Numericals on Unbalanced Assignment Model

(Unbalanced)

1.         A company has 4 machines to do 3 jobs. Each job can be assigned to only one machine. The cost of each job machine combination is given in the table below in Rupees. (MU, Apr. 2009)

Find an optimal assignment which will minimize the cost. Answer the following questions with justification.

(i)       Is there any alternate optimal assignment ?

(ii)      Is it possible to maintain the same minimum cost as per optimal solution if job B is assigned to either Machine II of Machine III ?

(iii)     Does any machine remain idle as per optimal assignment ?

           (Ans :  A ® I, B ® II, C ® III, D ® IV; Total Cost = 9 + 6 + 9 + 0 = 24

(i)       Alternate Soln. A ® I, B ® II, C ® III, D ® IV; Total Cost = 9 + 8 + 7 + 0 = 24

(ii)     Yes. From the above two alternate solutions it can be observed that Job B can be assigned to either Machine II or Machine III while maintaining an optimal cost of Rs. 24

(iii)    In both solutions, Job D (dummy) is assigned to Machine IV hence machine IV is idle.)

2.         Four major roads in a city badly need repairs. There are 5 contractors who are eager to do my job. The city corporation decides to give repair work of only one road to one contractor. The cost of repair is given below in thousand rupees. Find the assignment at minimum cost.

(Ans :  R₁ ® C₃, R₂ ® C₅, R₃ ® C₄, R₄ ® C₁, R₅ ® C₂

               Cost = 40 + 49 + 61 + 38 + 0 = ` 188)

3.         A finance faculty of management school decided to hold special seminars on 4 contemporary topics – leasing, portfolio management, private mutual funds and equity research. Such seminars should be held once per week in the afternoon, however scheduling these seminars (one for each topic and not more than one seminar afternoon) has to be done carefully so that the number of students unable to attend kept to the minimum, a carefully study indicates that the number of students who can not attend a particular on a specific day is as follows.

   (Ans :  Mon ® ER, Tues ® Dummy, Wed ® PM, Thurs ® PMF, Fri ® L; No. of
absentees = 20 + 0 + 20 + 20 + 10 = 70)