00:01
Hi, from the question given that consider the given constraints and the maximum c.
00:09
So, this is a linear programming problem maximum c will be minus x1 minus 2 x2 plus x3 instead of x y and z i here assumed as x1 x2 and x3.
00:28
So, and the subject to the constraints are given 2 x1 plus x2 plus x3 is less than or equal to 14 and 4 x1 plus 2 x2 plus 2 x3 which is less than or equal to 23 and 2 x1 plus 5 x2 plus 5 x3 that is also less than or equal to 30.
01:04
So, here we need to convert this into a canonical form by adding slack or surplus or artificial variable as an appropriate.
01:13
So, here the constraint 1 is in the type of less than or equal to.
01:17
So, here we need to add the slack variable constraint 2 and 3 all are less than or equal to symbol.
01:23
So, here we need to add the slack variable.
01:26
So, in maximize c is also we need to introduce the slack variable.
01:31
So, plus 0 is 1 plus 0 is 2 plus 0 is 3 here plus s1 plus s2 plus s3.
01:42
Now this symbol can be changes to equal and this is also equal and this is also equal.
01:50
Therefore x1 x2 x3 s1 s2 s3 all are greater than or equal to 0.
01:59
Now we move on to the iteration 1.
02:02
So, this is the iteration 1 in the first row we have cj that is the coefficient of x1 x2 x3 s1 s2 s3 here b cb xb x1 x2 x3 s1 s2 s3.
02:16
So, now we introduce the slack variable s1 s2 s3 and here the cb value will be 0 0 0 and xb is the equal the numbers that are present on the orange side that is 14 23 and the 13.
02:33
Here x1 is the coefficient of x1 for the constraint 1 and coefficient of x2 for the constraint 1 and this is coefficient of x3 for the constraint 1.
02:45
Similarly s1 is the coefficient of s1 in the constraint 1 here s2 s3 are 0.
02:52
In the similar manner we need to fill up the s2 and s3 values also and here first z equal to 0 therefore zj will be 0 0 0.
03:05
Now we need to find zj minus cj.
03:08
So, 0 minus 1 is 1 and 0 minus of minus 2 is 2 and 0 minus 1 is minus 1 0 0.
03:16
So, here the minimum value will be this one...