Linear Programming

Graphical Solution

Graphical Solution to an LP Problem

This is easiest way to solve a LP problem with two decision variables.

If there are more than two decision variables, it is not possible to plot the solution on two dimensional graph.

However, it provides us an useful insight how the other approaches work. So it is worthwhile to learn how it works.

To find the optimal solution to an LP problem we must first identify a set of feasible solution (a region of feasible solution)

First step in doing so to plot of the problem’s constraints on a graph.

X1 is usually plotted as the horizontal axis

X2 is usually plotted as the vertical axis

Because of the non-negativity constraint we are always working on the first quadrant

x1

x2This axis represents the constraint x2 ≥ 0

This axis represents the constraint x1 ≥ 0

Example 1

Giapetto’ Woodcarving, Inc.,

21max 23 xxZ

1002 21 xx

Subject to

8021 xx

401 x

0, 21 xx

The feasible region has been graphed. We may proceed to find the optimal solution to the problem.

The optimal solution is the point lying in the feasible region that produce the highest profit.

There are two different approaches to find it.

-Isoprofit line method (isocost line)

- Corner point method

Feasible Region

DG

F

EH x1

x2

Corner Point Solution Method

The second approach to find the optimal solution to LP problem is corner point method.

An optimal solution to LP problem lies at a corner point of (extreme point of) the feasible region.

Hence it is only necessary to find the value of them.

From the graph the problem is five sided polygon with five corner or extreme points.

These points are labeled H, E, F, G, D.

So we can find the coordinates of each corner and test the profit levels.

Feasible Region

DG

F

EH x1

x2

Carpentry Cons.

Demand Cons.

Finishing Cons.

Point H (0, 0) Z = 3(0) + 2(0) = 0

Point E (40, 0) Z = 3(40) + 2(0) = 120

Point F (40, 20) Z = 3(40) + 2(20) = 160

Note that: at Point F Cons.2 and Cons.3 intersect

So

2x1 + x2 = 100

x1 = 40

x1 = 40, x2 = 20

Point G (20, 60) Z = 3(20) + 2(60) = 180* optimal solution2x1 + x2 = 100 x1 + x2 = 80x1 = 20, x2 = 60

Point D (0, 80) Z = 3(0) + 2(80) = 160

The optimal solution to this problem (point G)

x1 = 20

x2 = 60

Z = 180

[ Z = 3x1 + 2x2 ⇒ 3(20) + 2(60) = $120]

If we substitute the optimal values of decision variables into the left hand side of the constraints.

2x1 + x2 ≤ 100 ⇒ 2(20) + 60 = 100

S1 = 100 – 100 = 0

x1 + x2 ≤ 80 ⇒ 20 + 60 = 80

S2 = 80 – 80 = 0

x1 ≤ 40 ⇒ 20

S3 = 40 – 20 = 20

S: Slack variable: represent the amount of resource unused

S1 = 0 means decision variables use up the resources completely

S2 = 0

S3 = 20 20 units of resource is left over.

Slack variable → [≤] less than or equal to

Once the optimal solution to the LP problem it is useful to classify the constraint.

A constraint is binding constraint if left hand side and right hand side of it are equal when the optimal values of decision variables are substituted into the constraint.

A constraint is nonbinding constraint if the left hand side and right hand side of constraint are not equal when the optimal values of decision variables are substituted in to the constraint.

Left hand side Right hand side

Constraint 1: 2x1 + x2 ≤ 100 2(20) + 60 = 100 100 Binding

Constraint 2: x1 + x2 ≤ 80 20 + 60 = 80 80 Binding

Constraint 3: x1 ≤ 40 20 40 Not binding

The other classification of resource

Resource Slack value Status

Finishing hours 0 Scarce

Carpentry hours 0 Scarce

Demand 20 Abundant

Example 2

Advertisement (Winston 3.2, p 61)

21min 10050 xxZ

2827 21 xx

Subject to

24122 21 xx

0, 21 xx

Example 3

Giapetto’ Woodcarving, Inc., (changed)

21max 24 xxZ

1002 21 xx

Subject to

8021 xx

401 x

0, 21 xx

Example 4

Giapetto’ Woodcarving, Inc., (changed)

21max 23 xxZ

1002 21 xx

Subject to

8021 xx

401 x

0, 21 xx202 x

301 x

Example 4

Giapetto’ Woodcarving, Inc., (changed)

21max 24 xxZ

Subject to

401 x

0, 21 xx202 x

301 x

Sensitivity Analysis

Shadow Prices

• Shadow price of resource i measures the marginal value of this resource.