UNIT 7 TRNSPORTTION PROLEM Structure 7.1 Introduction Objectives 2 asic Feasible Solution of a Transportation Problem 7. Modified Distribution (MODI) Method 7.4 Stepping Stone Method 7.5 Unbalanced Transportation Problem 7.6 Degenerate Transportation Problem 7.7 Transhipment Problem 7.8 Maximisation in a Transportation Problem 7.9 Summary 7.10 Key Words 7.1 1 nswers to SQs 7.1 INTRODUTION The transportation problem is a special type of linear programming problem where the objective is to minimise the wst of distributing a product from a number of sources or origins to a number of destinations. ecause of its special structural format, the usual simplex method is unsuitable for solving transportation problems. Tbese problems require a special method of solution.?be special features of a transportation problem are illustrated with the help of the examples. Objectives fter studying this unit, you should be able to Example 7.1 formulate a transportation problem, locate a basic feasible solution of a transportation problem by various methods, ascertain minimum transportation cost schedule by Modified Distribution (MODI) method and Stepping Stone Method, discuss appropriate method to convert an unbalanced transportation problems into a balanced transportation problem, deal with degenerate transportation problenk, formulate and solve transhipment problems, and discuss suitable method when the problem is to maximise the objective function instead of minimising it. onsider a manufacturer who operates three factories and despatches his products to five different retail shops. The Table 7.1 indicates the capacities of the three factories, the quantity of products required at the various retail shops and the cost of shipping one unit of product 6rom each of three factories to each of five retail shops. Table 7.1
'Opth.hrtim Tecbniqm-i The Table 7.1 is usually referred to as Trampolfation Table provides the basic data regarding the transportation problem The capacity of factories 1,2, is 50, 100 and 150 respectively. The requirement at retail shops 1.2..4.5 is 100,70.50,40 and 40 respectively. The quantities in the bordered rectangle are known as unlt transportation cost. The cost of transportation of one unit from factory 1 to retail shop 1 is 1, factory 1 to retail shop 2 is 9 and so on. transportation problem can be formulated as a linear programming problem using variables with two subscripts. Let xi 1 = mount to be transported from factory 1 to retail shop 1 112 = mount to be transpow from factory 1 to retail shop 2 x5 = mount to be transported from factory to retail shop 5. Let the unit transportation costs be denoted by i 1, 12,... 5, i.e. i 1 = 1, 12 = 9 and so on. Let the capacities of the three factories be denoted by a1 = 50, a2 = 100, as = 150. The requirement of the retail shops are bl = 100.62 = 70.6 = 50, b4 = 40 and bs = 40. Then, the transportation problem can be formulated as follows : Minirnise it xi1 + i~xi2 + + 5x5 Subject to :... XII +XIZ+XI~+X~~+XIJ = a1 Xll 2 0, x12 10... x5 2 0. Thus, the problem has 8 constraints and 15 variables. It will be unwise, if not impossible, to solve such a problem using simplex method. This is why, a spec computational procedure is necessary to solve transportation problem In the next section, a number of procedures have been presented to derive an in, basic feasible soiution of the problem. SQ 1 Fill up the blanks. (1) The objective of a transportation problem is to... the transportation t (2) The constraints of a transportation problem are... () If a transportation problem has m factories and n retail shops the number of variables is... and the number of constraints is... 7.2 SI FESILE SOLUTION OF TRNSPORTTION PROLEM We illustrate with the help of an example introduced in Section 7.1 the computation of m initial basic feasible solution of a transportation problem. lthough the problem has eight constraints and fifteen variables, one of the constraints can be eliminated since a1 + a2 + a = bl + b2 + b + b4 + b5. Thus, the problem has in fact seven constraints and fiftee~ variables. ny basic feasible solution thus has at most seven non-zero xk In general, any basic feasible solution of a transportation problem with rn origins (such as factories) and n destinations (such as retail shops) has at most m + n - 1 non-zero xq The following methods hre available for the computation of an initial basic feasible solution.
