Operations Research. Unbalanced transportation problem.

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1 Operations Research.

2 and properties of solutions In the previous lesson, the two special types of solutions of transportation problems (degenerate and alternative) mentioned. However, since both of these types of solutions have their own characteristics, deserves more attention.

3 Degenerate solution of transportation problem Basic solution of transportation problem is degenerate, if one of the basic variables of this solution is zero. Because m + n 1 basic variables corresponds exactly m + n 1 occupied boxes, it means that the degenerate solution find if some occupied box is occupied by zero. On the other hand, if in the table all occupied boxes contained a positive values, this is a non-degenerate solution. Test optima of a degenerate solution perform exactly the same way as in the case of non-degenerate solution. Similarly, the process of improving of a degenerate solution is no different from the process to improve the non-degenerate solutions.

4 Example 1. Determine the optimal solution of transportation problem for 3 suppliers and 4 consumers, the capacity of suppliers, consumer demands and the transport rates gives Table 1. c ij S 1 S 2 S 3 S 4 C D D D D Table 1: Entering of transportation problem from example 1.

5 Solution: Table 2 represents the initial basic solution of transportation problem. It was obtained by VAM method. S 1 S 2 S 3 S 4 C i u i D , D , D , D j 9,8 1,12 2,2 1, v j Table 2: Finding of the initial solution and test optima, example 1

6 Basic variable x 12 = 0, so the solution is degenerate. S 1 S 2 S 3 S 4 C u i D D D D v j Table 3: Optimal solution, example 1

7 Calculate the row numbers u i, i = 1,..., 3 and column numbers v j, j = 1,..., 4. Subsequently, in cross-boxes, we calculate indirect rates c ij. Test optima is negative because in the box D 3S 2 is c ij > c ij (9 > 7). This box is uniquely associated with a closed circuit, as shown in Table 2. One of -boxes of closed circuit is assigned values of zero (x 12 = 0), and hence improve the solution will not mean a reduction in total transportation cost. Only in the set of basic variables instead of x 12 appears variable x 32. The new solution, which displays tab. 3, is the only optimal solution of this problem and it is degenerate. As solution appropriate table 2 (ie initial), and solution from table 3 has the form 0 X Total transportation cost is z = CZK. 1 A.

8 If we find that the solution of the transportation problem is optimal (ie, for all cross-boxes paying c ij c ij) and at least for one this box paysc ij = c ij, there is another solution with the same value of the objective function, called alternative solution. Finding the alternative solution is similar to finding optimal solution. Box, where pay c ij = c ij, is the key box. We construct a closed circuit to this box by known way, we compute a new table.

9 Example 9. Determine all basic optimal solution of the transportation problem for 3 suppliers and 4 consumers. Capacity of suppliers, consumer demands and transportation rates shows Table 4. c ij S 1 S 2 S 3 S 4 C D D D D Table 4: Entering of transportation problem, example 9

10 Solution: At first find the initial basic feasible solution (using by VAM), see table 5. This solution is optimal, because for all cross-boxes paying c ij c ij. But there is cross-box (box D 2S 1), where this inequality satisfied as equality (15 = 15). Well, there is another basic solution with the same value of the objective function. The procedure of finding this alternative solution is the same as the procedure to improve the solution. For box D 2S 1 create a closed circuit, denote the vertices of the circuit by symbols + or. Minimum of -boxes is min{100, 500} = 100.

11 S 1 S 2 S 3 S 4 C i u i D ,7,1, D ,7,7, D , D j 11,1,1,1 5,7,7, 0,1,1,1 4,13, v j Table 5: Finding of the initial solution and test optima, example 9

12 S 1 S 2 S 3 S 4 K u i D D D P v j Table 6: Alternativ solution, example 9

13 A well-known way to create a new solution, see Table 6. It is necessary that this solution is again optimal and that there is also box D 1S 1, where the indirect rate equal to the rate direct. If we create a closed circuit for this box and accordingly created new solution, we would get exactly the table 5, ie. we would return to the first basic optimal solution of the transportation problem. Overall, we can noted that the solutions and 0 X 1 0 X are only two basic optimal solution of the problem. Total transportation costs are of course in both cases the same and equal to the value z = CZK. 1 A 1 A

