C. METODA TRANSPORTASIMetode transp

C. METODA TRANSPORTASI
Metode transportasi yang biasa digunakan untuk menyelesaikan masalah transportasi adalah suatu hasil modifikasi dari metode simpleks dengan memperhatikan pola khusus dari nilai koefisien pada fungsi pembatasnya. Seperti juga pada metode simpleks, langkah pertama dalam mnyelesaikan masalah transportasi adalah mencari solusi awal yang layak. Kemudian menguji apakan solusi awal tersebut tela optimal, bila belum dilakukan perbaikan-perbaikan yaitu dengan adanya leaving variable (perubahan variabel basis dan non basis) hingga diperoleh solusi yang optimal.

SOLUSI AWAL YANG LAYAK
Pada solusi awal yang layak harus dipenuhi kondisi berikut :
Jumlah variabel basis = m + n – 1
Dengan m SUMBER dan n TUJUAN
Jadi harus ada sejumlah (m + n -1) variabel yang bernilai nonnegatif.

PROSEDUR UMUM UNTUK MEMPEROLEH SOLUSI AWAL YANG LAYAK

Langkah awal : Semua baris sumber dan kolom tujuan pada tabel transportasi dapat dijadikan variabel basis (daerah pengalikasian = masih kosong)
Langkah I : Diantara baris dan kolom ayang masih dapat dijadikan variabel basis, pilihlah beberapa basis berikutnya berdasarkan beberapa kriteria (tergantung pada metode yang digunakan).
Langkah II : Pengalokasian dibuat sebanyak mungkin untuk memenuhi nilai penawaran atau permintaan (tergantung yang mana yang terkecil).
Langkah III : Menghilangkan baris atau kolom yang bisa menerima pengalokasian yang telah terpenuhi penawaran dan permintaannya dari pengaloksian berikutnya.
Langkah IV : Bila hanya tersisa satu baris atau kolom yang bisa menerima pengaloksian, maka pengalokasian ke kotak baris atau kolom tersebut, dan prosedur selesai; se;ain itu kembali ke langkah 1.




KRITERIA ALTERNATIF UNTUK LANGKAH 1

METODE NORTHWEST CORNER
Pengalokasian dimulai dari pojok barat laut (northwest corner). Selanjutnya pengalokasian dilakukan pada kotak Xij+1 bila permintaan ke j telah terpenuhi atau pada kotak Xi+1j bila penawara ke i telah terpenuhi.

METODE LEAST COST
Pengalokasian dimulai pada kotak variabel dengan biaya terendah. Selanjutnya pengalokasian dilakukan pada kotak variabel terendah berikutnya dengan memperhatikan nilai penawaran dan permintaan.

METODE APROKSIMASI VOGEL
Untuk setiap baris dan kolom dilakukan perhitungan nilai selisih antara kotak variabel dengan biaya terendah dengan kotak variabel biaya terendah berikutnya (smallest and next to smallest). Kemudian pilih baris atau kolom dengan nilai selisih terbesar. Pengalokasian dilakukan pada kotak variabel dengan biaya terendah pada baris atau kolom yang dipilih.

METODE APROKSIMASI RUSSELL
Untuk setiap baris ditentukan nilai ui yang merupakan biaya tertinggi pada baris tersebut. Sedangkan untuk setiap kolom ditentukan nilai vj yang merupakan biaya tertinggi pada kolom tersebut. Untuk stiap kotak variabel Xij dilakukan perhtiungan nilai Δij = cij –ui – vj , pengalokasian dilakukan pada kotak variabel dengan nilai Δij negatif terbesar.

UJI OPTIMALITAS
METODE STEPPING STONE
Uji Optimalitas: solusi awal yang layak disebut optimal jika dan hanya jika nilai perubahan biaya untuk setiap siklus yang dibuat ≥ 0.
Jadi pada metode stepping stone pengujian didasarkan pada hasil perhitungan perubahan biaya dari setiap siklus yang intinya adalah untuk mencoba mengalokasikan pada kotak kosong (variabel non basis). Prinsip nya sama dengan metode simpleks, yaitu menentukan variabel mana yang akan menjadi leaving variable dan mana yang akan menjadi entering variable.
Aturan penentuan siklus
1. Suatu siklus perubahan pengalokasian tidak boleh mengubah nilai penawaran dan permintaan.
2. Dalam satu siklus hanya boleh terdapat satu kotak kosong (variabel non basis) yang terlibat.
3. Suatu siklus berawal dan berakhir pada kotak yang sama
4. Hanya boleh ada 2 kotak yang berurutan yang terlibat dan yang terletak pada baris/kolom yang sama.

