- By N.W.CR:- Start allocating at north corner cell at the cell (1, 1): X11 = Min (a1 b1)
= min (5, 7) = 5
Here O1 is satisfied, next allocate at (2, 1) as:
X21 = Min (a2, b1 – X11) = Min (8, 7-5)
= Min (8, 2) = 2
D1 satisfied and next allocation is made at (2, 2):
X22 = Min (a2 – X21, b2)
= Min (8-2, 9)
= Min (6, 9) = 6
O2 availabilities are allocated, next allocate at (3, 2): X32
= Min (a3, b32 – X22)
= Min (7, 9 – 6) = Min (7, 3) = 3
D2 Satisfied, next allocate at (3, 3) as:
X33 = Min (a3 – X32, b3)
= Min (7-3, 18) = min (4, 18) = 4
O3 Satisfied, Allocate at (4, 3) as:
X43 = Min (a4, b3 – X33)
= Min (14, 18, -4) = Min (14,14) = 14
The Initial Basic feasible solution (I. B. F.*) is:
