TIME MINIMIZATION IN TRANSPORTATION PROBLEMS

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75 M O P vozmme 6 number J Jan^arz/ TIME MINIMIZATION IN TRANSPORTATION PROBLEMS J,K. SHARMA D.N. POST-GRADUATE COLLEGE, MEERUT. KANT I SWARUP INDIAN INSTITUTE OF PUBLIC ADMINISTRATION, NEW DELHI.

.Section if e x t e n d s t&is m e t ^ o d o Z o y y to m u l t i d i m e n s i o n a l t r a n s p o r t a t i o n p r o b Z e m s Tzauinp t^iree i ndices.

1 75 M O P vozmme 6 number J Jan^arz/ TIME MINIMIZATION IN TRANSPORTATION PROBLEMS J,K. SHARMA D.N. POST-GRADUATE COLLEGE, MEERUT. KANT I SWARUP INDIAN INSTITUTE OF PUBLIC ADMINISTRATION, NEW DELHI. Summary TTzis p a p e r c o n s i s t s o / t^o s e c t i o n s. /! t e c h n i q u e /or m i n i m i z i n g time in a t r a n s p o r t a t i o n p r o b Z e m is d e v e l o p e d in S e c t i o n J..Section if e x t e n d s t&is m e t ^ o d o Z o y y to m u l t i d i m e n s i o n a l t r a n s p o r t a t i o n p r o b Z e m s Tzauinp t^iree i ndices. T^e p r o c e d u r e s d e r e Z o p e d are / i n i t e a n d are b a s e d on m o v i n g / r o m one b a s i c s o Z x t i c n to a n o t h e r zzntiz o p t i m a Z i t ^ is r e ached. NzvmericaZ e x a m p Z e s izzmstratin<y t?ze t e c ^ n i q ^ e s are incz^ded. 1. Introduction The cost-minimizing transportation problem, popularly known as Hitchock- Koopmans problem, may be expressed as m n minimize Z = Y Y C.. x.- i = l j = l ^ ^ n (1) subject to x.. = a., i = 1,2,...,m j = l ^ ' m Y x..=b., j = l,2... n iii u J' ' For a feasible solution to exist it is necessary that total supply equals total demand. If in reality supply is greater than demand, a fictitious destination may be used to create the desired equality. If demand exceeds supply, a Manuscript received Oct. 26, 1976; revised May 1977.

76 fictitious source is introduced. Here the objective function is linear and convex. Algorithms, such as Dantzig's simplex method [21, provide efficient means of locating the optimal solution.

war time, fire services, ambulance services, etc.. In a time minimizing transportation problem, a time matrix {t..} is given where t.

2 76 fictitious source is introduced. Here the objective function is linear and convex. Algorithms, such as Dantzig's simplex method [21, provide efficient means of locating the optimal solution. But there may arise situations in which minimizing the time of tr a n s portation is of greater importance than cost, for example in the transportation of perishable goods, military equipment during war time, fire services, ambulance services, etc.. In a time minimizing transportation problem, a time matrix {t..} is given where t.. is the time of transporting ^ j th th ^ ^ goods from i origin to j destination. For any feasible solution X = (Xjj) satisfying (1), the time of transportation is the maximum of t^j's among the cells in which there are positive allocations, i.e. the time of transportation is [MAX t.. : x. - > 0]. The aim is to minimize this quantity, (i,j) ^ ^ i.e., find x^j, i = 1,2,...,m; j = 1,2,...,n that minimize [MAX t.. : x -. > 01 (PROBLEM I) subject to constraints (1). The time of transportation is independent of the amount of commodity shipped from suppliers to consumers while the cost of transportation is dependent on the variation in quantity of commodities shipped. Methods of minimizing transportation time for twodimensional problems have been given in [3,4,61. In Section 2 we present a new approach. The multidimensional transportation problem with respect to cost is studied by Corban [1]. Mathematically, the cost minimizing three-dimensional transportation problem is minimize subject to

origin i to destination j by means of t r a n s portation mode k; x^j^ is the amount of product transferred from origin i to destination j using transportation mode k; a.

