paper uas rekjas_rev 141213_12.28
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Professor
Prof. Ir. Isti Surjandari, M.T., M.A., Ph.D
Member of Group
[1206181244] Ibrahim Ali Marwan
[1206180336] Alan Dwi Wibowo
[1206180651] Billy Muhammad Iqbal
Due Date
December 16, 2013
Mid-Atlantic Bus Lines(Case 8.3)
Problem Case taken fromFitzsimmons, J. A. and Fitzsimmons, M. J. 2011. Service Management,
Operations, Strategy, Infromation technology 7th Edition. Mcgraw Hill. Singapore. p-210.
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A.BackgroundMid Atlantic Bus Lines, first-class express bus services operating between the major coastal
cities from Philadelphia, Pa. To Jacksonville, Fla. They established franchises in each city with
local enterpreneurs, who were authority to operate the pool stations. A percentage of the
passanger ticket sales and freight sales kept as cost and profit.
After several months some franchisees complained about inadequate profits. Because other
franchisees were pleased, a study was undertaken. Data outputs and inputs for Mid-Atlantic
Bus Lines as follows:
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Bus Depot City Served Ticket Sales Freight Sales Labor-Hours Facility Dollar
1 Philadelphia, Pa 700 300 40 500
2 Baltimore, Md. 300 600 50 500
3 Washington, D.C. 200 700 50 400
4 Richmond, Va. 400 600 50 500
5 Raleigh, N.C. 500 400 40 400
6 Charleston, S.C. 500 500 50 500
7 Savannah, Ga. 800 500 40 600
8 Jacksonville, Fla. 300 200 30 400
B.Problem Statements1. We have to identify bus terminals (pools) whether efficient or inefficient. The problem
as a linear programming model should have formulate and solve using computer
software (in this study, MS Excel Solver was used to solve the problem).
2. We have to describe our recommendations for changes in resource inputs for eachinefficient terminal include for the seriously inefficient terminal.
3. We have to discuss about shortcomings in using DEA of this case.
C. Solutions1. Solution 1 : A Linear Programming Model of Each Terminal & Productivity Frontier of
Mid-Atlantic Bus Lines
Data Envelopement Analysis (DEA) in Mid-Atlantic Bus Lines case conducted to determine
whether the terminal was efficient or inefficient. The first step, a linear programming model
was generated. Furthermore, the problem solved by Microsoft Excel (Solver Module Add-Ins).
The following is a linear programming model of each terminal.
Terminal 1 Model
Maximize E()= 700 300
Subject to
7 0 0 300 4 0 500 03 0 0 600 5 0 500 0
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2 0 0 700 5 0 400 04 0 0 600 5 0 500 05 0 0 400 4 0 400 05 0 0 500 5 0 500 08 0 0 500 4 0 600 03 0 0 200 3 0 400 0
4 0 500 1, , , 0
Terminal 2 Model
Maximize E()= 300 600
Subject to
7 0 0 300 4 0 500 03 0 0 600 5 0 500 02 0 0 700 5 0 400 04 0 0 600 5 0 500 05 0 0 400 4 0 400 05 0 0 500 5 0 500 08 0 0 500 4 0 600 03 0 0 200 3 0 400 0
5 0 500 1, , , 0
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Terminal 3 Model
Maximize E()= 200 700
Subject to
7 0 0 300 4 0 500 03 0 0 600 5 0 500 02 0 0 700 5 0 400 04 0 0 600 5 0 500 05 0 0 400 4 0 400 05 0 0 500 5 0 500 08 0 0 500 4 0 600 03 0 0 200 3 0 400 0
5 0 400 1, , , 0
Terminal 4 Model
Maximize E()= 400 600
Subject to
7 0 0 300 4 0 500 03 0 0 600 5 0 500 02 0 0 700 5 0 400 04 0 0 600 5 0 500 05 0 0 400 4 0 400 05 0 0 500 5 0 500 08 0 0 500 4 0 600 03 0 0 200 3 0 400 0
5 0 500 1, , , 0
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Terminal 5 Model
Maximize E()= 500 400
Subject to
7 0 0 300 4 0 500 03 0 0 600 5 0 500 02 0 0 700 5 0 400 04 0 0 600 5 0 500 05 0 0 400 4 0 400 05 0 0 500 5 0 500 08 0 0 500 4 0 600 03 0 0 200 3 0 400 0
4 0 400 1, , , 0
Terminal 6 Model
Maximize E()= 500 500
Subject to
7 0 0 300 4 0 500 03 0 0 600 5 0 500 02 0 0 700 5 0 400 04 0 0 600 5 0 500 05 0 0 400 4 0 400 05 0 0 500 5 0 500 08 0 0 500 4 0 600 03 0 0 200 3 0 400 0
5 0 500 1, , , 0
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Terminal 7 Model
