LEXI-SEARCH ALGORITHM USING PATTERN RECOGNITION TECHNIQUE FOR SOLVING A VARIANT CONSTRAINED BULK TRANSSHIPMENT PROBLEM

In this article we present a variant transportation problem called “A Variant Constrained Bulk Transshipment Problem”. The purpose of this paper is to propose an efficient Lexi-Search Algorithm using pattern recognition technique for solving “A Variant Constrained Bulk Transshipment Problem” on a scalable multicomputer platform and to obtain an optimal solution. Our results show that the proposed algorithm is highly competitive on a set of benchmark problems. The objective of the problem is to minimize the total bulk cost of supplying the required products to the destinations with the restriction that any destination should get its requirement from one source only, even when it gets from a destination. In the sequel we developed a Lexi-Search algorithm based “Pattern Recognition Technique” to solve this problem which takes care of simple combinatorial structure of the problem and computational results are reported. In this discussion we have studied a variation of the transportation problems called “A Variant Constrained Bulk Transshipment Problem”. It also comes under combinatorial programming problems.


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A VARIANT CONSTRAINED BULK TRANSSHIPMENT PROBLEM transshipment cost subjected to the availability and requirement constraints. The Classical Transshipment Problem is to minimize the total cost for shipping the various capacities of the goods on the requirement of destinations from the available sources. The usual transshipment consist a unit cost for supplied goods to destinations from the sources. But in bulk transshipment the cost is independent of number of goods supplied to destinations, it is practical. In this paper we investigated a "variant of constrained bulk transshipment problem". Let there are m-sources and n-destinations. The destinations can get its complete requirement from a source directly or through some destination. The practical constraint is considered as only fewer destinations are allowed to supply its availability to some limited destinations. The cost of transportation of products from the sources to destination and destinations to destination is given. In this problem we take care of the restriction of availability and requirement of product between source and destinations. i.e., the total availability of the product at the source is greater than or equal to the total requirement of the product at the destinations. Generally movement of a product from source to source or destinations to source is not natural or practical, hence these possibilities are avoided and movement from destinations to destination is only considered. This is more generalized problem and comes under combinatorial programming problem. Often, the model is expressed as a zero-one programming problem. The objective of the problem is minimize the total bulk cost of supplying the required products to the destinations with the restriction that any destination should get its requirement from one source only, even when it gets from a destination.
In this A Variant Constrained Bulk Transshipment Problem, in general any source i supply its product to destinations subject to its availability once. There is a set of S= {1, 2, 3,..........., m} sources which produces a particular product and set ofD = {1, 2, 3,.......,n} destinations. The requirement of place j є D is DR (j) and the capacity of the source i є S is SA (i). Let D 1 ={∝1, ∝2, ∝3,…......,∝k} where k<n(i.e.,D 1  D)be the sub set of destinations, which supply the product to destinations subject to its availability of product. Let S 1 be a set of effective sources including destinations (D 1 ) because we allowed supply the product from destination to destination also i.e., S 1 =SUD 1 ={1, 2, 3,……….,m 1 }, where m 1 =m+k. The cost of bulk transshipment from source i to destination j is C(i,j); i є S 1 ,jєD.The objective is to minimize the total bulk transshipment cost subjected to the availability and requirements of constraints. The model can be built as 0 or1 programming problem.

MATHEMATICAL FORMULATION
Subject to the constraints: The constraint (1) describes the minimization of the total bulk transshipment cost subjected to the constraints. The constraint (2) represents that a destination should get its complete requirement exactly once (i.e., from a source or via shipment node).The constraint (3) indicates the supply schedule to the n destinations. The constraint (4) represents that the sum of the requirements at different destinations should less than or equal to the sum of availabilities at various sources. The last constraint (5) indicates that if there is a transportation from i to j then X (i, j) = 1, Otherwise it is equals to 0.
In the sequel we developed a Lexi-Search algorithm based on the "Pattern Recognition Technique" to solve this problem which takes care of simple combinatorial structure of the problem.

NUMERICAL FORMULATION
The concepts and the algorithm developed will be illustrated by a numerical example. In which we have taken number of Sources as m=4 i.e., S={1,2,3,4}and number of Destinations as n=7 i.e., D={1,2,3,4,5,6,7}. Let SA is the availability of a product at sources and DR is the requirement of a product at the destinations. Let D 1 ={1,4} and S 1 be the number of sources including destinations (D 1 ) because the transshipment problem allow supplying product from destination to destination also. So the total number of sources will be increase to 6(4+2). But in this numerical example we consider, only two destinations can supply the product to the destinations subject to its availability of product. Let D 1 ={1, 4} be the set of destinations, which supply the product to the destinations subject to its availability of product. This is subset of set D. Hence the total number of sources increases from 4 to 6. Let S 1 be the total number of sources such that S 1 =SUD 1 = {1,2,3,4,5,6}.Where S 1 (5)=D 1 (1) and S 1 (6)=D 1 (4). Then the cost matrix D is given below in Table-1. (For convenience same notation D is taken for the distance matrix). source 3 to destination 4 is 5. Similarly D (6, 7) = 3 means that the cost of the product supply from source 6 (i.e., destination 4) to destination 7is 3.

FEASIBLE SOLUTION
Consider an ordered pair set {(1,1), (3,2), (4,7), (5,5), (3,4) In the above solution, source 1 supplies its product to destination 1. Source 5 (destination 1) supplies its product to the destinations 5 and 3. Source 3 supplies its product to the destinations 2 and 4. Source 4 supplies its product to the destinations 7 and source 2 supplies its product to the destinations 6.So the solution gives the feasible solution.
Then the total bulk transshipment cost from given 6 sources to 7 destinations with respective source is as follows.