3 77 (2 ) n 1 j = l m P 1 k=l X.-i ijk *ijk i = l m n 1 k = l j = l *ijk w m 1 i = l ^i ijk n = 1 j=l ^i' 'j' 'k > ijk One economic interpretation of this problem is the following: is the cost of transporting a unit of product from origin i to destination j by means of t r a n s portation mode k; x^j^ is the amount of product transferred from origin i to destination j using transportation mode k; a. is the supply available at the i ^ origin; b. is the th ^ amount required by j destination, and c^ is the total amount that can be transferred using the k ^ means of t r a n sportation. The three- dimensional time- minimizing transportation problem can be stated as minimize [Max t. : x. > 01 (PROBLEM II) (i,j,k) '3* ^ subject to (2), where is the time of transportation of goods from i ^ origin to j ^ destination using t r a n s p o r t a t ion mode k. It is further noted that this time of t r a n s portation is independent of the amount of goods shipped as long as x ^ ^ > 0, where it is assumed that all carrier start simultaneously. The three-dimensional time-minimizing transportation problem has been studied by Sharma 5 Swarup [7]. method presented here is a direct extension of the approach proposed for the two- dimensional case. The

78 2. The Two-dimensional Time Minimizing Transportation Problem The two-dimensional time minimizing transportation p r o b l e m was defined by PROBLEM I. PROBLEM I is not convex.

$fitj tmpravgj / e a s t M e sozmtton.- Let X. = tx^.l and 2 ^ i J X 2 = tx^jl be two feasible solutions of PROBLEM I, and let M^ = ((i,j) : x^. > 0} ; M^ = {(i,j) : x?j > 0} T^ = Max t.$

4 78 2. The Two-dimensional Time Minimizing Transportation Problem The two-dimensional time minimizing transportation p r o b l e m was defined by PROBLEM I. PROBLEM I is not convex. The objective function of We need the following definitions. CtJ 4 / e a s i M e aozmtton.- A set X = (x^j) of non-negative numbers satisfying (1) is called a feasible solution. fitj tmpravgj / e a s t M e sozmtton.- Let X. = tx^.l and 2 ^ i J X 2 = tx^jl be two feasible solutions of PROBLEM I, and let M^ = ((i,j) : x^. > 0} ; M^ = {(i,j) : x?j > 0} T^ = Max t.. : ; T- = Max t.. (i,j) e M ^ ^ ^ (i,j) e M ^ ^ = T^, (i,j) e M^} R 2 = H i J ) : = T^, (i,j) e M^} Pi = y x^. ; p. = y x?. (i,j) c R ^ ^ ^ 11 A solution X 2 is said to be better than solution X^ either when T^ < T^ or when 1*2 = T., Y x.. < Y ^ii' i*c*' P 2 ^ ^ (i,jl e R^ ^ (i,j) e R ^ 3 OptimaZ ^az^tion; A feasible solution for which CMax t.. : x.. >01 is minimal while no better feasible (i,j) ^ ^ solutions exist is said to be optimal. THEOREM 2.- If there is a feasible solution to a set of equations AX = b, X > 0, then there is a basic feasible solution. Hadley f5, p. 301 proves this theorem by reducing the number of positive variables in a given feasible solution one by one until the columns of A associated with positive variables are linearly independent. It may be noted that in

79 the proof of this theorem, the set of positive variables in the basic feasible solution is a subset of the positive variables in the given feasible solutions.

for the proof of this theorem see Hammer [4] and Swarc [ 6 ].

