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BMAN10960 Case Study 2(1)

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BMAN10960
Quantitative Methods for Business and
Management
Case Study 2
Dr Fanlin Meng
fanlin.meng@manchester.ac.uk
Office: AMBS 3.094
Problem description
Brief: SteelUK is a major steel producer in Northern England. The new CEO wants to implement a Social
Responsibility policy by committing SteelUK to reducing its annual emission of pollutants (see Table 1). The
company will invest in upgrading its technology regarding two aspects: better filters and better fuels. The
financial team has evaluated the annual cost of the investment in relation to the percentage of filters and
fuels upgraded to new technologies (Table 2). In particular, there is a basic (upkeep) cost that must be paid
independently of the upgrade and an additional cost that depends on the percentage of upgrade. For
example, if 10% of filters and 50% of fuels are upgraded, the total annual cost will be 0.15 · 10 + 2 + 0.05 ·
50 + 9. In addition, the engineering team has provided you with data estimating the reduction of pollutants as
a factor of the percentage of upgrade (Table 3). For example, if 10% of filters and 50% of fuels are upgraded,
the reduction in PM will be 0.3 · 10 + 0.09 · 50 = 7.5 tons ×1000, which is not enough to reach the target
reduction for PM of 10 tons ×1000 (Table 1).
The company has hired you to find out the percentage of its filters and fuels that must be upgraded in order
to achieve at least the target reduction (Table 1) for all pollutants while minimising the total annual cost.
BMAN10960 Case Study 2
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Problem description
Brief: SteelUK is a major steel producer in Northern England. The new CEO wants to implement a
Social Responsibility policy by committing SteelUK to reducing its annual emission of
pollutants (see Table 1).
BMAN10960 Case Study 2
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Problem description
The company will invest in upgrading its technology regarding two aspects: better filters and
better fuels. The financial team has evaluated the annual cost of the investment in relation to
the percentage of filters and fuels upgraded to new technologies (Table 2). In particular, there is a
basic (upkeep) cost that must be paid independently of the upgrade and an additional cost that depends on
the percentage of upgrade. For example, if 10% of filters and 50% of fuels are upgraded, the total annual
cost will be 0.15 · 10 + 2 + 0.05 · 50 + 9.
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Problem description
In addition, the engineering team has provided you with data estimating the reduction of pollutants
as a factor of the percentage of upgrade (Table 3). For example, if 10% of filters and 50% of fuels
are upgraded, the reduction in PM will be 0.3 · 10 + 0.09 · 50 = 7.5 tons ×1000, which is not enough to reach
the target reduction for PM of 10 tons ×1000 (Table 1).
BMAN10960 Case Study 2
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Problem description
Brief: SteelUK is a major steel producer in Northern England. The new CEO wants to implement a Social
Responsibility policy by committing SteelUK to reducing its annual emission of pollutants (see Table 1). The
company will invest in upgrading its technology regarding two aspects: better filters and better fuels. The
financial team has evaluated the annual cost of the investment in relation to the percentage of filters and
fuels upgraded to new technologies (Table 2). In particular, there is a basic (upkeep) cost that must be paid
independently of the upgrade and an additional cost that depends on the percentage of upgrade. For
example, if 10% of filters and 50% of fuels are upgraded, the total annual cost will be 0.15 · 10 + 2 + 0.05 ·
50 + 9. In addition, the engineering team has provided you with data estimating the reduction of pollutants as
a factor of the percentage of upgrade (Table 3). For example, if 10% of filters and 50% of fuels are upgraded,
the reduction in PM will be 0.3 · 10 + 0.09 · 50 = 7.5 tons ×1000, which is not enough to reach the target
reduction for PM of 10 tons ×1000 (Table 1).
percentage of its filters and fuels that must be
upgraded in order to achieve at least the target reduction (Table 1) for all pollutants while
minimising the total annual cost.
The company has hired you to find out the
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Q1
Question 1: Using the data in Tables 1, 2 and 3, formulate the problem as a linear programming (LP) model.
Label clearly the objective, constraints and decision variables.
