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
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.
Mar 31, 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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