Optimization ethods: Linear Programming Applications Transportation Problem odule 4 Lecture Notes Transportation Problem Introduction In the previous lectures, we discussed about the standard form of a LP and the commonly used methods of solving LPP. A key problem in many projects is the allocation of scarce resources among various activities. Transportation problem refers to a planning model that allocates resources, machines, materials, capital etc. in a best possible way so that the costs are minimized or profits are maimized. In this lecture, the common structure of a transportation problem TP and its solution using LP are discussed followed by a numerical eample. Structure of the Problem The classic transportation problem is concerned with the distribution of any commodity resource from any group of 'sources' to any group of destinations or 'sinks'. While solving this problem using LP, the amount of resources from source to sink will be the decision variables. The criterion for selecting the optimal values of the decision variables like minimization of costs or maimization of profits will be the objective function. And the limitation of resource availability from sources will constitute the constraint set. Consider a general transportation problem consisting of m origins sources O, O,, O m and n destinations sinks D, D,, D n. Let the amount of commodity available in i th source be a i i=,,.m and the demand in j th sink be b j j=,,.n. Let the cost of transportation of unit amount of material from i to j be c ij. Let the amount of commodity supplied from i to j be denoted as ij. Thus, the cost of transporting ij units of commodity from i to j is c ij ij. Now the objective of minimizing the total cost of transportation can be given as m n inimize = c ij ij i= j= f 4L
Optimization ethods: Linear Programming Applications Transportation Problem Generally, in transportation problems, the amount of commodity available in a particular source should be equal to the amount of commodity supplied from that source. Thus, the constraint can be epressed as n ij = a i, i=,,, m j= Also, the total amount supplied to a particular sink should be equal to the corresponding demand. Hence, m ij = b j, j =,,, n i= The set of constraints given by eqns and are consistent only if total supply and total demand are equal. m n a i = b j i= j= 4 But in real problems this condition may not be satisfied. Then, the problem is said to be unbalanced. However, the problem can be modified by adding a fictitious dummy source or destination which will provide surplus supply or demand respectively. The transportation costs from this dummy source to all destinations will be zero. Likewise, the transportation costs from all sources to a dummy destination will be zero. Thus, this restriction causes one of the constraints to be redundant. Thus the above problem have m n decision variables and m n - equality constraints. The non-negativity constraints can be epressed as ij, i=,,, m, j =,,, n 5 This problem formulation is elucidated through an eample given below. 4L
Optimization ethods: Linear Programming Applications Transportation Problem Eamples Problem Consider a transport company which has to supply 4 units of paper materials from each of the cities Faizabad and Lucknow to three cities. The material is to be supplied to Delhi, Ghaziabad and Bhopal with demands of four, one and three units respectively. Cost of transportation per unit of supply c ij is indicated below in the figure. Decide the pattern of transportation that minimizes the cost. Solution: Let the amount of material supplied from source i to sink j be ij. Here m =; n =. Total supply = 8 units and total demand = 4 = 8 units. Since both are equal, the problem is balanced. The objective function is to minimize the total cost of transportation from all combinations i.e. inimize m n f = i= j= c ij ij 4L
Optimization ethods: Linear Programming Applications Transportation Problem 4 inimize f = 5 8 4 7 6 subject to the constraints as eplained below: The total amount of material supplied from each source city should be equal to 4. ij = 4 i=, j= i.e. = 4 for i = 7 = 4 for i = 8 The total amount of material received by each destination city should be equal to the corresponding demand. i= = j =,, ij b j, i.e. = 4 for j = 9 = for j = = for j = Non negativity constraints ij i =, ; j=,, Thus, the optimization problem has 6 decision variables and 5 constraints. Since the optimization model consists of equality constraints, Big method is used to solve. The steps are shown below. 4L
Optimization ethods: Linear Programming Applications Transportation Problem 5 Since there are five equality constraints, introduce five artificial variables,,, 4 and 5. Thus, the objective function and the constraints can be epressed as inimize 5 4 7 4 8 5 f = subject to = 4 = 4 = 4 4 = 5 = odifying the objective function to make the coefficients of the artificial variable equal to zero, the final form objective function is 5 4 7 4 8 5 f The solution of the model using simple method is shown 4L
Optimization ethods: Linear Programming Applications Transportation Problem 6 Table First iteration Basic variables Variables 4 5 HS atio Z -5 - -8-4 - -7 6 4-4 4 4-4 5 - Table Second iteration Basic variables Variables 4 5 HS atio Z -5 - -8-4 -7-4 - 4 - - - 4 4 X - 5-4L
Optimization ethods: Linear Programming Applications Transportation Problem 7 Table Third iteration Basic variables Variables 4 5 HS atio Z -5-5 -8-7 4- - 8-4 4 X - - - - X - 5 - Table 4 Fourth iteration Basic variables Variables 4 5 HS atio Z -8-7 - 5- - 86 - - - - X - - - X - - X 5 epeating the same procedure, we get the final optimal solution f = 4 and the optimum decision variable values as : =.4, =., =.757, =.757, =., =.4. 4L
Optimization ethods: Linear Programming Applications Transportation Problem 8 Problem Consider three factories F located in three different cities, producing a particular chemical. The chemical is to be transported to four different warehouses Wh, from where it is supplied to the customers. The transportation cost per truck load from each factory to each warehouse is determined and are given in the table below. Production and demands are also given in the table below. Wh Wh Wh Wh4 Production F 5 68 458 85 6 F 4 4 6 79 F 67 558 895 695 5 Demand 65 85 8 7 Solution: Let the amount of chemical to be transported from factory i to warehouse j be ij. Total supply = 65 = and total demand = 658587 =. Since the total demand is less than total supply, add one fictitious ware house, Wh5 with a demand of. Thus, here m =; n = 5 Wh Wh Wh Wh4 Wh5 Production F 5 68 458 85 6 F 4 4 6 79 F 67 558 895 695 5 Demand 65 85 8 7 The objective function is to minimize the total cost of transportation from all combinations. inimize f = 5 68 458 85 4 5 4 4 6 79 4 5 67 558 895 695 4 5 subject to the constraints 4L
Optimization ethods: Linear Programming Applications Transportation Problem 9 4 5 = 6 for i = 4 5 = for i = 4 5 = 5 for i = = 65 for j = = 85 for j = =9 for j = 4 4 4 =8 for j = 4 5 5 5 = for j = 5 ij i =,,; j=,,,4 This optimization problem can be solved using the same procedure used for the previous problem. 4L