Maximize E()= 800 500
Subject to
7 0 0 300 4 0 500 03 0 0 600 5 0 500 02 0 0 700 5 0 400 04 0 0 600 5 0 500 05 0 0 400 4 0 400 05 0 0 500 5 0 500 08 0 0 500 4 0 600 03 0 0 200 3 0 400 0
4 0 600 1, , , 0
Terminal 8 Model
Maximize E()= 300 200
Subject to
7 0 0 300 4 0 500 03 0 0 600 5 0 500 02 0 0 700 5 0 400 04 0 0 600 5 0 500 05 0 0 400 4 0 400 05 0 0 500 5 0 500 08 0 0 500 4 0 600 03 0 0 200 3 0 400 0
3 0 400 1, , , 0
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The second step is run the model. LP solution for DEA study in Mid-Atlantic Bus Lines
(Appendix 1) show that terminal 1 (P1), terminal 3 (P3), terminal 5 (P5), and terminal 7 (P7)
were the most efficient than the other terminals. Productivity Frontier of Mid-Atlantic Bus
Lines terminal shows in Graph 1. Based on this graph and LP solution for DEA study in Mid-
Atlantic Bus Lines, opportunity cost as bases to give some recommendations was defined.
Productivity Frontier
Figure 1.Productivity Frontier of Mid-Atlantic Bus Lines
P(n)are the coordinates of the labor-hours and dollar facility. Each point represents the point of
productivity for each terminal Mid-Atlantic Bus Lines. Based on the results of the DEA
analysis of P1, P3, P5, dan P7coordinates represent the most efficient terminal productivity
levels compared to other terminals. All coordinates are connected by red lines as a frontier line(see Figure 1.).
Based on DEA analysis P2, P4, P6, dan P8 are inefficient. If each units are desired to be more
efficient there should be a correction to the nearest reference unit (the most efficient unit). To
calculate the potential of a benchmark unit which provides the best value in making corrections
to the unit that inefficient. To get the optimum inputs for inefficient units can be calculated by
intersection points and interpolation between these two benchmark units.
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P2, P4 dan P6 are units that have the same coordinates, so that the correction to the results are
equal for these units (terminals). The first step in making corrections (to defined opportunity
cost), needs to draw a line from the coordinate origin point (0,0) to P2, P4 dan P6 coordinate.
The line will intersect the frontier line (P7-P3) at coordinate C1 (46.7, 466.7). The same steps
are used to make corrections to the point P8, draw the line from the coordinate origin point(0,0) to the point P8. This line will intersect the frontier line P5-P7 at coordinate C2 (40,
533.33). C1 and C2 coordinates (see Appendix 2 for calculation measures the coordinates of
each point of intersection) obtained from equation (Ayres and Mendelson, 1999):
1
Both C1 and C2 are coordinates for calculating the correction for inefficient terminals based
on the benchmark terminal to get the relative weight to the benchmarking terminals, which is
represented as the opportunity cost. Opportunity cost calculation is done by interpolation
between the two closest points are used as a benchmarking terminals. In metallurgy sciencethis approach is also known as "The Lever Rule". Interpolation calculation (see Appendix 3)
can be solved by the following equation (Callister and Rethwisch, 2010):
2
Where,
WL = Relative Weight for Benchmark L
W = beban titik referensi Relative Weight for Benchmark
R = Distance from L and intersection point
S = Distance from and intersection point
Based on the results of the calculation of the importance of the opportunity cost of each of the
inefficient points to nearby benchmarking points. Point P2with 87.4% efficiency rate has the
opportunity cost of the P3at 0.6667 and P7at 0.3333. For terminal P4with 90.2% efficiency
rate and terminal P6with an efficiency of 89.4% has the same opportunity cost with terminal
P2. P8has the lowest efficiency levels between 57.1% other. Point P8has the opportunity cost
of P5at 0.3333 and P7at 0.6667 (see Appendix 1). The summary of DEA results are shown in
Table 1.