SOLUTION PROCEDURE
In the above figure-1, for the feasible solution we observe that 7 ordered pairs are taken along with the values from the cost/distance matrix for the numerical example in Table-
From the above figure-2, source 1 supplies its product to destination 1. The destination 2 gets its required product from source 3. Source 6 (i.e., destinsation4) supplies its product to destination 7.
Source 4 supplies its product to destination 2. But destination 2 already satisfied by source 3.
Source 5 (i.e., destination1) satisfies the requirement of destination 5. Source 3 supplies its product to destination 4. But the total sum of the requirement of the destinations 2, 4 and 7 is greater to the availability of the source 3, i.e., the total amount of supply is greater than actual amount of availability of source 3.

Definition of a Pattern
An indicator two dimensional arrays X which is associated with an allocation is called a "pattern". A pattern is said to be feasible if X is a feasible solution.
The value V(x) is gives the total cost of the tour for the solution represented by X. The pattern represented in the Table-2 Table-2 represents a feasible pattern of the feasible solution. In the above solution X (2, 6) = 1, represents that source 2 supplies its product to the destinations 6. In similar way X (4, 7) = 1, represents source 4 supplies its product to the destinations 7. Similarly all destinations can get its complete requirement from source directly or via some destination. So the above solution gives a feasible solution and it shown in figure-1. Table - Table-2, which is feasible solution and the ordered pair set {(1, 1), (3, 2), (6,7), (4, 2), (5,5), (3,4), (6, 6)} represents the pattern in Table -3, which is an infeasible solution . Table   There  For convenience same notation D is used for cost matrix and the increment array of cost in alphabet table. Then the alphabet of the given cost matrix ( Table -1) is as follows. From the above Table -2

Definition of a Word
Let SN = (1,2,…) be the set of indices, D be an array of corresponding distances of the ordered pairs and Cumulative sums of elements in D is represented as an array DC. Let arrays R, C be respectively, the row, column indices of the ordered pairs. Let Lk = {a1, a2, -----, ak}, ai∈ SN be an ordered sequence of k indices from SN. The pattern represented by the ordered pairs whose indices are given by Lk is independent of the order of ai in the sequence. Hence for uniqueness the indices are arranged in the increasing order such that ai< ai+1, i = 1, 2, ----, k-1.
The set SN is defined as the "Alphabet- Table" with alphabetic order as (1, 2, ----, 1 × ) and the ordered sequence Lk is defined as a "word" of length k. A word Lk is called a "sensible word".
If ai< ai+1, for i =1, 2, ----, k-1 and if this condition is not met it is called a "insensible word". A word Lk is said to be feasible if the corresponding pattern X is feasible and same is with the case of infeasible and partial feasible pattern. A Partial word Lk is said to be feasible if the block of words represented by Lk has at least one feasible word or, equivalently the partial pattern represented by Lk should not have any inconsistency. In the partial word Lk any of the letters in SN can occupy the first place. Since the words of length greater than n are necessarily infeasible, as any feasible pattern can have only n unit entries in it. Lk is called a partial word if k < n, and it is a full length word if k = n, or simply a word. A partial word Lk represents, a block of words with Lk as a leader i.e. as its first k letters. A leader is said to be feasible, if the block of word, defined by it has at least one feasible word.

Value of the Word
The value of the (partial) word Lk, V (Lk) is defined recursively as V (Lk) = V (Lk-1) + D (ak) with V (Lo) = 0 where D (ak) is the cost array arranged such that D (ak) < D (ak+1).V (Lk) and V(x) the values of the pattern X will be the same. Since X is the (partial) pattern represented by

Feasibility Criterion of a Partial Word
A feasibility criterion is developed, in order to check the feasibility of a partial word Lk+1 = (a1, a2… ak, ak+1) given that Lk is a feasible word. We will introduce some more notations which will be useful in the sequel.  Table -5 given below. We start with the partial word L1 = (a1) = (1). A partial word Lk is constructed as Lk = Lk-1 * (ak). Where * indicates chain formulation. We will calculate the values of V (Lk) and LB (Lk) simultaneously. Then two situations arises one for branching and other for continuing the search.
1. LB (Lk) < VT. Then we check whether Lk is feasible or not. If it is feasible we proceed to consider a partial word of under (k+1). Which represents a sub-block of the block of words represented by Lk? If Lk is not feasible then consider the next partial word p by taking another letter which succeeds ak in the position. If all the words of order p are exhausted then we consider the next partial word of order (k-1).
2. LB (Lk) > VT. In this case we reject the partial word Lk. We reject the block of word with Lk as leader as not having optimum feasible solution and also reject all partial words of order p that succeeds Lk.
To find Optimal Feasible word a Lexi-Search Algorithm using PRT is developed and is given in below.

SEARCH TABLE
The working details of getting an optimal word using the above algorithm for the illustrative numerical example is given in the following Table-

A Computer program for the proposed Lexi -Search Algorithm is written in C Language
and is tested at various hard instances. The experiments are carried out on a COMPAQ (dx2280 MT) system. The inputs like Cost Matrix D (i, j), source capacities (SA), and destination requirements (DR) are randomly generated for different instances. The cost values are uniformly generated in the interval [1,100]. The values for the availability and requirement of the product are also randomly generated between [1,500]. For different values of m, k, n, and Q a set of problems have been tested and their computational run time is recorded in seconds. The obtained results are tabulated in Table -9. It is observed that, the time required for the search of the optimal solution is fairly less. In the following table microseconds are represented by zero.
In Table- solution. It is seen that time required for the search of the optimal solution is moderately less.  In the above Table-