Unit 9.2
Analysing and Managing
Business Data
4BUS1149
Optimization: Formulating the Problem
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Introduction
• In the last sub-unit, we demonstrated a trial and error approach to
a typical optimisation problem. This was found to be
unsatisfactory even when only a small number of variables and
constraints are involved
• In this sub-unit we will formulate problems mathematically.
• Using this formulation, in the final sub-unit, we will the enter the
details into Excel and find the optimal solution using Excel’s
Solver.
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Steps involved in solving an optimisation problem
 Understand and summarise the problem
 Define the decision variables
 Write the objective function
 Write the constraints
We will then be in a position to:
 Develop the spreadsheet model and use Solver
 Analyse and interpret the results
Reminder: Optimisation Example 1
An office needs photocopiers in a print area covering 19.2 m2.
QuickCopy can make 15,000 copies per week
it costs £120 a week to rent and
takes 2.5 square metres of floor space.
FastPrint can make 18,500 copies per week
it costs £150 a week to rent and
takes 1.8 square metres of floor space.
Maximise the number of copies given a weekly budget of £1200.
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Example 1: Finding the optimal plan
Reminder: Summary of Example 1
QuickCopy
FastPrint
Number rented
?
?
Copies per unit
15000
18500
Maximise
Resource - cost
£120
£150
£1200 available
2.5 m2
1.8 m2
Resource - space
19.2 m2 available
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Example 1: Decision Variables
How many of each type of copier should we buy?
Let Q be the number of QuickCopy
and let F be the number of FastPrint.
Q and F are the
Q and F represent the number of copiers bought. decision variables
We might buy 0 or 1 or 2 or 3 or 4 or ... .
Q and F cannot
be negative
We have the two constraints Q ≥ 0
and F ≥ 0.
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Example 1: Objective Function
QuickCopy
FastPrint
Number rented
Q
F
Copies per unit
15000
18500
Maximise
Resource - cost
£120
£150
£1200 available
2.5 m2
1.8 m2
Resource - space
Q.F non-negative
19.2 m2 available
Objective
We want to maximise the number of copies, C.
The number of copies, C, is given by
C = 15000Q + 18500F
15000 copies on each QuickCopy.
Q are bought.
C is the
objective function
18500 sheets on each FastPrint.
F are bought.
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Example 1: Constraint - Cost
QuickCopy
FastPrint
Number rented
Q
F
Copies per unit
15000
18500
Maximise
Resource - cost
£120
£150
£1200 available
2.5 m2
1.8 m2
Resource - space
Q.F non-negative
19.2 m2 available
Constraints (i) cost
QuickCopy costs £120 per week and
FastPrint costs £150 per week.
The cost of rental, R, is given by
R = 120Q + 150F
The budget available for rental is £1200 so R ≤ 1200.
This constraint gives
120Q + 150F ≤ 1200
used ≤ available
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Example 1: Constraint - Space
QuickCopy
FastPrint
Number rented
Q
F
Copies per unit
15000
18500
Maximise
Resource - cost
£120
£150
£1200 available
2.5 m2
1.8 m2
Resource - space
Q.F non-negative
19.2 m2 available
Constraints (ii) space
QuickCopy needs 2.5 m2 of space and
FastPrint needs 1.8 m2 of space.
The space needed, S, is given by
S = 2.5Q + 1.8F
The space available is19.2 m2 so S ≤ 19.2.
This constraint gives
2.5Q + 1.8F ≤ 19.2
used ≤ available
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Example 1: Mathematical Formulation
Decision variables
Q = number of QuickCopy
F = number of FastPrint
Objective function
Maximize
C = 15000Q +18500F
Constraints
Q and F are non-negative Q ≥ 0, F ≥ 0
Cost
120Q + 150F ≤1200
Space
2.5Q + 1.8F ≤ 19.2
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Example 1: Feasible Region
The four constraints,
Q ≥ 0,
F ≥ 0, 120Q + 150F ≤ 1200,
2.5Q + 1.8F ≤ 19.2
form the feasible region.
F
Q
Only the points within the feasible region satisfy the constraints.
We want to find which point maximises the number of copies.
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Optimisation Example 2
An Indian clothing manufacturer makes 2 products:
– Kurta (Sherwani)
– Trousers
Both have a similar production requirements:
– Fabric
– Labour time.
