Formulating And Optimizing Multi-Objective Linear Programming Model For Garbage Recycling
Problem Description
As per given condition in problem, we have to solve the situation using multi objective linear programming (MOLP), In order to solve the problem using MOLP first we will solve the individual objective using solver, from this two-unit objective in which one increasing garbage capacity and another one is minimising transportation cost. The calculated result set as target for MOLP and we should provide weighted condition with few more constraints, so that we can achieve minimax condition of desired result.
The capacity of plant is decided according to their efficiency recorded for that plant. If plant is of 100 ton rated capacity and its efficiency is 60% then it should produce only 100*0.6 = 60 ton only. Similarly, the efficiency and rated capacity for garbage processing is given in first table we must derive actual capacity as per given production capacity and efficiency data.
|
Site location |
||||||
|
1 |
2 |
3 |
4 |
5 |
||
|
Capacity (Megaton) |
10 |
7 |
15 |
12 |
6 |
|
|
Efficiency |
0.35 |
0.45 |
0.25 |
0.75 |
0.55 |
|
|
Actual capacity |
3.5 |
3.15 |
3.75 |
9 |
3.3 |
Problem formulation
Suppose the section is denoted as I and site location is j, then Aij is the amount of garbage which is recycled at the site location j from the sector i.
The value of section I is in the range of
The value of site j is in the range of ,
In such scenario the distribution matrix can be given as
1 2 3 4 5
1 A11 A12 A13 A14 A15
2 A21 A22 A23 A24 A25
3 A31 A32 A33 A34 A35
4 A41 A42 A43 A44 A45
5 A51 A52 A53 A54 A55
6 A61 A62 A63 A64 A65
7 A71 A72 A73 A74 A75
8 A81 A82 A83 A84 A85
9 A91 A92 A93 A94 A95
10 A101 A102 A103 A104 A105
The given condition of maximum capacity of different site will be taken as constraint for the given distribution. The given constraint for location is as follows.
…………(i)
…………(ii)
…………(iii)
…………(iv)
…………(v)
…………(vi)
…………(vii)
…………(viii)
…………(ix)
……. (x)
The other five constraints will for site location, which is as follows as per their efficiency and capacity
………(xi)
………(xii)
………(xiii)
……….…(xiv)
……….…(xv)
The objective function for first optimisation i.e. maximising the garbage tonnage will be calculated as follows.
……. (xvi)
We must solve the given problem is excel solver, in We must arrange the following data as per given below
|
Site location |
||||||
|
1 |
2 |
3 |
4 |
5 |
||
|
garbage (Megaton) |
10 |
7 |
15 |
12 |
6 |
|
|
Efficiency |
0.35 |
0.45 |
0.25 |
0.75 |
0.55 |
|
|
Actual capacity |
3.5 |
3.15 |
3.75 |
9 |
3.3 |
The actual capacity is as calculated above for all 5 location. As per given table the recycling of garbage with their estimate is given below.
|
Different |
Recycling Plan |
|||||
|
1 |
24 |
10 |
34 |
52 |
65 |
4.6 |
|
2 |
17 |
15 |
58 |
64 |
62 |
4.6 |
|
3 |
10 |
20 |
26 |
66 |
60 |
4.7 |
|
4 |
18 |
25 |
32 |
57 |
62 |
4.2 |
|
5 |
11 |
22 |
15 |
55 |
62 |
3.8 |
|
6 |
29 |
34 |
46 |
54 |
43 |
3.9 |
|
7 |
34 |
43 |
69 |
43 |
40 |
3.4 |
|
8 |
38 |
42 |
36 |
53 |
34 |
3.3 |
|
9 |
22 |
29 |
46 |
53 |
50 |
3.9 |
|
10 |
22 |
46 |
50 |
42 |
58 |
4.1 |
In this condition, the objective function is being set at maximising the garbage tonnage and putt all the data in excel solver, which is shown below. There are 15 constraints for the given subject is set as per the following given details. The solving method is chosen as simplex LP, because it this problem show linear relationship. Click on the set objective to maximise the data and made all constraint non-negative.
We have ensured that all the cell value is given properly as per plan and run the solver. After running the solver, we have selected answer of the solution, which is given in excel sheet.
