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ON BICRITERIA LARGE SCALE TRANSSHIPMENT PROBLEMS Dr. Jasem M.S. Alrajhi, Dr. Hilal A. Abdelwali, Dr. Mohsen S. Al-Ardhi, Eng. Rafik El Shiaty In the following we will present an algorithm for determining all nondominated extreme points of the (BMTSP3) model from which the solution of the (BMTSP1) and the (BMTSP2) models can be deduced as special cases from it. Let: for independent constraints: D k , k= 1,2,…,N be the technological matrix of the k th stage activity, D k is (m k + n k ) * (m k + n k ) matrix, N is the number of stages, m k is the number of sources at k th stage, n k is the number of destinations. b k be the column vector consisting of the availabilities and requirements of the k th subproblem, b k is (m k + n k ) * 1 column vector. It follows that each set of independent constraints can be written as: D k x k = b k , k = 1,2,…,N x k represent the vector of the corresponding variables, x k is (m k + n k ) *1 column vector. Let: For the common constraints: A k be the technogical matrix of k th stage activity, A k is m 0 * (m k * n k ) matrix, m 0 be the number of common constraints. b 0 be its corresponding common resources vector, b 0 is (m 0 * 1) column vector. This gives: A 1 x 1 + A 2 x 2 + …….+ A k x k + ……. +A N x N = b 0 Let: for the objective functions: c K represent the vector of the first criterion coefficients for the k th stage activity, c k is 1*(m k *n k ) row vector. d k represent the vector of the second criterion coefficients for the k th stage activity, d k is 1*(m k *n k ) row vector. Let: For the master program: B be the basic matrix associated with the current basic solution, B is (m o *N) * (m o +N) matrix. C B be the row vector of the corresponding coefficients in the objective function, C B is 1*(m o +N) row vector. R o is the matrix of size (m o + N)*m o consisting of the first m o columns of B -1 , and v j is the (m o + j) th column of the same matrix B -1 The algorithm presented here is divided into two phases. Phase 1: Determine the nondominated extreme points in the objective space. And the algorithm is validated by the following theorem [1]. Theorem: A point z (q) q q = z1 , z2 is a nondominated extreme point is the objective space if and only if z(q) is recorded by the algorithm. Phase II: Is the decomposition algorithm which can be found in [7]. Since the special structure of the (BMTSP3) model may allow the determination of the optimal solution by, first decomposing the problem into small subproblems and then solving those subproblems almost independently, then the decomposition algorithm for solving large scale linear programming problems utilizing the special nature of transshipment problem can be used to solve it. Phase I: Step 1: Go to phase II, find z (1 ) Min z / x M 1 And find . 