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Problem in Transportation

Heena (Partner in Lokesh Khandelwal & co0) (204 Points)

28 March 2011

Hi,This is my question to CA final new course students.Kindly see illustration on page no.12.25 of Module of Advanced Management volume 1.On the last step in which a loop is to be formed because of negative value of non assigned cell.Loop is to be formed from the most negative value(there is printing mistake i suppose 10 (Dummy;B)is negative value.Loop has also involved e.Then in next step given in module why e is placed in Dummy;B and 10 is placed in 1;B?Please i am self studying and dont know how to solve my problem.Please Help.

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MOHIT MAHAJAN (LCS, ACA) (2050 Points)
Replied 28 March 2011

Hmmm...in my opinion -

loop will start from 100(5B) TO 10(1B) TO 0(1E) TO 100(5E). Now the -ve values will be at (1b) & (5E) i.e. -e & -500. Now we have to adjust the allocation having least -ve value which will be -e.

whenever e is reduced or added to any value the value will remain unchanged. Only difference will be that e-e will give nil and the nil column will be replaced by e.

In (1B) we will have e-e hence no allocation. In (1E) 500+E will remain 500. In (5E) 500-e will remain 500 nd in (5B) where there is no allocation e will be added hence e will come there.

I think in last table there should be no allocation at (1B), rest is ok.

I hope u understood wht i m trying to say, if there is any dount then do ask.

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