CMA Students Newsletter (For Intermediate Students)

Size: px

Start display at page:

Download "CMA Students Newsletter (For Intermediate Students)"

Roy Phillips
6 years ago
Views:

Special Edition on Assignment Problem An assignment problem is a special case of transportation problem, where the objective is to assign a number of resources to an equal number of activities so as

have varying degrees of efficiency for performing different activities such as job. Therefore the cost, profit or time of performing different activities is different.

1 Special Edition on Assignment Problem An assignment problem is a special case of transportation problem, where the objective is to assign a number of resources to an equal number of activities so as to minimize total cost or maximize total profit of allocation. The problem of assignment arises because available resources such as men, machines; etc. have varying degrees of efficiency for performing different activities such as job. Therefore the cost, profit or time of performing different activities is different. Thus, the problem is how the assignment should be made so as to optimize the given objective. Some of the problems where the assignment technique may be useful are: assignment of workers to machines; salesman to different sales areas; clerks to various checkout counters; classes to rooms, etc. Hungarian Method The Hungarian method (developed by Hungarian mathematician D. Konig) of assignment provides us with an efficient method of finding the optimal solution without having to make a direct comparison of every solution. It works on the principle of reducing the given cost matrix to a matrix of opportunity costs. Opportunity costs show the relative penalties associated with assigning resource to an activity as opposed to making the best or least cost assignment. If we can reduce the cost matrix to the extent of having at least one zero in each row and column, then it will be possible to make optimal assignments (opportunity costs are all zero). The procedure involves the following steps: Step 1: Subtract the smallest number in each row from all the numbers in that row. It is called Row Operation. Step 2: After step 1, from the reduced Matrix, Subtract the smallest numbers of each column from every number in that column. It is called Column Operation. Step 3: Draw the minimum number of vertical and horizontal (not diagonal) lines, to join maximum number of zeros in the reduced matrix. If the number of lines equals either the number of column or number of rows in the table, then the solution is optimal. If the number of lines drawn is smaller than number of rows and column, It is called Degeneracy. Proceeds to step 4 to resolve the degeneracy. 1

2 Step 4: Subtract the smallest number not covered by a line from every uncovered number and add the same number to the numbers which are placed at the intersections of the horizontal and vertical lines. Do not alter the elements through which only one line passes. Repeat step 3 and continue until an optimal assignment reached. Step 5: Given the optimal solution, make the project assignments (indicated by zero elements), as explained below: (a) Starting with first row, examine all rows of matrix in step 3 or 4 in turn until a row containing exactly one zero is found. Surround this zero by indication if an assignment there. Draw a vertical line through the column containing this zero. This eliminates any confusion of making any further assignments in that column. Process all the rows in this way. (b) Apply the same treatment to columns also. Starting with the first column, examine all columns until a column containing exactly one zero is found. Mark around this zero and draw a horizontal line trough the row-containing this marked zero. Repeat steps 5 a and b, until one of the following situations arises: (i) No unmarked (column) or uncovered (by a line) zero is left. (ii) There may be more than one unmarked zero in one column or row. In this case put around one of the unmarked zero arbitrarily and pass 3 lines in the cells of the reflecting zeros in its row and column. Repeat the process until no unmarked zero is left in the matrix. Problem 1: A solicitors firm employs typists on hourly piece-rate basis for their daily work. There are five typists for service and their charges and speeds are different. According to an earlier understanding only one job is given to one typist and the typist is paid for full hours even if he works for a fraction of an hour. Find the least cost allocation for the following data: Typist Rate per hour (`) No. of pages typed/hour A 5 12 B 6 14 C 3 8 D 4 10 E 4 11 P Q R S T No. of pages Solution: A B C D E

3 Row Operation A B C D E Column Operation A B C D E A B C D E As the minimum numbers of lines are not equal to order of matrix, let s take step to increase the number of zeros. A B C D E A B C D E

4 As the minimum numbers of lines are not equal to order of matrix, let s take step to increase the number of zeros. A B C D E A B C D E As the minimum number of lines are equal to order to matrix, optimal assignment should be made Optimal Assignment A B C D E Typist Job Cost (`) A T 75 B R 66 C Q 66 D P 80 E S 112 Total 399 Alternative Optimal assignment A B C D E

5 Typist Job Cost(`) A T 75 B R 66 C Q 114 D P 80 E S 64 Total 399 Unbalanced Assignment Problem In the number of jobs is not equal to number of persons the problem is said to be unbalanced. In this case we make it balanced by adding either dummy row or column with zero cost value of each element. Problem 2: Five swimmers are eligible to compete in a relay team which is to consist of four swimmers swimming four different swimming styles: back stroke, breast stroke, free style and butterfly. The time taken by the five swimmers Anand, Bhaskar, Chandru, Dorai and Easwar - to cover a distance of 100 meters in various swimming styles are given below in minutes: seconds. Anand swims the back stroke in 1:09, breast stroke in 1:15 and has never competed in free style or butterfly. Bhaskar is a free style specialist averaging 1.01 for the 100 meters but can also swim breast stroke in 1:16 and butterfly in 1:20. Chandru swims all four styles - back 1:10; butterfly 1:12, free style 1:05 and breast stroke 1:20. Dorai swims only the butterfly 1:11 while Easwar swims back stroke 1:20, the breast stroke 1:16, the free style 1:06 and the butterfly 1:10. Which swimmers should be assigned to which swimming style? Who will not be in relay? Solution: In this problem first of all we should introduce a dummy column, so that the above unbalanced matrix becomes a balanced matrix. Back stroke Breast stroke Free Style Butterfly Dummy Anand Bhaskar Chandru Dorai Easwar Row Operation is not required as there is zero in each row. Column Operation Back stroke Breast stroke Free Style Butterfly Dummy Anand Bhaskar Chandru Dorai Easwar