(1) The North West omer Rule Tr~llsporsoll ~oblem Inthe North West omer Rule, allocations are made starting from the north-west (ubper left) comer completely dlsregardlng the transportation cost. y applying north-west comer rule to the transportation problem of Section 7.1, we obtain xl I = 50 as the capacity of factory 1 is 50. Eliminating the first row as the fust factory is unable to'supply any more. The reduced transportation table becomes as Table 7.2. Table 72 Factories 2 Retail Shops 1 2 4 5. 24 12 16 20 1 14 1 2 26 50 70 50 40 40 apacity 100 150 250 The value of bt is reduced to 50 in the revised transportation table as 50 units have already been supplied in retail shop 1 from factory 1. We now allocate 50 units to the north west comer of the revised transportation table. Thus, x21= 50. Proceeding in this way, we obtain x22 = 50, ~ = 220, ~ = 50, x4 = 40, ~ = 540. The corresponding transportation cost is given by, 1x50+24x50+12x50+5x20+1x50+2x40+26x40 = 4520. It is clear that as soon as a value of xij is determined, a row or a column is eliminated from further consideration. Tlx last value of xij eliminates both a row and a column. Hence, a feasible solution computed by the north-west comer rule can have at most m + n - 1 positive no, if the transportation problem has m origins and n destinations. Thus, the obtained solution is a basic feasible solution. In the present problem, m = and n = 5. Hence, using the north west corner rule, we have derived a basic feasible solution with seven non-zero xij. (2) Matrix Minimum Method We look for the row arb the, column corresponding to which ij is minimum in the entire transportation table. If there are two or more minimum costs then we should select the row and the column corresponding to the lower numbered row. If they appear in the same row we should select the lower numbered column. We choose the value of the corresponding xrj as much as possible subject to capaclty and requirement constraints. row or a column is dropped and the same procedure is repeated with the reduced transportation cost matrix. The method is illustrated with the help of the transportation problem presented in Section 7.1. We observe that i 1 = 1 which is the minimum transportation cost in the entire transportation Table 7.1. Hence, xt t = 50 and the first row is eliminated from any further allocation. The reduced Transportation Matrix is given in Table 7.. 25 = 1 is the minimum transportation cost in the reduced transportation table. So x25 = 40. Proceeding in this way we observe that x = 50, x22 = 60, x1= 50, x2 = 10, x4 = 40. The basic feasible solution developed by the matrix minimum method has a transportation cost, 1x50+1x40+1x50+12x60+14x50+x10'+2x40 = 2810. Table 7 Retail Shops Factories '- - 1 2 4 5 apacity 2 24 12 16 20 1 100 14 1 2 26 150 50, 70 50 40 40 250 \ ' \ p~dtion cost obtained by using matrix method is much lower idb cdst of the solution derived by using north west comer rule. This is to be expected as the matrix minimum method takes into account the unit transportation cost while choosing the values of the basic variables.
qtimizntion Techniques-I () Vogel pproximation Method (VM) Vogel pproximation method (VM) for finding an initial basic feasible solution involves the following steps : (i) From the transportation table, we determine the penalty for each row and column. The penalties are calculated for each row (column) by subtracting the lowest cost element in that row (column) from the next lowest cost element in the same row (column). (ii) We identify the row or column with the largest penalty among all the rows and columns. If the penalties corresponding to two or more rows or columns are equal, we select the topmost row and the extreme left column. (iii) We select xij as a basic variable if ij is the minimum cost in the row or column with largest penalty. We choose the numerical value of xq as high as possible subject to the row and the column constraints. Depending upon whether ai or bj is the smaller of the two, ith row or jth column is eliminated. Example 7.2 (iv) The step (ii) is now performed on the reduced matrix until all the basic variables have been identified. consider the transportation problem presented in the Table 7.4. Table 7.4 Solution Origin 1 2 4 Destination 1 2 4 20 22 17 4 24 7 9 7 2 7 20 15 60 40 0 110 The Table 7.5 shows the computation of penalty for various rows and columns. Table 7.5 : omputation of Penalty for VM ai 1 20 70 50 240 Destination Origin 1 2 4 ai Penalty 1 20 22 17 4 1 20 1 2 24 7 9 7 70 2 2 7 20 15 50 5 bi Penalty 60 40 0 110 4 15 8 240 The highest penalty occurs in the second column. The minimum ij in this column is 12 = 22. Hence, xi2 = 40 and the second column is eliminated. Proceeding in this way, we get xi4 = 80, xu = 0, xa = 0, mi = 10, mi = 50. The transportation cost corresponding to this choice of basic variables is 22x40+4x80+9x0+7x0+24x10+2x50 = 520. Generally, the VM provides a basic feasible solution whose associated cost is quite closer to the minimum transportation cost. SQ 2 Four factories (,,, D) supply the requirements of three warehouses (E, F, G) The availability at the factories, the requirement of the warehouses and the various associated unit transportation costs are presented in Table 7.6. Find an initial basic feasible solution of the transportation problem by using (a) North-west comer rule (b) Matrix minimum method (c) Vogel approximation method
