Transportation Modelling - Quantitative Analysis and Discrete Maths
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Transportation Modelling - Quantitative Analysis and Discrete Maths

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Transportation Modelling - Quantitative Analysis and Discrete Maths Transportation Modelling - Quantitative Analysis and Discrete Maths Presentation Transcript

  • Transportation Model Subject – QADM Professor Kazi Batch 97, Group 5 • Atul Sande, Roll no. 04 • Krupesh Shah, Roll no. 07 • Vengadeshwaran Perumal, Roll no. 19 • Pooja Chilap, Roll no. 32 • Sagar Kuckian, Roll no. 36
  • Outline O Transportation Modeling O Benefits and Applications O Methods Of Transportation O Obtain the Initial Feasible Solution using O Northwest - Corner Rule O Intuitive Least Cost Method / Minimization Method O Vogel’s Approximation Method O Obtain Feasible Solution for Optimality using O Modified Distribution Method O Conclusion
  • Transportation Modeling O What is Transportation Model? A Transportation Model (TP) consists of determining how to route products in a situation where there are several destinations in order that the total cost of Transportation is minimised O Can be used to help resolve distribution and location decisions O Need to know: O The origin points and the capacity or supply per period at each O The destination points and the demand per period at each O The cost of shipping one unit from each origin to each destination
  • Transportation Problem (TP) SUPPLY DEMAND A D B E C F
  • Transportation Problem To From Andheri Bandra Chandivali Dadar 5 4 3 Elphinston 8 4 3 Fort 9 7 5
  • Transportation Matrix To Andheri Bandra From 5 Dadar 8 Elphinston 9 Fort Warehouse requirement 300 Cost of shipping 1 unit from factory to Bandra warehouse 4 Factory Chandivali capacit y 3 100 Dadar capacity constraint Cell representi 4 3 ng a 300 possible source-todestinatio 7 5 n shipping 300 assignme nt (Elphinsto 200 200 700 n to Chandivali ) Total demand Chandivali and total supply warehouse demand
  • DEVELOPING AN INITIAL SOLUTION— THE NORTHWEST CORNER RULE Once the data have been arranged in tabular form, we must establish an initial feasible solution to the problem. One systematic procedure, known as the northwest corner rule, requires that we start in the upper left hand cell (or northwest corner) of the table and allocate units to shipping routes as follows •Exhaust the supply ( factory supply) at each row before moving down to the next row •Exhaust the (warehouse) requirements of each column before moving to the next column, on the right •Check that all the supply and demands are met.
  • Demand Not Equal To Supply A situation occurring quite frequently in real-world problems is the case where total demand is not equal to total supply. These unbalanced problems can be handled easily by introducing a dummy Demand or dummy Supply. In the event that total supply is greater than total demand, a dummy destination, with demand exactly equal to the surplus, is created. If total demand is greater than total supply, we introduce a dummy source (factory) with a supply equal to the excess of demand over supply. In either ease, cost coefficients of zero are assigned to each dummy location.