Decision variables
π‘₯1 = % π‘’π‘π‘”π‘Ÿπ‘Žπ‘‘π‘’ π‘“π‘–π‘™π‘‘π‘’π‘Ÿπ‘ 
π‘₯2 = % π‘’π‘π‘”π‘Ÿπ‘Žπ‘‘π‘’ 𝑓𝑒𝑒𝑙𝑠
Objective function
𝑀𝑖𝑛 𝑧 = 0.15π‘₯1 + 0.05π‘₯2 + 2 + 9
Constraints
𝑃𝑀: 0.3π‘₯1 + 0.09π‘₯2 ≥ 10
𝑆02 : 0.4π‘₯1 + 0.01π‘₯2 ≥ 34
𝐢𝑂: 0.2π‘₯1 + 0.5π‘₯2 ≥ 50
π‘₯1 , π‘₯2 ≥ 0
π‘₯1 , π‘₯2 ≤ 100
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Q2
Question 2: Solve the LP problem from Question 1 graphically and using
Excel solver, respectively.
For the graphical solution, draw a coordinate system with all the constraints,
indicate where the feasible region is, and draw the objective function going
through the optimal solution. Provide the coordinates of the optimal solution
(approximately).
Compare your graphical solution with the solution obtained from Excel Solver.
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Q2 Graphical Solution
Constraints:
𝑃𝑀: 0.3π‘₯1 + 0.09π‘₯2 ≥ 10; 𝑆02 : 0.4π‘₯1 + 0.01π‘₯2 ≥ 34; 𝐢𝑂: 0.2π‘₯1 + 0.5π‘₯2 ≥ 50
π‘₯1 , π‘₯2 ≥ 0; π‘₯1 , π‘₯2 ≤ 100
Feasible Region
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Q2 Graphical Solution
Objective function:
𝑀𝑖𝑛 𝑧 = 0.15π‘₯1 + 0.05π‘₯2 + 2 + 9
BMAN10960 Case Study 2
π’™πŸ ≈ πŸ–πŸ”
π’™πŸ ≈ πŸ”πŸ”
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Q3
Question 3: Two other types of technologies are also considered for upgrade: Furnaces and
smokestacks. Data regarding the financial cost and pollution reduction is given in Table 4 as a linear
factor (i.e., the slope of the linear function) of the percentage of upgrade.
Extend the Linear Programming (LP) model from Q2 to include these two additional decision
variables. Note that the inclusion of these two decision variables will require you to modify both the
objective and the constraints. Solve the resulting LP problem by means of the Solver tool in Excel.
What is the percentage of filters, fuels, furnaces and smokestacks that should be upgraded to the
new technologies? Compare your results to Q2 and analyse the reason(s) for a potential shift in the
optimal solution (compared to Q2).
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Q3
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Q3
Decision variables
π‘₯1 = % π‘’π‘π‘”π‘Ÿπ‘Žπ‘‘π‘’ π‘“π‘–π‘™π‘‘π‘’π‘Ÿπ‘ 
π‘₯2 = % π‘’π‘π‘”π‘Ÿπ‘Žπ‘‘π‘’ 𝑓𝑒𝑒𝑙𝑠
π’™πŸ‘ = % π’–π’‘π’ˆπ’“π’‚π’…π’† 𝑭𝒖𝒓𝒏𝒂𝒄𝒆𝒔
π’™πŸ’ = % π’–π’‘π’ˆπ’“π’‚π’…π’† π‘Ίπ’Žπ’π’Œπ’†π’”π’•π’‚π’„π’Œπ’”
Objective function
𝑀𝑖𝑛 𝑧 = 0.15π‘₯1 + 0.05π‘₯2 + 𝟎. πŸπ’™πŸ‘ + 𝟎. πŸŽπŸ“π’™πŸ’ + 2 + 9
Constraints
𝑃𝑀: 0.3π‘₯1 + 0.09π‘₯2 + 𝟎. πŸŽπ’™πŸ‘ + 𝟎. πŸπ’™πŸ’ ≥ 10
𝑆02 : 0.4π‘₯1 + 0.01π‘₯2 + 𝟎. πŸŽπŸ‘π’™πŸ‘ + 𝟎. πŸπŸ“π’™πŸ’ ≥ 34
𝐢𝑂: 0.2π‘₯1 + 0.5π‘₯2 + 𝟎. πŸπŸ“π’™πŸ‘ + 𝟎. πŸŽπŸ–π’™πŸ’ ≥ 50
π‘₯1 , π‘₯2 , π’™πŸ‘ , π’™πŸ’ ≥ 0
π‘₯1 , π‘₯2 , π’™πŸ‘ , π’™πŸ’ ≤ 100
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