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OMGT2087 Supply Chain Modelling And Design
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OMGT2087 Supply Chain Modelling And Design
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Course Code: OMGT2087
University: Royal Melbourne Institute Of Technology
MyAssignmentHelp.com is not sponsored or endorsed by this college or university
Country: Australia
Question:
Recycling is an important and complex activity in Country A. To enable timely operations, the country is divided into 10 sectors and recycling operations are commenced simultaneously in each sector. The recyclable garbage is collected from public bins, loaded into trucks, and transported to recycling sites. Each site can accommodate different amounts of recyclable garbage because of its available land size at the facility. The annual capacities for five recycling sites are given in the table below (in megatonnes):
Each recycling site is installed with facilities that have different recycling efficiencies which are summarised in the table below (in percentages):
The cost of collecting and transporting recyclable garbage primarily depends on the distance between the sectors and the recycling sites. The following table summarises the distances between each sector and each recycling site (in kilometres):
Formulate a multiple-objective linear programming (MOLP) model for this problem in a Word file with a brief description of an equation, and implement the MOLP model in an Excel spreadsheet.
Determine the optimal value for each objective in the problem.
Suppose the management considers maximising the amount of recycled garbage to be three times as important as minimising the transportation cost. Formulate a GP model to optimise both objectives simultaneously with a brief description of anequation in a Word file, and implement the MOLP model in an Excel spreadsheet. What do the results suggest?
Answer:
The formulation of multi objective linear programming start with solving the linear programming from each variable and combine to make a common objective with and solve this including he two optimised result obtained from individual objective solving,
The given problem will also work like this, for which first I must understand and identify he given data in the question,
Recycling Site
1
2
3
4
5
Capacity (Megaton)
10
8
15
11
10
Efficiency
0.35
0.42
0.39
0.75
0.6
Actual capacity
3.5
3.36
5.85
8.25
6
The recycling site from five individual location is given with capacity, which is given above, but as per the efficiency the actual capacity of all five site is calculated as given above. For calculation purpose we will consider the actual capacity not the rated capacity, because its efficiency is given. Now the problem can be formulated as below. The site of the location with section wise is given as follows.
1
2
3
4
5
1
S11
S12
S13
S14
S15
2
S21
S22
S23
S24
S25
3
S31
S32
S33
S34
S35
4
S41
S42
S43
S44
S45
5
S51
S52
S53
S54
S55
6
S61
S62
S63
S64
S65
7
S71
S72
S73
S74
S75
8
S81
S82
S83
S84
S85
9
S91
S92
S93
S94
S95
10
S101
S102
S103
S104
S105
Now for solution we must adjust all the data in above given section and site.
The available constraints for given condition can be formulated as follows
S11+ S21+S31+S41+S51+S61+S71+S81+S91+S101 3.50…… (11)
S12+ S22+S32+S42+S52+S62+S72+S82+S92+S102 3.36…… (12)
S13+ S23+S33+S43+S53+S63+S73+S83+S92+S103 5.85 …… (13)
S14+ S24+S34+S44+S54+S64+S74+S84+S94+S104 8.25 . …. (14)
S15+ S25+S35+S45+S55+S65+S75+S85+S95+S105 6.0 .…… (15)
After complete ting the decision variable we must define the objective function for the given problem, which will be as follows
This is the formulation of the linear programming for first problem. We have to arrange all the data in excel solver which is given as a screen short.
We have inserted all the necessary data in solver as given above and execute it.
The result came from the execution is given below.
SUM
Recycling Plan
Sector
1
0
0
0
0
5
5
5
2
0
0
0
5
1
6
6
3
0
0
4.75
3.25
0
8
8
4
3.5
3.36
1.1
0
0
7.96
9
5
0
0
0
0
0
0
3
6
0
0
0
0
0
0
4
7
0
0
0
0
0
0
7
8
0
0
0
0
0
0
3
9
0
0
0
0
0
0
4
10
0
0
0
0
0
0
2
SUM
3.5
3.36
5.85
8.25
6
26.96
51
Objective function 1
26.96
Maximise
The results show that, the maximum garbage that can be handled is 26.96 megaton, which is equal to the efficient capacity of all plant. The distribution of garbage consumption in plant is given above in yellow. This shows that maximum garbage that can be handled from sector 3 of site 3 and 4. This reached to maximum estimated target.
Solution1(b)
The problem statement for second part means that we must reduce the cost of transportation cost from the given condition. The required objective for this solution will be minimising, the SUMPRODCT of objecting function, which is given below
The arrangement of solver is given in print screen, which is given below.
The result came from running the solver shows that the minimum cost which occurred in minimising the transportation cost is around $ 146964507. The distribution of consumption of garbage, shown in excel sheet. The overall consumption is 26.96 megaton. Based on the above two solution we formulate the MOLP problem, which will provide the minmax value of cost as well as maximum tonnage.