Table 7.6 ÿ or tat ion Problem D Required Warehouses E F G 10 8 9 5 2 6 7 4 7 6 8 25 26 49 vailable 15 20 0 5 100 7. MODIFIED DISTRIUTION (MODI) METHOD The modified distribution method, also known as MODI method or u - v method provides a minimum cost solution to the tiansportation problem. The steps involved in the Modified Distribution Method are as follows : (1) Find out an initial basic feasible solution of the transportation problem using one of the three methods described in the Section 7.2. (2) We introduce dual variable corresponding to the row constraints and the column constraints. If there are m origins and n destinations, then there will be m + n dual variables. The dual variables corresponding to the row constraints are denoted by ui ( i = 1,2,..., m ) while the dual variables corresponding to column constraints are ¬ed by Vj ( j = 1.2,..., n ). The values of the dual variables should be determined from the following equations. () ny basic feasible solution of the transportation problem has (m + n - 1) non-zero xij. Thus, there will be m + n - 1 equations to determine m + n dual variables. One of the dual variables can be chosen arbitrarily. It is also to be noted that as the primal constraints are equations, the dual variables are unrestricted in sign. (4) If xij = 0, the dual variables computed in stop are compared with the ij values of this allocation as ij- Ui - Vj. If all ij - ui - vj 2 0, then by an application of complementary slackness theorem, it can be shown that the corresponding solution of the transportation problem is optimum. If one or more of ij - ui - Vj < 0, we choose the cell with least value of ij - u; - vj and allocate as much as possible subject to the row and the column constraints. The allocation of a number of adjacent cell are adjusted se&at a basic variable becomes non-basic. (5) fresh set of dual variables are computed and entire procedure is repeated. Let us consider the transportation problem of Example 7.1 given in Table 7.7 with a basic feasible solution computed by Matrix Minimum method. Table 7.7 orkin 1 2 - Destination 1 2 4 5 1 9 1 6 51 24 12 16 20 1 vailability 50 100 14 1 2 26 100 70 50 40 40 150 00 (1) Tbe initial basic feasible solution by matrix minimum method is
~ = 2 10, X = 50. X4 = 40. (2) The dual variablgs ui, u2, u and vi, tr, m, v4, vs can be computed from the corresponding ij values () Since one of the dual variables can be chosen arbitrarily, we take.us = 0 is it occurs most often in the equations. The values of the dual variables aie u1=- 1,~2=-21,ug=O,vl=14,v2=,v=1,~4=2,~5=22. (4) We now oompute ij - ui - vj values for all the cells where xij = 0. ll the ij-ui-vj r Oexceptforcell(1,2) where 12-~1 -n = -11. Thus, in the next iterationxi2 will be a basic variable changing one of the present basic variables non-basic. We also observe that for allocating one unit in cell (1,2), we have to reduce one unit in cells (.2) and (1, 1) and increase unit in cell (.1). The net reduction in the transportation cost for each unit of such reallocation is The maximum that can be allocated to cell (1.2) is 10, otherwise the allocation in cell (.2) will be negative. The revised basic feasible solution is It can be verified that the new set of dual variables satisfy the optimality condition. Thus, the minimum cost transportation schedule is xi1 = 40, ~ 1= 2 10, ~ 2= 2 60, xz = 40, ~ 1 = 60. x = 50, ~ = 4 40. The corresponding transportation cost is 2700 which is about 4% less than the transportation cost arrived at by matrix minimum method. SQ ompute the dual variables of the second iteration in the above example and verify that the solution presented is the optimum solution. 7.4 STEPPING STONE METHOD - Stepping Stone Method is another method for finding the optimum solution of the transportation problem. me steps necessary in the stepping stone method are given below : (1) Find an initial basic feasible solution of the transportation problem. (2) Next check for degeneracy. basic feasible solution with m origins and n destinations is said to be degenerate if the number of non-zero basic variables is less than m + n - 1. When a transportation problem is degenerate it has to be properly modified. This has been included in Section 7.6. () Each empty (non-allocated) cell is now examined for a possible decrease in the transportation cost. One unit is allocated to an empty cell. number of adjacent cells are balanced so that the row and the column constraints are not violated. If the net result of such re-adjustment is a decrease in the transportation cost, we include as many units as possible in the selected empty cell and carry out the necessary re-adjustment with other cells.