1 (1) (1) z 2 Min . z 2 / z 1 z 1 and x M . Step 2: (1) (1) Record ( z 1 , z 2 ) and set q = 1. Similarly, go to phase II, find z ( 2 ) 2 And find Min . z 2 / x ( 2 ) ( 2 ) z 1 Min . z 1 / z 2 z 2 and x M . ( 2 ) ( 2 ) (1) (1) ( z 1 , z 2 ) ( z 1 , z 2 ), stop M If . (2) (2) Otherwise record ( z 1 , z 2 ) and set q = q+1 Defines sets L = {(1,2)} and E = , and go to step 2. Choose an element (r,s) L and set ( r , s ) ( s ) ( r ) a z z and a 1 ( r , s ) 2 z 2 ( s ) 1 z 2 ( r ) 1 k Go to phase II to obtain the optimal solution ( x , k=1,2,..,N) to the multistage transshipment problem. Minimize x k N k 1 ik , jk and ( r, s) k ( r, s) k (e 1 cik j a d k 2 ik j ) k Subject to k M , x o , k 1,2 ,.., N x k ik jk If there are alternative optima, choose an optimal solution k=1,2,.,N, for which N k 1 ik , jk Let z 1 = z 2 = N k1 ( c N k1 k ik jk i , j k x ik , jk k k ik jk d c k ik jk k i j k k min.) x x k ik jk k i j k k ; and ( r ) ( r ) If ( z 1 , 2 ( z 1 , z 2 ) Set E = E {(r,s)} and go to step 3. ( q) ( q) Otherwise record ( z 1 , z 2 ) such that z ( s ) ( s ) z ) is equal to or ( z , z ) ( q ) ( q ) 1 z1 , z 2 z 2 and set q q Step 3: L = L {(r,q)}, (q,s)} and go to step 3. Set L = L - {(r-s)}. If L = , stop. Otherwise go to step 2. 1 1, 2 k x , Phase II: Step 1: Reduce the original problem to the modified form in terms of the new variables k Step 2: Find an initial basic feasible solution to the modified problem. Step 3: Solve the subproblems k k k k k w ( c OR d c B R o A ) x Subject to k k k D x b , x k o , k 1,2,..., N . AcademyPublish.org – Journal of Engineering and Technology Vol.2, No.2 24
ON BICRITERIA LARGE SCALE TRANSSHIPMENT PROBLEMS Dr. Jasem M.S. Alrajhi, Dr. Hilal A. Abdelwali, Dr. Mohsen S. Al-Ardhi, Eng. Rafik El Shiaty Note: c k will be used with first criteria, and d k will be used with the second criteria. Obtaining xˆ k l and optimal objective values transportation technique, Go to step 4. Step 4: For the current iteration, find * k k k w c B v , k k Then determine Min ( ) k 1,2,..., N , k w * , by using the If o , the current solution is optimal and the process is terminated, the optimal solution to multistage transportation problem is: x k L K L K xl , k 1,2 ,..., L 1 Otherwise, go to step 5. k Step 5: Introduce the variable L corresponding to into the basic solution. Determine the leaving variable using the feasibility condition and compute the next B -1 using the revised simplex method technique, go to step 3. Illustrative Example The suggested algorithm for solving problem of the type BMTSP 3 will be illustrated in the following example: Consider the following bicriteria two stage transshipment problem. For each stage the availabilities, requirements, costs and deteriorations for each stage are given as: 1 a 1 = 6, 2 b 1 = 6, 1 a 2 = 4, a = 2, 1 3 2 2 b 2 = 2, b 3 = 4 1 b 1 = N 2 a 1 = 9, 1 b 2 = Table 1. Transportation cost at stages (1) and (2). D 1 1 D 1 2 S 1 1 S 1 2 S 1 3 S 1 1 5 4 0 2 1 S 1 2 10 8 1 0 4 S 1 3 9 9 3 2 0 D 1 1 0 1 5 9 9 D 1 2 3 0 4 6 7 D 2 1 D 2 2 D 2 3 S 2 1 S 2 2 S 2 1 4 3 3 0 3 S 2 2 8 4 7 2 0 D 2 1 0 2 4 8 7 D 2 2 4 0 3 3 5 D 2 3 3 4 0 4 9 Table 2. Deterioration cost at stages (1) and (2). D 1 1 D 1 2 S 1 1 S 1 2 S 1 3 S 1 1 3 6 0 1 4 S 1 2 