INTRODUCTION The transportation problems are one of the types of the LLP (Linear Programming Problems), In which objective is to transport various quantities of a single homogenous commodity, to different destinations in such a way that the total transportation cost is minimum. Transportation problems gave direct relevance to decisions in the area of distribution policy making, where the objective is minimisation of transportation cost . Here the availability as well as requirements of the various centres are finite and constitute the limited resources. It is also assumed that cost of shipping is linear. Thus these problems could also be solved by “SIMPLEX METHOD”.

Terminology used in transportation model FEASIBLE SOLUTION (FS) : Non negative values of xij where i = 1,2,……m and j = 1,2 ……n which satisfy the constraints of availability (supply) and requirement (demand) is called the feasible solution to the transportation problem. BASIC FEASIBLE SOLUTION (BFS) : A feasible solution to a m-origin, n-destination problem is said to be basic if the number of positive allocations are m+n-1 i.e. one less than the sum of rows and columns. OPTIMAL SOLUTION : A feasible solution is said to be optimal if it minimizes the total transportation cost. The optimal solution itself may or may not be a basic solution. This is done through successive improvements to the initial basic feasible solution until no further decrease in transportation cost is possible.

BALANCED TRANSPORTATION PROBLEM : A transportation problem in which the total supply from all the sources equals the total demand in all the destinations . UNBALANCED TRANSPORTATION PROBLEM : Such problems which are not balanced are called unbalanced .

MATRIX TERMINOLOGY : In the matrix used in transportation problem, the squares are called ‘cells’ . These cells form ‘columns’ vertically and ‘row’ horizontally. Unit costs are written in the cells . WAREHOUSES 1 2 3 4 PLANTS 4 2 10 3 6 8 7 5 15 7 8 12 A B DEMAND The cell located at intersection of row B and column 4 is the one in which the unit cost 5 is written as (B,4). DEGENERATE BASIC FEASIBLE SOLUTION (DBFS) : If the number of allocation in a basic feasible solutions are less than (m+n-1), it is called degenerate basic feasible solution (DBFS).

BASIC ASSUMPTION OF THE MODEL AVAIABLITY OF THE QUANTITY : Quantity available for distribution at different sources or depots is equal to total requirement of different consumption centres. 2. TRANSPORTATION OF ITEMS : Items can be conveniently transported from every production centre to every consumption centre.

3. COST PER UNIT : The per unit transportation cost of items from one production centre to another consumption centre is certain. INDEPENDENT COST : The per unit cost of transportation is independent of the quantity despatched 5. OBJECTIVE : The objective of such an arrangement is to minimise the total cost of transportation for the organisation as a whole

OF THE TRANSPORTATION PROBLEM OPTIMAL SOLUTION OF THE TRANSPORTATION PROBLEM There are 2 steps to find the optimal solution of the transportation problems : find an initial basic feasible solution. obtain an optimal solution by making successive improvements to initial basic feasible solution until no further decrease in the transportation cost is possible.

METHODS FOR INITIAL BASIC FEASIBLE SOLUTIONS Following are the important methods of developing an initial feasible solution : North West Corner Methods (NWCM). (b) Lowest Cost Entry Method (LCEM or Matrix Minima Method). (c) Vogel’s Approximation Method (VAM).

(A) NORTH-WEST CORNER METHOD (NWCM) This is the most systematic and easiest method for obtaining initial feasible solution : Steps involved in this method are stated as : Step 1. Construct an empty m*n matrix, completed with rows and columns. Step 2. Indicate the rows totals and columns totals at the end. Step 3. starting with (1,1) cells at the North-West Corner of the matrix, allocate maximum possible quantity keeping in view that allocation can neither be more than the quantity required by the respective warehouses nor more than the quantity available at each supply centre.

Step 4. Adjust the supply and demand numbers in the respective rows and columns allocations. Step 5. If the supply for the first row is exhausted then move down to the first cell in the second row and first column and go to step 4. Step 6. If the demand for the first column is satisfied, then move to the next cell in the second column and first row and go to step 4. Step 7. If for any cell, supply equals demand then the next allocation can be made in cell either in the next row or column. Step 8. Continue the procedure until the total available quantity is fully allocated to the cells as required.

