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प्रश्न
Give mathematical form of Assignment problem
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उत्तर
Consider the problem of assigning n jobs to n machines (one job to one machine).
Let Cij be the cost of assigning ith job to the jth machine and xij represents the assignment of ith job to the jth machine.
Then, xij = `{{:(1",", "if" "i"^"th" "job is assigned to" "j"^"th" "machine"),(0",", "if" "i"^"th" "job is assigned to" "j"^"th" "machine"):}`
| Machines | ||||||
| 1 | 2 | … | n | Supply | ||
| 1 | `""^((x_11))"C"_11` | `""^((x_12))"C"_12` | … | `""^((x_(1n)))("C"_(1n))` | 1 | |
| 2 | `""^((x_21))"C"_21` | `""^((x_22))"C"_22` | … | `""^((x_(2n)))("C"_(2n))` | 1 | |
| Jobs | : | : | : | : | 1 | |
| m | `""^((x_"ij"))"C"_("n"1)` | `""^((x_(m2)))"C"_("n"1)` | … | `""^((x_"ij"))("C"_"nn")` | 1 | |
| Demand | b1 | b2 | … | bn | ||
xij is missing in any cell means that no assignment is made between the pair of job and machine.
i.e xij = 0.
xij is presents in any cell means that an assignment is made their.
In such cases xij = 1
The assignment model can written in LPP as follows:
Minimize Z = `sum_("i" = 1)^"m", sum_("j" = 1)^"n" "C"_"ij" "X"_"ij"`
Subject to the constrains
`sum_("i" = 1)^"n" "X"_"ij"` = 1, j = 1, 2, …. n
`sum_("i" = 1)^"n" "X"_"ij"` = 1, i = 1, 2, …. n and xij =0 or 1 for all i, j
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| P | Q | R | S | |
| A | 31 | 25 | 33 | 29 |
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| Job | Machines (Processing cost in ₹) |
|||
| I | II | III | IV | |
| P | 31 | 25 | 33 | 29 |
| Q | 25 | 24 | 23 | 21 |
| R | 19 | 21 | 23 | 24 |
| S | 38 | 36 | 34 | 40 |
Complete the following activity to find the optimal assignment to minimize the total processing cost.
Solution:
Step 1: Subtract the smallest element in each row from every element of it. New assignment matrix is obtained as follows :
| Job | Machines (Processing cost in ₹) |
|||
| I | II | III | IV | |
| P | 6 | 0 | 8 | 4 |
| Q | 4 | 3 | 2 | 0 |
| R | 0 | 2 | 4 | 5 |
| S | 4 | 2 | 0 | 6 |
Step 2: Subtract the smallest element in each column from every element of it. New assignment matrix is obtained as above, because each column in it contains one zero.
Step 3: Draw minimum number of vertical and horizontal lines to cover all zeros:
| Job | Machines (Processing cost in ₹) |
|||
| I | II | III | IV | |
| P | 6 | 0 | 8 | 4 |
| Q | 4 | 3 | 2 | 0 |
| R | 0 | 2 | 4 | 5 |
| S | 4 | 2 | 0 | 6 |
Step 4: From step 3, as the minimum number of straight lines required to cover all zeros in the assignment matrix equals the number of rows/columns. Optimal solution has reached.
Examine the rows one by one starting with the first row with exactly one zero is found. Mark the zero by enclosing it in (`square`), indicating assignment of the job. Cross all the zeros in the same column. This step is shown in the following table :
| Job | Machines (Processing cost in ₹) |
|||
| I | II | III | IV | |
| P | 6 | 0 | 8 | 4 |
| Q | 4 | 3 | 2 | 0 |
| R | 0 | 2 | 4 | 5 |
| S | 4 | 2 | 0 | 6 |
Step 5: It is observed that all the zeros are assigned and each row and each column contains exactly one assignment. Hence, the optimal (minimum) assignment schedule is :
| Job | Machine | Min.cost |
| P | II | `square` |
| Q | `square` | 21 |
| R | I | `square` |
| S | III | 34 |
Hence, total (minimum) processing cost = 25 + 21 + 19 + 34 = ₹`square`