METODE MODIFIED DISTRIBUTION (MODI)
Uji optimalitas : solusi awal yang layak disebut optimal jika dan hanya jika (cij – ui - vj ) ≥ 0 untuk setiap (i,j) dimana Xij nya adalah variabel non basis.
Jadi, pada uji optimalitas perlu dilakukan penentuan nilai ui dan vj pada solusi yang layak yang diperoleh, kemudian baru dilakukan perhitungan nilai (cij – ui – vj ).
Dikaranakan nilai (cij – ui – vj) = 0 untuk Xij adalah varibel basis, maka nilai ui dan vj dapat ditentukan dengan persamaan berikut :
Cij = ui + vj
Untuk setiap (i,j) dimana Xij adalah variabel basis.
Dikarenakan adanya sejumlah (m + n – 1) variabel basis, maka akan ada sejumlah (m + n – 1) persamaan tersebut diatas. Dan dikarenakan jumlah ui dan vj yang tidak diketahui ada (m + n), maka salah satu variabel tersebut diberi suatu nilai yang tidak mempengaruhi persamaan tersebut di atas. Pemberian nilai tersebut dilakukan dengan cara memilih ui dengan alokasi terbesar dan memberinya nilai nol.

Results (English) 1:

C. THE METHOD OF TRANSPORTATIONThe transportation method used to solve the transportation problem is a result of the modification of simplex method with attention to specific patterns of coefficients in the function pembatasnya. As well as in the simplex method, the first step in the mnyelesaikan transportation problem is to find a proper solution. Then test whether the tela solution is optimal, if not yet done the repair-repair IE with the leaving variable (variable base changes and non-base) to the optimal solution obtained.A DECENT SOLUTIONAt the beginning of a proper solution must meet the following conditions:The number of variables base = m + n – 1With m and n SOURCE DESTINATIONSo there has to be a number of (m + n-1) variable that is worth nonnegatif.THE GENERAL PROCEDURE IS TO OBTAIN A DECENT SOLUTIONInitial steps: all lines of source and destination columns in the table can be used as a transport base variables (area = pengalikasian still empty)Step i: between the rows and columns can still be used as my father learned a variable base, choose some the next base based on several criteria (depending on the method used).Step II: Assignment made as much as possible to meet the value offer or request (depending on which is the smallest).Step III: Remove rows or columns that can receive the assignment that has fulfilled the supply and demand of the next pengaloksian.Step IV: when only one row or column, then the pengaloksian can accept assignment to the row or column, and the procedure is completed; SE; ain back to step 1.ALTERNATIVE CRITERIA for step 1 NORTHWEST CORNER METHODAllocation starts from the North-West corner (northwest corner). Next the assignment is done in the Xij + 1 when the request to j have been met or the Xi + 1j when penawara to i have met.LEAST COST METHODS Assignment begins on the variable with the lowest cost. Next the assignment is done on the next lowest variable by observing the value of supply and demand.THE METHOD OF APPROXIMATION VOGEL For each row and column is done the calculation value difference between the box variable with the lowest cost with the next lowest cost variable box (smallest and next to smallest). Then select the row or column with the largest difference value. Assignment is done at the lowest cost with the variables box on the row or column is selected.THE METHOD OF APPROXIMATION RUSSELL For each row of the specified values of the ui which is the highest fee on those lines. While prescribed for each column value vj who is the highest fee on that column. For the variables Xij box stiap done perhtiungan the value cij = Δij – ui – vj, assignment is done on the value of the variable with the largest negative Δij.TEST OPTIMALITASSTONE STEPPING VALUE METHOD Optimalitas: test the solution earlier worth mentioning is optimal if and only if the value of the change fee for each cycle made ≥ 0. So on the method based on testing stone stepping value calculation result changes the cost of each cycle that the point is to try to allocate on the empty boxes (non variable base). His principles are the same as the simplex method, that is, determine which variables will be leaving variable and which ones will be the entering variable.The rules determining the cycle1. A cycle of change of assignment should not change the value of supply and demand.2. in one cycle there should only be one empty boxes (non variable base) are involved.3. A cycle begins and ends on the same box4. There should only be 2 sequential boxes involved and located on the row/column.THE METHOD OF MODIFIED DISTRIBUTION (MODI)Optimalitas: test the solution earlier worth mentioning is optimal if and only if (cij – ui-vj) ≥ 0 for every (i, j) where the Xij are non variable base.So, in a test of optimalitas to do the determination of the value of the ui and vj on a decent solution is obtained, then recently done the calculation value (cij – ui – vj).Dikaranakan value (cij – ui – vj) = 0 to Xij is varibel a ui, the value base and the vj can be determined by the following equation:Cij = ui + vjFor each (i, j) where Xij is a variable base.Due to the existence of a number of (m + n – 1) variable base, then there will be a number of (m + n – 1) the equation above. And because of the number of ui and vj unknown there is (m + n), then one of these variables are given a value that does not affect the equation above. The granting of such values is done by selecting the ui with the biggest allocation and give him a zero value.