Table 1. Summary of DEA Results
Service UnitEfficiency
Rating (E)
Efficiency Reference
Set
Relative Labor-
Hour Value (V1)
Relative Facility
Dollars Value (V2)
P1 1,000 N.A. 0,0000 0,0020
P2 0,874P3 (0,6667)
P7 (0,3333)0,0200 0,0000
P3 1,000 N.A. 0,0144 0,0007
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P4 0,902P3 (0,6667)
P7 (0,3333)0,0026 0,0017
P5 1,000 N.A. 0,0032 0,0022
P6 0,894 P3 (0,6667)P7 (0,3333)
0,0026 0,0017
P7 1,000 N.A. 0,0022 0,0015
P8 0,571P5 (0,3333)
P7 (0,6667)0,0033 0,0023
Conclusion 1
1. Efficient terminals (Benchmark) : P1 (Philadelphia, Pa), P3 (Washington, D.C.), P5(Raleigh, N.C.) dan P7(Savannah, Ga).
2. Inefficient terminals: P2(Baltimore, Md.), P4(Richmond, Va.), P6(Charleston, S.C.)dan P8(Jacksonville, Fla.).
2. Solution 2 : Recommendations For Changes In Resource Inputs For Each InefficientTerminal
DEA offer opportunities for an inefficient terminal to become efficient terminal regarding its
reference set of efficient terminals. The calculation of excess inputs used by each of inefficent
terminal shown in Table 2. The aim of this calculation is to generate any recommendations for
inefficient terminals to become efficient terminal.
a. Calculation of Excess Input Used by P2(Baltimore, Md.)Table 2. contains the calculation for a hypothetical Unit C1, which is composite
reference unit defined by the weighted input of the reference P3and P7. Composite unit
C1is located in intersection of the productivity frontier and the dashed line drawn from
the origin to unit P2. Thus compared with this reference unit C1, inefficient unit P2is
using excess inputs in the amounts of 3,33 labor-hours and 33,33 facility dollars.
So in order to make efficient for P2 unit, inputs must be reduced to a number of labor-
hours of 3.33 and 33.33 dollars for the facility. Unit P2 inputs are excess of its business
activities, thus resulting in a loss of 100 for ticket sales and 33.33 for freight sales.
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Table 2. Calculation of Excess Input Used by P2(Baltimore, Md.)
Outputs and
Inputs
Reference Set Composite
Reference
Unit C1
Excess
Input
UsedP3 P7 P2
Ticket Sales (0.6667) x 200 + (0.3333) x 800 400 300 -100
Freight Sales (0.6667) x 700 + (0.3333) x 500 633,33 600 -33,33
Labor-Hours (0.6667) x 700 + (0.3333) x 40 46,67 50 3,33
Facility Dollar (0.6667) x 400 + (0.3333) x 600 466,67 500 33,33
b. Calculation of Excess Input Used by P4(Richmond, Va.)Table 3. contains the calculation for a hypothetical Unit C1, which is composite
reference unit defined by the weighted input of the reference P3and P7. Composite unit
C1is located in intersection of the productivity frontier and the dashed line drawn from
the origin to unit P4. Thus compared with this reference unit C1, inefficient unit P4is
using excess inputs in the amounts of 3,33 labor-hours and 33,33 facility dollars.
So in order to make efficient the unit P4inputs should be reduced to a number of labor-
hours of 3.33 and 33.33 dollars for the facility. Unit P4 input excess in business
activities, thus resulting in a loss of 33.33 for freight sales.
Table 3.Calculation of Excess Input Used by P4(Richmond, Va.)
Outputs and
Inputs
Reference Set Composite
ReferenceUnit C1
Excess
InputUsedP3 P7 P4
Ticket Sales (0.6667) x 200 + (0.3333) x 800 400 400 0
Freight Sales (0.6667) x 700 + (0.3333) x 500 633,33 600 -33,33
Labor-Hours (0.6667) x 50 + (0.3333) x 40 46,67 50 3,33
Facility Dollar (0.6667) x 400 + (0.3333) x 600 466,67 500 33,33
c. Calculation of Excess Input Used by P6(Charleston, S.C.)Table 4. contains the calculation for a hypothetical Unit C1, which is composite
reference unit defined by the weighted input of the reference P3and P7. Composite unit
C1is located in intersection of the productivity frontier and the dashed line drawn from
the origin to unit P6. Thus compared with this reference unit C1, inefficient unit P6is
using excess inputs in the amounts of 3,33 labor-hours and 33,33 facility dollars.
