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Question 24 of 32
Q.

Three new machines M1M_1, M2M_2, M3M_3 are to be installed in a machine shop. There are four vacant places A, B, C, D. Due to limited space, machine M2M_2 can not be placed at B. The cost matrix (in hundred rupees) is as follows:

MachinesPlaces
ABCD
M1M_113101211
M2M_215-1320
M3M_357106
Determine the optimum assignment schedule and find the minimum cost.
Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2024Subjective· 3mImportance★★★★★
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Balance with a dummy machine and block M2–BM_2\text{–}B; the Hungarian method gives M1→B, M2→C, M3→AM_1\to B,\ M_2\to C,\ M_3\to A with minimum cost 2828 hundred =₹2800=₹2800.

Step 1 — set up: the problem is unbalanced (3 machines, 4 places) and restricted (M2M_2 cannot go to BB). Add a dummy machine M4M_4 with all costs 00, and put a prohibitive cost ∞\infty at M2–BM_2\text{–}B:

ABCD
M1M_113101211
M2M_215∞\infty1320
M3M_357106
M4M_40000

Step 2 — row reduction (subtract row minima 10,13,5,010,13,5,0):

ABCD
M1M_13021
M2M_22∞\infty07
M3M_30251
M4M_40000

Every column already contains a 00, so no column reduction is needed.

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