14 Transportation problem can have a single optimal solution. In this case is certainly that the solution is basic. If the problem has alternative solutions, ie more (at least two) basic optimal solutions, then the problem may have nonbasic optimal solution. Every feasible solution, which is a convex linear combination of basic optimal solutions, is also an optimal solution. Practically we can obtaine nonbasic solutions by conversion table, where we will not perform with a minimum of -boxes, but the value that we add to the boxes labeled + and subtracted from the boxes labeled must «be between zero and this minimum ie from the interval 0, min. -boxes {xij} Whether the number of optimal non-basic solutions will be finitely many or infinitely many depends on whether the problem is an integer (x ij N 0, transported goods is counted to pieces, is an indivisible), or whether variables can be any real value from a interval (x ij R + 0, transported goods is the nature of the fabric is freely divisible).

15 Consider therefore unbalanced transportation problem, ie let mx a i i=1 nx b j. j=1, we will always solve by transferring to balance problem. This is achieved by introducing a fictional factor supplier or customer. We differentiate two types of unbalanced transportation problems, depending on whether it exceeds the total capacity of goods by the suppliers of the total requirements of all consumers, and vice versa. In accordance with this we will lead next interpretation.

16 In case, that P a) m P a i > n b j, i=1 j=1 it is necessary to introduce a fictitious consumer S f with P demand b f = m P a i n b j. i=1 j=1 This give a transportation problem, which has against the original one consumer more, but it is the balanced problem. Transportation costs along the way D is f are zero, ie P b) m P a i < n i=1 b j j=1 c if = 0, i = 1, 2,..., m. it is necessary to introduce a fictitious supplier D f with a P capacity a f = n P b j m a i. j=1 i=1 This give a transportation problem, which has against the original one supplier more, but it is the balanced problem. Transportation costs along the way D f S j are zero, ie c fj = 0, j = 1, 2,..., n.

17 Example 17. Determine the optimal solution of transportation problem for 3 suppliers and 5 consumers, the capacity of suppliers, consumer demands and transportation rates shows table 7. c ij S 1 S 2 S 3 S 4 S 5 K D D D P Table 7: Entering of transportation problem, example 17

18 Solution: At first find out whether it is a balanced transportation problem. c ij S 1 S 2 S 3 S 4 S 5 K D D D D f P Table 8: Entering already balanced transportation problem, example 17

19 P The sum of capacities of all suppliers is 3 a i = and the sum of demands P of all consumers is 5 b j = Thus, the problem is unbalanced, type of j=1 b), so we introduce a fictitious supplier D f with capacity a f = = 500. Transportation rates from this supplier to all consumers have zero value. Thus transformed transportation problem is already balanced, see tab. 8 Finding the initial basic solution of the problem by VAM is done in Table 9. Then, still on the table 9, was carried test optima of this solution. Founding basic solutions is non-degenerate, because all m + n 1 = 8 basic variables (occupied boxes) takes positive values. It represents the transport of goods with a the total transportation cost i=1 z 1 = = CZK.

20 S 1 S 2 S 3 S 4 S 5 C i u i D ,1, D ,1,1,1,1, D ,3,3,3,3, D f , D j 2,5,4,4, 3,2,2,2,2, 1,8,1,1,1,1 4,2,2,2,2,2 7,5,5, v j Table 9: Finding of initial solution and test optima, example 17 9

21 The solution is not optimal because for cross-box D f S 3 indirect rate is greater than direct (1 > 0). Minimum of -boxes, ie the value of min{200, 1500, 500} = 200, subtract of -boxes and added to + -boxes. The obtained improved solutions is shown in Table 10.

22 S 1 S 2 S 3 S 4 S 5 C u i D D D D f D v j Table 10: Improved solutions and test optima, example 17

23 Solution of Table 10 is already optimal. The solutions shows that the consumer demands S 3 and S 5 not fully satisfied. Total transportation cost is z 2 = = CZK.

24 Operations Research.

Study support. Course Name: Operations Research. Supervisor: doc. RNDr. Jiří Moučka, Ph.D. Author: RNDr. Michal Šmerek, Ph.D.

Study support. Course Name: Operations Research. Supervisor: doc. RNDr. Jiří Moučka, Ph.D. Author: RNDr. Michal Šmerek, Ph.D. Study support Course Name: Operations Research Supervisor: doc. RNDr. Jiří Moučka, Ph.D. Author: RNDr. Michal Šmerek, Ph.D. 2015, University of Defence 1 Contents 1 Introduction 3 1.1 Phases of solution