(4) Step is performed with all the empty cells till no further reduction in the -m transportation cost is possible. If there is another allocation with zero increase or decrease in the transportation cost than the transportation problem has multiple solutions. Example 7. onsider the transportation problem given in Table 7.8 (cost in Rupees). Table 7.8 Depot D E F G 4 6 8 6 5 2 5 9 6 5 400 450 50 500 apacity 700 400 600 1700 (1) We compute an initial basic feasible solution of the problem by North-West orner Rule as presented in Table 7.9. Table 7.9 Requhment Depot D E F G 4(400) 6(00) 8 6 5(150) 2(250) 5 9 6(100) 5(500) 400 450 50 500 apacity 700 400 600 1700 The figures in the parenthesis indicate the allocation in the corresponding cells. (2) The solution is not degenerate as the number of non-zero basic variables is min-1 = 6. () The cell D is empty. The result of allocating one unit along with the necessary adjustment in the adjacent cells is indicated in Table 7.10. Table 7.10 Depot D E F G 4(99) 6(01) 8 6 (+1) 5(149) 2(250) 5 9 6(100) S(500) 400 450 50 500 apacity 700 400 600 1700 The net increase in the transportation cost per unit quantity of reallocation in celldis+6-4-5 = 0. This indicates that every unit allocated to route D will neither increase nor decrease the transportation cost. Thus, such a reallocation is unnecessary. (4) The result of reallocating one unit to cell D is indicated in Table 7.1 1. Table 7.11 Depot D E F G 4(99) 6(01) 8 6 5(149) 2(251) 5 (+1) 9 6(99) 5(500) 400 450 50 500 apacity 700 400 600 1700 The net increase in the transportation cost per unit quantity of reallocation in celldis+6+2-4-5-6 = -4.
Tbus, the new route would be beneficial to the company. The maximum amount that can be allocated jn D is 100 and this will make the current basic variable corresponding to cetl F non basic. Table 7.12 shows the transportation table after the reallocation. Table 7.12 Factoty Depot D E F G 400) 6(400) 8 6 5(50) 2(50) 5 (100) 9 6. s(s00) 400 450 50 500 apacity 700 400 600 1700 'Ihis procedure is repeated with remaining empty cells E, F, F, G and G. Tbe results are summarised in the Table 7.1. Unoccupied ell E M F G G Table 7.1 Incw In coat Per udt of realloclrtlon 9+4-6- = 4 8+5-6-2 = 5 6-2+5-6+4- = 4 6-5+-4 = 0 5-5+6-4+-5 = O Since reallocation in any other unoccupiedcell cannot further decrease the transportation cost, the present allocation is optimum. Tbus, we get, Tbe minimum transportation cost is 4x00+6x400+5x50+2x50+x100+5x500 = 750. The transportation schedule is, however, not unique as there are a number of unoccupied cells with zero increase in transportation cost. SQ 4 Solve the transportation problem given in Table 7.14 by stepping stone method. Table 7.14 1 2 Order Dktrlbutor 1 2 2 1 5 7 4 6 5 15 22 18, Inventory 10 25 20 55 7.5 ' UNLNED TRNSPORTTION PROLEM We solved the various tramportation problems with the assumption that the total supply at the origins is equal to the total requirement at the destinations. If they are unequal the problem is known as an unbalanced transportation problem. If the total supply is more than the total demand, we introduce an additional column which will indicate the surplus supply with transportation cost zero. Likewise, if the total demand is more than the total
supply an additional row is introduced in the table which represents unsatisfied demand with transportation cost zero. The balancing of an unbalanced transportation problem is illustrated in the following example. Example 7.4 Table 7.15 Plant Warehouses WI w2 w 25 17 25 I5 10 18 00 00 500 vailable 00 500 In Table 7.15, the total requirement is 1100 whereas the total supply 800. Thus, we introduce an additional row with transportation cost zero indicating the unsatisfied demand as shown in Table 7.16. Table 7.16 Plant Unsatisfied demand Warehouses w1 WZ w 25 17 25 15 10 18 0 0 0 00 00 500 vahble 00 500 00 1100 Now the problem can be worked out as discussed in previous sections. 