Download Transportation slides and more Slides Mathematics in PDF only on Docsity!
- THE TRANSPORTATION PROBLEM AND THE AS- SIGNMENT PROBLEM
- The transportation problem
- The matrix format for the transportation problem
- Formulating transportation problems
- Theorems and definitions
- Finding an initial basic feasible solution 5.1 The Northwest Corner method 5.2 Vogel’s approximation method
- Improvement of a basic feasible solution 6.1 Selection of the entering vector 6.2 Selection of the leaving vector
- The transportation tableau
- The transportation algorithm
- Degeneracy
- The assignment problem 10.1 The Hungarian method 10.2 The Hungarian algorithm 10.3 Maximization problems
1. The transportation problem
The transportation problem deals with the transporta- tion of any product from m origins, O 1 ,... , O (^) m , to n destinations, D 1 ,... , D (^) n , where:
- The origin O (^) i has a supply of a (^) i units, i = 1,... , m.
- The destination D (^) j has a demand for b (^) j units to be delivered from the origins, j = 1,... , n.
- c (^) ij is the cost per unit distributed from the origin O (^) i to the destination D (^) j , i = 1,... , m, j = 1,... , n.
In mathematical terms, the above problem can be ex- pressed as finding a set of x (^) ij ’s, i = 1,... , m, j = 1 ,... , n, to meet supply and demand requirements.
The aim: to minimize the total distribution cost.
Linear model:
min z =
�^ m
i=
�^ n
j=
c (^) ij x (^) ij
subject to �^ n
j=
x (^) ij ≤ a (^) i , i = 1,... , m
�m
i=
x (^) ij ≥ b (^) j , j = 1,... , n
x (^) ij ≥ 0 , i = 1,... , m, j = 1,... , n
We can write the constraints in equation form, be- cause the total supply is equal to the total demand.
The model in matrix form:
min z = (8 , 6 , 10 , 10 , 4 , 9)
x (^11) x (^12)
x (^13) x (^21)
x (^22) x (^23)
subject to
1 1 1 0 0 0
0 0 0 1 1 1 1 0 0 1 0 0
0 1 0 0 1 0 0 0 1 0 0 1
x (^11)
x (^12) x (^13)
x (^21) x (^22)
x (^23)
x 11 , x 12 , x 13 , x 21 , x 22 , x 23 ≥ 0
2. The matrix format for the transporta-
tion problem
The relevant data for any transportation problem can be summarized in a matrix format using a tableau called the transportation costs tableau.
D 1 D 2 · · · D (^) n Supply
O 1 c 11 c 12 · · · c (^1) n a (^1)
O 2 c 21 c 22 · · · c (^2) n a (^2) ... ... ...... ... ...
O (^) m c (^) m 1 c (^) m 2 · · · c (^) mn a (^) m
Demand b 1 b 2 · · · b (^) n
Example.
D 1 D 2 D 3 Supply
O 1 8 6 10 2000
O 2 10 4 9 2500
Demand 1500 2000 1000
5. Finding an initial basic feasible solu-
tion
To compute an initial basic feasible solution we will use a tableau of the same dimensions as the transportation costs tableau; the transportation solution tableau.
Each position (i, j) is associated with the decision vari- able x (^) ij , that is, the number of units of product to be transported from origin O (^) i to destination D (^) j.
The transportation solution tableau:
D 1 D 2 · · · D (^) n Supply
O 1 x 11 x 12 · · · x (^1) n a (^1)
O 2 x 21 x 22 · · · x (^2) n a (^2) ... ... ...... ... ...
O (^) m x (^) m 1 x (^) m 2 · · · x (^) mn a (^) m
Demand b 1 b 2 · · · b (^) n
The main difference between the Northwest Corner method and Vogel’s approximation method lays in the criteria used to select a cell in the solution tableau.
We have to balance the transportation problem before solving it.
5.1 The Northwest Corner method
Step 1. In the rows and columns under consider- ation, select the cell (i, j) in the northwest corner of the solution tableau.
Step 2. Assign to the variable x (^) ij the maximum feasible amount consistent with the row and the column requirements of that cell, x (^) ij = min{a (^) i , b (^) j }. Adjust the supply a (^) i and the demand b (^) j as follows:
- If a (^) i happens to be the minimum, then the supply of the origin O (^) i becomes zero, and the row i is eliminated from further consideration. The demand b (^) j is replaced by b (^) j − a (^) i.
