Question

Consider the following problem: Maximize 2 = S.t. 321 -2.22 2:01 +02 21 +22 21,02 > 0 < < . 30 10 The simplex method yields the following final simplex tableau: BV || 2 || 2172 || s1 s2 || RHS | 2 ||1|| 05 | 03 | 30 si || 0 || 0 -1 || 1 -2 || 10 21 || 0 || 1 1 | o 1 || 10 | (a) (9 points) Determine the allowable ranges for C and C2, independently of one another. (b) (6 points) Determine the allowable ranges for bi and b2, independently of one another.

I need help solving this in a step by step manner. I have been following along with the videos and along with the notes, however this is not making much sense and I really want to learn how to do this. So if you can do a step by step for both part A and B, I would greatly appreciate it.

Answer 1

Given Lpp is max z = 3%, -222 constraints subject to The 2x + x = 30 a, tx = 10 and Simplen Now we solve this app by methad The probleon is converted to Canonical form by adding slacks and entiticiel Vareasles as appropriate ". As the constraint 0 is of type , we should add slack Variable s... 2. As the constraint - ① is of type £, we should add slacce Variable sa. After introduing slack Variables max z = 30,- 2 da to sto se subject to 28, +f S = 30 4td2 + sq = 10 and 11, 42, 5x>0.

Heration rej 3 man Ratio 100 = 15 10 roli loll 10 = 102 -7;oooo -C) -3 11 2 Ketw ZO Key column) Negative minimum z; - cj is -3 and its Column inden is 1. so, the entering variable is si. minimum ratio is 10 and its rove index in a. so, the leaving baric variable in Sa. The pirot element is "li Entering - Deporting = sa, key element - 1 Iteration L e; 3 1- 20 lol 30

since all z. - c; so. Hence, optional solution is arrived with Valee of variables is a, = 10, 22=0. Now we final max z . max z = 3x - 222 . 3 (10) + 2(0) 30. - Hence the optimal solution of the is attained . max z = 30 and ai:10, 7,=0 Lop