  • Northwest Corner rule To (A) (B) (C) Dummy From (E) 250 4 3 0 4 3 0 50 200 9 (F) Warehouse requirement 5 8 (D) 300 7 200 50 150 200 5 150 150 0 Factory capacity 250 300 300 850
  • Northwest – Corner Rule To From (D) Dadar (E) Elphinston (A) Andheri 100 200 (C) Chandivali 5 4 3 8 4 3 9 (F) Fort Warehouse requirement (B) Bandra 300 100 100 200 7 200 200 5 Factory capacity 100 300 300 700 Means that the firm is shipping 100 bathtubs from Fort to Bandra
  • Northwest Corner Computed Shipping Cost Route From To D E E F F A A B B C Tubs Shipped Cost per Unit 100 200 100 100 200 5 8 4 7 5 Total Cost 500 1,600 400 700 1,000 Total: 4,200 This is a feasible solution but not necessarily the lowest cost alternative
  • Drawbacks O The Northwest- corner rule is easy to use, but O O O O this approach totally ignores the costs Demand not equal to supply Called an unbalanced problem Resolved by introducing dummy source or dummy destination where in the aim of transportation model is minimization of cost so introducing a dummy source is not a good solution. Missing out the best cost effective path
  • Intuitive Lowest-Cost Method O Identify the cell with the lowest cost O Allocate as many units as possible to that cell without exceeding supply or demand; then cross out the row or column (or both) that is exhausted by this assignment O Find the cell with the lowest cost from the remaining cells O Repeat steps 2 and 3 until all units have been allocated
  • Lowest - Cost Method To From (A) Andheri Warehouse requirement 300 3 4 3 9 (F) Fort 4 8 (E) Elphinston (C) Chandivali 5 (D) Dadar (B) Bandra 7 5 200 200 Factory capacity 100 300 300 700 First, 3 is the lowest cost cell so ship 100 units from Dadar to Chandivali and cross off the first row as Dadar is satisfied
  • Lowest Cost Method To From (A) Andheri 5 (F) Fort 300 4 9 (E) Elphinston 4 8 (D) Dadar Warehouse requirement (B) Bandra (C) Chandivali 7 200 100 100 3 3 5 200 Factory capacity 100 300 300 700 Second, 3 is again the lowest cost cell so ship 100 units from Elphinston to Chandivali and cross off column C as Chandivali is satisfied
  • Lowest Cost Method To From (A) Andheri 3 4 200 9 (F) Fort Warehouse requirement 4 8 (E) Elphinston (C) Chandivali 5 (D) Dadar (B) Bandra 300 100 100 7 200 3 5 200 Factory capacity 100 300 300 700 Third, 4 is the lowest cost cell so ship 200 units from Elphinston to Bandra and cross off column B and row E as Elphinston and Bandra are satisfied
  • Lowest Cost Method To From (A) Andheri 5 (E) Elphinston Warehouse requirement 300 300 4 8 (D) Dadar (F) Fort (B) Bandra 4 200 9 (C) Chandivali 100 100 7 200 3 3 5 200 Factory capacity 100 300 300 700 Finally, ship 300 units from Andheri to Fort as this is the only remaining cell to complete the allocations
  • Lowest Cost Method To From (A) Andheri 8 (E) Elphinston Warehouse requirement Total Cost (C) Chandivali 4 3 5 (D) Dadar (F) Fort (B) Bandra 300 300 200 9 4 100 100 7 200 3 5 200 = 3(100) + 3(100) + 4(200) + 9(300) = 4,100 Factory capacity 100 300 300 700
  • Lowest Cost Method To From (A) Andheri (B) Bandra This is a feasible solution, and 5 an (D) Dadar improvement over the previous solution, but not necessarily the lowest 8 cost alternative (E) Elphinston 200 (F) Fort Warehouse requirement Total Cost 300 300 9 4 4 (C) Chandivali 100 100 7 200 3 3 5 200 = 3(100) + 3(100) + 4(200) + 9(300) = 4,100 Factory capacity 100 300 300 700
  • Vogel’s Approximation Method O Calculate penalty for each row and column by taking the difference between the two smallest unit costs. This penalty or extra cost has to be paid if one fails to allocate the minimum unit transportation cost. O Select the row or column with the highest penalty and select the minimum unit cost of that row or column. Then, allocate the minimum of supply or demand values in that cell. If there is a tie, then select the cell where maximum allocation could be made. O Adjust the supply and demand and eliminate the satisfied row or column. If a row and column are satisfied simultaneously, only of them is eliminated and the other one is assigned a zero value.Any row or column having zero supply or demand, can not be used in calculating future penalties. O Repeat the process until all the supply sources and demand destinations are satisfied
  • Vogel’s Approximation Method D1 S1 S2 S3 D2 D3 Supply Row Diff. D4 19 30 50 10 70 30 40 60 40 8 70 20 7 9 Demand Col.Diff. 5 21 8 22 D1 S1 S2 S3 Demand Col.Diff. 7 10 D3 19 50 10 70 40 60 40 70 20 7 10 Supply Row Diff. 7 20 10 14 10 9 9 5 5 21 12 34 14 10 D4 10 18 8 9 20 34
  • Vogel’s Approximation Method D3 50 S3 60 70 S2 10 40 S1 20 S1 S2 Demand Col.Diff. 50 7 10 7 10 9 4 50 20 10 50 34 14 10 2 40 9 2 60 40 Supply Row Diff. D4 10 40 2 10 Demand Col.Diff. D3 Supply Row Diff. D4 20 34 D3 S2 Demand Col.Diff. Supply Row Diff. D4 40 60 7 7 2 2 9 34 20
  • Vogel’s Approximation Method O The total transportation cost obtained by this method = 8*8+19*5+20*10+10*2+40*7+60*2 = Rs.4055 O Here, we can see that Vogel’s Approximation Method involves the lowest cost than North-West Corner Method and Least Cost Method and hence is the most preferred method of finding initial basic feasible solution.
