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

Shokoufeh Mirzaei Thank you very much..

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).

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.

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

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

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

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

+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

ok. Thanks

please watch 4:21-6:49

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

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

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

+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

+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

+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 thank you so much .. :)

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