Formulating the MOLP Model
After running the solver, the maximum value of garbage recycling occurred is 22.7 megaton. The result obtained from running the solver illustrates that, the sector number. The capacity is reached at its optimum level without taking the section 7, 8 ,9, and 10. This show that, the capacity of recycling site is very low as compared to sum of garbage of 10 sector. We have to either increase the efficiency of given recycling site or increase the recycling plant for given capacity of different sector.
We can see that, the second objective function is related with cost of transportation of garbage from different location, the given garbage cost is $ 109603 megaton per kilometre, which is to be multiplied with given distance and tonnage. The second objective function w will be set as follows
……(xvii)
After running the solver, the result is as above. We must set again the solver parameter with some minor changes which is given below
|
SUM |
Recycling Plan |
Deviation |
|||||||
|
Sector |
1 |
0.690833 |
0.308333 |
0.375833 |
0.900833 |
0.330833 |
2.6066667 |
4.6 |
0.433333 |
|
2 |
0.008333 |
0.308333 |
0.375833 |
0.900833 |
0.330833 |
1.9241667 |
4.6 |
0.581703 |
|
|
3 |
2.7425 |
0.375 |
0.3675 |
0.8925 |
0.3225 |
4.7 |
4.7 |
0 |
|
|
4 |
0.008333 |
0.308333 |
0.375833 |
0.900833 |
0.330833 |
1.9241667 |
4.2 |
0.541865 |
|
|
5 |
0.008333 |
0.308333 |
0.375833 |
0.900833 |
0.330833 |
1.9241667 |
3.8 |
0.49364 |
|
|
6 |
0.008333 |
0.308333 |
0.375833 |
0.900833 |
0.330833 |
1.9241667 |
3.9 |
0.506624 |
|
|
7 |
0.008333 |
0.308333 |
0.375833 |
0.900833 |
0.330833 |
1.9241667 |
3.4 |
0.434069 |
|
|
8 |
0.008333 |
0.308333 |
0.375833 |
0.900833 |
0.330833 |
1.9241667 |
3.3 |
0.416919 |
|
|
9 |
0.008333 |
0.308333 |
0.375833 |
0.900833 |
0.330833 |
1.9241667 |
3.9 |
0.506624 |
|
|
10 |
0.008333 |
0.308333 |
0.375833 |
0.900833 |
0.330833 |
1.9241667 |
4.1 |
0.530691 |
|
|
SUM |
3.5 |
3.15 |
3.75 |
9 |
3.3 |
22.7 |
40.5 |
0.439506 |
|
|
Objective function 2 |
104294470 |
Minimize |
The manipulation in sheet done by solve is given above. The minimised cost of running and transporting the garbage to the plant is calculated as $ 104294470. The overall average deviation from estimated plan is around 44%, same data for previous case was around 47%.
Now based on the following data we must mover forward foe MOLP problem which is as follows
MOLP
For setting the problem in MOLP we must fine the deviation from actual target. The deviation is calculated as given below
The % deviation will be calculated as and The weightage for MOLP problem is amount of recycled garbage to be three times as important as minimising the transportation cost. In this condition, we must add three new constraints in addition to above 15 constraints.
The new constraints will The new table for deviation calculated is set, which is given below
|
Site location |
|||||||||||
|
1 |
2 |
3 |
4 |
5 |
Required |
Target |
% Dev |
Weight |
% wt Dev |
||
|
Goal 1: Tonnage |
3.5 |
3.15 |
3.75 |
9 |
3.3 |
22.7 |
22.7 |
0.00% |
3 |
0.00% |
|
|
Goal 2: Cost |
24660675 |
31346458 |
45156436 |
59076017 |
58747208 |
104294470 |
104294470 |
0.00% |
1 |
0.00% |
|
|
0 |
|||||||||||
|
Minmax |
0.00% |
After running the solver as per given screen
After running the solver, we got the following changes in table and given deviation parameter.