5 79 the proof of this theorem, the set of positive variables in the basic feasible solution is a subset of the positive variables in the given feasible solutions. The values of the variables in the two sets, however, may be different. TNEJ/frW A locally optimal solution to problem (I) is also a globally optimal solution. for the proof of this theorem see Hammer [4] and Swarc [ 6 ]. On the basis of theorems 1 and 2, we develop the following three-step new algorithm: Step 1: Determine an initial basic feasible solution by any available method for the cost m i n i m i z ing transportation problem. Step 2: 1-ind an adjacent better basic feasible s o l u t i o n. Step 3: Perform step 2 until no better adjacent basic feasible solution can be obtained. Step 2 deals with the determination of a cell (i,j) not in the basis which if introduced into the basis will reduce the time of transportation or reduce the allocation x ^ for at least one (h,k) er. This is done in such a way that the allocation is reeuced in that cell belonging to R in which the allocation is maximum. Hence step 2 consists of the following substeps: 2.1 Find the greatest x ^, (h,k) cr. 2.2 Identify the set S ^ of all cells (i,j) not in the basis, such that if one of these cells is i n t r o duced into the basis it will eliminate the cell (h,k) or reduce the amount x ^. 2.3 Choose among the elements of S ^ the one, say, (p,q) for which t is minimum. pq 2.4 Introduce the cell (p,q) into the basis.

Cells eligible for entry into the basis are those for which d < 0 which rs implies u^+v^ = 1 and C ^ = 0. Thus = ((r,s) : (r,s)^:m, u +v r s 1, C ^ = 0, t. / < T). If 4), the process terminates.

6 80 m a t r i x. The set is determined as follows: Define a cost 'ij if if > T < T Using C = fc..1 as the cost matrix associated with the. 1 initial feasible solution = (x^j}, determine u^ for rows and Vj for columns by the (u-v) method f21 and d^.g = ( C ^ - (u^.+vg) : (r,s) ^ M). Cells eligible for entry into the basis are those for which d < 0 which rs implies u^+v^ = 1 and C ^ = 0. Thus = ((r,s) : (r,s)^:m, u +v r s 1, C ^ = 0, t. / < T). If 4), the process terminates. If <(), go to step 2.3. Note that there is no loss of generality by assuming that the problem is non-degenerate, since degeneracy can be handled in the same manner as in the ordinary transportation problem. Numerical E x a m p l e : Let there be four producers supplying 37, 22, 32 and 14 units, respectively, with six consumers demanding 15, 20, 15, 25, 20 and 10 units, respectively, with the following transportation time, t.., i = 1,2,...,4, j = 1,2,...,6 : b. J The North-west Corner rule gives the following initial feasible solution X ^ :

$= Max t.. = 45, R. = {(3,5)}, hence * (i,j) c M i ^ ^ (h,k) = (3,5) Define Cost Matrix tc..l as follows H fl if t. - > 45 C.$ $= { ^ " (0 if t-j < 45 We deduce that = {(1,6),(2,5),(2,6),(3,6)}.$

7 , Then Ml = {(i,j) : x > 0) = {(1,1),(1,2),(1,3),(2,3),(2,4),(3,4),(3,5),(4,5), (4,6)) Ti = Max t.. = 45, R. = {(3,5)}, hence * (i,j) c M i ^ ^ (h,k) = (3,5) Define Cost Matrix tc..l as follows H fl if t. - > 45 C. = { ^ " (0 if t-j < 45 We deduce that = {(1,6),(2,5),(2,6),(3,6)}. The 1 smallest t^j with (i,j) being t 2^ = 20, we bring this cell into the basis; this reduces the allocation of (3,5) = (h,k). Thus the new solution X-, is given by u. l

82 Proceeding in the same manner we get an optimal solution (with time T = 40) given below 15 13 9 0-6 6 10 0 7 25 0 14-1 v. J 0 0 0 0 1 0 3.

We use the same definitions and notation as in Section 2 (for details see reference [7]). THEOREM 3 ; A locally optimal solution to PROBLEM II is also globally optimal.