Solution1(c)
The multi objective linear programming, dictates the overall optimisation for any problem. At the given scenario, we have to bring both the solution in same platform, and % deviation is the tool by which we can do this. Therefore, we must calculate the % deviation and multiply it with given weight and feed into the solver, which is as follows.
The deviation in % will be calculated as follows.as
As given in question, the weightage is triple of cost to the tonnage.
The new constraints will
After feeding the value in the solver, we got the following result.
The minimax value comes around 0, which means that there is no change in MOLP solution, as it is optimised in the previous state.
Solution 2
The basic objective in this problem is to find the location of warehouses in such way that the distance will be minimum,
We know that minimum distance can be calculated from distance formula
where x and y are the coordinates given in the problem.
Suburb
X
Y
Warehous-1
Warehous-2
Warehous-3
Distance
Ascot Vale
25
13.8
1
0
0
3.00
1
Avondale Heights
19.7
14.2
1
0
0
5.06
1
Brooklyn
18.2
9.4
1
0
0
5.76
1
Burnside
10.7
16.2
0
1
0
4.93
1
Caroline Springs
9.7
16.8
0
1
0
5.82
1
Derrimut
10.7
10.2
0
1
0
1.71
1
Flemington
24.3
11.8
1
0
0
0.92
1
Footscray
22.4
11
1
0
0
1.30
1
Footscray
23.7
11.1
1
0
0
0.00
1
Hopper Crossing
6.3
4.7
0
0
1
0.94
1
Laverton north
13.5
7.2
0
1
0
4.51
1
Melbourne
28.6
8.9
1
0
0
5.37
1
Seabrook
10.9
2.3
0
0
1
6.12
1
Southbank
29.8
8.1
1
0
0
6.80
1
ST kilda
30.4
3.4
1
0
0
10.21
1
Sunshine west
16.6
10.2
0
1
0
4.86
1
Tarneit
5.2
8.1
0
0
1
3.09
1
tarneit
5.1
6.6
0
0
1
1.62
1
Werribee
0.5
0
0
0
1
7.02
1
Wyndham Vale
0
2
0
0
1
6.20
1
The other decision variable that can be considered for equation is the calculation of distances for ware houses, which is given below
Warehous-1
23.70
11.10
Warehous-2
11.90
11.41
Warehous-3
5.42
5.01
The result above calculated indicates that the suburb will be supplied by following warehouses. In this condition all the minimum distance will be covered. After feeding the data in solver and running it, it was found that, total distance calculated is around 85.2243
Solution 3
As given, the use of Analytical hierarchy process is done where lot of criteria is given, and we must decide which one is best. In the given problem, Mr Jones is surrounded with Four different criteria and we must solve the problem by AHP process,
The tabulated criteria is as follows
We have additionally calculated, SUMPV, CI, CR and λ for the given value.
The AHP process starts with normalising the table by dividing the value with their sum, after normalising the value will be as given below.
Price
Safety
Economy
Comfort
Average
Price
0.0667
0.0870
0.0400
0.0566
0.0626
Safety
0.4000
0.5217
0.4800
0.5660
0.4919
Economy
0.2000
0.1304
0.1200
0.0943
0.1362
Comfort
0.3333
0.2609
0.3600
0.2830
0.3093
Sum
1.0000
From the similar fashion normalising all the criteria given in the question
PRICE
X
Y
Z
Average
X
0.2353
0.2258
0.3333
0.2648
Y
0.7059
0.6774
0.5833
0.6555
Z
0.0588
0.0968
0.0833
0.0796
SUM
1.0000
1.0000
1.0000
1.0000
Safety
X
Y
Z
Average
X
0.2222
0.2286
0.1818
0.2109
Y
0.6667
0.6857
0.7273
0.6932
Z
0.1111
0.0857
0.0909
0.0959
Sum
1.0000
Economy
X
Y
Z
Average
X
0.1000
0.0400
0.2000
0.1133
Y
0.6000
0.2400
0.2000
0.3467
Z
0.3000
0.7200
0.6000
0.5400
Sum
1.0000
Comfort
X
Y
Z
Average
X
0.0769
0.0303
0.1579
0.0884
Y
0.6154
0.2424
0.2105
0.3561
Z
0.3077
0.7273
0.6316
0.5555
Sum
1.0000
1.0000
1.0000
1.0000
From the above four calculated table, we can summarise the data which is as follows
The AHP process in excel s quite easier, we will use the function which is known as MMULT with two columns, ranking each of the criteria
Final Score
Rank
X
0.1631
3
Y
0.5394
1
X
0.2975
2
The above table describes that, the maximum score is achieved by Y, it means that, Mr jones will go for Y van which, is superior in all the criteria.
Solution 4
As given in the question, we must decide that which utility car is best as per given criteria.