6 Back stroke Breast stroke Free Style Butterfly Dummy Anand Bhaskar Chandru Dorai Easwar As the minimum numbers of lines are not equal to order of matrix, let s take step to increase the number of zeros. Back stroke Breast stroke Free Style Butterfly Dummy Anand Bhaskar Chandru Dorai Easwar Back stroke Breast stroke Free Style Butterfly Dummy Anand Bhaskar Chandru Dorai Easwar As the minimum number of lines are equal to order of matrix, optimal assignment should be made. Optimal Assignment Back stroke Breast stroke Free style Butterfly Dummy Anand Bhaskar Chandru Dorai Easwar Swimmer Activity Assigned Duration (seconds) Anand Best stroke 75 Bhaskar Free style 61 Chandru Back stroke 70 Dorai Dummy 0 Easwar Butterfly 70 Total 276 6

7 Maximization case in Assignment Problem: In this case profits/productions are given instead of costs. In this case, the given matrix is converted into an opportunity loss matrix by subtracting all the elements of the matrix from the largest element. Problem 3: Solve the following Assignment problem. The data in the table refers to production in units: Operators A B C D Solution: In this problem first of all we should introduce a dummy column, so that the above unbalanced matrix becomes a balanced matrix. Operators A B C D Dummy Opportunity Loss matrix Operators A B C D Dummy Row Operation Operators A B C D Dummy Column Operation Operators A B C D Dummy

8 Operators A B C D Dummy As the minimum number of lines are equal to order of matrix, optimal assignment should be made. Optimal Assignment Operators A B C D Dummy Computing maximum production: Operators Production (units) 1 A 10 2 B 10 3 C 9 4 Dummy 0 5 B 9 Total 38 For Advanced Level Learning: Problem 4: Air-pacific airways operating 7 days a week has given the following time-table. Crews must have a minimum layover of 5 hours between flights. Obtain the pairing flight and minimizes layover time away from home. For any given pairings the crew will be based at the city that results in the smaller layover. Kolkata-Bangkok Bangkok-Kolkata Flight Number Department Arrive Flight Number Department Arrive KX1 05:15 AM 07:15 AM BX1 07:15 AM 09:15 AM KX2 07:15 AM 09:15 AM BX2 08:15 AM 10:15 AM KX3 01:15 AM 03:15 AM BX3 01:15 AM 03:15 AM KX4 07:15 AM 09:15 AM BX4 06:15 AM 08:15 AM 8

9 Solution: Let us first assume that the crew is based at Kolkata. The flight KX1, which starts from Kolkata at 05:15 AM reaches Bangkok is 07:15 AM. The Schedule time for the flight at Bangkok is 7:15 AM. Since the Minimum layover time for crew is 5 hours, this flight can depart only on the next day i.e. the layover time will be 24 hours similarly layover times for other flights are also calculated and given in the following table; Crew based at Kolkata Flight No. BX1 BX2 BX3 BX4 KX KX KX KX The layover time for various flight connections when crew is assumed to be based at Bangkok are similarly calculated in the following table. Crew based at Bangkok Flight No. BX1 BX2 BX3 BX4 KX KX KX KX Now since the crew can be based at either of the laces, minimum layover can be obtained for different flight numbers by selecting the corresponding lower Value out of the above two tables. The resulting table is as given below: Flight No. BX1 BX2 BX3 BX4 KX1 20* 19* 6 9* KX2 22* 21* 16* 9 KX * 17* KX4 10* 9* (*) with an entry in the above table indicates that the corresponds to layover time when the crew is based at Bangkok. Now we can apply the assignment algorithm to find optimal solution. Subtracting the minimum element of each row from all the elements of that row, we get the following matrix. Flight No. BX1 BX2 BX3 BX4 KX KX KX KX Since there is a zero in each column, there is no need to perform column reduction. The minimum number of lines to cover all zeros is four which is equal to the order of the matrix. Hence, the above table will given the optimal solution. The assignment is made below: Flight No. BX1 BX2 BX3 BX4 KX KX KX KX

10 The optimal assignment is From Flight No. To Flight No. City at which crew is Layover(hrs) based KX1 BX3 Kolkata 6 KX2 BX4 Kolkata 9 KX3 BX1 Kolkata 16 KX4 BX2 Bangkok 9 40 hours 10

The Assignment Problem

CHAPTER 12 The Assignment Problem Basic Concepts Assignment Algorithm The Assignment Problem is another special case of LPP. It occurs when m jobs are to be assigned to n facilities on a one-to-one basis