7.6 DEGENERTE TRNSPORTTION PROLEM If a basic feasible solution of a transportation problem with m origins and n destinations has fewer than m + n - 1 positive xu (occupied cells) the problem is said to be a degenerate transportation problem. While in the simple computation, degeneracy does not cause any serious difficulty, it can cause computational difficulty in a transportation problem If we apply modified distribution method, then the dual variables ui and vj are obtained from the ii values of the basic variables. If the problem is degenerate, you will be unable to locate one or more ij value which should be equated to corresponding Ui + vi omputational difficulty will also arise while applying stepping stone method to a degenerate transportation problem. It is thus necessary to identify a degenerate transportation problem at the very beginning and take appropriate step to avoid any computational difficulty. The degeneracy in a transportation problem can be identified through the following result (Must&, 1988) : degenerate basic feasible solution in a transportation problem exists if and only If some partial sum of availabilities (row) is equal to a partial sum of requirements (column). s for illustration the transportation problem presented in Section 7.5 is degenerate as az+a = 800 = h+b. Perturbation Technique The degenerate basic solutions of the transportation problem can be avoided if we ensure that no partial sum of ai and bj are the same. We set up. a new problem where
-QTcrhniqlrcs-l This modified problem has been constructed in such a manner that no partial sum of af is equal to a partial sum of bj. fter the problem is solved, we put d = 0 leading to the optimum solution of the original problem. For illustration, consider the transportation problem presented in Section 7.5 (after the introduction of the additional row). Here, m =, n =. 'Ihe perturbed problemhas been presented in Table 7.17.. Table 7.17 Phnt Unsstlsfled demand Warehouses w1 wz w 25 17 25 15 10 18 0 0 0 00 00 500 + d vailable 00+d 500 +d 00 +d ll00+d The problem can be solved using any of the methods described before. The& we take d = 0 to obtain the solution of the original problem. SQ 5 Solve the perturbed problem given in Table 7.17. 7.7 TRNSHIPMENT PROLEM In a transportation problem consignments are always transported from an origin to a destination. There could be a situation where it might be economical to transport items in several stages : First within certain origins and destinations and fmally the reform to the ultimate receipt points. It is not uncommon to maintain dumps for central storage of certain bulk material. Similarly, movement of material involving two different modes of transport -road and railways or between stations connected by broad gauge and meter gauge lines will necessarily require transhipment. Thus, for the purpose of transhipment, the distinction between an origin and destinations is dropped so that from a transportation problem with m origins and n destinations, we obtain a banshipment problem with m + n origins and m + n destinations. The formulation and solution of a transhipment problem is illustrated with the help of the following example. Example 7.5 onsider a transportation problem where the origins are plants and destinations are depots. The unit transportation costs; capacity at the plants and the requirements at the depots are indicated in Table 7.18 Table 7.18 Plant Depot X Y Z 1 15 5 25 150 150 150 150 00 450 When each plant is also considexed a destination and each depot is also considered an origin, there are altogether five origins dnd five destinations. Some additional cost data are also necessary. These are presented in the following Tables.
Table 7.19 Unit Transportation ost from Plant to Plant Tramportation Problem From Plant To To Plant Plant 0 65 1 0 Table 7.20 Unit Transportation ost From Depot to Depot From Plant X Y Z To Depot X Depot Y Depot Z 0 2 1 1 0 65 0 Table 7.21 Unit Transportation ost fkom Depot to Plant Depot X Y z 25 45 Plant 15 55 From Table 7.18, Table 7.19, Table 7.20 and Table 7.21, we obtain the transportation formulation of the transhipment problem. Table 7.22 Transbipment Table buffer stock of 450 which is the total capacity and total requirement in the original transportation problem is added to each row and column of the transhipment problem The resulting transportation problem has m + n = 5 origins and m + n = 5 destinations. On solving the transportation problem presented in Table 7.22. we obtain,xll= 150, x1 = 00; x14 = 150, x21 = 00, x22 = 450, x = 00, x5 = 150, m = 450, x55 = 450. The description of the transhipment problem is given below: (1) Transport x21 = 00 units from plant to plant. This increase the availability at plant to 450 units including the 150 unik originally available from. (2) From plant transport xi = 00 to depot X and X14 = 150 to depot Y. () From 00 units available at depot X transport x5 = 150 units to depot Z. The total tianshipment cost is 1 x 00 + x 150+ 1 x00 + 1 x 150 = 1200 If, however, the consignments are transported from plants, to depots X, Y, Z only according to the transportation Table 7.18, the minimum transportation cost schedule is x1 = 150 x21= 150 x22 = 150 with a minimum cost of 450. Thus, transhipment reduces the cost of cargo movement in this-case.