The overall criteria matrix is given as follows
Criteria
Weight
Nissan Navara
Mazda BT-50
Mitsubishi Triton
Toyota Hilux
Objective
Price
8
32990
24950
14999
11450
Min
Odometer(km)
5
26555
83427
316000
246000
Min
Automatic transmission
4
1
0
1
0
Max
Fuel Consumption (1/100 km)
3
7
8.4
9.9
10.9
Min
Fuel Average distance (km)
7
1143
952
758
606
Max
CO2 Emission
2
186
222
261
290
Min
Payload
7
982
1371
927
1017
Max
Build Year
5
2016
2012
2007
2004
Max
Fuel Cost per fill
6
127
119
119
105
Min
We must transpose the table to calculate further
Criteria
Price
Odometer(km)
Automatic transmission
Fuel Consumption (1/100 km)
Fuel Average distance (km)
CO2 Emission
Payload
Build Year
Fuel Cost per fill
Weight
8
5
4
3
7
2
7
5
6
Nissan Navara
32990
26555
1
7
1143
186
982
2016
127
Mazda BT-50
24950
83427
0
8.4
952
222
1371
2012
119
Mitsubishi Triton
14999
316000
1
9.9
758
261
927
2007
119
Toyota Hilux
11450
246000
0
10.9
606
290
1017
2004
105
Objective
Min
Min
Max
Min
Max
Min
Max
Max
Min
The choice of ideal and worst attributed is done through mx and if function in excel
Ideal
11450
26555
1
7
1143
186
1371
2016
105
Worst
32990
316000
0
10.9
606
290
927
2004
127
Further the steps take is as follows as well as in excel sheet
Weighted Normed matrix
Criteria
Price
Odometer(km)
Automatic transmission
Fuel Consumption (1/100 km)
Fuel Average distance (km)
CO2 Emission
Payload
Build Year
Fuel Cost per fill
Nissan Navara
0
2.88472E-05
0.707106781
0.117560572
0.000563029
0.004382076
0.000459432
0.00024879
0
Mazda BT-50
1.1038E-05
7.37796E-06
0
0.062799451
0.000563029
0.002400576
0.000459432
0.00024879
0.002717486
Mitsubishi Triton
4.10865E-05
0
0.707106781
0.021313753
0.000563029
0.000870797
0.000459432
0.00024879
0.002717486
Toyota Hilux
6.44387E-05
7.53089E-07
0
0
0.000563029
0
0.000459432
0.00024879
0.008469498
Ideal
6.44387E-05
2.88472E-05
0.707106781
0.117560572
0.000563029
0.004382076
0.000459432
0.00024879
0.008469498
Worst
0
0
0
0
0.000563029
0
0.000459432
0.00024879
0
For Ideal Condition
Criteria
Price
Odometer(km)
Automatic transmission
Fuel Consumption (1/100 km)
Fuel Average distance (km)
CO2 Emission
Payload
Build Year
Fuel Cost per fill
Nissan Navara
6.44387E-05
0
0
0
0
0
0
0
0.008469498
Mazda BT-50
5.34006E-05
2.14692E-05
0.707106781
0.054761121
0
0.0019815
0
0
0.005752012
Mitsubishi Triton
2.33522E-05
2.88472E-05
0
0.096246819
0
0.003511278
0
0
0.005752012
Toyota Hilux
0
2.80941E-05
0.707106781
0.117560572
0
0.004382076
0
0
0
For Worst Condition
Criteria
Price
Odometer(km)
Automatic transmission
Fuel Consumption (1/100 km)
Fuel Average distance (km)
CO2 Emission
Payload
Build Year
Fuel Cost per fill
Nissan Navara
0
2.88472E-05
0.707106781
0.117560572
0
0.004382076
0
0
0
Mazda BT-50
1.1038E-05
7.37796E-06
0
0.062799451
0
0.002400576
0
0
0.002717486
Mitsubishi Triton
4.10865E-05
0
0.707106781
0.021313753
0
0.000870797
0
0
0.002717486
Toyota Hilux
6.44387E-05
7.53089E-07
0
0
0
0
0
0
0.008469498
For Ideal Condition
Criteria
Value
RANK
Nissan Navara
0.008469743
4
Mazda BT-50
0.709250164
2
Mitsubishi Triton
0.096482466
3
Toyota Hilux
0.716826124
1
For Worst Condition
Criteria
Value
RANK
Nissan Navara
0.716826124
1
Mazda BT-50
0.062904044
3
Mitsubishi Triton
0.707433686
2
Toyota Hilux
0.008469743
4
From Ideal and worst condition ranking for Ideal and worst condition is maximum for Toyota Hilux. Therefore, the trader must select Toyota Hilux.
References
Bolumole, K, L, 2009, The Customer Service Management Process, The International Journal of Logistics Management, 14(2), pp, 1-33.
David, Z,, Gilbert , N, N, & Gary, Y, 2016, Supply chain risk management and hospital inventory: Effects of system affiliation, Journal of Operations Management, 44(1), pp, 30-47.
Gomes, R, 2011, Physical Distribution Service, Jams, 1(1), pp, 1-10.
Gulc, A,, 2017, Models and Methods of Measuring the Quality of Logistic Service, Project, and Production Management, 255(264), pp, 1-10.
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