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costing query- transportation

(Guest)

i have a query from the transportation chapter of cost management. while solving a transportation problem, suppose we have the initial basic feasible solution. but there is degeneracy so we need to allocate 'e' to least cost independent cell. so my question is, what do we mean by 'independent cell'? i came across this transportation question from may 07 and the solution stated that the least cost cell was not 'independent cell' therefore 'e' was allocated to the next least cost and independent cell. anyone?

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12 Replies

CA CS Bhumika Thakkar (CA,CS) (1618 Points)
Replied 22 October 2009

werever u place d e,it shd nt form a rectangular or square loop in d intial BFS..

1 Like

LAKSHMI NARAYAN (CA Final Student) (904 Points)
Replied 22 October 2009

As per my knowledge Independent cell represents

a) an unallocated cell and

b) a cell which should fall in the regular row or column i.e, it should not fall in the dummy row or column if inserted.

Please let me know if i am correct

1 Like

(Guest)

Originally posted by :LAKSHMI NARAYAN
"	As per my knowledge Independent cell represents a) an unallocated cell and b) a cell which should fall in the regular row or column i.e, it should not fall in the dummy row or column if inserted. Please let me know if i am correct	"

both these conditions are satisfied. however i still don't understand what bhumika means by forming a loop. i have given the IBFS below. allocation is as follows-

row 1- columns 2,4,5

row 2 - column 1

row 3- column 3

row 4- column 1 & 4

11	2	8	6	2
9	9	12	9	6
7	6	3	7	7
9	3	5	6	11

now can u explain why '5' ( row 4 column 3) is chosen to allocate 'e' and not '3' (row 4 column 2)

mallika bansal (na) (120 Points)
Replied 22 October 2009

as bhumika rightly said, independent cell means it should not form a loop with the allocated cell. the cell which has least cost of 3 forms a loop with row 1 colum 2 and column 4 and row 4 column 4 whereas the next least cost 5 is unable to form any loop whatsoever, hence 5 is selected

2 Like

Ashish M (Chartered Accountant) (2731 Points)
Replied 22 October 2009

Mallika's and bhumika's answer is right and perfect......

BY USING '3', WE ARE ABLE TO MAKE A LOOP, BUT FROM THE NEXT LEAST NO. I.E. '5', LOOP CAN'T BE MADE, SO WE WILL PUT 'e' TO IT........

AND 'e' JUST MEANS 0... i.e. allocate 0 units to that cell to satisfy the equation 'm+n-1'.,.............

Christina (CA in practice) (26 Points)
Replied 22 October 2009

When allocating u and v values for the matrix u+v=Cij for allocated cells, it is not possible to allocate u & v values for column 3 & row3

e is to be allocated to the least cost unallocated cell in this column 3 & row3

evenif 3 is the least cost unallocated cell in the matrix, it is not in this column3 or row3

Therefore 5 is selected- which is in the column 3

(Guest)

ok got it now. thanks a lot everyone.

work is worship (Service) (2728 Points)
Replied 26 October 2009

Hey one question. how is loop formed.. is it formed using the already located cell.

CA Punit (CA) (1775 Points)
Replied 28 October 2009

i think d final allocation will be same even if u choose least cost independent cell as 3' (row 4 column 2)

kaka (none) (21 Points)
Replied 17 September 2013

Hey guys so if this is the initial basic solution can we insert 'e' in any of the 400 cell(i.e; row 1 column 2 or row 3 column 3)? The cells in colour are the allocated cells.