00:01
High from the constraint given that consider the given simplex problem.
00:07
Minimized z is equal to w is equal to 6 x1 plus 4 x2.
00:16
So, in the question it is given that y1 and y2.
00:19
So, instead of y1 and y2 i have taken here it as x1 and x2.
00:24
Then the subject to the constraints will be 6 y1 plus y2 which is greater than or equal to 60 and 15 y1 plus y2 which is greater than or equal to 105.
00:37
So, here y1 and y2 all are greater than or equal to 0.
00:41
So, here we need to find the solution of y1 and y2 and x1 and x2.
00:47
So, here instead of y1 and y2 i have taken here it as x1 and x2.
00:52
So, change it to x1 and x2.
00:53
So, here minimized z is equal to 6 x1 plus 4 x2 and subject to the constraints or greater than or equal to symbol.
01:09
So, we should add the artificial variable and subtract the surplus variable.
01:14
So, that the minimized z is plus 0 s1 plus 0 s2 minus m a1 minus m a2.
01:29
So, here this can be written as 6 x1 plus x2 plus a1 minus s1 which is equal to 60 and 15 x1 plus x2 minus s2 plus a2 is equal to 105.
01:51
So, here all the values are greater than or equal to 0.
01:56
Now, let us move on to the first iteration.
01:58
So, here b pb xb x1 x2 s1 s2 a1 a2 and for the iteration 1 we should add the artificial variable force a1 a2 cj is coefficient of the minimized z value.
02:12
So, 6 4 0 0 minus m minus m.
02:15
Here z is equal to 1 minus 165 m and let it be cb is the minus m minus m that is the coefficient of a1 and a2 in the minimized z equation.
02:31
So, xb is nothing but 60 and 105.
02:35
So, a variable coefficient of the variable x1 x2 s1 s2 a1 a2 are represented here.
02:42
So, zj is minus 21 m minus 2 m m m minus m minus m.
02:51
Now zj minus cj when we subtract zj from cj so we obtain minus 21 m minus 6 minus 2 m minus 4 m m 0 0.
03:01
So, here negative minimum that is zj minus cj is negative minimum in minus 21 m minus 6 and its column index is 1.
03:15
Therefore, so from this we need to pick up the greater value though that is 15.
03:22
So, we need to find the minimum ratio.
03:25
So, here 60 by 6 and 105 by 15.
03:29
So, we obtain the minimum ratio is 7.
03:31
So, we have pick up the 15.
03:34
So, here we need to entering the variable x1 since the minimum ratio is 7 and its row index is 2.
03:41
So, leaving the basic variable a2.
03:44
So, that here the pivot element is 15...