Linear Programming

Linear Programming provides methods for allocating limited resources among competing activities in an optimal way.

Any problem whose model fits the format for the linear programming model is a linear programming problem.

Wyndor Glass Co. example Two variables – Graphical method Maximize profit

Mary diagnosed with cancer of the bladder → needs radiation therapy

Radiation therapy Involves using an external beam to pass radiation

through the patient’s body Damages both cancerous and healthy tissue Goal of therapy design is to select the number,

direction and intensity of beams to generate best possible dose distribution

Doctors have already selected the number (2) and direction of the beams to be used

Goal: Optimize intensity (measured in kilorads) referred to as the dose

Area

Fraction of Entry Dose

Absorbed byArea (Average)

Restriction on TotalAverage Dosage,

KiloradsBeam1 Beam2

Healthy Anatomy 0.4 0.5 minimizeCritical Tissues 0.3 0.1 ≤ 2.7Tumor Region 0.5 0.5 = 6.0Tumor Center 0.6 0.4 ≥ 6.0

Graph the equations to determine relationships

Minimize Z = 0.4x1 + 0.5x2

Subject to:0.3x1 + 0.1x2 ≤ 2.7

0.5x1 + 0.5x2 = 6

0.6x1 + 0.4x2 ≥ 18

x1 ≥ 0, x2 ≥ 0

In order to ensure optimal health (and thus accurate test results), a lab technician needs to feed the rabbits a daily diet containing a minimum of 24 grams (g) of fat, 36 g of carbohydrates, and 4 g of protein. But the rabbits should be fed no more than five ounces of food a day.

Rather than order rabbit food that is custom-blended, it is cheaper to order Food X and Food Y, and blend them for an optimal mix. Food X contains 8 g of fat, 12 g of carbohydrates, and 2 g of

protein per ounce, and costs $0.20 per ounce. Food Y contains 12 g of fat, 12 g of carbohydrates, and 1 g of

protein per ounce, at a cost of $0.30 per ounce.

What is the optimal blend?

Daily Amount

Food TypeDaily Requirements

(grams)X Y

Fat 8 12 ≥ 24Carbohydrates 12 12 ≥ 36

Protein 2 1 ≥ 4

maximum weight of the food is five ounces:

X + Y ≤ 5

Minimize the cost:

Z = 0.2X + 0.3Y

Graph the equations to determine relationships

Minimize Z = 0.2x + 0.3y

Subject to:fat: 8x + 12y ≥ 24 carbs: 12x + 12y ≥ 36 protein: 2x + 1y ≥ 4 weight: x + y ≤ 5

x ≥ 0, y ≥ 0

When you test the corners at: (0, 4), (0, 5), (3, 0), (5, 0), and (1,

2) you get a minimum cost of sixty cents per daily serving, using three ounces of Food X only.

Only need to buy Food X

You have $12,000 to invest, and three different funds from which to choose.

Municipal bond: 7% return

CDs: 8% returnHigh-risk acct: 12% return (expected)

To minimize risk, you decide not to invest any more than $2,000 in the high-risk account.

For tax reasons, you need to invest at least three times as much in the municipal bonds as in the bank CDs.

Assuming the year-end yields are as expected, what are the optimal investment amounts?

Bonds (in thousands): xCDs (in thousands): yHigh Risk: z

Um... now what? I have three variables for a two-dimensional linear plot Use the "how much is left" concept Since $12,000 is invested, then the high

risk account can be represented as z = 12 – x – y

Constraints:Amounts are non-negative:

x ≥ 0

y ≥ 0z ≥ 0 12 – x – y ≥ 0 y ≤ –x + 12

High risk has upper limit z ≤ 2 12 – x – y ≤ 2

y ≤ –x + 10

Taxes: 3y ≤ x y ≤ 1/3 x

Objective to maximize the return:

Z = 0.07x + 0.08y + 0.12z

Z = 1.44 - 0.05x – 0.04y

When you test the corner points at (9, 3), (12, 0), (10, 0), and (7.5, 2.5), you should get an optimal return of $965 when you invest $7,500 in municipal bonds, $2,500 in CDs, and the remaining $2,000 in the high-risk account.

Machine data

Product dataP Q R

Revenue per unit $90 100 $70

Material cost per unit $45 $40 $20

Profit per unit $45 $60 $50

Maximum sales 100 40 60

Unit processing times(min)

Availability(min)

Machine \ Product P Q RA 20 10 10 2400B 12 28 16 2400C 15 6 16 2400D 10 15 0 2400

Total processing time 57 59 42

Revenue: $70/unit

Max sales: 60 units/week

Component 3

Revenue: $90/unit

Max sales: 100 units/week

P Q

D 10 min/unit

D 15 min/unit

C 9 min/unit

C 6 min/unit

B 16 min/unit

A 20 min/unit

B 12 min/unit

A 10 min/unit

RM1 $20/unit

RM2 $20/unit

RM3 $20/unit

Purchased part:

$5/unit

Machines A,B,C,D Available time: 2400 min/week Operating expenses: $6000/week

RRevenue: $100/unit

Max sales: 40 units/week

Component 1 Component 2

C 16 min/unit

max 45 xP + 60 xQ Objective Function

s.t. 20 xP + 1800

12 xP + 28 xQ 144015 xP

+ 6 xQ 204010 xP

+ 15 xQ 2400

demand

Are we done? nonnegativity

Are the LP assumptionsvalid for this problem?

Optimal solution x *P = 81.82 x*

Q = 16.36

Structuralconstraints

xP ≥ 0, xQ ≥ 0

xP 100, xQ 40

10 xQ

40

80

120

160

200

240

00 40 80 120 160 200 240 280 320 360

P

Q

C B

D

A

Max Q

Max P

B0

20

40

60

80

100

120

0 10 20 30 40 50

A

Max Q

P

Q60

Z = $3600

Z = $4664

Optimal solution = (16.36, 81.82)

Optimal objective value is $4664 but when we subtract the weekly operating expenses of $3000 we obtain a weekly profit of $1664.

Machines A & B are being used at maximum level and are bottlenecks.

There is slack production capacity in Machines C & D.