How to Solve a Linear Programming Problem Using the Big M Method

In this lesson we learn how to solve a linear programming problem using the big M method. Change the setting of your youTube to HD for the best quality.

[@Jedilance8]

Definitely the best tutorial in between those bunch of big m videos. Thank you so much.

[@mariamilingou8217]

Fantastic tutorial!Thank you very much for your videos!I will hopefully pass my final exams thanks to you!

[@emptycl0ud9]

Thanks for the wonderful tutorials. Best ones I've used to learn.

[@pradeepgaur5362]

this will definately help me to clear my CT......thank you ma'm

[@ashiqahamed9775]

Thanks Great Teaching...................... Best In KZbin For BIG M method :)

[@herryfrd2740]

Very good job! Best tutorial of big M on youtube

[@shauryamalik3455]

i have my Operation Research exam in next 2 hrs and here I am watching this video

[@tejasvi0claw]

Thank you SO much! This was perfect.

[@nerdumdigitalis]

Brilliant, Thank you very much Shokoufeh :)

[@perkelele]

I'm used to reading cursive, so i find your handwriting legible. But I think it may indeed help others who aren't accustomed to cursive if you were to write in capitals. Nice handwriting though. Great videos too, it was a clear explanation

[@MuhammadSalman-om4rn]

Very helpful, thanks!

[@cansengul5915]

you are really good, thank you professor

[@khaledgh2464]

thank you so much you did great job

[@luckymact1085]

thank you ma'am, you are awesome.

[@lillianreyes5465]

Great Video!

[@andreluisal]

Excellent video!

[@poorianorouzi214]

Thanks professor, please upload some videos about sensitivity and shadow price, kheylii khub boodm mamnun ,goftam farsi am tashakor karde basham ;)

[@sankarkumar7898]

Mirzaei, this session was good and very useful. Did you posted any video relating Gomory's Cutting plane method and Branch & Bound method problems. Kindly give the link. Thanks

[@99pedropotter]

thank you very much

[@aysanrafiei5791]

hello it was a great video thanks

[@sirousyousefnejad1658]

Thank u, very helpfull

[@srikar720]

excellent professor

[@saihehehe]

thank you this is helpful 😘😗

[@waraymohamedamine696]

Thank you for the videos, they are great. I have two questions : Do we always add -M ? or we add -M for maximization problem and M for minimization problem ? Same question for the artificial variable (a) ?

[@sxmirzaei]

we add -M for a Max and +M for a Min problem. Also, for constraints with >= signs we always add +a regardless of the objective function direction (max or min).

[@manishjoshi539]

great video !!

[@mahmodel-attar2856]

thanks too much u r awesome :)

[@MerveNoyan]

You helped a lot but I didn't understand how and why you switched from s2 to e4

[@PassiveInnovation]

Great tutorial

[@mariamilingou8217]

So if I get it right..if the initial problem was a minimization problem then we would have to add e's to the

[@sxmirzaei]

regardless of your objective function direction (Max or Min) when you have

[@logan89ism]

Hey Prof So do we use artificial variables in Simplex too in case we have a perfect equality constraint?

[@sxmirzaei]

+Ashwin Nair yes, in that case you only add an artificial variable and you don't add any excess variable: x1+x2=1 --> x1+x2+a1=1

[@logan89ism]

ok. Thanks

[@harun5407]

Thank you professor but i have one question. What we supposed to do when there is more than one > or = sign ?

[@sxmirzaei]

let's say you have 2 constraints with = and >= signs. Then, you have Max z=3x1+4x2 x1+x2=3 --> x1+x2+a1=3 x1+2x2+3x3>=1 --->x1+2x2+3x3-e2+a2=1 Note that for constraints with "=" sign we don't add any excess variable (e.g. the first constraint in the above example). You need to add both a1 and a2 with big M to the objective function. for example, in this case you have max z=3x1+4x2 ---> z=3x1+4x2-Ma1-Ma2

[@harun5407]

Shokoufeh Mirzaei Thank you very much..

[@camilot9501]

so if the artificial variable (either a4 or 4e) are another number other than 0, does that mean that there is no optimal solution, because there is no feasible region?

[@sxmirzaei]

no! only "a" variables are artificial variables. You have a feasible solution when e.g. e4=4. However, if any of your artificial variables (a variables) in the final optimal table are positive then you don't have a feasible solution

[@LuKaS2oo9]

What if all z row elements are positive but you still have M at the final value of z??

[@sxmirzaei]

if there is M in the final value of Z, it means at least one of the artificial variables is still positive (basic). If you are in the optimal table, this is a sign that shows the problem does not have a feasible solution.

[@farazraza4054]

can you tell me how can we differ Big M method by simplex method..? by checking the constraint of inequalities ? if differ we use big M otherwise simplex . please comment.

[@sxmirzaei]

If all constraints in an LP are in form of = or = signs we should use big M or two-phase method.

[@kikkasluca]

what if I decide to solve the min LP with the dual simplex ? do I still have to add an artificial variable in the

[@leddel746]

the information is hard to read, please choose other font

[@sxmirzaei]

That is my handwriting. I try to make it more legible next time. thanks for your feedback led del

[@imedhosting]

+led del if you consider her handwriting as a font it means that her handwriting is beautiful hahahaha

[@Shumayal]

Why can't we add a positive M? I understand that we want to avoid a positive M as it's a maximization problem. But if we do add a +M to the objective function, after standardizing, it becomes -M. So by normal procedure we would automatically make that column as the pivot column because it's the MOST NEGATIVE number there?

[@sxmirzaei]

please watch 4:21-6:49

[@Shumayal]

Yes, I have. I understood that but I don't get why we can't do it this way? Thanks ma'am.

[@malayait]

nice tutorial but what if we have two Ms in cost functions? how to get the new Z or cost function ?

[@sxmirzaei]

+malayait if you have two Ms you have to add both Ms to your objective function. Example (note: this is a minimization problem): Min Z=x1+2x2 Z=X1+2X2+M1a1+M2a2 ---> Z-X1-2X2-M1a1-M2a2=0 X1>=2 X1-e1+a1=2 X1-X2>=3 X1-X2-e2+a2=3 X1,X2>=0 X1,X2>=0

[@malayait]

+Shokoufeh Mirzaei thank you lecturer but i mean new z . mean the one which we have to use with. like you multiply M with r3 and then added to r4 to get new Z to use. sorry for bothering and last uncleare question

[@sxmirzaei]

+malayait I see. you have to do it for both columns of a1 and a2. you can do it in two steps. first, use one of the rows to get rid of the M under a1, then use another row to get rid of the other M left in the row of Z. Note that when you are in the second step, you have to use the new row of z that you have created.

[@malayait]

+malayait thank you so much .. :)

[@sxmirzaei]

+malayait if you find a feasible solution, Ms will be eliminated from your objective function.

[@methemamacita]

In some parts it's hard to see what you wrote, wish you wrote them bigger. But thanks a lot for the video! :)

[@MW2KidGun]

thanks so much