8 82 Proceeding in the same manner we get an optimal solution (with time T = 40) given below v. J The Three-Dimensional Time-Minimizing Transportation Problem This problem was mathematically defined as PROBLEM II. Again, the objective function is not convex. We use the same definitions and notation as in Section 2 (for details see reference [7]). THEOREM 3 ; A locally optimal solution to PROBLEM II is also globally optimal. Proof; Consider a locally optimal solution X^ = to PROBLEM II. Assume there exists a better solution 2 ^2 ' ^ i j k ^ * Consider a maximization problem defined by constraints (2) and the matrix [C...1 as follows: ^ ^ ljk Cijk 1 if t... < T. ljk 1 0 if t. - i > T, l jk 1 C^ = return of solution X^ I C. x?- = ^ x^ ( i, j, k ) E M ^ J * ^ ( i, j, k ) e ( M ^ R ^ ^ * = K - p.

$Therefore, there exists an adjacent basic feasible solution X^ which is better than X^ such that > C^, where is the return of solution X^. Hence, T^ jl T\. Let T^ > T^.$

9 83 m n p where K = ^ a- = [ b. = c, i = l j=l ^ k=l ^ C 7 = return of solution X-, = Y C. - x^.. ^ (i,j,k) c Since X 2 is a better solution than X^, then either T^ < T^ or T^ = T^, p-, < p^. When T^ < T^, = 1 for occupied cells of X^. Therefore C-, = Y C..- xt.. = Y x^., = K. (i,j,k) (i,j,k) Since K > K-p^ it follows that When T-, = P 7 < P^^ then > C^. C ^ = Y C..- x... = Y x.^.. 2 (i,j,ib E - K - p^. Since K-p^ > K-p^, it follows that > C^. Thus, X^ cannot be the optimal solution to the maximizing problem. Therefore, there exists an adjacent basic feasible solution X^ which is better than X^ such that > C^, where is the return of solution X^. Hence, T^ jl T\. Let T^ > T^. Now there may exist a possibility in which X^ contains all the cells of X^ with time T^ and the allocations are the same as in X^. Then (\ = I C. x*? ^ r ^ - I T K Ijk Ijk (1,j,k) C v 3 v 1 x...=k-p--p-. (i,j,k) e (M^-R^ U R 3) ^ ^ As = K'P^ and K-p^ > K-p^-p^, it follows that < C^. This contradicts the fact that X^ is better than X^. Therefore T^ ^ T^, hence T^ T^.

Hence a locally optimal solution is also globally optimal. The following solution procedure will converge to the optimal solution to PROBLEM II in a finite number of i t e r a t i o n s : Let X = tx^.

10 84 When Tg = T^, C? = I C... x^.. = I x?.. = K P? Hence C- > C. since K-p^ > K-p- and p^. < p-. Thus either T 3 ' T i r T ^, p ^. This implies that X^ is a better basic feasible solution with respect to time, and that X^ is not locally optimal. This contradicts the hypothesis. Hence a locally optimal solution is also globally optimal. The following solution procedure will converge to the optimal solution to PROBLEM II in a finite number of i t e r a t i o n s : Let X = tx^.^1 be any basic feasible solution to PROBLEM II yielding time T of transportation which can be obtained in the same way as in [7]. Let M = t(i,j,k) : * ijk ^ ^ ^ ^ em}. Define a return matrix tc^.^1 as in Theorem 3, i.e. fl if t.., < T C- - = ^ ^0 if t... > T ijk - For this return maximizing problem, determine dual variables u-, v., w, [11. Since the return is to be 1 j K maximized, the cells eligible for entry into the basis are those for which d ghk > 0, where ^ghk = ^ g h k " ^ g ^ ^ h ^ k ^ ' ^ u "**v^+w^ = 1 for (g,h,k) e (M-R), all Vg, u^, w^ are either 0 or 1. For cells (g,h,k) with t g ^ T' ^ghk ' ^ therefore d g ^ will never be positive. For cells (g,h,k) with tg^^ < T, C g ^ = 1, and therefore d g ^ is positive only if Ug+v^+w^. = 0. Ug+v^+w^ = 0, C g ^ = 1}. Thus S g ^ = {(g,h,k) : (g,h,k) ^M, The procedure is repeated until

Note again that degeneracy can be Let us consider the following example with four origins, having a^ = 24, a 2 = 8, a^ = 18, a^ = 10; four

11 85 S = 4*- ghk The procedure is bound to converge for it involves movement from one basic feasible solution to another, which are finite in number. handled as explained by Corban [1]. Note again that degeneracy can be Let us consider the following example with four origins, having a^ = 24, a 2 = 8, a^ = 18, a^ = 10; four independent destinations, having b^ = 11, b^ = 19, b^ = 21, b^ = 9; and three means of transport with c^ = 17, c^ = 31, Cg = 12, as depicted in Figure 1. Figure 1 : Three-dimensional time-minimizing transportation problem. is shown in Figure 2.

