How to Solve a Linear Programming Problem Using the Two Phase Method

[@superchore1]

This was one of the best video on 2 phase method, in fact best series addressing the lpp problems... Thanks a lot :D

[@alejandracasas9621]

You made this method a lot less scary. Thank you so much

[@Saqer36]

4:34 very important to know.

[@isiyaramuthumuni4676]

Thank you so much for this tutorial, your approach makes the two-phase method a lot easier to understand!

[@aiolosdaftsis]

I love you so much. You helped me pass this course. I really dont know what I would have done had you not uploaded this. Thank you

[@NejiraTrako]

I love your tutorials, they are so helpful. thank you

[@secludedone3781]

Thank you very much Dr. Shokoufeh , it was helpful for me. My answer also is 900 (The value of the objective function). But, in my solution, all values of C - Z row are negative or zero.

[@sxmirzaei]

Its because we don't calculate C-Z, our row is Z-C which is the negative of (C-Z), so your answer is correct as well!

[@secludedone3781]

Shokoufeh Mirzaei Yeah, I see ! thank you again doctor :)

[@chanthawitanuntasethakul8289]

Thanks. You really help me in the exam.

[@muaazhammad4446]

thanxs. (ur student from Pakistan)

[@erfaaalii1440]

Obviously, by looking at your name, I can bet that you're Iranian(I'm Iranian as well) Your tutorial is excellent

[@pren5948]

thanks, it is very useful for me.

[@mahbubehparhizkar862]

دستتون درد نکنه. خیلی خوب بود

[@sxmirzaei]

مرسی‌

[@fatihyalcn8796]

Hi Professor. What about if we obtain equal two ratios on the Ratio Test ? How are we able to select which one to use ?

[@manoharreddy8760]

Thank you very much. Please upload a vedio on graphical method for Theory of games

[@syshYarak]

10:15 why change the value of Z from (-4 )to 0 ???

[@sxmirzaei]

+HAFIDI SOUHAIB because X2 is a basic variable, meaning that it is one of the variables for which we are solving the system of equations. Therefore, its coefficient in the row of z must be zero.

[@qoribmunajat]

thanks a lot :) I understand it well from your lecture. Please give another video of explanation, if you don't mind, regarding with unbounded negative constraint. Thanks :)

[@syshYarak]

+qorib munajat why change the value of Z from (-4 )to 0 ???

[@riyoua]

superb video .. thanks

[@Karan-zm3qp]

Great video, however I think you may want to change the font you use, cursive can be a bit less organized than regular text and mistakes can more easily happen when the problem is in an unused font.

[@sxmirzaei]

+Karan Kadaru that is my hand writing. I ll try to make it more legible next time!

[@masoudaghaie4435]

عالی واقعا مشکلم رو حل کرد ممنون

[@liwaanaim9142]

hi do you have an example where you use revised simplex method

nicely done

[@Asmodath]

Thank, you for this tutorial, it saved my arse...

[@tszchungpoon5798]

If the question is minimize, We do maximize approach in phase 2?

[@sxmirzaei]

Tsz Chung Poon no, for phase 2 you always use your original objective function. So , if your problem is a max, your second phase will have a max OF. However, your phase I objective is always a min problem , no matter if your OF is a max or min.

[@syshYarak]

+Shokoufeh Mirzaei why change the value of Z from (-4 )to 0 ???

[@FPrimeHD1618]

So this basically helps you find a basis?

[@valeriebureau]

Hello, For the phase 1 , we minimize a1 , what do you mean by we need to standardize the Z function so a1 becomes negative ? Why is a1 negative? In the simplex method I saw that to be optimal in a minimization the variable in the Z function need to be positive or 0, why is it different in that case ? (We minimize a1 but we want the Z function to have negative variables) Thank you so much!

[@sxmirzaei]

+Valérie Bureau 1- in the first phase your objective function is always minimization; bcz you want to eliminate the artificial variables (or hopefully reduce them to zero ). by standardization, I mean bring all decision variables in the left hand side and constant values in the right hand side. like when you solve a systems of equations, if you do this the coefficient of a1 (not a1) becomes negative. you need to do this step to initialize the simplex table and solve it. your second question: suppose you have an objective function of max z=2x1+3x2 when you standardize (z-2x1-3x2=0) negative coeffs in the row of Z shows the variables that if we increase their values, Z will improve. For instance, in this case adding one unit of x1 will increase z by 2 units and adding 1 unit of x3 will increase z by 3 units. now if I am in a table where all z coeffs are positive or zero , (e.g. row of z: 2 3 0 0 | 120 --> you can write z+2x1+3x2=120 --> z=-2x1-3x2) this implies that there is no variable that if we increase its value, Z will improve. because positive values in the row of z are in fact negative contribution in the original objective function. you can use the same logic for minimization. hope it helps!

[@sxmirzaei]

+Valérie Bureau 1) when you standardize, it means you put all the variables in the LHS and all the constant values in the RHS. For example, for z=x1+x2, the standardized form is z-x1-x2=0 in a system of equations. 2) instead of confusing yourself with negatives and positives learn how the optimization works. if you want to minimize Z= - x1 + x2, increasing which decision variable would decrease the value of objective function? obviously X1. why? bcz it has a negative contribution to the objective function. For example if x1=10 and x2=0 , z=-10 . So, its in favor to increase x1 as much as we can to minimize the value of z. The negative coefficient appears as positive when you standardize the equation to z+x1-x2=0. in this form, when you see a positive coeff, it implies that increase in the variable will result to decrease in the value of Z. Also, the bigger the coeff the higher the contribution and that's why we choose the most positive value in the row of z for minimization. you can apply the same reasoning for a max problem.

[@syshYarak]

+Shokoufeh Mirzaei 10:15 why change the value of Z from (-4 )to 0 ???

[@db2079]

why you don't take the 4th row when you look for smallest value? 75 the one that corresponds to x2 value you only take the minimum of the division between S1 S2 S3 values and the the column correspondent to the smallest value of z

[@chiharunamine]

Why can we eliminate both the row and column of the artificial variable if it is a basic variable with value 0?

[@sxmirzaei]

+Sara Zayed correct, if the table is in the optimal state.

[@syshYarak]

+Shokoufeh Mirzaei why change the value of Z from (-4 )to 0 ???

[@PK-zi3nn]

do we always get the minimum* for the first phase?

[@sxmirzaei]

yes

[@mohamedboudouar]

thank u so much (y)

[@hassanhashemi6478]

Thanks.

[@ToStand2]

thanks

[@syshYarak]

+ToStand why change the value of Z from (-4 )to 0 ???

[@ToStand2]

+HAFIDI SOUHAIB where do you see that

[@syshYarak]

10:15 why she change the value Z = -4 to a new line with Z= 0 ???

[@syshYarak]

the intersection with X2 and (Z = -4 )

[@ToStand2]

HAFIDI SOUHAIB oh yes i see thx ; it's because x2 is a base variable (the 4 variables on the left are the base variables) and these variables can't have a negative value in the z row so she combinates z with another row to have the coeffficient >= 0 if you take a close look, you'll see that at the end of the algorithm, there is no negative value in the columns that correspond to the base variables that's what i noticed

[@amineberrada5276]

i love you.

[@Rezamath]

Thank you so much.

[@PK-zi3nn]

akh dastet dard nakone :* :* :*