MIT and James Orlin © The Geometry of Linear Programs –the geometry of LPs illustrated on GTC

MIT and James Orlin © Data for the GTC Problem Want to determine the number of wrenches and pliers to produce given the available raw materials, machine hours and demand.

MIT and James Orlin © Formulating the GTC Problem P = number of pliers manufactured W = number of wrenches manufactured Maximize Profit = Steel: Molding: Assembly: Pliers Demand: Wrench Demand: P,W  0 Non-negativity: 1.5 W + P  15,000 W + P  12, W P  5,000 P  10,000  W  8,000.4 W +.3 P

MIT and James Orlin © Reformulation P = number of 1000s of pliers manufactured W = number of 1000s of wrenches manufactured Maximize Profit = Steel: Molding: Assembly: Pliers Demand: Wrench Demand: P,W  0 Non-negativity: 1.5 W + P  15 W + P  W P  5 P  10  W  W P

MIT and James Orlin © Finding an optimal solution Introduce yourself to your partner Try to find an optimal solution to the linear program, without looking ahead.

Graphing the Feasible Region 2 W P We will construct and shade the feasible region one or two constraints at a time. 6

Graphing the Feasible Region W P Graph the Constraint: 1.5 W + P  15 7

Graphing the Feasible Region W P Graph the Constraint: W + P  12

Graphing the Feasible Region W P Graph the Constraint: 0.4 W P  5 What happened to the constraint : W + P  12?

Graphing the Feasible Region W P Graph the Constraints:  W  8 P  10

How do we find an optimal solution? W P Maximize z = 400W + 300P It is the largest value of q such that 400W + 300P = q has a feasible solution 11

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

W P How do we find an optimal solution? Maximize z = 400W + 300P Is there a feasible solution with z = 2400 ? Is there a feasible solution with z = 3600 ? 13 z = 2400 z=3600

W P Can you see what the optimal solution will be? z = 2400 z = 3600 How do we find an optimal solution? Maximize z = 400W + 300P 14

W P What characterizes the optimal solution? What is the optimal solution vector? W = ? P = ? What is its solution value? z = ? How do we find an optimal solution? Maximize z = 400W + 300P 15

Optimal Solution Structure W P Maximize z = 400W + 300P 1.5 W + P  15.4W +.5P  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 Binding constraints 16

How do we find an optimal solution? W P Maximize z = 400W + 300P 1.5W + P = 15.4W +.5P = 5 Solution:.7W = 5, W = 50/7 P = /7 = 30/7 z = 29,000/7 = 4,142 6/7 Optimal solutions occur at corner points. In two dimensions, this is the intersection of 2 lines. 17

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

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

Preview of the Simplex Method W 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. Maximize z = 400W + 300P 20

Preview of Sensitivity Analysis W P Suppose the pliers demand is decreased to 10 - . What is the impact on the optimal solution value? The shadow price of a constraint is the unit increase in the optimal objective value per unit increase in the RHS of the constraint. 21 Changing the RHS of a non- binding constraint by a small amount has no impact. The shadow price of the constraint is 0.

Preview of Sensitivity Analysis W P Suppose slightly more steel is available? 1.5W + P  15 +  What is the impact on the optimal solution value? 22

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

Shifting a Constraint W P Steel is increased to 15 + . What happens to the optimal solution? What happens to the optimal solution value? 24

MIT and James Orlin © Finding the New Optimum Solution Maximize z = 400W + 300P Binding Constraints: Solution: Thus the shadow price of steel is 1,600/7 = 228 4/7. 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. 1.5W + P = 15 + .4W +.5P = 5 W = 50/7 +(10/7)  P= 30/7 -8/7  z = 29,000/7 +(1,600/7) 

MIT and James Orlin © Some Questions on Shadow Prices Suppose the amount of steel was decreased by  units. What is the impact on the optimum objective value? How large can the increase in steel availability be so that the shadow price remains as 228 4/7? Suppose that steel becomes available at $1200 per ton. Should you purchase the steel? 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.)

MIT and James Orlin © Bounds on RHS coefficients in Sensitivity Analysis Recall that the optimum solution is a corner point, which in 2 dimensions is the solution of 2 equations in 2 variables, and the equations are the binding constraints. Compute the largest changes in the RHS coefficient so that all constraints remain satisfied.

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

MIT and James Orlin © Changing the RHS coefficient P,W  W + P  15 W + P  W P  5 P  10 W  8 Binding Constraint 1.5 W + P  15 +  Binding Constraint 0.4 W P  5 Increase steel from 15 to 15 +  W = 50/7 +(10/7)  ; P= 30/7 –(8/7) 

MIT and James Orlin © Changing the RHS coefficient P,W  0 W + P  W P  5 P  10 W  W + P  15 +  0.4 W P  5 W = 50/7 +(10/7)  ; P= 30/7 –(8/7)  W + P = 80/7 + (2/7)   12 W = 50/7 + (10/7)   8 P = 30/7 – (8/7)   10 50/7 +(10/7)   0; 30/7 –(8/7)   0 Compute changes in the LHS of remaining constraints

MIT and James Orlin © Changing the RHS coefficient P,W  W P  5 P  10 W  W + P  15 +  0.4 W P  5 30/7 –(8/7)   0 80/7 + (2/7)   12 50/7 + (10/7)   8 30/7 – (8/7)   10 50/7 +(10/7)   0; Compute upper and lower bounds on    2   3/5   -5    -5   15/4 So, -5   3/5

MIT and James Orlin © Summary for changes in RHS coefficients Determine the binding constraints Determine the change in the “corner point solution” as a function of . Compute the largest and smallest values of  so that the solution stays feasible. The shadow price is valid so long as the “corner point solution” remains optimal, which is so long as it is feasible. If there are three binding constraints, then choose two of these to get the two equations to solve, and the technique still works. (But the change in the solution as a function of  depends on which two constraints are chosen.)

MIT and James Orlin © Bounds on Cost coefficients in Sensitivity Analysis Recall that the optimum solution is a corner point, which in 2 dimensions is the solution of 2 equations in 2 variables, and the equations are the binding constraints. The solution has two neighboring corner point solutions Compute the largest changes in the cost coefficient so that the current corner point solution has a better objective value than its neighboring corner point solutions.

Shifting a Cost Coefficient The objective is: Maximize z = 400W + 300P What happens to the optimal solution if 300P is replaced by (300+  )P How large can  be for your answer to stay correct? 34 GTC W P W +.5P = 5

MIT and James Orlin © Determining Bounds on Cost Coefficients W P z = 400W + (300+  ) P W = 50/7 ; P= 30/7; z = 29,000/  /7 W = 0 ; P= 10; z =  W = 8 ; P= 3; z = 

MIT and James Orlin © Determining Bounds on Cost Coefficients W P W = 50/7 ; P= 30/7; z = 29,000/  /7 W = 0 ; P= 10; z =  W = 8 ; P= 3; z =  z = 29,000/  /7      -100/3 z = 29,000/  /7     200

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