativity constraints =5 Each time we solve two equations and two unknowns. Select two out of five (m+n)!/m!n! = COMBIN(5,2)= 10 How many feasible corner points: 4

x 2 x 1 Problem 2 : Original version Max Z = 2x 1 + 3x 2 Subject to x1 +3x2  6 4x1 +3x2  12 4x1 +x2  8 x 1  0, x 2  0 2x1+3x2 = 6 x2=0 x1=3 x1=0 x2= 2