• A Kurta requires 2.5 metres of fabric and take 3 hours to make
• Trousers require 2 metres of fabric and take 1.5 hours to make
In the next week there are 300 metres of fabric and 275 hours of labour
time available.
Each Kurta yields a profit of £7 and each pair of Trousers £5.
Maximise profit.
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Example 2: Summary
It may be helpful to summarise the data given.
Kurta
Trousers
Number made
?
?
Profit per unit
7
5
Maximise
2.5
2
300 available
3
1.5
275 available
Resource - Fabric
Resource - labour time
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Example 2: Decision Variables
How many of each type should we make?
Let K be the number of Kurta (Sherwani )
and let T be the number of Trousers
.
K and T are the
K and T represent the number of each item made. decision variables
We might make 0 or 1 or 2 or 3 or 4 or ... .
K and T cannot
be negative
We have the two constraints K ≥ 0 and T ≥ 0.
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Example 2: Objective Function
Kurta
Trousers
Number made
K
T
K, T non negative
Profit per unit (£)
7
5
Maximise
Resource - Fabric
2.5
2
300 available
3
1.5
275 available
Resource - labour time
Objective
We want to maximise profit, P.
P is the
objective function
The profit, P, is given by
A.
B.
C.
D.
P = 7K + 5T
P = 12(K + T)
P = 12K + 12T
P = (7+K) + (5+T)
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Example 2: Objective Function
We want to maximise profit, P.
P is the
objective function
The profit, P, is given by
P = 7K + 5T
£7 profit on each kurta.
x are made.
£5 profit on each pair of trousers.
y are made.
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Example 2: Constraint – Fabric
Kurta
Trousers
Number made
K
T
K, T non negative
Profit per unit (£)
7
5
Maximise
2.5
2
300 available
3
1.5
275 available
Resource - Fabric
Resource - labour time
Resources (i) Fabric
Kurta use 2.5 metres of fabric
K are made.
2.5K
Trousers use 2 metres of fabric
T are made.
2T
The fabric used, F, is given by
F = 2.5K + 2T
The fabric available is 300 metres so F ≤ 300
giving the constraint
2.5K + 2T ≤ 300
used ≤ available
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Example 2: Constraint – Labour Time
Kurta
Trousers
Number made
K
T
K, T non negative
Profit per unit (£)
7
5
Maximise
Resource - Fabric
2.5
2
300 available
3
1.5
275 available
Resource - labour time
Resources (ii) labour time
The constraint for labour time is:
A. 3K + 1.5T ≥ 275
B. 3K + 1.5T ≤ 275
C. 3K + 1.5T = 275
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Example 2: Constraint – Labour
Kurta
Trousers
Number made
K
T
K, T non negative
Profit per unit (£)
7
5
Maximise
Resource - Fabric
2.5
2
300 available
3
1.5
275 available
Resource - labour time
Resources (ii) labour time
A kurta uses 3 hours of labour time.
K are made.
3K
Trousers use 1.5 hour of labour time.
T are made.
1.5T
The total labour time used, L, is given by
L = 3K + 1.5T
The labour time available is 275 hours so L ≤ 275
giving the constraint
3K + 1.5T ≤ 275
used ≤ available
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Example 2: Mathematical Formulation
Kurta
Trousers
Number made
K
T
K, T non negative
Profit per unit (£)
7
5
Maximise
Resource - Fabric
2.5
2
300 available
3
1.5
275 available
Resource - labour time
Decision variables
K = number of Kurta (Sherwani)
T = number of Trousers
Objective function
Maximize
P = 7K +5T
Constraints
K and T are non-negative
Fabric
Labour time
K ≥ 0,T ≥ 0
2.5K + 2T ≤ 300
3K + 1.5T ≤ 275
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Example 2: Feasible Region
The four constraints,
K ≥ 0,
T ≥ 0,
4K + 3T ≤ 240,
2K + T ≤ 100
form the feasible region.
T
K
Only the points within the feasible region satisfy the constraints.
We want to find which point maximises the profit.
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Optimisation Example 3
A company makes (window) frames and doors. It makes a
profit of £20 on a frame and £25 on a door.
Each frame require 6 units of materials, 3 units of machine
time and 2 units of labour.