|
SUM |
Recycling Plan |
Daviation |
|||||||
|
Sector |
1 |
1.431418 |
1.323139 |
0.454812 |
0.299317 |
1.091314 |
4.6 |
4.6 |
1.93E-16 |
|
2 |
0.448159 |
0.20979 |
0.251432 |
0.89666 |
0.294085 |
2.10012444 |
4.6 |
0.543451 |
|
|
3 |
0.265121 |
0.942066 |
0.510643 |
0.887122 |
0.50125 |
3.10620062 |
4.7 |
0.339106 |
|
|
4 |
0.410376 |
0.152992 |
0.445298 |
1.111549 |
0.391721 |
2.51193553 |
4.2 |
0.40192 |
|
|
5 |
0.36094 |
0.175161 |
0.711868 |
1.036708 |
0.391721 |
2.67639808 |
3.8 |
0.295685 |
|
|
6 |
0.209151 |
0.077886 |
0.01273 |
1.223593 |
0.006104 |
1.52946406 |
3.9 |
0.60783 |
|
|
7 |
0.060629 |
0.00772 |
0.657372 |
0.706848 |
0.032456 |
1.46502475 |
3.4 |
0.56911 |
|
|
8 |
0 |
0 |
0.527407 |
0.898235 |
0.302792 |
1.72843391 |
3.3 |
0.476232 |
|
|
9 |
0.156556 |
0.226407 |
0.01273 |
1.015131 |
0.070101 |
1.48092556 |
3.9 |
0.620275 |
|
|
10 |
0.157652 |
0.03484 |
0.16571 |
0.924837 |
0.218454 |
1.50149305 |
4.1 |
0.633782 |
|
|
SUM |
3.5 |
3.15 |
3.75 |
9 |
3.3 |
22.7 |
40.5 |
0.439506 |
|
Site location |
|||||||||||
|
1 |
2 |
3 |
4 |
5 |
Required |
Target |
% Dev |
Weight |
% wt Dev |
||
|
Goal 1: Tonnage |
3.5 |
3.15 |
3.75 |
9 |
3.3 |
22.7 |
22.7 |
0.00% |
3 |
0.00% |
|
|
Goal 2: Cost |
24660675 |
31346458 |
45156436 |
59076017 |
58747208 |
104294470 |
104294470 |
0.00% |
1 |
0.00% |
|
|
0 |
|||||||||||
|
Minmax |
0.00% |
The calculated parameter by MOLP solver suggests that, the optimisation done in previous condition individually has the most optimum solution for given condition and the is not change in parameter after giving the weightage and deviation condition. So, the suggestion given for first objective function is still viable for this condition also. Similarly, the cost objective is also remain unchanged due to plant capacity problem.
As per given question, we must minimise the distance between the different location by provide as warehouse as such place that, the total distance will be minimum for planned warehouse. First, we must set the data in excel sheet in such as way that, we can calculate the distance between given coordinate with proposed warehouse, with the formula given below..
The sum of the distance will be given below, and we will apply it as a objective function for rung GRG non-linear solver. Excel sheet will be set as given below
|
Location of Warehouses |
||||||||
|
Suburb |
X |
Y |
Warehous-1 |
Warehous-2 |
Warehous-3 |
Distance |
||
|
Ascot Vale |
25 |
13.8 |
1 |
0 |
0 |
2.51 |
1 |
|
|
Avondale |
19.7 |
14.2 |
1 |
0 |
0 |
5.36 |
1 |
|
|
Brooklyn |
18.2 |
9.4 |
0 |
1 |
0 |
5.00 |
1 |
|
|
Burnside |
10.7 |
16.2 |
0 |
1 |
0 |
5.98 |
1 |
|
|
Caroline |
9.7 |
16.8 |
0 |
1 |
0 |
7.00 |
1 |
|
|
Derrimut |
10.7 |
10.2 |
0 |
1 |
0 |
2.80 |
1 |
|
|
Flemington |
24.3 |
11.8 |
1 |
0 |
0 |
0.41 |
1 |
|
|
Footscray |
22.4 |
11 |
1 |
0 |
0 |
1.91 |
1 |
|
|
Footscray |
23.7 |
11.1 |
1 |
0 |
0 |
0.64 |
1 |
|
|
Hopper |
6.3 |
4.7 |
0 |
0 |
1 |
0.94 |
1 |
|
|
Laverton north |
13.5 |
7.2 |
0 |
1 |
0 |
3.67 |
1 |
|
|
Melbourne |
28.6 |
8.9 |
1 |
0 |
0 |
5.00 |
1 |
|
|
Seabrook |
10.9 |
2.3 |
0 |
0 |
1 |
6.12 |
1 |
|
|
Southbank |
29.8 |
8.1 |
1 |
0 |
0 |
6.44 |
1 |
|
|
ST kilda |
30.4 |
3.4 |
1 |
0 |
0 |
10.08 |
1 |
|
|
Sunshine |
16.6 |
10.2 |
0 |
1 |
0 |
3.25 |
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 |
We must set objective for minimum with constraint as shown in the screen shot
The screen shot of solver look like this,
Implementing the MOLP Model in Excel
After running the solver, we get the following result as shown below.
|
Warehous-1 |
24.26 |
11.40 |
Distance |
85.0386528 |
||
|
Warehous-2 |
13.42 |
10.87 |
||||
|
Warehous-3 |
5.42 |
5.01 |
We can see that the location of warehouse is given as coordinate and total distance which can be covered is 85.039 km.