OpthnizPtlm Techniques-I SQ 6 Solve the transportation problem given in Table 7.22 (Transhipment problem) using the modified distribution method. 7.8 MXIMISTION IN TRNSPORTTION PROLEM There are certain types of transportation problems where the objective function is to be maximised instead of being minimised. These problems can be solved by converting the maximisation problem into a minimisation problem. The formulation and solution of this class of problems are illustrated with the help of the following example. Example 7.6 firm has three factories located in ity, ity and ity and supplies goods to four dealers spread all over the country. The production capacities of these factories are 1000,700 and 900 units per month respectively. The net return per unit product varies for different combinations of dealers and factories which is given in Table 7.2. Table 7.2 ity ity ity Dealer Dealers 1 2 4 6 6 6 4 4 2 4 5 5 6 7 8 900 800 500 ' 400 GP~~Y lo00 700 900 2600 Determine a suitable allocation to miximise the total net return. If xij denotes the number of units to be despatched from the ith city to the jth dealer rij be the corresponding return, then the objective function is Maximise rll xll + rlz xlz + r1 xlg +... + r4 x4 The value of decision variables xij which maximise the objective function are also the values where - rll xli - r12 x12 -... cr114 x4 is minimised. In order the express the objective function in a more convenient form we observe that the per unit return is maximum corresponding to m with a value 8. If we add and subtract 8 x 2600, the midmisation problem will remain unchanged. Hence, the function to be minimised is ut 2600 = xll + x12 +... + m + m. Hence, the function to be minimised is (8-rll )xll+(8-rlz)x1~+...+( 8-r )x+( 8 - f 1x4-2600x8 ~ This is identical to minimise, the objective function ( 8 - rll )xll+ ( 8 - rlz )x12 +... + ( 8 - r )x t ( 8 - r4 )XM Hence, we have a revised transportation problem given in Table 7.24. Table 7.24 Dealers itg 1 2 4 2 2 2 4 4 6 4 2 1 0 900 = bl 800 = 500 = 400 = b4 ai 1000 = a1 700 = 9F = a 2600
s the partial sum a = b- + b4, the problem is degenerate. We consider the corresponding perturbed problem in Table 7.25. Table 7.25 Dealers ity 1 2 4 2 2 2 4 4 6 4. 2 1 0 900 800 500 400 + d ai 1000 + d 700 + d 900 + d 2600 + d The initial basic feasible solution by matrix minimum method is, xi1 900, xi2 = 100 + d, xz = 700 - d, xz = 26, x = 500-26, X- = 400 + d. Using modified distribution method, we obtaih, ui + vi = 2, ui + vz = 2, ~~+~=6,~2+~=4,~+v=1,~+~4=0. Taking ul = 0 arbitrarily, we obtain, ui = 0,142 = 4, u = 1, vl = 2, v2 =, v = 0. On verifying the optimality condition, we observe that 12-u1-v* < o,c2-u-v2 < 0. We allocate n2 = 700 - d and make re-adjustment in some of the other basic variables. The revised values are as follows : xll = 200+d,x12 = 800,x21 = 700-d,x~ = 26,x = 500-d,and x4 = 400 + d. The dual variables must satisfy the conditions, ul+vl = 2,u1+vz = 2,u2+vi = 4,m+v = 4,u+v = 1,u+v4 = 0. Taking ul = 0, arbitrarily u1 = 0, U2 = 2. U = -1, V1 = 2, V2 = 2, V = 2, V4 = 1. The optimality conditions are now satisfied. Taking d = 0, the optimum solution of the problem is xll = 200, x12 = 800, x21 = 700, x = 500, x- = 400. The maximum net return is 6~200+6~800+4~700+7~500+8~400 = 15500. 7.9 SUMMRY lkansportation Problem is a special type of linear programming problem Simplex method is not suitable for the solution of a transportation problem. On the other hand the transportation problem has a special structure which may be utilised to develop efficient computational techniques for its solution. In the most general form, a hmqmtation problem has a number of origins and a number of destinations. certain amount of a particular consignment is available in each origin Likewise, each destination has a certain requirement. The transportation problem indicates the amount of consignment to be transported from various origins to different destinations so that the total transportation cost k minimised without violating the availability constraints and the requirement constraints. The decision variables x~ of a tmqofhtion problem indicate the amount to be transported from the ith origin to the jth destination. Tho subscripts are necessary to describe these decision variables. number of techniques are available for computing an initial basic feasible solution of a transportation problem. These are the North West orner Rule, Math Minimum method and Vogel's pproximation Method (VM). Generally, the Vogel's pproximation Method (VM) provides an initial basic feasible solution whose assbciated cost is quite closer to the minimum transportation cost. Further, optimum solution of a transportation problem can be obtained from Modified Distribution (MODI) Method or Stepping Stone Method.
