i think at the end when we were proceeding for the last step. we had value of zj-cj as (1,0,2,1) for (x1,x2,s1,s2) but at the last we have (x1,x2) as basic variables and you said in middle of the video that the net evaluation (zj-cj) for basic variables are always zero so accordingly we must have values of zj-cj as (0,0,2,1) and not (1,0,2,1). i think you just had a writing error.
Because zj-cj for x1 variable is more -ve (or we may say minimum), so the value of z is improved the most. If we choose x2 here then increment in z will be less in comparison with x1.
Not the coefficient matrix A has order 2*4, and there are total four decision variables X1, X2, S1, s2. Now as coefficient matrix has order 2*4, the rank of this matrix cannot exceed 2. Also notice there is a submatrix of order 2*2in A which is an identity matrix so this gives us information that rank of A is 2 and so this becomes the order of Basis matrix 2*2. Now to apply simplex, we usually make first simplex table with identity column, which corresponds to S1, S2. I hope above explanation clear your doubts. You can look at Basic Solutions of LPP videos for more background.
I think we can also get a 0 in R3 by R3^-R2^. However i am getting 3 in place of 1 by doing Z1- C1. Rest my entire result is correct. Don't know why???
For maximization, enter a variable with most -ve zj-cj. For Minimization, enter a variable with most +ve zj-cj value. For the proof of this you can watch some last videos on mathematical background for simplex method.
@@MathPod ok mam, but as u said most -ve. so here most -ve will be -3 rather than -4. isn't it? so, according to that x2 should be the entering variable. isn't it?
@@LUCKYSINGH-vz2pd Well, here 'most' stand for maximum value, and then attached a negative sign. It's one of the standard ways of saying in optimization techniques literature. So most -ve is -4.
![Noob To Max With DRAGON REWORK In Blox Fruits [FULL MOVIE]](http://i.ytimg.com/vi/LnBMOinoOvA/mqdefault.jpg)
22:02
it should be
Z 0
S1 0
X1 1
And not
1
0
1
Good catch!
i think at the end when we were proceeding for the last step. we had value of zj-cj as (1,0,2,1) for (x1,x2,s1,s2) but at the last we have (x1,x2) as basic variables and you said in middle of the video that the net evaluation (zj-cj) for basic variables are always zero so accordingly we must have values of zj-cj as (0,0,2,1) and not (1,0,2,1). i think you just had a writing error.
Maximization? More like "Magnificent demonstration!" 👍
At 13:46, why did we replace s2 with x1, why didn't we replace s1 with x1?
The value of z1 - c1 = 0, not 1.
Thanks for the great effort.
22:42 how can I identify which biases to put in table ?
the same here 14:14
bro did you get the solution i'm stuck here too
i know the one she wrote at 14:14 but not at 22:42
Can you please explain why we have taken next bv as (S1, X1). I mean why specifically X1 & not X2? Thanks
Because zj-cj for x1 variable is more -ve (or we may say minimum), so the value of z is improved the most. If we choose x2 here then increment in z will be less in comparison with x1.
🙏
I cannot understand the last step of solutions
To find R1,R2,R3 is difficult to understand me.
Mam i have'nt understand : BXb=b , (B)2*2 *=b, B=(1 0 01) how u did this and why this is 2*2 only.. does it depends on something. plz help me out
Not the coefficient matrix A has order 2*4, and there are total four decision variables X1, X2, S1, s2.
Now as coefficient matrix has order 2*4, the rank of this matrix cannot exceed 2. Also notice there is a submatrix of order 2*2in A which is an identity matrix so this gives us information that rank of A is 2 and so this becomes the order of Basis matrix 2*2. Now to apply simplex, we usually make first simplex table with identity column, which corresponds to S1, S2.
I hope above explanation clear your doubts. You can look at Basic Solutions of LPP videos for more background.
Please correct first word as Note*
I think we can also get a 0 in R3 by R3^-R2^. However i am getting 3 in place of 1 by doing Z1- C1. Rest my entire result is correct. Don't know why???
how are you finding entering variable?
For maximization, enter a variable with most -ve zj-cj. For Minimization, enter a variable with most +ve zj-cj value. For the proof of this you can watch some last videos on mathematical background for simplex method.
@@MathPod ok mam, but as u said most -ve. so here most -ve will be -3 rather than -4. isn't it? so, according to that x2 should be the entering variable. isn't it?
@@LUCKYSINGH-vz2pd Well, here 'most' stand for maximum value, and then attached a negative sign. It's one of the standard ways of saying in optimization techniques literature. So most -ve is -4.
@@MathPod Oooo, I got it. thank you, ma'am
Thank you ma'am
stop explaining 5 different methods in one video of the same thing just stick to one method and teach that is all you have to do
Yes
Thanks Mam :-)