Ques : find the initial basic feasible solution by at least three different methods for the following transportation problem : From\To D1 D2 D3 D4 Available F1 10 7 3 6 3 F2 1 6 7 3 5 F3 7 4 5 6 7 Demand 3 2 6 4 15\15

Solution : (a) Initial basic feasible solution by NWCM : From\To D1 D2 D3 D4 Available F1 3 F2 5 F3 7 Demand 3 2 6 4 15\15 10 7 3 6 3 1 6 7 3 2 3 7 4 5 6 3 4

TOTAL COST IS : F1-D1 = 10*3 =30 F2-D2 = 6*2 =12 F2-D3 = 7*3 =21 F3-D3 = 5*3 =15 F3-D4 = 6*4 =24 TOTAL COST Rs. 102

METHOD OR MATRIX MINIMA METHOD (B) LOWEST COST ENTRY METHOD OR MATRIX MINIMA METHOD This method takes into consideration the lowest cost and therefore takes less time to solve the problem; various steps of this method can be summarized as under: Step1. select the cell with the lowest transportation cost among al the rows or columns of the transportation table. If the minimum cost is not unique then select arbitrarily any cell with the lowest cost. Step2. Allocate as many units as possible to the cell determined in step1 and eliminate that row in which either capacity or requirement is exhausted.

Step 3. Adjust the capacity and requirement for the next allocations. Step 4. Repeat steps 1 to 3 for the reduced table until the entire capacities are exhausted to fill the requirement at different destinations.

From/To D1 D2 D3 D4 Available 5 F2 7 F3 Demand 3 2 6 4 15/15 10 7 3 6 3 1 6 7 3 3 2 7 4 5 6 2 3 2

TOTAL COST IS : F1-D3 = 3*3 = 9 F2-D1 = 1*3 = 3 F2-D4 = 3*2 = 6 F3-D2 = 4*2 = 8 F3-D3 = 5*3 = 15 F3-D4 = 6*2= 12 TOTAL COST Rs. 53

(C) INITIAL BASIC FEASIBLE SOLUTION BY VOGEL’S APPROXIMATION METHOD (IBFS BY VAM) This method is preferred over the other two methods because the initial basic feasible solution obtained with VAM is either optimal or very close to the optimal solution. Therefore, the amount of time required to calculate the optimum solution is reduced. In Vogel’s approximation method the basis of allocation is unit cost penalty i.e., that column or row which has the highest unit cost penalty (difference between the lowest and the next higher cost) is selected first for allocation and the subsequent allocation in cells are also done keeping in view the highest unit cost penalty.

The steps are as follows : Step 1. For each row of the table identify the lowest and the next lowest cost cell. Find their differences and place it to the right of that row. In case two cells contains the same least cost then the difference shall be zero. Step 2. Similarly find the differences of each column and place it below each column. These differences found in steps (1) and (2) are also called penalties. Step 3. Looking at all the penalties. Identify highest of them and the row or column relative to that penalty. Allocate the maximum units to the least cost cell in the selected row or column. Ties should be broken in this order Maximum difference Least Cost Cell. Maximum difference Least Cost Cell. Maximum Units allocation Arbitrary Tie Tie

Step 4. Adjust the supply and demand and cross the satisfied row or column. Step 5. Recompute the column and row difference ignoring deleted rows/columns and go to step number (3) Repeat the procedure until all the column and rows totals are satisfied.

OPTIMALITY TEST After computation of an initial basic feasible solution, we can proceed to know, whether the solution so obtained is optimum or not. Solution so obtained may be optimal, so it becomes essential for us to test for optimization. For this purpose we first define non-degenerate basic feasible solution. Basic feasible solution of an (m*n) transportation problem is said to be non-degenerate if it has following two properties : Initial basic feasible solution must contain exactly m+n-1 number of individual allocations. (b) These allocations must be in independent positions. Independent position of a set of allocations means that it is always impossible to form any closed loop through these allocations. See figures given below :

For this purpose two methods are followed. Stepping Stone Method. Modified Distribution Method (MODI) These methods are followed to discuss the optimality of an initial basic feasible solution. The stepping stone method. The stepping stones method is an interactive technique from moving an initial feasible solution. In order to apply the Stepping Stone Method, to transportation problem, one rule about the number of shipping routes being used must first be observed. In this rule, the number of occupied routes (or square) must always be equal to one less than the sum of the number of rows plus the number of columns. (B) Testing the solution for possible improvement. How does the stepping stone method work? Its approach is to evaluate the cost effectiveness of shipping goods via transportation routes not currently in the solution.

Stepping Stone Method The stepping stone method for testing the optimality can be summarized in the following steps : Step 1. Prepare a transportation table with a given unit cost of transportation along with the Rim Requirements. Step 2. Determine an initial basic feasible solution using any method (preferably VAM). Step 3. Evaluate all unoccupied cell for the effect of transferring one unit from an occupied cell. This transfer is made by forming a closed path (loop) that retains the supply and demand condition of the problem. The evaluation is conducted as follows. Select an unused square to be evaluated. Beginning with the selected unused square trace a closed path (or loop) through atleast three occupied cell. (it is called home square). The direction of the movement taken is immaterial because the result will be same in either case. In the closed path formulation of only right angle turn is allowed, and therefore skip all other cells which are not at the turning points.