Results (English) 2:

C. METHOD OF TRANSPORTATION
Transportation method used to solve the transport problem is a modified version of the simplex method with special attention to the pattern of the coefficient on the function of its boundary. As well as the simplex method, the first step in mnyelesaikan transport problems is seeking solutions decent start. Then test whether the initial solution of the tela optimal, if not already done repairs, with the occurrence leaving variable (change of variable basis and non base) to obtain the optimal solution. SOLUTION AT BEGINNING OF LIVING At the initial solution worth having fulfilled the following conditions: Number of variables base = m + n - 1 with m and n sOURCE gOAL So there must be a number (m + n -1) variable that is worth nonnegative. PROCEDURE PRELIMINARY GENERAL SOLUTIONS TO OBTAIN ADEQUATE first step: Each line source and destination columns in the table can transport used as a variable base (area pengalikasian = blank) Step I: Among the rows and columns ayang can still be used as a variable basis, choose some of the next base is based on several criteria (depending on the method used). Step II: the allocation is made ​​as much as possible to meet the bid value or request (whichever is the smallest). Step III: Eliminate rows or columns that can receive allocations have been met supply and demand of pengaloksian next. Step IV: If only the remaining one row or column can accept pengaloksian, the allocation to the box row or column, and the procedure is completed; se; ain it back to step 1. CRITERIA FOR STEP 1 ALTERNATIVE METHOD OF NORTHWEST CORNER allocation starts from the northwest corner (the northwest corner). Furthermore, the allocation is done on the box Xij + 1 to j when the demand has been met or the box Xi + 1h when penawara to i have been met. METHOD OF LEAST COST Allocation beginning on variable box at the lowest cost. Furthermore, the allocation is done on the box next lowest variable with regard to the value of supply and demand. METHOD VOGEL APPROXIMATION For each row and column calculating the difference between the value of the variable boxes at the lowest cost with the next lowest cost variable box (smallest and next to smallest). Then select the row or column with the largest difference value. The allocation is done on the box the variable with the lowest cost in the selected row or column. METHOD APPROXIMATION RUSSELL For each row is determined ui value which is the highest cost on that line. As for any given column value vj which is the highest cost in that column. To stiap box Xij variables do perhtiungan Δij value = ij -ui - vj, allocation is done on the box the variable with the largest negative Δij. TEST optimality METHOD OF STEPPING STONE Test Optimality: initial solution worth mentioning optimal if and only if the value changes in the cost for each cycle created ≥ 0. So the stepping stone test methods are based on the calculation result of changes in costs of each cycle that the point is to try to allocate the empty box (non-variable basis). The principle is the same as the simplex method, which was to determine which variable will be leaving variable and which will be entering variable. The rule of determining the cycle 1. A cycle of allocation changes should not alter the value of supply and demand. 2. In one cycle can only be one empty box (non-variable basis) involved. 3. A cycle begins and ends on the same box 4. There should only be two boxes sequential involved and which is located on the row / column of the same. METHOD MODIFIED DISTRIBUTION (MODI) Test optimality: initial solution worth mentioning optimal if and only if (ij - ui - vj) ≥ 0 for every ( i, j) where Xij it is a variable non-base. So, the test optimality is necessary to determine the value ui and vj at a feasible solution is obtained, then a new calculation of the value (ij - ui - vj). Dikaranakan value (ij - ui - vj) = 0 for Xij is a variable basis, the value ui and vj can be determined by the following equation: Cij = ui + vj for each (i, j) where Xij is a variable basis. Due to the number of (m + n - 1) variable basis, then there will be a number of (m + n - 1) of the above equation. And because the number of ui and vj are not known to exist (m + n), then one variable is given a value that does not affect the above equation. Scoring is done by selecting ui with the largest allocations and gave him a zero value.