A factory produces two types of drink, an ‘energy’ drink and a ‘refresher’ drink. The day’s output is to be planned. Each drink requires syrup, vitamin.

Post on 31-Mar-2015

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A factory produces two types of drink, an ‘energy’ drink and a ‘refresher’ drink. The day’s output is to be planned. Each drink requires syrup, vitamin supplement and concentrated flavouring, as shown in the table.

The last row in the table shows how much of each ingredient is available for the day’s production.

How can the factory manager decide how much of each drink to make?

THE PROBLEM

Linear Programming : Introductory Example

SyrupVitamin

supplement

Concentrated

flavouring

5 litres of energy drink

1.25 litres 2 units 30 cc

5 litres of refresher

drink1.25 litres 1 unit 20 cc

Availabilities 250 litres 300 units 4.8 litres

Energy drink sells at £1 per litre

Refresher drink sells at 80 p per litre

THE PROBLEM

Syrup constraint:

Let x represent number of litres of energy drink

Let y represent number of litres of refresher drink

0.25x + 0.25y 250

x + y 1000

FORMULATION

Vitamin supplement constraint:

Let x represent number of litres of energy drink

Let y represent number of litres of refresher drink

0.4x + 0.2y 300

2x + y 1500

FORMULATION

Concentrated flavouring constraint:

Let x represent number of litres of energy drink

Let y represent number of litres of refresher drink

6x + 4y 4800

3x + 2y 2400

FORMULATION

Objective function:

Let x represent number of litres of energy drink

• Energy drink sells for £1 per litre

Let y represent number of litres of refresher drink

• Refresher drink sells for 80 pence per litre

Maximise x + 0.8y

FORMULATION

- 200 200 400 600 800 1000 1200

- 200

200

400

600

800

1000

1200

1400

1600

x

y

Empty grid to accommodate the 3 inequalities

SOLUTION

- 200 200 400 600 800 1000 1200

- 200

200

400

600

800

1000

1200

1400

1600

x

y

1st constraint

Draw boundary line:

x + y = 1000x y

0100

0

1000

0

SOLUTION

- 200 200 400 600 800 1000 1200

- 200

200

400

600

800

1000

1200

1400

1600

x

y

1st constraint

Shade out unwanted region:

x + y 1000

SOLUTION

- 200 200 400 600 800 1000 1200

- 200

200

400

600

800

1000

1200

1400

1600

x

y

Empty grid to accommodate the 3 inequalities

SOLUTION

- 200 200 400 600 800 1000 1200

- 200

200

400

600

800

1000

1200

1400

1600

x

y

2nd constraint

Draw boundary line:

2x + y = 1500x y

0150

0

750 0

SOLUTION

- 200 200 400 600 800 1000 1200

- 200

200

400

600

800

1000

1200

1400

1600

x

y

2nd constraint

Shade out unwanted region:

2x + y 1500

SOLUTION

- 200 200 400 600 800 1000 1200

- 200

200

400

600

800

1000

1200

1400

1600

x

y

Empty grid to accommodate the 3 inequalities

SOLUTION

- 200 200 400 600 800 1000 1200

- 200

200

400

600

800

1000

1200

1400

1600

x

y

3rd constraint

Draw boundary line:

3x + 2y = 2400

x y

0120

0

800 0

SOLUTION

- 200 200 400 600 800 1000 1200

- 200

200

400

600

800

1000

1200

1400

1600

x

y

3rd constraint

Shade out unwanted region:

3x + 2y 2400

SOLUTION

- 200 200 400 600 800 1000 1200

- 200

200

400

600

800

1000

1200

1400

1600

x

y

All three constraints:

First:

x + y 1000

SOLUTION

- 200 200 400 600 800 1000 1200

- 200

200

400

600

800

1000

1200

1400

1600

x

y

All three constraints:

First:

x + y 1000

Second:

2x + y 1500

SOLUTION

- 200 200 400 600 800 1000 1200

- 200

200

400

600

800

1000

1200

1400

1600

x

y

All three constraints:

First:

x + y 1000

Second:

2x + y 1500

Third:

3x + 2y 2400

SOLUTION

- 200 200 400 600 800 1000 1200

- 200

200

400

600

800

1000

1200

1400

1600

x

y

All three constraints:

First:

x + y 1000

Second:

2x + y 1500

Third:

3x + 2y 2400

Adding:

x 0 and y 0

SOLUTION

- 200 200 400 600 800 1000 1200

- 200

200

400

600

800

1000

1200

1400

1600

x

y

Feasible region is the unshaded area and satisfies:

x + y 1000

2x + y 1500

3x + 2y 2400

x 0 and y 0

SOLUTION

- 200 200 400 600 800 1000 1200

- 200

200

400

600

800

1000

1200

1400

1600

x

y

Evaluate the objective function

x + 0.8yat vertices of the feasible region:

O: 0 + 0 = 0

A: 0 + 0.8x1000 = 800

B: 400 + 0.8x600 = 880

C: 600 + 0.8x300= 840

D: 750 + 0 = 750

O

A

B

C

D

Maximum income = £800 at (400, 600)

SOLUTION

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