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Question 27 of 32
Q.

Solve the following assignment problem to maximization:

IIIIIIIVV
11824192023
21921201822
32223202123
42018211919
51822232221
Step - I
Subtract the Smallest element of each row from every element of that row
IIIIIIIVV
------------------
106124
213203
323013
420311
504543
Step - II
Subtract the smallest element of each column from every element of that column:
IIIIIIIVV
------------------
106124
213203
323012
420310
504542
Step - III
Draw minimum number of lines covering all zeros.
IIIIIIIVV
------------------
106124
213203
323012
420310
504542
Step - 1V
The smallest uncovered element is 1, which is to be subtracted from all uncovered elements and add it to all elements which lie at the intersection of two lines:
IIIIIIIVV
------------------
1050□\square3
223203
3330□\square2
430310
50343□\square
Step - V
Draw minimum number of lines that are required to cover all zeros:
IIIIIIIVV
------------------
105013
223203
333012
430310
503431
Here minimum number of Lines ≠ order of matrix.
Step - VI
Find smallest uncovered element (1). Subtract this number from all uncovered elements and add it to all elements which lie at the intersection of two lines:
IIIIIIIVV
------------------
104002
233303
33□\square00□\square
430310
502430
Now minimium number of lines = order of matrix.
The optimal assignment can be made.
Optimal solution is
1 → I
2 → IV
3 → □\square
4 → □\square
5 → V
Minimum value = □\square
Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2025Subjective· 4mImportance★★★★★
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Because the objective is to maximize, convert the profit matrix into a loss matrix using 24−aij24-a_{ij} (subtract from the largest entry), then apply row reduction, column reduction and the Hungarian method. A complete set of five independent zeros gives the optimal assignment 1→II, 2→V, 3→I, 4→III, 5→IV1\to\text{II},\,2\to\text{V},\,3\to\text{I},\,4\to\text{III},\,5\to\text{IV} with maximum profit 111111.

Given profit matrix (rows == jobs 11–55, columns == machines I–V):

IIIIIIIVV
11824192023
21921201822
32223202123
42018211919
51822232221

Step 1 — Convert to a minimization problem. The largest entry is 2424. Replace each element by 24−aij24-a_{ij} to get the opportunity-loss matrix:

IIIIIIIVV
160541
253462
321431
446355
562123

Step 2 — Row reduction. Subtract the smallest entry of each row (row minima 0,2,1,3,10,2,1,3,1):

IIIIIIIVV
160541
231240
310320
413022
551012

Step 3 — Column reduction. Subtract the smallest entry of each column (column minima 1,0,0,1,01,0,0,1,0):

IIIIIIIVV
150531
221230
300310
403012
541002

Step 4 — Make the assignment (Hungarian method). Assign one zero per row and column:

  • Row 1 has a single zero at II ⇒1→II\Rightarrow 1\to\text{II}. …

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