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4. The president of a consulting firm wants to minimize the total number of hours it will take to complete four projects for
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Answer #1

Solution:

As no specific information is mentioned in the question, we will solve the given assignment problem by following the steps of the Hungarian Method of Task (Job) Assignment as mentioned below:

We are given the following problem-table:

A B C D Ch 15 | 16 | 12 | 18 Ge 13 13 | 15 16 | 12 Jo 15 21 20 15 Ri 17 19 12 22

(Here, we have denoted the consultants by the first two initials of their names)

Step 1:

Pick a minimum element from each row and subtract it from that row:

Hence, we get:

AB CD Ch 3 4 06 (-12) Ge 13 4 0 (-12) Jo 0 6 5 0 (-15) Ri 5 5 70 010 (-12)

Step 2:

Now, find out each column minimum element and subtract it from that column:

Hence, we get the opportunity cost table as mentioned below:

A B С D 1 se / 1 Ch 3 1 06 Ge 0 4 0 Jo 03 5 0 Ri 5 4 010 (-0) (-3) (-0) (-0)

Step 3:

Make assignment in the opportunity cost table, by following the below-mentioned substeps:

a. Identify rows with exactly one unmarked 0. Make an assignment to this single 0 by making a square ( [0] ) around it and cross off all other 0 in the same column.

b. Identify columns with exactly one unmarked 0. Make an assignment to this single 0 by making a square ( [0] ) around it and cross off all other 0 in the same rows.

c. If a row and/or column has two or more unmarked 0 and one cannot be chosen by inspection, then choose the cell arbitrarily.

d. Continue this process until all 0 in rows/columns are either assigned or cross off

Thus,

(1) Rowwise cell (Ch,C) is assigned, so columnwise cell (Ri,C) crossed off.

(2) Columnwise cell (Jo,A) is assigned, so rowwise cell (Jo,D) crossed off.

(3) Columnwise cell (Ge,B) is assigned, so rowwise cell (Ge,D) crossed off.

A D Bc [O] 6 Ch 3 1 Ge 1 [0] 4 5 Jo [0] 3 Ri | 5 4 10

Here, Number of assignments = 3, number of rows = 4
Which is not equal, so the solution is not optimal.

Step 4:

Draw a set of horizontal and vertical lines by following the below-mentioned substeps:

a. Tick(✓) mark all the rows in which no assigned 0.

b. Examine Tick(✓) marked rows, If any 0 cell occurs in that row, then tick(✓) mark that column.

c. Examine Tick(✓) marked columns, If any assigned 0 exists in that columns, then tick(✓) mark that row.

d. Repeat this process until no more rows or columns can be marked.

e. Draw a straight line for each unmarked rows and marked columns.

Thus,

(1) Mark(✓) row Ri since it has no assignment

(2) Mark(✓) column C since row Ri has 0 in this column

(3) Mark(✓) row Ch since column C has an assignment in this row Ch.

(4) Since no other rows or columns can be marked, therefore draw straight lines through the unmarked rows Ge,Jo and marked columns C

bo C D Ch 3 1 roj [0] 6 Ge 1014 Jo 1013 $ o Ri 5 4 | 10

Step 5:

Develop the new revised table by selecting the smallest element, among the cells not covered by any line (say k = 1) Subtract k = 1 from every element in the cell not covered by a line.
Add k = 1 to every element in the intersection cell of two lines.

Ge AB C D Ch 2 2 0 0 0 5 10 5 0 Jo 03 6 0 Ri 4 309 1

Step 6:

Here, follow step 3 in order to make the assignments:

(1) Rowwise cell (Ri,C) is assigned, so columnwise cell (Ch,C) crossed off.

(2) Rowwise cell (Ch,B) is assigned, so columnwise cell (Ge,B) crossed off.

(3) Rowwise cell (Ge,D) is assigned, so columnwise cell (Jo,D) crossed off.

(4) Rowwise cell (Jo,A) is assigned

B C D Ch 2 roj 5 Ge 1 5 [0] Jo [0] 3 60 Ri 4 | 3 [O] 9


Here, Number of assignments = 4, number of rows = 4
Which is equal, so the solution is optimal:

12 Consultant Project Time (Hours) Charlie B 16 Gerald D Johny А 15 Rick С 12 Min Time (Hours) 55

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