Linear Programming

Consider the following problem:  A tailor has the following materials available in stock: 16 m2 of cotton, 11 m2 of silk and 15 m2 of wool. A suit requires 2 m2 of cotton, 1 m2 of silk and 1 m2 of wool to produce, while a gown requires 1 m2 of cotton, 2 m2 of silk and 3 m2 of wool. A clothing outlet will pay the tailor $150 for each suit and $250 for each gown he produces. How many suits and how many gowns should the tailor prepare in order to maximize his revenue from the sale of these articles of clothing?

First we must consider some mathematical theory and review

 y = mx + b

x

y

 y > mx + b lies above the line

x

y

1

 b = –4  m = 3

test point (0, 0)

LS = y = 0

RS = 3(0) – 4 RS = –4 LS > RS

(0, 0) is in the region

 y ≥ 3x – 4

 y < mx + b lies below the line

The line is in the region as well

RS = 3(x) – 4

–4

3

If we solve the inequality for y we get y > mx + b (above the line)

x

y

–2

 a = 5  b = –2

test point (0, 0)

LS = 0

RS = 10 LS < RS

(0, 0) is in the region

 2x – 5y < 10

5

The line is not in the region

 x > a lies to the right of the line

x

y

2

 a = 2

vertical line

test point (0, 0) LS = 0

RS = 6

LS < RS

(0, 0) is not in the region

3x > 6

 x < a lies to the left of the line

x

y

–2

x – y 2

2x + y –2

y 1

2

1

–1

Above this line

Below this line

Above this lineThe boundary lines are included

y

x

PointIntercept Position

SlopeInclinationDirection

Consider the straight line on the x-y plane

 y = mx + b

 y = m(x a)

 Ax + By + C = 0

y

x

The family of lines with slope of 2

 y = 2x + b

y

x

The family of lines with y-intercept 2

 y = mx + 2

2

m is called a parameter (or free variable)

 y = 3x + b  y = a(x – h)2 + k

 x2 + y2 = r2

Here are some additional examples:

 y = asin(x – p) + q

 y = mx + 2

 y = logbx + c

y

x

The parameter k is part of an expression that determines the y-intercept (or the x-intercept).

3x – 4y = k

The slope is always ¾

Definitions:The objective function:

The feasible region:

The corner points:

A linear function (consisting of two variables if a plane is used) which is to be maximized or minimized.

The set of interior and boundary points of a polygon defined by the inequality relations which specify the restrictions (the constraints) on the variables used.

The vertices of the feasible region (the set of possible solutions to the problem).

The theory of linear programming states that, to find the optimum value of f, where  f = Ax + By, we need only test the values of f at the corner points of the feasible region R. This assumes that region R is a convex set of points determined by the intersection of 3 or more linear inequalities. The region R may be open (also called bounded) or closed (also called unbounded). The region is usually open when we are testing for a minimum value of f.

C

B

Q

P

DO

A

y

x

B

Maximum

The Objective Function

C

B

Q

P

DO

A

y

x

B

C

Maximums

The Objective Function

A

C

B

Q

P

DO

y

x

B

Minimum

The Objective Function

Maximize: 150x + 250yR =

Products Resources

Resources Suits (x) Gowns (y) Available

Cotton 2 1 16

Silk 1 2 11

Wool 1 3 15

Product - Resource Chart (figures in m2)

Subject to: 2x + 1y 16

1x + 2y 11

1x + 3y 15

x 0

y 0

The quantities must be positive

y

x

16

15

5

5.5

811

A (0, 5)

B (3, 4)

E (6.6, 2.8)

D (8, 0)

C (7, 2)

O (0, 0)

2x + y = 16 [1]

x + 3y = 15 [2]

6x + 3y = 48 [3]3 [1]

[3] [2] 5x = 33

x = 6.6

y = 2.8 Substituting in [1]

Divide by 5

Thus, (x, y) = (6.6, 2.8)

Point E in previous diagram

Conclusion: The maximum revenue can be generated by producing 7 suits and 2 gowns

Table for Testing the Revenue Function

Corner Point Objective Function: 150x + 250y

A ( 0 , 5 ) R = 150(0) + 250(5) B ( 3 , 4 ) R = 150(3) + 250(4) = 1450 C ( 7 , 2 ) R = 150(7) + 250(2) = 1550 D ( 8 , 0 ) R = 150(8) + 250(0) = 1200 O (0, 0) R = 150(0) + 250(0) = 0

= 1250

= 1550