- If b (^) j happens to be the minimum, then the demand of the destination D (^) j becomes zero, and the column j is eliminated from further consideration. The supply a (^) i is replaced by a (^) i − b (^) j.
- If a (^) i = b (^) j , then the adjusted values for the supply a (^) i and the demand b (^) j become both zero. The row i and the column j are eliminated from further consideration.
- Step 3. Two cases may arise:
- If only one row or only one column remains under consideration, then any remaining cells (i, j), that is, variables x (^) ij associated with these cells, are selected and the remaining supplies are assigned to them. Stop.
- Otherwise, go to Step 1.
- Improvement of a basic feasible solution
The dual transportation problem is used to find an improved basic feasible solution.
Balanced transportation problem:
min z =
�^ m
i=
�^ n
j=
c (^) ij x (^) ij
subject to
�^ n
j=
x (^) ij = a (^) i , i = 1,... , m
�m
i=
x (^) ij = b (^) j , j = 1,... , n
x (^) ij ≥ 0 , i = 1,... , m, j = 1,... , n
We denote by u 1 ,... , u (^) m and v 1 ,... , v (^) n the dual vari- ables,
The dual model:
max G =
�^ m
i=
a (^) i u (^) i +
�^ n
j=
b (^) j v (^) j
subject to
u (^) i + v (^) j ≤ c (^) ij , i = 1,... , m, j = 1,... , n
u (^) i , v (^) j : unrestricted, i = 1,... , m, j = 1,... , n
Example. A balanced transportation problem:
min z = 8x 11 + 6x 12 + 10x 13 + 10x 21 + 4x 22 + 9x (^23)
subject to
x 11 +x 12 +x 13 = 2000 x 21 +x 22 +x 23 = 2500
x 11 +x 21 = 1500 x 12 +x 22 = 2000
x 13 +x 23 = 1000
x 11 , x 12 , x 13 , x 21 , x 21 , x 23 ≥ 0
The dual variables: u 1 , u 2 , v 1 , v 2 , v 3. The dual model:
max G = 2000u 1 + 2500u 2 + 1500v 1 + 2000v 2 + 1000v (^3)
subject to
u 1 +v 1 ≤ 8 u 1 +v 2 ≤ 6
u 1 +v 3 ≤ 10 u 2 +v 1 ≤ 10
u 2 +v 2 ≤ 4 u 2 +v 3 ≤ 9
u (^) i , v (^) j : unrestricted
6.2 Selection of the leaving vector
We need to take into account the following:
- The m + n − 1 basic cells corresponding to any basic solution never contain a cycle.
- The m + n − 1 basic variables together with the entering variable contain a unique cycle.
To find such a unique cycle: The cell that corresponds to the entering variable is assumed to be basic and is marked with the symbol ↑. We cross out all the rows and columns which contain only one basic cell: we first cross out the rows, for instance, and afterwards the columns, then we check the rows again and so on. At the end of this process, the basic cells that have not been crossed out as well as the cell that corresponds to the entering variable contain the unique cycle.
To adjust the entries in the cycle: since the entering variable will be assigned a positive value, the basic cells in the cycle that are in the same row or in the same column are marked by the symbol ↓, the basic cells in the cycle that are in the same row or in the same column as the ones previously marked with the symbol ↓ are marked with the symbol ↑ and so on. Stop when all the cells in the cycle are marked.
The first basic variable in the cycle decreased to zero among those marked by the symbol ↓ becomes the leaving variable. All the basic variables in the cycle are recomputed. The basic variables that do not belong to the cycle remain unchanged. The entering variable is assigned the value the leaving variable had before being adjusted.
7. The transportation tableau
Up to now, we have carried out all the calculations in two different tableaux: the transportation costs tableau and the transportation solution tableau.
To improve a solution, we have computed the values of the dual variables u (^) i , v (^) j and the reduced cost coef- ficients z (^) ij − c (^) ij outside the two tableaux.
