Foundations for LP

Sean Eom

In this course, LP is the foundation of all other quantitative models.

Study of LP and related techniques (Integer LP and Goal Programming) requires 2/3 of our time in the semester.

It is critical to understand some building blocks of LP conceps.

Foundational Concepts/tools

Drawing lines on a two-dimensional space

Identifying the feasible area Identifying extreme points and their

values Understanding the properties of the

objective function line

Drawing lines on a two-dimensional space

An LP model consists of the following: Max 3x1 + 3x2

s.t.(subject to)2x1 + 4x2 =< 12

6x1 + 4x2 =< 24x1 , x2 >= 0

2x1 + 4x2 =< 12

LP constraints must be drawn on a 2D space.

To do so, at minimum two sets of values that are on the constraint line (2x1 + 4x2 = 12) must be found.

For example, x1 = 4, x2 = 1 or x1 = 2, x2 = 2 x1 = 6, x2 = 0 or x1 = 0, x2 = 3

2x1+4x2=<12

Variations of linear constraints Draw a graph for x1 => .5 (x1 + x2 )

You need to rearrange the constraint. x1 => .5 (x1 + x2 ) x1 -.5 (x1 + x2 ) =>0 x1 -.5x1 -.5x2 ) =>0 .5x1 -.5x2 ) =>0

6x1 + 4x2 =< 24

At minimum two sets of values that are on the constraint line (6x1 + 4x2 = 24) must be found.

For example, x1 = 2, x2 = 3 or x1 = 4, x2 = 0 x1 = 0, x2 = 6

It is easier for you to find values with (x1 = 0, x2 =? Or x1 = ?, x2 = 0 )

6x1 + 4x2 =< 24

All LP constraints on one space

The Objective Function (Max 3x1 + 3x2 )

The difference between the objective function and constraint lines is the absence of the right-hand side value.

Arbitrary value may be added before drawing the line.

You can use any numbers, but add the smallest number that can be dividable by both coefficients of x1 and x2

In this case, 3.

3x1 + 3x2 = 3

All lines on one space

About the object function line (3x1 + 3x2 = 3)

About the object function line (3x1 + 3x2 = 3) The objective function line indicates

the $ amount of profits/costs “3x1 + 3x2 = 3” can be interpreted

this way: Any combination of production of the two products along the objective function line produces same profit, $3.

Therefore, it is often called “ISO- profit/cost line. Iso=same.

3x1 + 3x2 = 6

3x1 + 3x2 = 60

Interpretation of the objective function line

3x1 + 3x2 = 6 3x1 + 3x2 = 60

The larger right hand side value indicates the larger profit ($)

Figure 1

Figure 2

Figure 3

Feasible area & Extreme points in figure 3

The green colored area is called the feasible area.

There are 4 extreme points in the feasible area.

Extreme points are “corners” (vertices) of the feasible area.

The optimal solution = one of the extreme points

Computing the optimal answer

In figure 3, the optimal point is the intersection of an extreme point and the objective function line.

What is the values of the optimal point (x1=? and x2=?)?

How do you compute them?

Computing the optimal answer Identify the two constraints that

intersect.2x1 + 4x2 =< 12

6x1 + 4x2 =< 24 Ignore in-equality signs

2x1 + 4x2 = 12 (1) 6x1 + 4x2 = 24 (2)

Computing the optimal answer Eliminate either x1 or x2. In this case, equations 1 and 2

have same objective function coefficient of x2, let’s eliminate x2 first.

2x1 + 4x2 = 12 (1)- ( 6x1 + 4x2 = 24) (2) -4x1 = -12 x1= 3

Use x1= 3 and substitute this value to find x2 using either one of the two equation.2*3 + 4x2 = 126 + 4x2 = 124x2 = 12-6=6x2 = 1.5

The final answer is

x1 = 3x2 = 1.5

Z= 3x1 + 3x2 = 3*3 + 3*1.5 =9 + 4.5 =13.5