So in order to make efficient the unit P6input must be reduced to a number 3.33 of
labor-hours and 33.33 dollars for the facility. Unit P6 input excess in its business
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activities, thus resulting in a loss of 133.33 for freight sales and 100 for the ticket sales.
However, if the aggregate income units P6count is still a loss for ticket sales and freight
sales are loss reached 33.33.
Table 4.Calculation of Excess Input Used by P6(Charleston, S.C.)
Outputs and
Inputs
Reference Set Composite
Reference
Unit C1
Excess
Input
UsedP3 P7 P6
Ticket Sales (0.6667) x 200 + (0.3333) x 800 400 500 100
Freight Sales (0.6667) x 700 + (0.3333) x 500 633,33 500 -133,33
Labor-Hours (0.6667) x 50 + (0.3333) x 40 46,67 50 3,33
Facility Dollar (0.6667) x 400 + (0.3333) x 600 466,67 500 33,33
d. Calculation of Excess Input Used by P8(Jacksonville, Fla.)Table 5. contains the calculation for a hypothetical Unit C2, which is composite
reference unit defined by the weighted input of the reference P5and P7. Composite unit
C2is located in intersection of the productivity frontier and the dashed line drawn from
the origin to unit P8. Thus compared with this reference unit C2, inefficient unit P8is
using excess inputs in the amounts of 10 labor-hours and 133,33 facility dollars.
So in order to make efficient the P8unit input must be to increase 10 labor-hours and
133.33 dollars for the facility. P8lack of input unit in business activities, thus resulting
in a loss of 400 for ticket sales and 266.67 for freight sales.
Table 5.Calculation of Excess Input Used by P8(Jacksonville, Fla.)
Outputs and
Inputs
Reference Set Composite
Reference
Unit C2
Excess
Input
UsedP5 P7 P8
Ticket Sales (0.3333) x 500 + (0.6667) x 800 700 300 -400
Freight Sales (0.3333) x 400 + (0.6667) x 500 466,67 200 -266,67
Labor-Hours (0.3333) x 40 + (0.6667) x 40 40 30 -10
Facility Dollar (0.3333) x 400 + (0.6667) x 600 533,33 400 -133,33
3. Solution 3 : Recommendations For The One Seriously Inefficient In Regard to IncreasingIts Outputs.
Based on Table 1, unit P8 (Jacksonville, Fla.) is the location of the most inefficient terminal.
The level of efficiency at the point P8 based DEA analysis is 57.1%. Based on Table 5,
calculation of ticket sales and freight sales had the largest losses compared to the other terminal
locations, 400 for the number of ticket sales and 266.67 for freight sales. Thus it is clear that
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the point P8is the most serious inefficient terminals. This situation happened because P 8lack
of input unit in business activities. So, first recommendation in order to increase efficiency the
P8unit input must be to increase 10 labor-hours and 133.33 dollars for the facility.
When combined with profitablility, DEA efficiency analysis can be useful in strategic planningfor service of each Mid-Atlantic Bus Lines terminal. Four possibilities of strategic planning
that arise from combining efficiency and profitability presents in Figure 2. Four possibilities of
strategic planning describes as follows:
a. Benchmark group Quadrant 1: Efficiency and profit high.b. Underperforming potential stars Quadrant 2 : Profit highh and efficiency lowc. Candidates for divesture Quadrant 3: Efficiency high and profit lowd. Problem branches Quadrant 4: Efficiency and profit low
Figure 2.DEA Matrix
To determine the second recommendation which is based on the strategic planning of the most
inefficient terminal P8, plotting between the efficiency of the profit obtained for each location
on the X axis and Y-axis Plotting the results are presented in Figure 3.