Each door requires 5 units of materials, 4 units of machine
time and 1 unit of labour.
The resources available are 4000 units of materials, 2600
units of machine time and 2000 units of labour.
Find the production plan to maximise profit.
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Example 3: Summary
Frame
Door
Number made
?
?
Profit per unit
20
25
Maximise
Resource - materials
6
5
4000 available
Resource – machine time
3
4
2600 available
Resource - labour
2
1
2000 available
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Example 3: Decision Variables
How many of each type should we make?
Let F be the number of frames
and let D be the number of doors.
F and D represent the number of each item made.
We might make 0 or 1 or 2 or 3 or 4 or ... .
F and D are the
decision variables
F and D cannot
be negative
This gives two constraints. What are these constraints?
A.
B.
C.
D.
F ≤ 0, D ≤ 0
F ≥ 0, D ≥ 0
F+D≥0
F = 0, D = 0
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Example 3: Objective Function
Frame
Door
Number made
?
?
Profit per unit
20
25
Maximise
Resource - materials
6
5
4000 available
Resource – machine time
3
4
2600 available
Resource - labour
2
1
2000 available
Objective
We want to maximise profit, P.
P is the
objective function
The profit, P, is given by
P = 20F + 25D
£20 profit on each frame.
F are made.
£25 profit on each door.
D are made.
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Example 2: Constraint – Materials
Frame
Door
Number made
?
?
Profit per unit
20
25
Maximise
Resource - materials
6
5
4000 available
Resource – machine time
3
4
2600 available
Resource - labour
2
1
2000 available
Resources (i) materials
The constraint for materials is of the form:
aF + bD ≤ c
The values of a, b and c are:
A.
B.
C.
D.
a = 11,
a = 6,
a = 5,
a = 11,
b = 11,
b = 5,
b = 6,
b = 10,
c = 8600
c = 4000
c = 4000
c = 8600
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Example 3: Constraint – Materials
Frame
Door
Number made
?
?
Profit per unit
20
25
Maximise
Resource - materials
6
5
4000 available
Resource – machine time
3
4
2600 available
Resource - labour
2
1
2000 available
Resources (i) materials
A frame uses 6 units of materials. F are made.
A door uses 5 units of materials. D are made.
6F
5D
The total of materials used, M, is given by M = 6F + 5D
The materials available is 4000 units so M ≤ 4000
giving the constraint
6F + 5D ≤ 4000
used ≤ available
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Example 3: Mathematical Formulation
Frame
Door
Number made
?
?
Profit per unit
20
25
Maximise
Resource - materials
6
5
4000 available
Resource – machine time
3
4
2600 available
Resource - labour
2
1
2000 available
Exercise: Complete the mathematical formulation
Decision variables:
F = number of frames
D = number of doors
Objective function:
Maximize
Constraints:
F and D are non-negative
materials
machine time
labour
………………
………………
………………
………………
………………
Example 3: Mathematical Formulation
Frame
Door
Number made
?
?
Profit per unit
20
25
Maximise
Resource - materials
6
5
4000 available
Resource – machine time
3
4
2600 available
Resource - labour
2
1
2000 available
Exercise Answers: Complete the mathematical formulation
Decision variables:
F = number of frames
D = number of doors
Objective function:
Maximize
Constraints:
x and y are non-negative
materials
machine time
labour
P = 20F +25D
F ≥ 0, D ≥ 0
6F + 5D ≤ 4000
3F + 4D ≤ 2600
2F + D ≤ 200029
Example 3: Feasible Region
The five constraints,
F ≥ 0, D ≥ 0, 6F + 5D ≤ 4000, 3F + 4D ≤ 2600, 2F + D ≤ 2000
form the feasible region.
D
F
Only the points within the feasible region satisfy the constraints.
We want to find which point maximises the profit.
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Solving an optimisation problem using Excel
• Now that we have formulated the problem
mathematically, we can build the model in Excel.
• We can then use the Solver Add-In to find the optimal
solution. We will look at this in the next sub-unit.
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Student Tasks
• Try the ‘Unit 9.2 Formulating Optimisation
Problems Quiz’ to check your understanding and
follow the feedback if you make any mistakes.
• When you feel confident with this topic,
go on to the next session in the unit.
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