Operations Management MBA 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 1.Formulate the problem and arrange the data in matrix form 2.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

Initial basic solution by North west corner method

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.

D1D2D3D4CAPACI TY S S S Demand

The no. of positive allocations (occupied cells) = m + n -1 = 6 D1D2D3D4CAPACI TY S S S Demand

Total cost = 5 * * * * * * 20 = Rs. 1,015

Initial basic solution by VAM

Choose largest value  22 in columnD 2 Choose cell with lowest cost Satisfy it D1D1 D2D2 D3D3 D4D4 capacityRow differences S1S S2S S3S Demand Column differences ghj

D1D1 D2D2 D3D3 D4D4 capRow diff S1S S2S S3S Demand Column difference s

D1D1 D2D2 D3D3 D4D4 capac ity Row diff S1S S2S S3S Demand Column differences

D1D1 D2D2 D3D3 D4D4 capa city Row diff S1S S2S S3S Demand Column difference s

D1D1 D2D2 D3D3 D4D4 capa city Row diff S1S S2S S3S Demand Column differences

Total cost : 5*19 + 2* *40 + 2* * *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 1.NWCM 2.VAM Find the optimal soln (after NWCM) using UV method

Dest source 1234supply Demand

By VAM

Dest source 1234supply Demand

m+n-1 = 6 Hence this is a feasible soln

Total Cost = Rs. 2850

By NWCM Dest source 1234supply Demand

Total cost = Rs. 4400

Optimal soln Row 1,2,3 are assigned values U1,U2,U3 AND Col 1, col2, col3 and col 4 are assigned variables V 1,V 2,V 3,V 4 FOR BASIC cells Ui + Vj = c ij Take U1 = 0

V1=3V2=1V3=0V4= supply U1= U2= U3=

If all pij<=0; optimality is reached Compute Pij (penalties) for the non basic cells by using the formula : Pij = Ui + Vj - cij

V1=3V2=1V3=0V4= supply U1= U2= U3= ve 1

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.

V1=3V2=1V3=0V4= supply U1= U2=522 +  U3= ve ve ve 200

V1=-3V2=1V3=0V4= supply U1= U2= U3= X ve 300 -ve ve

V1=-2V2=1V3=1V4=0 1234supply U1= U2= U3= ve 300 -ve ve 50 -ve

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 1.NWCM 2.VAM APPLY UV method to find the optimal solution.

12345supply Deman d

The given problem is unbalanced because the total demand(1550) < total supply (2000) Convert to a balanced transportation problem.

123456Supply De man d

Initial basic soln using NWCM

123456supply Dem and

Total cost = Rs. 19,700

Initial basic solution using VAM

123456Supply Demand

Total cost = Rs Asssign Ui & Vj

V1=6V2=2V3=12V4= 10 V5= 8 V6= Supply U1= U2= U3= U4= D ve 0 2

V1=6V2=2V3=12V4=10 V5=8V6= S U1= U2= U3= U4= D ve

V1=6V2=2V3=12V4= 6 V5=8V6= S U1= U2= U3= U4= D ve

Total cost = Rs. 11,500