The result obtained in warehouse 1, 2, and 3 in excel column indicate that, the suburb is supplied with that ware house. All the data are calculated on minimum distance covered.
As per given problem, we must apply the analytical hierarchy process and help David decide which Van is suitable to buy. The given criteria tabulated below for further processing of data to make decision through analytical hierarchy process.
|
Price |
Safety |
Economy |
Comfort |
G.M. |
|
|
Price |
1 |
0.166667 |
0.333333 |
0.2 |
0.324667915 |
|
Safety |
6 |
1 |
4 |
2 |
2.632148026 |
|
Economy |
3 |
0.25 |
1 |
0.333333 |
0.707106781 |
|
Comfort |
5 |
0.5 |
3 |
1 |
1.65487546 |
The first step is to normalise the value dividing by sum of each column to each entity, the result is as follows
|
Normalised Value |
|||||
|
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 |
In the similar way, we must normalise all the criteria such as Price, Safety, economy, comfort which is as follows
|
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 |
Now we have again summarised all the normalised data for further calculation which as follows
|
Price |
Safety |
Economy |
Comfort |
Overall |
|
|
X |
0.2648 |
0.2109 |
0.1133 |
0.0884 |
0.0626 |
|
Y |
0.6555 |
0.6932 |
0.3467 |
0.3561 |
0.4919 |
|
Z |
0.0796 |
0.0959 |
0.5400 |
0.5555 |
0.1362 |
|
1.0000 |
1.0000 |
1.0000 |
1.0000 |
0.3093 |
Now have calculated the final score with use of MMULT in excel sheet, which provide the matrix product of two array, in this condition this provide final score for each parameter, which is numerically equalised and ranked thereafter.
|
Final Score |
Rank |
|
|
X |
0.1631 |
3 |
|
Y |
0.5394 |
1 |
|
X |
0.2975 |
2 |
After using the rank function it was found that, the highest score occurred for ‘Y’ Van, i.e. this Van is suitable for David jones.
In this problem we have to decide best location of out of the three warehouses on the basis of TOPSIS decision making tool. For this problem, fist we have to revese the row which is given below
|
Criteria |
Weight |
Smithfield |
Eagle Farm |
Derrimut |
Objective |
|
Land Cost |
4 |
5 |
7 |
6 |
Min |
|
Labour Cost |
6 |
7 |
6 |
6 |
Min |
|
Labour availability |
8 |
6 |
7 |
5 |
Max |
|
Construction cost |
5 |
5 |
6 |
7 |
Min |
|
Transportation |
5 |
6 |
6 |
7 |
Max |
|
Access to customers |
9 |
8 |
7 |
7 |
Max |
|
Long Range goals |
7 |
5 |
7 |
6 |
Max |
After reversing the row and column we the table is look like this
|
Criteria |
Land Cost |
Labour Cost |
Labour availability |
Construction cost |
Transportation |
Access to customers |
Long Range goals |
|
Weight |
4 |
6 |
8 |
5 |
5 |
9 |
7 |
|
Smithfield |
5 |
7 |
6 |
5 |
6 |
8 |
5 |
|
Eagle Farm |
7 |
6 |
7 |
6 |
6 |
7 |
7 |
|
Derrimut |
6 |
6 |
5 |
7 |
7 |
7 |
6 |
|
Objective |
Min |
Min |
Max |
Min |
Max |
Max |
Max |
The ideal attribute is maximised by choosing maximum for ideal and minimum value for worst
|
Ideal |
5 |
6 |
7 |
5 |
7 |
8 |
7 |
|
Worst |
7 |
7 |
5 |
7 |
6 |
7 |
5 |
Now the minimisation criteria has to be converted by subtracted the respective attributes
|
Minimising criteria must be converted |
|||||||
|
Criteria |
Land Cost |
Labour Cost |
Labour availability |
Construction cost |
Transportation |
Access to customers |
Long Range goals |
|
Smithfield |
2 |
0 |
6 |
2 |
6 |
8 |
5 |
|
Eagle Farm |
0 |
1 |
7 |
1 |
6 |
7 |
7 |
|
Derrimut, |
1 |
1 |
5 |
0 |
7 |
7 |
6 |
Further we must normalise the above by dividing the square root of sum of square of all the cells as given below.