$86 Figure 2 : Initial solution 11 19 21 9 Here T^ = 15, R^ = {(2,3,2)}. Hence (g,h,k) = (2,3,2), = 2. Define a return matrix as = (1 if < 15; 0 if >_ 15}.$ Therefore the cell with minimum time of transportation belonging to S 232 ^ (1,3,2), and is introduced into the basis. obtain a new basic feasible solution X 2 shown in Figure 3.

Therefore the cell with minimum time of transportation belonging to S 232 ^ (1,3,2), and is introduced into the basis. obtain a new basic feasible solution X 2 shown in Figure 3.

12 86 Figure 2 : Initial solution Here T^ = 15, R^ = {(2,3,2)}. Hence (g,h,k) = (2,3,2), = 2. Define a return matrix as = (1 if < 15; 0 if >_ 15}. We deduce that $ 2^2 ' {(1,3,1),(1,3,2), (1,3,3),(1,4,1),(1,4,2),(1,4,3),(2,3,3),(2,4,3)}. Therefore the cell with minimum time of transportation belonging to S 232 ^ (1,3,2), and is introduced into the basis. obtain a new basic feasible solution X 2 shown in Figure 3. Proceeding in the same manner we get an optimal solution (with time T = 10), as shown in Figure 4. We Acknowledgement We are thankful to Mr A.K. Gupta, D.N. College, Meerut, for his co-operation in this research and to the Council of Scientific and Industrial Research for its financial assistance.

87 Figure 3 : New solution LI 19 21 9 RF.l FRFNCHS [1] Corban, A., "A Multidimensional Transportation Problem", Fev.

, Linear Programming and Extensions^ Princeton University Press, 1963. [31 Garfinkel, R.

13 87 Figure 3 : New solution LI RF.l FRFNCHS [1] Corban, A., "A Multidimensional Transportation Problem", Fev. Fo^m. Math. P^res et /IppZ.^ 9, 8 (1964). [2] Dantzig, C. B., Linear Programming and Extensions^ Princeton University Press, [31 Garfinkel, R.S. and M.R. Rao, "The Bottleneck T r a n s portation Problem", JVaf. Fes. Log. $Kart., 18, (1971). [41 Hammer, P.L., "Time Minimizing Transportation Problems" Afav. Fes. Log. $^art.^ 16, (1969).

88 Figure 4 : Optimal solution 11 13 21 9 [5] Hadley, G., Linear Programming, Addison- Wesley, 1962.

18, 473-484 (1971). [7] Sharma, J.K. and K.

14 88 Figure 4 : Optimal solution [5] Hadley, G., Linear Programming, Addison- Wesley, [6 ] Swarc, W., "Some Remarks on the time Transportation Problem", JVau. Fes. Zoy. <$Mart. 18, (1971). [7] Sharma, J.K. and K. Swarup, "The Time Minimizing Multidimensional Transportation Problem", Presented at the 9th annual conference of ORSI, December 1976, Calcutta, and Journal o/ Engineering Production, Vol. 1, 1977.

THREE-DIMENS IONAL TRANSPORTATI ON PROBLEM WITH CAPACITY RESTRICTION *

NZOR volume 9 number 1 January 1981 THREE-DIMENS IONAL TRANSPORTATI ON PROBLEM WITH CAPACITY RESTRICTION * S. MISRA AND C. DAS DEPARTMENT OF MATHEMATICS, REGIONAL ENGINEERING COLLEGE, ROURKELA - 769008,