Linear Programming: Model Formulation and Graphic Solution Lecture 2

A Maximization Model Example The Beaver Creek Pottery Company Given these limited resources, the company desires to know how many bowls and mugs to produce each day in order to maximize profit. There are 40 hours of labor and 120 kg of clay available each day for production.

Resource Requirements Beaver Creek Pottery Resource Requirements Product Labor (hr/unit) Clay (kg/unit) Profit ($/unit) Bowl 1 4 40 Mug 2 3 50

Maximize Z = 40 x1 + 50 x2 Modeling Decision Variables Objective Function x1 : number of bowls to produce x2 : number of mugs to produce Total Profit = 40 x1 + 50 x2 40 x1 = profit from bowls 50 x2 = profit from mugs Maximize Z = 40 x1 + 50 x2

Non-negativity Constraint Model Constraints Labor Constraint Total Labor used in production = 1 x1 + 2 x2 1 x1 + 2 x2 ≤ 40 hr Clay Constraint Total Clay used in production = 4 x1 + 3 x2 4 x1 + 3 x2 ≤ 120 lb Non-negativity Constraint x1 ≥ 0, x2 ≥ 0

Linear Programming Model Maximize Z = 40 x1 + 50 x2 Subject to: 1 x1 + 2 x2 ≤ 40 4 x1 + 3 x2 ≤ 120 x1, x2 ≥ 0

Feasible / Infeasible If x1 = 5, x2 = 10 Z = 40 . 5 + 50 . 10 = 700

Graphical Solution of Linear Programming Models 10 20 30 40 50 60 Coordinates for graphical analysis

Graph of the labor constraint line x2 10 20 30 40 50 60 x1 + 2 x2 = 40 x1

The labor constraint area x2 10 20 30 40 50 60 M L x1 + 2 x2 ≤ 40 K x1

Graph of the labor constraint line x2 10 20 30 40 50 60 4 x1 + 3 x2 = 120 x1

The clay constraint area x2 10 20 30 40 50 60 M L 4 x1 + 3 x2 ≤ 120 K x1

Graph of both model constraints x2 10 20 30 40 50 60 4 x1 + 3 x2 = 120 x1 + 2 x2 = 40 x1

The feasible solution area constraints x2 10 20 30 40 50 60 T: Infeasible S: Infeasible R: Feasible 4 x1 + 3 x2 = 120 T S R x1 + 2 x2 = 40 x1

Objective function line for Z = $ 800 x2 10 20 30 40 50 60 800 = 40 x1 + 50 x2 x1

Alternative objective function lines for profits, Z, of $ 800, $ 1200, $ 1600 x2 40 800 = 40 x1 + 50 x2 30 1200 = 40 x1 + 50 x2 20 1600 = 40 x1 + 50 x2 10 10 20 30 40 x1

Identification of optimal solution point x2 10 20 30 40 50 60 800 = 40 x1 + 50 x2 Optimal solution point B x1

Optimal solution coordinates x2 40 35 4 x1 + 3 x2 = 120 30 25 A 20 15 10 B 8 x1 + 2 x2 = 40 5 C Prof. M.A.Shouman 5 10 15 20 24 25 30 35 40 x1

Solutions at all corners points x2 40 35 4 x1 + 3 x2 = 120 x1 = 0 bowls x2 = 20 mugs Z = $ 1,000 30 25 x1 = 24 bowls x2 = 8 mugs Z = $ 1,360 A 20 x1 = 30 bowls x2 = 0 mugs Z = $ 1,200 15 10 B 8 5 x1 + 2 x2 = 40 C Prof. M.A.Shouman 5 10 15 20 24 25 30 35 40 x1

Optimal solution point The optimal solution with Z = 70 x1 + 20 x2 x2 40 35 4 x1 + 3 x2 = 120 30 25 A 20 Optimal solution point x1 = 30 bowls x2 = 0 mugs Z = $ 1,200 15 10 B 5 x1 + 2 x2 = 40 C Prof. M.A.Shouman 5 10 15 20 25 30 35 40 x1

A Minimization Model Example The Farmer’s Field The farmer’s field requires at least 16 kg. of nitrogen and 24 kg. of phosphate. Super-gro costs $6 per bag, and Crop-quick costs $3. The farmer wants to know how many bags of each brand to purchase in order to minimize the total cost of fertilizing.

Chemical Contribution The Farmer’s Field Chemical Contribution Brand Nitrogen (kg./bag) Phosphate Super-gro 4 2 Crop-quick 3

Minimize Z = 6 x1 + 3 x2 Subject to: 2 x1 + 4 x2 ≥ 16 4 x1 + 3 x2 ≥ 24 Linear Programming Model Minimize Z = 6 x1 + 3 x2 Subject to: 2 x1 + 4 x2 ≥ 16 4 x1 + 3 x2 ≥ 24 x1, x2 ≥ 0

Constraint lines for fertilizer model x2 12 10 8 4 x1 + 3 x2 = 24 6 4 2 2 x1 + 4 x2 = 16 Prof. M.A.Shouman 2 4 6 8 10 12 14 16 x1

Feasible solution area x2 12 10 8 4 x1 + 3 x2 = 24 6 Feasible solution area 4 2 2 x1 + 4 x2 = 16 x1 2 4 6 8 10 12

Optimal solution point The optimal solution point x2 Optimal solution point 12 x1 = 0 bags of Super-gro x2 = 8 bags of Crop-quick Z = $ 24 10 A 8 6 Z = 6 x1 + 3 x2 4 B 2 C x1 2 4 6 8 10 12

Irregular Types of Linear Programming Problems

Multiple Optimal Solutions x2 40 35 30 Point B Point C 25 x1 = 24 x2 = 8 Z = $ 1,200 x1 = 30 bowls x2 = 0 mugs Z = $ 1,200 A 20 15 10 B 5 C Prof. M.A.Shouman 5 10 15 20 25 30 35 40 x1

An Infeasible Problem C B A x1 = 4 x2 = 6 4 x1 + 2 x2 = 8 x2 12 10 8 6 2 4 6 8 10 12

An Unbounded Problem x1 = 4 Minimize Z = 4 x1 + 2 x2 Subject to: 12 10 8 Z = 4 x1 + 2 x2 6 x2 = 6 4 2 x1 2 4 6 8 10 12