(c) At each corner of the closed path assign plus (+) and minus (-) sign alternatively beginning with plus sign for the unoccupied square (water square) to be evaluated. The +ve and –ve signs can be assigned either in a clockwise or counter clockwise direction. (d) Compute the net change in cost along the closed path by adding together the unit transportation costs associated with each of the cell traced in the closed path. Comparing the addition to cost with decreases, will be the improvement index. (e) Repeat steps 3 (a) to 3 (d) until net change in cost has been calculated for all unoccupied cells. Step 4. Check the sign of each of the net change in the unit transportation costs. If all net changes are plus (+) or zero, then we have obtained an optimal solution, otherwise go to step 5.

Step 5. Select the unoccupied cell with most negative net change among all unoccupied cells. If two minus values are equal, select that one which will result in moving as many units as possible into the selected unoccupied cell with the minimum cost. Step 6. Assign as many units as possible to unoccupied cell satisfying rim conditions. The maximum number of units to be assigned are equal to the smaller circled number ignoring sign among the occupied cells with minus value in the closed path. Step 7. Go to step 3 and repeat the procedure until all unoccupied cells are evaluated and net change is positive or zero values.

(B) MODIFIED DISTRIBUTION METHOD (MODI METHOD) The MODI (modified distribution) method allow us to compute improvement indices quickly for each unbiased square without drawing all of the closed paths (loops). Because of this it can often provide considerable time savings over the stepping stone method for solving transportation problems. MODI provide a new means of finding the unused route with the largest negative improvement index. Once the largest index is identified, we are required to trace only one closed path, just as with the stepping stone approach, this path helps to determine the maximum number of units that can be shipped via the best unused route.

The following steps are followed to determine the optimality. Step 1. from the given data construct a transportation table with the given cost of transportation and rim conditions. Step 2. Determine in initial basic feasible solution using a suitable method (i.e. NWCM, LCEM or VAM). Step 3. For the current basic feasible solution with m+n-1 occupied cells, calculate index numbers (dual variables) Ri = (i = 1,2,....m) and Kj = (j=1,21.....,n) for rows and columns, respectively. For calculating values of Ri and Kj, the following relationship (formula) for occupied cells is used, cij = Ri + kj for all i,j Step 4. For occupied cells, the opportunity cost by using the formula. dij = cij – Ri + Kj for all i,j Step 5. Now the opportunity cost of an unoccupied call is determined by using the formula : Opportunity cost = Actual cost- Implied cost dij = cij – (Ri + kj)

Step 6. Examine unoccupied cells evaluation for Dij If dij > 0, then the cost of transportation will increase, i.e., and optimal solution has been arrived at. If dij = 0, then the cost of transportation will remain unchanged. But there exists an alternative solution. If dij < 0, then an improved solution can be obtained by introducing cell (i,j) in the basis and go to step 7. (i.e. Next step) Step 7. Select an unoccupied cell (i.e. Water square) with largest negative opportunity cost among all unoccupied cells. Step 8. Construct a closed path for the unoccupied cell determined in step 7 and assign plus (+) and minus (-) sign alternatively beginning with plus sign for the selected unoccupied cell in clockwise or other direction. Step 9. Assign as many units as possible to the unoccupied cell satisfying rim conditions. The smallest allocation in a cell with negative sign on the closed path indicated the number of units that can be transported to the unoccupied cells. This quantity is added to all the unoccupied cells on the path marked with plus sign and subtracted from the those occupied cells on the path marked with minus signs.

Step 10. Go to step 4 and repeat procedure until all dij > 0, i. e Step 10. Go to step 4 and repeat procedure until all dij > 0, i.e., an optimal solution is reached. Calculate the associated total transportation cost. Ques : A product is manufactured by four factories A,B,C and D. The unit production cost in them are Rs.2, Rs.3,Rs.1 and Rs.5 respectively. Their production capacities are 50, 70,30 and 50 units respectively. These factories supply the product to four stories, demands of which are 25,35,105 and 20 units respectively. Unit transportation cost in rupees from each factory to each store is given in the table below : STORES 1 2 3 4 A B C D 2 4 6 11 10 8 7 5 13 3 9 12 4 6 8 3 FACTORIES Determine the transportation plan to minimize the total production – cum-transportation cost.

