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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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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