Objective and Constraints
Formulate an 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.
Objective
10X1 + 7X2 + 15X3 + 12X4 + 6X5
Constraints
24*$109,603X1 + 10*$109,603X2 + 34*$109,603X3 +52*$109,603 X4 + 65*$109,603X5 4.6
17*$109,603X1 +15*$109,603X2 + 58*$109,603X3 + 64*$109,603X4 + 62*$109,603X5 4.6
10*$109,603X1 + 20*$109,603X2 + 26*$109,603X3 + 66*$109,603X4 + 60*$109,603X5 4.7
18*$109,603X1 + 25*$109,603X2 + 32*$109,603X3 + 57*$109,603X4 + 62*$109,603X5 4.2
11*$109,603X1 + 22*$109,603X2 + 15*$109,603X3 + 55*$109,603X4 + 62*$109,603X5 3.8
29*$109,603X1 + 34*$109,603X2 + 46*$109,603X3 + 54*$109,603X4 + 43*$109,603X5 3.9
34*$109,603X1 + 43*$109,603X2 + 69*$109,603X3 + 43*$109,603X4 + 40*$109,603X5 3.4
38*$109,603X1 + 42*$109,603X2 + 36*$109,603X3 + 53*$109,603X4 + 34*$109,603X5 3.3
22*$109,603X1 + 29*$109,603X2 + 46*$109,603X3 + 53*$109,603X4 + 50*$109,603X5 3.9
22*$109,603X1 + 46*$109,603X2 + 50*$109,603X3 + 42*$109,603X4 + 58*$109,603X5 4.1
- Determine the optimal value for each objective in the problem.
X1 = 1.28 * 10-7
X2 = 8.93 * 10-8
X3 = 1.91 * 10-7
X4 = 1.53 * 10-7
X5 = 7.65 * 10-8
- 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 an equation in a Word file, and implement the MOLP model in an Excel spreadsheet. What do the results suggest?
Objective
10X1 + 7X2 + 15X3 + 12X4 + 6X5
Constraints
24*$109,603X1 + 10*$109,603X2 + 34*$109,603X3 +52*$109,603 X4 + 65*$109,603X5 13.8
17*$109,603X1 +15*$109,603X2 + 58*$109,603X3 + 64*$109,603X4 + 62*$109,603X5 13.8
10*$109,603X1 + 20*$109,603X2 + 26*$109,603X3 + 66*$109,603X4 + 60*$109,603X5 14.1
18*$109,603X1 + 25*$109,603X2 + 32*$109,603X3 + 57*$109,603X4 + 62*$109,603X5 12.6
11*$109,603X1 + 22*$109,603X2 + 15*$109,603X3 + 55*$109,603X4 + 62*$109,603X5 11.4
29*$109,603X1 + 34*$109,603X2 + 46*$109,603X3 + 54*$109,603X4 + 43*$109,603X5 11.7
34*$109,603X1 + 43*$109,603X2 + 69*$109,603X3 + 43*$109,603X4 + 40*$109,603X5 10.2
38*$109,603X1 + 42*$109,603X2 + 36*$109,603X3 + 53*$109,603X4 + 34*$109,603X5 9.9
22*$109,603X1 + 29*$109,603X2 + 46*$109,603X3 + 53*$109,603X4 + 50*$109,603X5 11.7
22*$109,603X1 + 46*$109,603X2 + 50*$109,603X3 + 42*$109,603X4 + 58*$109,603X5 12.3
It is recommended that for effective maximization of the amount of recycled garbage to be three times as important as minimization of the cost transportation the total capacity should be 2.13476 *10-5 megatonnes, the achievement of this capacity is directly influenced by a reduction in the cost of transport for each of the 10 sectors, which in perspective should be less than the estimated recyclable garbage, with respective objective of each site to be X1 = 0, X2 = 2.68 * 10-7 ,X3 = 4.93 * 10-7 , X4 = 8.91 * 10-7 and X5 = 2.3 * 10-7
The same scenario is experience when the management intended to maximise the amount of recycled garbage and minimise the transportation cost without altering the either the amount of the of recycle garbage or the cost of transportation, the maximized capacity at this scenario was determined to be 7.06647*10–6 megatonnes, and this achievement was resulted from minimizing the cost to be less than the amount of the estimated recycled garbage, with respective objective of each site to be X1 = 1.28 * 10-7, X2 = 8.93 * 10-8 ,X3 = 1.91 * 10-7 , X4 = 1.53 * 10-7 and X5 = 7.65 * 10-8
Q2)
The NLP spreadsheet model is following:
Formulas:
G2 =D2*((B2-$B$23)^2+(C2-$C$23)^2)^0.5+E2*((B2-$B$24)^2+(C2-$C$24)^2)^0.5+F2*((B2-$B$25)^2+(C2-$C$25)^2)^0.5 copy to G2:G21
H2 =SUM(D2:F2) copy to H2:H21
G23 =SUM(G2:G21)
The locations of the three warehouses are following
|
X |
Y |
|
|
Wh 1: |
13.1 |
10.4 |
|
Wh 2: |
23.9 |
10.8 |
|
Wh 3: |
5.0 |
4.6 |
Recommendation
The subrubs are supplied by the warehouses as indicated in the matrix (D2:F21). Value 1 indicates that particular Suburb is supplied by that warehouse.
