Advertisements
Advertisements
Question
Suggest optimum solution to the following assignment. Problem, also find the total minimum service time.
Service Time ( in hrs.)
| Counters | Salesmen | |||
| A | B | C | D | |
| W | 41 | 72 | 39 | 52 |
| X | 22 | 29 | 49 | 65 |
| Y | 27 | 39 | 60 | 51 |
| Z | 45 | 50 | 48 | 52 |
Advertisements
Solution
This problem is already 4 x 4.
Select the smallest element in each row and subtract it from every element in each row :
Select the smallest element in each column of the above matrix and subtract it form every element in that column.
Draw the minimum lines covering all zeros.
Minimum lines covering all zeros is not equal to the order of the matrix.
Minimum uncovered value 2 is subtracted from uncovered values and added to values at intersection of the lines.
Draw minimum lines covering all the zeros.
Minimum lines covering all the zeros equal to the order of the matrix.
∴ Allocation of counters can be done.
The allocation of the counters to the salesmen is
W → C, X → B, Y → A, Z → D
The minimum time = 39 + 29 + 27 +52 = 147 Hrs.
APPEARS IN
RELATED QUESTIONS
A job production unit has four jobs A, B, C, D which can be manufactured on each of the four machines P, Q, R and S. The processing cost of each job is given in the following table:
|
Jobs
|
Machines |
|||
|
P |
Q |
R |
S |
|
|
Processing Cost (Rs.)
|
||||
|
A |
31 |
25 |
33 |
29 |
|
B |
25 |
24 |
23 |
21 |
|
C |
19 |
21 |
23 |
24 |
|
D |
38 |
36 |
34 |
40 |
How should the jobs be assigned to the four machines so that the total processing cost is minimum?
Solve the following minimal assignment problem and hence find the minimum value :
| I | II | III | IV | |
| A | 2 | 10 | 9 | 7 |
| B | 13 | 2 | 12 | 2 |
| C | 3 | 4 | 6 | 1 |
| D | 4 | 15 | 4 | 9 |
Solve the following minimal assignment problem :
| Machines | A | B | C | D | E |
| M1 | 27 | 18 | ∞ | 20 | 21 |
| M2 | 31 | 24 | 21 | 12 | 17 |
| M3 | 20 | 17 | 20 | ∞ | 16 |
| M4 | 21 | 28 | 20 | 16 | 27 |
Determine `l_92 and l_93, "given that" l_91 = 97, d_91 = 38 and q_92 = 27/59`
In a factory there are six jobs to be performed each of which should go through two machines A and B in the order A - B. The processing timing (in hours) for the jobs arc given here. You are required to determine the sequence for performing the jobs that would minimize the total elapsed time T. What is the value of T? Also find the idle time for machines · A and B.
| Jobs | J1 | J2 | J3 | J4 | J5 | J6 |
| Machine A | 1 | 3 | 8 | 5 | 6 | 3 |
| MAchine B | 5 | 6 | 3 | 2 | 2 | 10 |
The assignment problem is said to be balanced if ______.
The objective of an assignment problem is to assign ______.
Fill in the blank :
When an assignment problem has more than one solution, then it is _______ optimal solution.
Choose the correct alternative:
The assignment problem is generally defined as a problem of ______
Choose the correct alternative:
When an assignment problem has more than one solution, then it is ______
Choose the correct alternative:
The assignment problem is said to be balanced if ______
State whether the following statement is True or False:
In assignment problem each worker or machine is assigned only one job
Three jobs A, B and C one to be assigned to three machines U, V and W. The processing cost for each job machine combination is shown in the matrix given below. Determine the allocation that minimizes the overall processing cost.
| Machine | ||||
| U | V | W | ||
| Jobs | A | 17 | 25 | 31 |
| B | 10 | 25 | 16 | |
| C | 12 | 14 | 11 | |
(cost is in ₹ per unit)
A computer centre has got three expert programmers. The centre needs three application programmes to be developed. The head of the computer centre, after studying carefully the programmes to be developed, estimates the computer time in minitues required by the experts to the application programme as follows.
| Programmers | ||||
| P | Q | R | ||
| Programmers | 1 | 120 | 100 | 80 |
| 2 | 80 | 90 | 110 | |
| 3 | 110 | 140 | 120 | |
Assign the programmers to the programme in such a way that the total computer time is least.
Choose the correct alternative:
The purpose of a dummy row or column in an assignment problem is to
A job production unit has four jobs P, Q, R, and S which can be manufactured on each of the four machines I, II, III, and IV. The processing cost of each job for each machine is given in the following table:
| 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 |
Find the optimal assignment to minimize the total processing cost.
Five wagons are available at stations 1, 2, 3, 4 and 5. These are required at 5 stations I, II, III, IV and V. The mileage between various stations are given in the table below. How should the wagons be transported so as to minimize the mileage covered?
| I | II | III | IV | V | |
| 1 | 10 | 5 | 9 | 18 | 11 |
| 2 | 13 | 9 | 6 | 12 | 14 |
| 3 | 7 | 2 | 4 | 4 | 5 |
| 4 | 18 | 9 | 12 | 17 | 15 |
| 5 | 11 | 6 | 14 | 19 | 10 |
A job production unit has four jobs P, Q, R, S which can be manufactured on each of the four machines I, II, III and IV. The processing cost of each job for each machine is given in the following table :
| 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`
A plant manager has four subordinates and four tasks to perform. The subordinates differ in efficiency and task differ in their intrinsic difficulty. Estimates of the time subordinate would take to perform tasks are given in the following table:
| I | II | III | IV | |
| A | 3 | 11 | 10 | 8 |
| B | 13 | 2 | 12 | 2 |
| C | 3 | 4 | 6 | 1 |
| D | 4 | 15 | 4 | 9 |
Complete the following activity to allocate tasks to subordinates to minimize total time.
Solution:
Step I: Subtract the smallest element of each row from every element of that row:
| I | II | III | IV | |
| A | 0 | 8 | 7 | 5 |
| B | 11 | 0 | 10 | 0 |
| C | 2 | 3 | 5 | 0 |
| D | 0 | 11 | 0 | 5 |
Step II: Since all column minimums are zero, no need to subtract anything from columns.
Step III: Draw the minimum number of lines to cover all zeros.
| I | II | III | IV | |
| A | 0 | 8 | 7 | 5 |
| B | 11 | 0 | 10 | 0 |
| C | 2 | 3 | 5 | 0 |
| D | 0 | 11 | 0 | 5 |
Since minimum number of lines = order of matrix, optimal solution has been reached
Optimal assignment is A →`square` B →`square`
C →IV D →`square`
Total minimum time = `square` hours.