1 2 3 4 5 CAP. UP1 UP2 UP3 UP4 UP5 UP6 UP7 A 50 2 2 2 2 2 6 6 B 70 5 2 2 2 1 7 7 C 30 3 6 6 - - - - D 50 3 1 1 3 2 8 - DEM.25 35 105 20 15 200 UP1 2 1 1 2 0 UP2 2 1 1 2 - UP3 2 2 1 2 - UP4 - 2 1 2 - UP5 - 2 2 - - UP6 - - 1 - - UP7 - - 1 - - 2 4 6 11 10 8 7 5 13 3 9 12 4 6 8 3

Since Rim Requirement = m+n-1 = 4+5-1 = 8 Rim Requirement = number of stone squares ; 8 = 8 The above solution is non-degenerate so minimum transportation cost we get : A-1 = 2*25 = 50 A-2 = 4*5 = 20 A-3 = 6*20 = 120 B-3 = 7*55 = 385 B-3 = 0*15 = 0 C-2 = 3*30 = 90 D-3 = 8*30 = 240 D-4 = 3*20 = 60 TOTAL COST Rs.965 Now, we proceed to test its optimality by MODI Method.

Calculation of opportunity cost of water squares is as below : Unused Square Cij – Ri - Kj Improvement index A- 4 11-0-1 10 A- 5 0-0+1 1 B - 1 10-1-2 7 B-2 8-1-4 3 B-4 5-1-1 C-1 13+1-2 12 C-3 9+1-6 4 C-4 12+1-1 C-5 0+1+1 2 D-1 4-2-2 D-2 6-2-4 D- 5 0-2+1 -1

DEGENERACY IN THE TRANSPORTATION PROBLEM It may be noted that total number of stone squares must be equal to the m+n-1 i.e. rim requirement . If this condition is not satisfied the solution is degenerate. Such situation occurs in the following two cases : (a) There may be excessive number of stone squares in a solution than the rim requirement that is m+n-1. This type of degeneracy arises only in developing the initial solution and is caused by an improper assignment or an error in formulating the problem. In such cases there is a need to modify the initial solution so as to satisfy the rule of m+n-1 (rim requirements). (b) The second situation occurs when there may be insufficient number of stone squares in a solution. Degeneracy of this type may occur either in the initial solution or in subsequent table .

Stage 1 : Degeneracy occur at the initial solution This particular case of degeneracy arises when both a column requirement and row requirement are satisfied simultaneously. We introduce Greek letter epsilon (e) in a water square with the lowest transportation cost and by this process the number of stone squares becomes equal to the rim requirements and we can proceed to solve the solution in an ordinary way. The value of Epsilon (e) is assumed to be something greater than 0 but which does not affect our mathematical calculation in anyway. Once (e) is introduced into the solution, it will remain there until degeneracy is removed or a final solution is arrived at, whichever occurs first. The use of (e) is illustrated in the following example. : Ques : The sunrise transport company ships truck loads of grain from silos to four mills. The supply (in truck loads) and the demand (also in truck loads) together with the unit transportation cost per truckload on the different routes are summarized in the following table-unit transportation costs. Cij are in the hundreds of dollars . Determine the minimum cost shipping schedule between the silos and mills (By using Vogel's approximation Method for I.B.F.S. is obtained as below ).

MILLS SILOS 1 2 3 4 5 15 25 10 Require- ment 5 15 15 15 50\50 10 2 20 11 15 12 7 9 20 15 10 4 14 16 18 5 5

Solution : Since Rim Requirement = m+n-1 = 3+4-1 = 6 But stone square = 5 This is a case of degeneracy because Rim Requirement Stone Squares We have introduced and epsilon in the least cost independent cell

MILLS SILOS 1 2 3 4 SUPPLY 15 2 25 10 Require- Ment 5 15 15 15 50\50 10 2 20 11 15 12 7 9 20 e 15 10 4 14 16 18 5 5

We calculated the values of unused square by using stepping stone method 1 – 1 = 10-4+18-20+7-2 = 9 1 – 3 = 20-9+7-2 = 16 1 – 4 = 11-20+7-2 = -4 2 – 1 = 12-4+18-20+6 = 12 3 – 2 = 14-18+20-7 = 9 3 – 3 = 16-18+20-9 = 9 We make a closed loop path from the highest negative value i.e. -4 and proceed further. Since the value of cell 1 – 4 is least –ve and the improved solution is shown below :

MILLS SILOS 1 2 3 4 SUPPLY 1 15 2 25 3 10 REQUIR 5 15 15 15 50\50 -EMENT 2 20 11 5 10 7 9 20 10 15 14 16 18 5 5

Value of unused square by using stepping stone method. 1 – 1 = 10-4+8-11 = 13 1 – 3 = 20-9+7-2 = 16 2 – 1 = 12-7+2-11+18-4 = 10 2 – 4 = 20-7+2-11 = 4 3 – 2 = 14-2+11-18 = 5 3 – 3 = 16-9+7-2+11-18 = 5 Since the values of all opportunity cost of all squares cells is +ve. So we get the optimum solution and rim requirement = stone squares i.e.6 = 6 Total transportation cost = 2*5+7*10+9*15+11*10+18*5+4*5 = 10+70+135+110+90+20 = Rs. 435

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