  • Benefits of Vogel’s Approximation Method O VAM is an improved version of the least- cost method that generally, but not always, produces better starting solutions. O This method is preferred over the other methods because it generally yields, an optimum, or close to optimum, starting solutions.
  • Obtain Feasible solution for Optimality using MODI Method The modified distribution method, MODI for short , is an improvement over the stepping stone method for testing and finding optimal solutions
  • Steps Involved in MODI Method O Find a basic solution by any standard method. If supply and demand are equal then it is a balanced transportation problem. O Test for optimality. The number of occupied cells should equal to m + n -1. If the initial basic feasible solution does not satisfy this rule, then optimal solution cannot be obtained. Such solution is a degenerate solution O Set up a cost matrix for allocated cells only.
  • Steps Involved in MODI Method O Determine a set of number Ui for each row and a set of number Vj on the bottom of the matrix. O Compute the value of Ui and Vj with the formula Ui + Vj = Cij to all basic(occupied) cells. O Calculate the water value of of non-basic ( unoccupied) cells using the relation Ui+Vj=Cij. O Compute the penalties for each unoccupied cell by using the formula Dij=Pij=Ui+Vj-Cij. Examine whether all Pij ≤ 0. If all Pij < 0, then the solution is optimal and unique. If all Pij ≤ 0, then the solution is optimal and an alternative solution exists. If at least one Pij > 0, then the solution is not optimal. At the end, prepare the optimum solution table and calculate the optimum/minimum transportation cost.
  • Optimum Solution Using Modi Method CAPACITY A D B 8 C 8 15 120 E 15 10 17 80 3 9 10 80 F REQUIREMENT 150 80 50
  • W1 F1 W2 W3 8 8 120 15 120 F2 15 F3 3 30 17 10 50 9 150 80 30 80 10 50 IBFS= 120(8)+30(15)+50(10)+30(9)+50(10) IBFS=960+450+500+270+500=2680 50 80
  • OCCUPIED MATRIX UNOCCUPIED MATRIX -5 Ui 8 U1=0 -11 4 11 7 -15 -6 U2=7 10 0 3 -8 15 Ui Ui -17 11 9 6 14 10 U3=6 -3 Vj V1=8 V2=3 V3=4 Ui + Vj = Cij Vj 4 8 Pij=(Ui+Vj)-Cij 3
  • LOOP CONSTRUCT 8 8 15 8 120 15 _ 10 + 9 17 _ 15 Si 120 E 10 15 17 80 80 3 10 30 + 120 50 30 3 8 50 1 30 0 9 10 50 80 DEGENERACY OCCUAR Dj 150 80 50 280 Value of O is equal to the minimum of the existing allocation among the signed cells on STONE SEQUARE=RIM REQUIREMENT m+n-1=5 the loop.
  • OPTIMUM SOLUTION TABLE OPTIMUM COST 120 15 8 10 17 80 3 30 9 10 50 8*E 10*80 =800 • F3 W1 3*30= 90 • F3 W3 E 8*120 =960 • F1 W2 15 • F1 W1 • F2 W2 8 10*50 =500 = _ ____________ 2350
  • Why MODI Method? O It Is the simplex method O It is a minimum cost solution to the transportation problem. O All the drawbacks which were in all the three methods is covered in this modi method.
  • Thank You!!