It is recommended that for a company to build its warehouses in locations that minimise the distances to each of the stations it serves, the location of the three warehouse from the suburbs X and Y for warehouse 1 should be 13.1 and 10.4, similarly warehouse 2 should be 23.9 and 10.8 and finally the warehouse 3 location should be 5.0 and 4.6 respectively, therefore to determine the respective distance from the respective petrol station there location will be based on the respective reference for of the position of the three warehouse which will result to a total distance which is equal 86, this is the best minimum total distance from all the station that should be considered as far as the position of the warehouse is concern
Step 1. Determine the priority of the factors which are taken into the consideration. The factors taken into consideration for the van here are – price, safety, economy, and comfort.
How to find the priority-
- Find the geometric mean of all the factors, of their rank factors. GM = (factor 1*factor 2*..factor n)1/n
- Find the priority vector for each factor, priority vector = GM of the factor/ (Sum of all GMs). This givesthe actual priority values of all the factors. But now we also need to check the consistency of the data, consistency means that the data is valid to use, for this we need to find the Consistency Ratio
- Finding the CR-Find the sum of all columns as below-
- Now we multiply the sum of columns of the factorsinto their respective PVs. See below
- Now we find the sum of all the Sum PV values, this sum will be the value called lambda max
- Now we find Consistency Index; CI; CI= (lamda max- n)/(n-1); n= number of factors; 4 here. As shown below. CI=0.024
- There is a term random index (RI) corresponding to number of fators; for n=4; RI=0.9. CR=CI/RI/ See below. If CR is below 0.1; then data is consistent and can be used. The CR value is .0264; hence the data is consistent
- Now, we find the same CR for all the other 4 matrices; that is Van A,B,C against Price, safety, economy and comfort. Find the PVs of all the Vans against all the factors, using the same procedure described above. Check their consistency at all levels. Use RI for n=3 as 0.58.As shown below
You see that for comfort and economy, the CR is above 0.1; hence the data is inconsistent and should be disregarded, but here we can continue to solve the question.
Now we make the final matrix, to find the final rank. For all the Van, A,B,C, write their PV values against the corresponding 4 factors; which we found above. Also write the PV values we found of the four factors.
Now, multiply the PV of the factor, to the corresponding PV of the respective Van, and find the final score of each van. For eg; for Van A; score = D40*D39 + E40*E39 + F40*F39 + G40*G39; and do so for all the Vans. See below
It is recommended that David purchases Van B, since it is clearly seen from the calculation that final score for Van B is max.
The determination of purchasing Van B is build up through first formation of geometric mean which will help to determine the priority vector to demonstrate an actual priority values of all the factors that will help in eliminating other factors, the priority vector helps determine the consistency ratio where consistency ratio will show that the value is valid to use, based on the several values of priority values determined from the safety of van A, B and C, the maximum value which shows a greater priority value of 0.536877059 is determined on van B, which is recommended to be bought
NLP Spreadsheet Model
Q4)
Here we have 3 alternatives for locations and 7 criteria to select the location. We can form the alternate-criterion matrix which will look as shown below:
Let J be the set of benefit attributes (which needs to be maximized) and K be the set of negative attributes.
J = {Labour Availability, Transportation, Access to Customers, Long-range Goals}
K = {Land Cost, Labour Cost, Construction Cost}
We now normalize the matrix by dividing it by the square root of sum of squares of all the cell values in the matrix. We then get the following matrix:
Now we multiply each of the columns by the respective weights given to the criterion. This yields the following matrix. Now get the ideal and negative ideal solution. You get the ideal solution by picking the max value for set J (calculated above) and min value for set K. You do the opposite for negative ideal solution.
Ideal Solution = {0.69, 1.45, 1.93, 0.86, 1.2, 2.48, 1.69}
Negative Ideal Solution = {0.96, 1.45, 1.38, 1.2, 1.03, 2.17, 1.20}
Now for each Location we’ll calculate the distance from these solutions. These distances are calculated using the distance formula or the L2 norm. The distances from ideal and negative ideal solutions are as follows:
Now we calculate the relative closeness to the ideal solution dividing the distance from ideal solution by the sum of distances for every alternate. This gives the following closeness to ideal solution:
It is recommended that in order to determine the best location then a consideration based on considering the location whose closeness is least distance from 1. It will be recommended that the best location will be Derrimut, VIC.
The location whose closeness is least is determine through first creation of an evaluation matrix which consists of alternatives and criteria, {displaystyle (x_{ij})_{mtimes n}}the matrix {displaystyle (x_{ij})_{mtimes n}} is then normalised to form the matrix, this followed by calculation of the weighted normalised decision matrix, after which this will help determine the worst alternative {displaystyle (A_{w})} and the best alternative{displaystyle (A_{b})}, from determined worst alternative and best alternative the calculation of the L2-distance between the target alternative {displaystyle i}and the worst condition is determined, which is similar to the distance between the alternative {displaystyle i} and the best condition, the calculation of the similarity to the worst condition is also determined and finally the alternatives are ranked accordingly, and in this case the greater closeness is a value of 0.81 therefore best location will be Derrimut, VIC.
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