Figure 3 shows that for the location for terminal P1, P2, P3, P4, P5, P6and P7plotted in thequadrant 1 (Benchmark Group) while the terminal P8plotted in quadrant 3 (Candidates for
divesture). Location P8(Jacksonville) has performed quite efficiently but profit is very low.
This point has very limited profits that may result from the location is lack of the customer.
Based on DEA recommendation matrix can be used as references in preparing the strategic
plan is the location P8terminal divesture. Divesture of this terminal could be done by selling
the facilities to get capital that can be used for development in other prospective area.
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Figure 3. DEA Matrix for each of Mid-Atlantic Bus Lines terminal
Conclusion 3 : P8terminal location (Jacksonville, Fla.) is the lowest efficiency with the most
serious problems. There are two recommendations can be given as follows:
1. Inputs must be to increase which are 10 labor-hours and 133.33 dollars for the facility.2. Selling the facilities to get capital that can be used for development in other prospective
area.
4. Solution 4 : Shortcomings in the implementation of DEA to Mid-Atlantic Bus LinesIn practical use, especially in the case of the Mid-Atlantic Bus Lines, felt there are some
shortcomings of the use of the DEA method as mention below:
1. DEA did modeling LP one by one for each terminal, this has resulted in a verynumerous model that should be made and error probability in modeling is very high.
2. The use of LP approach which is a non-heuristic approach for optimization has aweakness if the model or unit that must be solved in very large number, so it can make
the process of calculation or iteration process takes a very long time.
3. The use of statistical hypothesis is difficult to do, as in this case is at the terminal 2, 4,and 6 where we difficult to provide a significant difference if the terminal has the same
input that result in overlapping cases.
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Reference
Callister, W. D. and Rethwisch, D. G. 2010.Material Science and Engineering An Introduction
Eighth Edition. John Wiley & Sons, Inc. United States of America.
Ayres, F. and Mendelson, E. 1999.Schaums Outlines of Theory and Problems of Calculus
Fourth Edition. McGraw-Hill.
Fitzsimmons, J. A. and Fitzsimmons, M. J. 2011. Service Management, Operations, Strategy,
Infromation Technology 7th Edition. Mcgraw Hill. Singapore.
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Appendix 1Appendix 1 : LP solution for DEA study in Mid-Atlantic Bus Lines
Summarized Results For Pool 1
VariablesSolutions Opportunity Cost
VariablesSolutions Opportunity Cost
No Names No Names
1 U1 0,001273 0 8 P4 0,272727 0
2 U2 0,000364 0 9 P5 0,018182 0
3 V1 0 0 10 P6 0,181818 0
4 V2 0,002 0 11 P7 0 0
5 P1 0 1,0000000 12 P8 0,345455 0
6 P2 0,4 0 13 I9 0 100,00000
7 P3 0,290909 0
Maximize Objective Function = 100
Summarized Results For Pool 2
VariablesSolutions Opportunity Cost
VariablesSolutions Opportunity Cost
No Names No Names
1 U1 0.00013 0 8 P4 0,11304 0
2 U2 0,001391 0 9 P5 0,17826 0
3 V1 0,002 0 10 P6 0,23913 0
4 V2 0 0 11 P7 0 0,33333333
5 P1 0,291305 0 12 P8 0,28261 0
6 P2 0,126087 0 13 I9 0 87,39139
7 P3 0 0,6666667
Maximize Objective Function = 87,39139
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Summarized Results For Pool 3
VariablesSolutions Opportunity Cost
VariablesSolutions Opportunity Cost
No Names No Names
1 U1 0,000427 0 8 P4 0,11486 0
2 U2 0,001307 0 9 P5 0,11954 0
3 V1 0,014431 0 10 P6 0,20281 0
4 V2 0,000696 0 11 P7 0 0
5 P1 0,2344 0 12 P8 0,32196 0
