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Operations Management MBA Sem II. Module IV Transportation. …the physical distribution of goods and services from several supply centers to several demand centres. Transportation.
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Operations ManagementMBA Sem II Module IV Transportation
…the physical distribution of goods and services from several supply centers to several demand centres.
Transportation The structure of transportation problem involves a large no. of shipping routes from several supply origins to several demand centres.
Objective : To determine the number of units of an item that should be shipped from an origin to a destination in order to satisfy the required quantity of goods or services at each destination centre.
Method • Formulate the problem and arrange the data in matrix form • Obtain an initial basic feasible solution by - North West Corner Method - Least Cost method - Vogel’s Approximation Method
The solution must satisfy all the supply and demand constraints The number of positive allocations must be m+n-1, where m is the no. of rows and n is the no. of columns 3. Test the initial solution for optimality (MODI) 4. Update the solution
Q A company has three production facilities S1,S2, S3 with production capacity of 7,9 and 18 units (in 100’s) per week of a product, respectively. These units are to be shipped to four warehouses D1, D2, D3 and D4 with requirement of 5,6,7 and 14 units (in 100’s) per week, respectively.
The transportation costs(in rupees) per unit between factories to warehouses are given in the table in the next slide. Minimize the total transportation cost.
The no. of positive allocations (occupied cells) = m + n -1 = 6
Total cost = 5 *19 + 2 * 30 + 6*30 + 3 * 40 + 4 * 70 + 14* 20 = Rs. 1,015
Choose largest value 22 in columnD2 Choose cell with lowest cost Satisfy it ghj
5 8
5 10 8
5 2 10 8
5 2 2 7 8 10
Total cost : 5*19 + 2*10 + 7 *40 + 2*60 + 8 *8 + 10 *20 = Rs 779
Q consider the following transportation problem involving 3 sources and four destinations. The cell entries represent the cost of transportation per unit. Obtain the initial basic feasible soln using • NWCM • VAM Find the optimal soln (after NWCM) using UV method
300 250 150 50 250 200
m+n-1 = 6 Hence this is a feasible soln
By NWCM 250 50 300 100 300 200
Optimal soln Row 1,2,3 are assigned values U1,U2,U3 AND Col 1, col2, col3 and col 4 are assigned variables V1,V2,V3,V4 • FOR BASIC cells Ui + Vj = cij • Take U1 = 0
250 50 100 300 200 300
If all pij<=0; optimality is reached Compute Pij (penalties) for the non basic cells by using the formula : Pij = Ui + Vj - cij
50 -ve -ve 250 6 100 -ve 300 -ve 1 200 300
If all Pij are <= 0; then optimality is reached; Cell (2,1) with penalty 6, has the most positive penalty. So, there is scope for improving soln. Construct a loop using one basic cell and other non basic cells. Assign + and – signs alternatively to the basic cells.
Choose the minimum among the –vely assigned basic cells = 250 Add to all +vely cells. Subtract from all –vely assigned cells.
-ve 250 -ve 50 6 300 -ve 100 -ve 1 300 200
300 -ve -ve -ve 250 50 100 -ve 300 200 -ve 1
300 -ve -ve -ve 150 250 -ve -ve 250 -ve 50 200
All penalties <=0; Hence optimum solution is reached Total = 2850
Q Consider the transportation problem as shown in the table. Find the initial basic solution using • NWCM • VAM APPLY UV method to find the optimal solution.
The given problem is unbalanced because the total demand(1550) < total supply (2000) Convert to a balanced transportation problem.
300 50 400 50 200 150 75 400 375
300 75 350 75 100 175 150 400 375
Total cost = Rs. 12250 Asssign Ui & Vj
-ve 300 - ve -ve -ve 0 350 -ve 75 75 -ve -ve 100 175 -ve 150 400 2 -ve 375 -ve 2 -ve -ve
300 -ve -ve -ve -ve -ve 150 350 -ve -ve -ve -ve 0 150 75 100 100 400 4 375 -ve -ve -ve -ve
300 -ve -ve -ve -ve -ve 350 -ve -ve -ve 150 -ve -ve 0 100 100 225 400 -ve -ve 150 -ve -ve 225