Optimiz~~*e~q~~-I Sometimes the total available consignment at the origins is different from the total requirement at the destinations. Such a transportation problem is said to be unbalanced. n unbalanced transportation problem can be made balanced where the total available consignment at the origins is equal to the total requirement at the destinations by introducing an additional row or column with zero transportation cost. The basic feasible solutions of a transportation problem with m origins and n destinations should have m + n - 1 positive basic variables. However, if one or more basic variables are zero the solution is said to be degenerate. degenerate transportation problem can be suitably modified by the perturbation method so that the problem can be solved without any difficulty. problem having a structure similar to that of a transportation problem where the objective function is tqbe maxhnised can also be solved by the techniques developed in this unit with the slight modifications. Transportation problem can be generalised into a transhipment pmbiem where shipment is possible from origin to origin or destination as well as from destination to origin or destination. This may result in an economy of transportation in some cases. transhipment problem can be formulated as a transportation problem with an increased number of origin and destinations. 7.10 KEY WORDS The Origin The Destination Unit Transportation ast North West orner Ruie : The origin of a transportation problem is the location from which the shipments are despatched to the destination. : The destinationof a transportation problem is the location to which the shipments are transported from origin. : The unit transportation cost is the cost of transporting one unit of the consignment from an origin to a destination. This has been represented by ij : The north-west comer rule is a method of computing an initial basic feasible solution of a transportation problem where basic variables are selected from the north-west comer, i.e. top left corner. The Matrix Minimum : The matrix minimum method is a method of computing an Method initial basic feasible solution of a transportation problem where the basic variables are chosen according to the unit cost of transportation. Vogel's pproximation: Method (VM) Stepping Stone Method: The Vogel's pproximation Method (VM) is an iterative procedure of computing an initial basic feasible solution of the transportation problem. Stepping Stone Method is a method of computing optimum solution of a transportation problem. Modified Distribution : The Modified Distribution Method also known as u - v (MODI) Method method is a method of computing optimum solution of a transportation problem. Unbalanced Transportation Problem Degenerate Transportation Problem Perturbation Technique Teanshipment Problem : The unbalanced transportation problem is a transportation problem where the total availability at the origin is different from the total requirement at the destinations. : degenerate transportation problem with m origins and n destinations has a basic feasible solution with fewer than m + n - 1 positive basic variables. : The perturbation technique is a method of modifying a degenerate transportation problem s-o that the degeneracy can be resolved. : The transhipmept problem is a transportation problem where shipment is possible from an origin to an origin or a destination as well as from a destination to an origin or a destination.
7.11 NSWERSTOSOs Transportation Problem SQ 1 SQ 2 SQ (1) minimise (2) equations () mn, m + n. North West omer Rule xi i = 15, x21= 10, n2 = 10, x2 = 16, m = 14, m = 5. Minimum ost = 668. Matrix Minimum Method xi = 15, x22 = 20, x = 0; mi = 25, ~ 4 2 Minimum ost = 58. = 6, ~4 Vogel pproximation Method xi = 15, x22 = 20, x = 0, mi = 25, m2 = 6, m = 4. Minimum ost = 58. = 4. SQ 4 ll ij - ui - vj 2 0. The solution is optimum. SQ 5 = 10, x22 = 22, X =, mi = 5, m + 15 f Minimum Transportation ost is 17. xi2 = 00, ni = 00, x2 = 200, m= 00 Minimum Transportation ost is 1200.