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    Final Exam Service EngineeringPostgarduate Program of Industrial Engineering Universitas Indonesia

    1

    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.

    ----------------------------------------------------------------------------------------------------------------

    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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    Final Exam Service EngineeringPostgarduate Program of Industrial Engineering Universitas Indonesia

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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.