6 P2 0,157566 0 13 I9 0 100,00000
7 P3 0 1,0000000
Maximize Objective Function = 100
Summarized Results For Pool 4
VariablesSolutions Opportunity Cost
VariablesSolutions Opportunity Cost
No Names No Names
1 U1 0,000851 0 8 P4 0,097872 0
2 U2 0,000936 0 9 P5 0 0
3 V1 0,002553 0 10 P6 0,106383 0
4 V2 0,001745 0 11 P7 0 0,3333333
5 P1 0,0097872 0 12 P8 0,331915 0
6 P2 0,182979 0 13 I9 0 90,21285
7 P3 0 0,6666667
Maximize Objective Function = 90,21285
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Summarized Results For Pool 5
VariablesSolutions Opportunity Cost
VariablesSolutions Opportunity Cost
No Names No Names
1 U1 0,001064 0 8 P4 0,122341 0
2 U2 0,00117 0 9 P5 0 1,000000
3 V1 0,003191 0 10 P6 0,132979 0
4 V2 0,002181 0 11 P7 0 0
5 P1 0,122341 0 12 P8 0,414894 0
6 P2 0,228724 0 13 I9 0 100,00000
7 P3 0 0
Maximize Objective Function = 100
Summarized Results For Pool 6
VariablesSolutions Opportunity Cost
VariablesSolutions Opportunity Cost
No Names No Names
1 U1 0,000851 0 8 P4 0,097872 0
2 U2 0,000936 0 9 P5 0 0
3 V1 0,002553 0 10 P6 0,106383 0
4 V2 0,001745 0 11 P7 0 0,333333
5 P1 0,097872 0 12 P8 0,331915 0
6 P2 0,182979 0 13 I9 0 89,36170
7 P3 0 0,66666667
Maximize Objective Function = 89,36170
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Summarized Results For Pool 7
VariablesSolutions Opportunity Cost
VariablesSolutions Opportunity Cost
No Names No Names
1 U1 0,000741 0 8 P4 0,085185 0
2 U2 0,000815 0 9 P5 0 0
3 V1 0,002222 0 10 P6 0,092593 0
4 V2 0,001519 0 11 P7 0 1,000000
5 P1 0,085185 0 12 P8 0,288889 -
6 P2 0,159259 0 13 I9 0 100,00000
7 P3 0 0
Maximize Objective Function = 100
Summarized Results For Pool 8
VariablesSolutions Opportunity Cost
VariablesSolutions Opportunity Cost
No Names No Names
1 U1 0,001099 0 8 P4 0,126374 0
2 U2 0,001209 0 9 P5 0 0,33333333
3 V1 0,003297 0 10 P6 0,137363 0
4 V2 0,002253 0 11 P7 0 0,66666667
5 P1 0,126374 0 12 P8 0,428572 0
6 P2 0,236264 0 13 I9 0 57,14291
7 P3 0 0
Maximize Objective Function = 57,14291
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Appendix 2Appendix 2 : Calculation of the point of intersection
Based on Graph 1.
1. C1IntersectionLine : P1(40,600) and P3(50,400)
600
400 600
40
5 0 4 0
600
200
40
10
10 6000 200 8000
10 200 14000
20 1400 1
Line : O (0,0) and P2, P4, P6(50,500)
0
5000
0
5 0 0
50 500
10 0 2
Subtitute (1) and (2)
20 1400
10 0
,
Furthermore, y was defined:
10 .140
3
,
So that C1(46.7,466.7)
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2. C2IntersectionLine : P5(40,400) and P7(40,600)
400
600 400
40
4 0 4 0
0 200 8000
40 1
Line : O (0,0) and P8(30,400)
0
400 0
0
3 0 0
30 400
3 40 0 2
Subtitute (1) and (2)
3 4040 0
3 1600
,
So that C2(40,533.333)
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Appendix 3Appendix 3 : Calculation of the opportunity cost
1. Opportunity cost for P2, P4, and P6: intersection C1Line : P1(40,600) and P3(50,400)
Intersection : C1(46.7,466.7)
Based on formula (2) we can calculate that,
50 46,7
50 40
3,3
10 0,33
Because WL= 0,33, we can conclude that opportunity cost for P2, P4, and P6by P3 as
reference is 1 - 0,3333 = 0,66667. So that, we can calculate the opportunity cost for P2,
P4, and P6by P1 as reference is 1 0,66667 = 0,33333.
2. Opportunity cost for P8: intersection C2Line : P5(40,400) and P7(40,600)
Intersection : C2(40,533.333)
Based on formula (2) we can calculate that,
600 533,3333
600 400
66,66667
200 0,333333
Because WL= 0,33333, we can conclude that opportunity cost for P8by P5 as reference
is 0,33333. So that, we can calculate the opportunity cost for P8by P7 as reference is 1
0,33333 = 0,66667.