|
Normalising |
2.236068 |
1.414213562 |
10.48808848 |
2.236067977 |
11 |
12.72792206 |
10.48808848 |
After normalization the data looks like as follows.
|
Normed matrix |
|||||||
|
Criteria |
Land Cost |
Labour Cost |
Labour availability |
Construction cost |
Transportation |
Access to customers |
Long Range goals |
|
Smithfield |
0.894427 |
0 |
0.572077554 |
0.894427191 |
0.545454545 |
0.628539361 |
0.476731295 |
|
Eagle Farm |
0 |
0.707106781 |
0.667423812 |
0.447213595 |
0.545454545 |
0.549971941 |
0.667423812 |
|
Derrimu |
0.447214 |
0.707106781 |
0.476731295 |
0 |
0.636363636 |
0.549971941 |
0.572077554 |
To find the weighted normalised data we must multiply the above result with weight given in the question. Further outcome is as follows.
|
Weighted Normed matrix |
|||||||
|
Criteria |
Land Cost |
Labour Cost |
Labour availability |
Construction cost |
Transportation |
Access to customers |
Long Range goals |
|
Smithfield, (NSW) |
3.577709 |
0 |
4.576620428 |
4.472135955 |
2.727272727 |
5.656854249 |
3.337119062 |
|
Eagle Farm, QLD |
0 |
4.242640687 |
5.3393905 |
2.236067977 |
2.727272727 |
4.949747468 |
4.671966687 |
|
Derrimut, VIC |
1.788854 |
4.242640687 |
3.813850357 |
0 |
3.181818182 |
4.949747468 |
4.004542875 |
|
Ideal |
3.577709 |
4.242640687 |
5.3393905 |
4.472135955 |
3.181818182 |
5.656854249 |
4.671966687 |
|
Worst |
0 |
0 |
3.813850357 |
0 |
2.727272727 |
4.949747468 |
3.337119062 |
For Ideal condition we must reduce the weighted normalised data with data obtained for ideal condition. Similar for worst condition also
|
From Ideal Condition |
|||||||
|
Criteria |
Land Cost |
Labour Cost |
Labour availability |
Construction cost |
Transportation |
Access to customers |
Long Range goals |
|
Smithfield, (NSW) |
0 |
4.242640687 |
0.762770071 |
0 |
0.454545455 |
0 |
1.334847625 |
|
Eagle Farm, QLD |
3.577709 |
0 |
0 |
2.236067977 |
0.454545455 |
0.707106781 |
0 |
|
Derrimut, VIC |
1.788854 |
0 |
1.525540143 |
4.472135955 |
0 |
0.707106781 |
0.667423812 |
|
From worst condition |
|||||||
|
Criteria |
Land Cost |
Labour Cost |
Labour availability |
Construction cost |
Transportation |
Access to customers |
Long Range goals |
|
Smithfield, (NSW) |
3.577709 |
0 |
0.762770071 |
4.472135955 |
0 |
0.707106781 |
0 |
|
Eagle Farm, QLD |
0 |
4.242640687 |
1.525540143 |
2.236067977 |
0 |
0 |
1.334847625 |
|
Derrimut, VIC |
1.788854 |
4.242640687 |
0 |
0 |
0.454545455 |
0 |
0.667423812 |
Now by using square root of sum of all squares for each location and ranked them by using rank formula in excel. The result is as follows.
|
Ideal |
||
|
Criteria |
Ideal |
Rank |
|
Smithfield, |
4.535444 |
2 |
|
Eagle Farm, |
4.301931 |
3 |
|
Derrimut, |
5.145165 |
1 |
|
Non-Ideal |
||
|
Criteria |
Non-Ideal |
Rank |
|
Smithfield, |
5.820809 |
3 |
|
Eagle Farm, |
5.206639 |
2 |
|
Derrimut |
4.674619 |
1 |
From Ideal ranking it was found that, the highest point got by Derrimut,VIC, Similarly from non-ideal point of view the lowest score is for Derrimute, VIC, it suggests that this location is best for providing all the criteria given in question.
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