The transportation tableau puts together all the cal- culations needed to solve a transportation problem.
v 1 v 2 · · · v (^) n z 11 − c 11 c 11 z 12 − c 12 c 12 · · · z (^1) n − c (^1) n c (^1) n u 1 x 11 x 12 x (^1) n a (^1) z 21 − c 21 c 21 z 22 − c 22 c 22 · · · z (^2) n − c (^2) n c (^2) n u 2 x 21 x 22 x (^2) n a (^2)
...... ...
z (^) m 1 − c (^) m 1 c (^) m 1 z (^) m 2 − c (^) m 2 c (^) m 2 · · · z (^) mn − c (^) mn c (^) mn
u (^) m x (^) m 1 x (^) m 2 x (^) mn a (^) m
b 1 b 2 · · · b (^) n
- Degeneracy
If a solution to a given balanced transportation prob- lem with m origins and n destinations has less than m + n − 1 positive variables, that is, if at least one of the basic variables has the value zero, then it is said to be degenerate.
Degeneracy may occur in the following two cases:
- While computing an initial basic feasible solution, either applying Vogel’s approximation method or using the northwest corner method. If at Step 2 of either algorithm we find that when assigning to a variable the maximum feasible amount consis- tent with the row and the column requirements, both the supply and the demand are equal, then they become both zero; a row and a column are eliminated at the same time.
- While applying the transportation algorithm at Step 5, once we have identified the entering vari- able and found the cycle. If two decreasing vari- ables in the cycle (two cells marked by the symbol ↓) tie for the minimum, then after recomputing the values in the cycle, they both become zero. However, only one of them can be selected as the leaving variable.
If degeneracy is encountered, we have to distinguish the zero-valued basic variable from the nonbasic ones.
10. The assignment problem
It is a special case of the transportation problem.
It deals with assigning n origins O (^) i to n destinations D (^) j with the goal of determining the minimum cost assignment. Each origin must be assigned to one and only one destination, and each destination must have assigned one and only one origin. c (^) ij represents the cost of assigning the origin O (^) i to the destination D (^) j.
Decision variables are defined like this:
x (^) ij =
1 if O (^) i is assigned to D (^) j
0 otherwise
A linear model in standard form for the assignment problem:
min z =
�^ n
i=
�^ n
j=
c (^) ij x (^) ij
subject to �^ n
j=
x (^) ij = 1, i = 1,... , n
�n
i=
x (^) ij = 1, j = 1,... , n
x (^) ij = 0, 1 , i, j = 1,... , n If the assignment problem has the same number of origins and destinations, it is balanced.
If it is not balanced, we can always add as many dummy origins or dummy destinations as necessary.
Theorem 5 Given c (^) ij ≥ 0 and a solution x (^) ij to the
assignment problem, if z =
�^ n
j=
�^ n
i=
c (^) ij x (^) ij = 0 holds,
then the values of x (^) ij are optimal, i, j = 1,... , n.
If a constant is subtracted from each cost in a row or column, the optimal solution to the problem remains unchanged.
The solution of an assignment problem proceeds by transforming the assignment costs tableau to create zero assignment-costs. Then, if a feasible assignment is found in which all the x (^) ij ’s that equal 1 have zero costs and thus z = 0, the assignment is an optimal solution to the problem.
10.2 The Hungarian algorithm
The objective is to minimize.
Step 1. Balance the assignment problem.
Step 2. Create zero entries in the rows. For each row in the assignment costs tableau, subtract the row minimum u (^) i from each element in the row, u (^) i = min j
{c (^) ij }. The new entries in the resulting
tableau are c � ij = c (^) ij − u (^) i , i, j = 1,... , n.
- Step 3. Create zero entries in the columns. For each column in the resulting tableau, subtract the column minimum v (^) j from each element in the column, v (^) j = min i
{c � ij }. The new entries are c �� ij = c � ij − v (^) j , i, j = 1,... , n.
- Step 4. Choose independent zeros. Find the row or column with the smallest number of zero entries. Choose one of its zeros, and cross out all the zeros in the same row or column. Proceed to choose more zeros among the ones that have not been crossed out, starting at the row or column with the smallest number of them, until all zeros are either chosen or crossed out.
- If n independent zeros have been chosen, an optimal solution is available. Stop.
- If less than n independent zeros have been chosen, then go to Step 5.
