Assignment Problem
Minimization or maximization of the cost of transporting goods from one source to another
Maximization of the total profit or minimization of the total cost in assigning tasks to individuals
Nature of problem
Involves transporting goods from sources to destinations
Involves assigning tasks to individuals
Number of sources and destinations
Multiple sources and destinations
An equal number of sources and destinations
Availability and demand
Each source and destination have a supply or demand value
Each task has only one individual who can perform it
Decision variables
Amount of goods transported from each source to each destination
Binary variables indicate whether an individual is assigned a task or not
Constraints
Capacity constraints on sources and demand constraints on destinations
Each individual can only perform one task
Solution method
Transportation simplex method, northwest corner rule, Vogel’s approximation method
Hungarian algorithm, brute force method
Example
Transporting goods from factories to warehouses
Assigning tasks to employees or jobs to machines
Decision Variables:
In a transportation problem, the decision variables represent the flow of goods from sources to destinations. Each variable represents the quantity of goods transported from a source to a destination.
In contrast, in an assignment problem, the decision variables represent the assignment of agents to tasks. Each variable represents whether an agent is assigned to a particular task or not.
Constraints:
In a transportation problem, the constraints ensure that the supply from each source matches the demand at each destination and that the total flow of goods does not exceed the capacity of each source and destination.
In contrast, in an assignment problem, the constraints ensure that each task is assigned to exactly one agent and that each agent is assigned to at most one task.
Objective function:
The objective function in a transportation problem typically involves minimizing the total cost of transportation or maximizing the total profit of transportation.
In an assignment problem, the objective function typically involves minimizing the total cost or maximizing the total benefit of assigning agents to tasks.
In summary,
The transportation problem is concerned with finding the optimal way to transport goods from sources to destinations,
while the assignment problem is concerned with finding the optimal way to assign agents to tasks.
Both problems are important in operations research and have numerous practical applications.
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There are two phases to solve the transportation problem. In the first phase, the initial basic feasible solution has to be found and the second phase involves optimization of the initial basic feasible solution that was obtained in the first phase. There are three methods for finding an initial basic feasible solution,
NorthWest Corner Method
This article will discuss how to optimize the initial basic feasible solution through an explained example. Consider the below transportation problem.
Check whether the problem is balanced or not. If the total sum of all the supply from sources
is equal to the total sum of all the demands for destinations
then the transportation problem is a balanced transportation problem.
If the problem is not unbalanced then the concept of a dummy row or a dummy column to transform the unbalanced problem to balanced can be followed as discussed in
Finding the initial basic feasible solution. Any of the three aforementioned methods can be used to find the initial basic feasible solution. Here,
will be used. And according to the NorthWest Corner Method this is the final initial basic feasible solution:
Now, the total cost of transportation will be
(200 * 3) + (50 * 1) + (250 * 6) + (100 * 5) + (250 * 3) + (150 * 2) = 3700
U-V method to optimize the initial basic feasible solution. The following is the initial basic feasible solution:
– For U-V method the values
have to be found for the rows and the columns respectively. As there are three rows so three
values have to be found i.e.
for the first row,
for the second row and
for the third row. Similarly, for four columns four
. Check the image below:
There is a separate formula to find
is the cost value only for the allocated cell. Read more about it
. Before applying the above formula we need to check whether
m + n – 1 is equal to the total number of allocated cells
or not where
is the total number of rows and
is the total number of columns. In this case m = 3, n = 4 and total number of allocated cells is 6 so m + n – 1 = 6. The case when m + n – 1 is not equal to the total number of allocated cells will be discussed in the later posts. Now to find the value for u and v we assign any of the three u or any of the four v as 0. Let we assign
in this case. Then using the above formula we will get
). Similarly, we have got the value for
so we get the value for
which implies
. From the value of
. See the image below:
Now, compute penalties using the formula
only for unallocated cells. We have two unallocated cells in the first row, two in the second row and two in the third row. Lets compute this one by one.
If we get all the penalties value as zero or negative values that mean the optimality is reached and this answer is the final answer. But if we get any positive value means we need to proceed with the sum in the next step. Now find the maximum positive penalty. Here the maximum value is 6 which corresponds to
cell. Now this cell is new basic cell. This cell will also be included in the solution.
The rule for drawing closed-path or loop.
Starting from the new basic cell draw a closed-path in such a way that the right angle turn is done only at the allocated cell or at the new basic cell. See the below images:
Assign alternate plus-minus sign to all the cells with right angle turn (or the corner) in the loop with plus sign assigned at the new basic cell.
Consider the cells with a negative sign. Compare the allocated value (i.e. 200 and 250 in this case) and select the minimum (i.e. select 200 in this case). Now subtract 200 from the cells with a minus sign and add 200 to the cells with a plus sign. And draw a new iteration. The work of the loop is over and the new solution looks as shown below.
Check the total number of allocated cells is equal to (m + n – 1). Again find u values and v values using the formula
is the cost value only for allocated cell. Assign
then we get
. Similarly, we will get following values for
Find the penalties for all the unallocated cells using the formula
There is one positive value i.e. 1 for
. Now this cell becomes new basic cell.
Now draw a loop starting from the new basic cell. Assign alternate plus and minus sign with new basic cell assigned as a plus sign.
Select the minimum value from allocated values to the cell with a minus sign. Subtract this value from the cell with a minus sign and add to the cell with a plus sign. Now the solution looks as shown in the image below:
Check if the total number of allocated cells is equal to (m + n – 1). Find u and v values as above.
Now again find the penalties for the unallocated cells as above.
All the penalty values are negative values. So the optimality is reached. Now, find the total cost i.e.
(250 * 1) + (200 * 2) + (150 * 5) + (50 * 3) + (200 * 3) + (150 * 2) = 2450
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