7 9 3 0 6 S 1 3 12 11 4 6 0 D 1 1 0 3 7 11 12 D 1 2 5 0 7 8 8 2 a 2 = 3, D 2 1 D 2 2 D 2 3 S 2 1 S 2 2 S 2 1 6 5 5 0 6 S 2 2 11 6 9 5 0 D 2 1 0 4 6 11 9 D 2 2 6 0 5 4 7 D 2 3 5 7 0 6 11 One requirement is added to the above problem: It is required that the quantity shipped from the first source to the first destination in the first stage is equal to the quantity shipped from the first source to the first destination in the second stage. The mathematical model is given as follows: Minimize z 1 = 5x 1 11 + 4x 1 12 + 0x 1 13 + 2x 1 14 + x 1 15 + 10x 1 21 + 8x 1 22 + x 1 23 + 0x 1 24 + 4x 1 25 + 9x 1 31 + 9x 1 32 + 3x 1 33 + 2x 1 34 + 0x 1 35 + 0x 1 41 + x 1 42 + 5x 1 43 + 9x 1 44 + 9x 1 45 + 3x 1 51 + 0x 1 52 + 4x 1 53 + 6x 1 54 + 7x 1 55 + 4x 2 11 + 3x 2 12 + 2x 2 13 + 0x 2 14 + 3x 2 15 + 8x 2 21 + 4x 2 22 + 7x 2 23 + 2x 2 24 + 0x 2 25 + 0x 2 31 + 2x 2 32 + 4x 2 33 + 8x 2 34 + 7x 2 35 + 4x 2 41 + 0x 2 42 + 3x 2 43 + 3x 2 44 + 5x 2 45 + 3x 2 51 + 4x 2 52 + 0x 2 53 + 4x 2 54 + 9x 2 55 Subject to: Z 2 = 3x 1 11 + 6x 1 12 + 0x 1 13 + 1x 1 14 + 4x 1 15 + 7x 1 21 + 9x 1 22 + 3x 1 23 + 0x 1 24 + 6x 1 25 + 12x 1 31 + 11x 1 32 + 4x 1 33 + 6x 1 34 + 0x 1 35 + 0x 1 41 + 3x 1 42 + 7x 1 43 + 11x 1 44 + 12x 1 45 + 5x 1 51 + 0x 1 52 + 7x 1 53 + 8x 1 54 + 8x 1 55 + 6x 2 11 + 5x 2 12 + 5x 2 13 + 0x 2 14 + 6x 2 15 + 11x 2 21 + 6x 2 22 + 9x 2 23 + 5x 2 24 + 0x 2 25 + 0x 2 31 + 4x 2 32 + 6x 2 33 + 11x 2 34 + 9x 2 35 + 6x 2 41 + 0x 2 42 + 5x 2 43 + 4x 2 44 + 7x 2 45 + 5x 2 51 + 7x 2 52 + 0x 2 53 + 6x 2 54 + 11x 2 55 x 1 11 = x 2 11 x 1 11 + x 1 12 + x 1 13 + x 1 14 + x 1 15 = 18 x 1 21 + x 1 22 + x 1 23 + x 1 24 + x 1 25 = 16 x 1 31 + x 1 32 + x 1 33 + x 1 34 + x 1 35 = 14 x 1 41 + x 1 42 + x 1 43 + x 1 44 + x 1 45 = 12 x 1 51 + x 1 52 + x 1 53 + x 1 54 + x 1 55 = 12 x 1 11 + x 1 21 + x 1 31 + x 1 41 + x 1 51 = 21 x 1 12 + x 1 22 + x 1 32 + x 1 42 + x 1 52 = 15 x 1 13 + x 1 23 + x 1 33 + x 1 43 + x 1 53 = 12 x 1 14 + x 1 24 + x 1 34 + x 1 44 + x 1 54 = 12 x 1 15 + x 1 25 + x 1 35 + x 1 45 + x 1 55 = 12 x 2 11 + x 2 12 + x 2 13 + x 2 14 + x 2 15 = 21 x 2 21 + x 2 22 + x 2 23 + x 2 24 + x 2 25 = 15 x 2 31 + x 2 32 + x 2 33 + x 2 34 + x 2 35 = 12 x 2 41 + x 2 42 + x 2 43 + x 2 44 + x 2 45 = 12 x 2 51 + x 2 52 + x 2 53 + x 2 54 + x 2 55 = 12 x 2 11 + x 2 21 + x 2 31 + x 2 41 + x 2 51 = 18 x 2 12 + x 2 22 + x 2 32 + x 2 42 + x 2 52 = 14 x 2 13 + x 2 23 + x 2 33 + x 2 43 + x 2 53 = 16 x 2 14 + x 2 24 + x 2 34 + x 2 44 + x 2 54 = 12 x 2 15 + x 2 25 + x 2 35 + x 2 45 + x 2 55 = 12 AcademyPublish.org – Journal of Engineering and Technology Vol.2, No.2 25
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Page 3 and 4: PARAMETRIC STUDY OF SANDWICH PANEL
Page 17 and 18: A STUDY OF SIMPLIFIED SHALLOW WATER
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Page 21 and 22: ON BICRITERIA LARGE SCALE TRANSSHIP
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Page 27 and 28: TRIBOLOGY OF HIGH SPEED MOVING MECH
Page 35 and 36: THE USE OF IRON IN PEAT WATER FOR F
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