15 053 February 5 2002 l The Geometry

15. 053 February 5, 2002 l The Geometry of Linear Programs – the geometry of LPs illustrated on GTC Handouts: Lecture Notes 1

But first, the pigskin problem (from Practical Management Science) l Pigskin company makes footballs l All data below is for 1000 s of footballs l Forecast demand for next 6 months – 10, 15, 30, 35, 25 and 10 l Current inventory of footballs: 5 l Max Production capacity: 30 per month l Max Storage capacity: 10 per month l Production Cost per football for next 6 months: – $12: 50, $12. 55, $12. 70, $12. 85, $12. 95 l Holding cost: $. 60 per football per month l With your partner: write an LP to describe the problem 2

On the formulation l Choose decision variables. – Let xj= the number of footballs produced in month j (in 1000 s) – Let yj= the number of footballs held in inventory from year j to year j + 1. (in 1000 s) – y 0= 5 l Then write the constraints and the objective. Pigskin Spreadsheet 3

Data for the GTC Problem Wrenches Pliers Available Steel 1. 5 1. 0 15, 000 tons Molding Machine 1. 0 12, 000 hrs Assembly Machine . 4 . 5 5, 000 hrs Demand Limit 8, 000 10, 000 Contribution ($ per item) $. 40 $. 30 Want to determine the number of wrenches and pliers to produce given the available raw materials, machine hours and demand. 4

Formulating the GTC Problem P= number of pliers manufactured (in thousands) W= number of wrenches manufactured (in thousands) Maximize Profit = Steel: Molding: Assembly: Plier Demand: Wrench Demand: Non-negativity: 300 P + 400 W P + 1. 5 W ≤ 15 P + W ≤ 12 0. 5 P + 0. 4 W ≤ 5 P ≤ 10 W ≤ 8 P, W ≥ 0 5

Graphing the Feasible Region We will construct and shade the feasible region one or two constraints at a time. 6

Graphing the Feasible Region Graph the Constraint: P + 1. 5 W ≤ 15 7

Graphing the Feasible Region Graph the Constraint: W + P ≤ 12 8

Graphing the Feasible Region Graph the Constraint: . 4 W +. 5 P ≤ 5 What happened to the constraint : W + P ≤ 12? 9

Graphing the Feasible Region Graph the Constraint: W ≤ 8, and P≤ 10 10

How do we find an optimal solution? Maximize z = 400 W + 300 P Is there a feasible solution with z = 1200? 11

How do we find an optimal solution? Maximize z = 400 W + 300 P Is there a feasible solution with z = 1200? Is there a feasible solution with z = 3600? 12

How do we find an optimal solution? Maximize z = 400 W + 300 P Can you see what the optimal solution will be? 13

How do we find an optimal solution? Maximize z = 400 W + 300 P What characterizes the optimal solution? What is the optimal solution vector? W = ? P = ? What is its solution value? z=? 14

Optimal Solution S Binding constraints Maximize z = 400 W + 300 P 1. 5 W + P ≤ 15. 4 W +. 5 P ≤ 5 plus other constraints A constraint is said to be binding if it holds with equality at the optimum solution. Other constraints are non-binding 15

How do we find an optimal solution? Optimal solutions occur at extreme points. In two dimensions, this is the intersection of 2 lines. Maximize z = 400 W + 300 P 1. 5 W + P ≤ 15. 4 W +. 5 P ≤ 5 Solution: . 7 W = 5, W = 50/7 P = 15 -75/7 = 30/7 z = 29, 000/7 = 4, 142 6/7 16

Finding an optimal solution in two dimensions: Summary l The optimal solution (if one exists) occurs at a “corner point” of the feasible region. l In two dimensions with all inequality constraints, a corner point is a solution at which two (or more) constraints are binding. l There is always an optimal solution that is a corner point solution (if a feasible solution exists). l More than one solution may be optimal in some situations 17

Preview of the Simplex Algorithm l In n dimensions, one cannot evaluate the solution value of every extreme point efficiently. (There are too many. ) l The simplex method finds the best solution by a neighborhood search technique. l Two feasible corner points are said to be “adjacent” if they have one binding constraint in common. 18

Preview of the Simplex Method Maximize z = 400 W + 300 P Start at any feasible extreme point. Move to an adjacent extreme point with better objective value. Continue until no adjacent extreme point has a better objective value. 19

Preview of Sensitivity Analysis Suppose the plier demand is decreased to 10 -∆. What is the impact on the optimal solution value? Theshadow price of a constraint is the unit increase in the optimal objective value per unit increase in the RHS of the constraint. Changing the RHS of a nonbinding constraint by a small amount has no impact. The shadow price of the constraint is 0. 20

Preview of Sensitivity Analysis Suppose slightly more steel is available? 1. 5 W + P ≤ 15 +∆ What is the impact on the optimal solution value? 21

Shifting a Constraint Steel is increased to 15 + ∆. What happens to the optimal solution? What happens to the optimal solution value? 22

Shifting a Constraint Steel is increased to 15 + ∆. What happens to the optimal solution? What happens to the optimal solution value? 23

Finding the New Optimum Solution Maximize z = 400 W + 300 P Binding Constraints: 1. 5 W + P = 15 + ∆. 4 W +. 5 P = 5 W = 50/7 +(10/7)∆ Solution: P= 30/7 -8/7∆ z = 29, 000/7 +(1, 600/7)∆ Conclusion: If the amount of steel increases by ∆units (for sufficiently small∆) then the optimal objective value increases by(1, 600/7)∆. The shadow price of a constraint is the unit increase in the optimal objective value per unit increase in the RHS of the constraint. Thus the shadow price of steel is 1, 600/7 = 228 4/7. 24

Some Questions on Shadow Prices l Suppose the amount of steel was decreased by ∆units. What is the impact on the optimum objective value? l How large can the increase in steel availability be so that the shadow price remains as 228 4/7? l Suppose that steel becomes available at $1200 per ton. Should you purchase the steel? l Suppose that you could purchase 1 ton of steel for $450. Should you purchase the steel? (Assume here that this is the correct market value for steel. ) 25

Shifting a Constraint Steel is increased to 15 + ∆. What happens to the optimal solution? Recall that W <= 8. The structure of the optimum solution changes when ∆=. 6, and W is increased to 8 26

Shifting a Cost Coefficient The objective is: Maximize z = 400 W + 300 P What happens to the optimal solution if 300 P is replaced by (300+δ )P How large can δ be for your answer to stay correct? . 4 W +. 5 P = 5 27

Summary: 2 D Geometry helps guide the intuition l The Geometry of the Feasible Region – Graphing the constraints l Finding an optimal solution – Graphical method – Searching all the extreme points – Simplex Method l Sensitivity Analysis – Changing the RHS – Changing the Cost Coefficients 28