the geometry of linear programs the geometry of lps illustrated on gtc handouts: lecture notes
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15.053. February 5, 2002. The Geometry of Linear Programs the geometry of LPs illustrated on GTC Handouts: Lecture Notes. But first, the pigskin problem (from Practical Management Science). Pigskin company makes footballs All data below is for 1000s of footballs - PowerPoint PPT PresentationTRANSCRIPT
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The Geometry of Linear Programs– the geometry of LPs illustrated on GTC
Handouts: Lecture Notes
15.053 February 5, 2002
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But first, the pigskin problem (from Practical Management Science)
Pigskin company makes footballs All data below is for 1000s of footballs Forecast demand for next 6 months
– 10, 15, 30, 35, 25 and 10
Current inventory of footballs: 5 Max Production capacity: 30 per month Max Storage capacity: 10 per month Production Cost per football for next 6 months:
– $12:50, $12.55, $12.70, $12.80, $12.85, $12.95
Holding cost: $.60 per football per month With your partner: write an LP to describe the proble
m
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On the formulation
Choose decision variables. – Let xj= the number of footballs produced in m
onth j (in 1000s)
– Let yj= the number of footballs held in inventory from year j to year j + 1. (in 1000s)
– y0= 5
Then write the constraints and the objective.
Pigskin Spreadsheet
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Data for the GTC Problem
Wrenches Pliers Available
Steel 1.5 1.0 15,000 tons
Molding Machine
1.0 1.0 12,000 hrs
Assembly Machine
.4 .5 5,000 hrs
Demand Limit 8,000 10,000
Contribution
($ per item)
$.40 $.30
Want to determine the number of wrenches and pliersto produce given the available raw materials, machinehours and demand.
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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
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Graphing the Feasible Region
We will construct and shade the feasible region one or two constraints at a time.
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Graphing the Feasible Region
Graph the Constraint:
P + 1.5W ≤15
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Graphing the Feasible Region
W + P ≤12Graph the Constraint:
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Graphing the Feasible Region
.4W + .5P ≤5Graph the Constraint:
What happened to the constraint :W + P ≤12?
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Graphing the Feasible Region
W ≤8, and P≤10Graph the Constraint:
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How do we find an optimal solution?
Maximize z = 400W + 300P
Is there a feasible solution with z = 1200?
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How do we find an optimal solution?
Maximize z = 400W + 300P
Is there a feasible solution with z = 1200?
Is there a feasible solution with z = 3600?
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How do we find an optimal solution?
Maximize z = 400W + 300P
Can you see what the optimal solution will be?
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How do we find an optimal solution?
Maximize z = 400W + 300P
What characterizes the optimal solution?
What is the optimal solution vector? W = ? P = ?
What is its solution value? z = ?
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Optimal Solution SBinding constraints
Maximize z = 400W + 300P
1.5W + P ≤15
.4W + .5P ≤5
plus other constraints
A constraint is said to be bindingif it holds with equality at the optimum solution.
Other constraints are non-binding
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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 = 400W + 300P
1.5W + P ≤15
.4W + .5P ≤5
Solution:
.7W = 5, W = 50/7
P = 15 -75/7 = 30/7
z = 29,000/7 = 4,142 6/7
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Finding an optimal solution in two dimensions: SummaryThe optimal solution (if one exists) occurs at
a “corner point” of the feasible region.In two dimensions with all inequality constrai
nts, 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
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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.
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Preview of the Simplex Method
Maximize z = 400W + 300P
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.
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Preview of Sensitivity AnalysisSuppose 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 non-binding constraint by a small amount has no impact. The shadow price of the constraint is 0.
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Preview of Sensitivity Analysis
Suppose slightly more steel is available? 1.5W + P ≤15 +∆
What is the impact on the optimal solution value?
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Shifting a Constraint
Steel is increased to 15 + ∆.
What happens to the optimal solution?
What happens to the optimal solution value?
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Shifting a Constraint
Steel is increased to 15 + ∆.
What happens to the optimal solution?
What happens to the optimal solution value?
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Finding the New Optimum SolutionMaximize z = 400W + 300P
Binding Constraints:
Solution:
1.5W + P = 15 + ∆
.4W + .5P = 5
W = 50/7 +(10/7)∆
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.
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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.)
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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
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Shifting a Cost Coefficient
The objective is:
Maximize z = 400W + 300P
What happens to the optimal solution if 300P is replaced by (300+δ)P
How large can δbe for your answer to stay correct?
.4W + .5P = 5
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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