(3.5 # 1) Each of the following tableaux corresponds to a LP problem in canonica
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Question
(3.5 # 1) Each of the following tableaux corresponds to a LP problem in canonical form with 3 equality constraints, an objective function to be minimized, 7 nonnegative variables (i) The solution of the problem can be determined from the given tableau, which ii) One or more iterations of the simplex algorithm are necessary to complete the ri, .., and with variables ai,r, s serving as basic variables. For each, either case state the complete resolution of the problem, or problem, in which case determine all valid pivot points for the tableau. You do NOT need to carry out any more iterations zs 050 3-1839 (a) 1310 6 1-1 0 0-6110 1 9 08 0-3 4 88 0 6 04 2 3-75+3Explanation / Answer
To obtain the solution of LP from the simplex table, thereare two basic rules.
Rule-1..If all the coefficients in the last row(objective function) are positive, then optimal solution is reached.
Rule-2..If not optimal as in Rule-1, choose the entering variable from the last row with the most negative coefficient and then enter the basic variables rows(pivots) where the ratio of RHS/Coefficient of entering variable is least.
a) Rule-1 is violated hence not optimal. Entering variable is x4 with negative coef. The pivot is x2 as among(39/3,-10/1,88/8). x2 ratio is least.
b)Rule-1 is satisfied hence optimal solution. with x5=39, x3=10 and x1=88. and z=75
c)Rule-1 is violated hence not optimal. Entering variable is x2 with negative coef. The pivot is x1 as among(3/5,2/6,0/9). x1 ratio is least.
d)Rule-1 is violated hence not optimal. Entering variable can be x2 or x6 with negative coef. The pivot for x2 is x1 as among(3/5,2/6,1/9). x1 ratio is least. The pivot for x6 is x1 as among(-3/1,2/0,-1/3) x1 ratio is least.
e)Rule-1 is violated hence not optimal. Entering variable can be x2 ,x6 or x7 with negative coef. The pivot for x2 is x3 as among(60/5,30/6,50/9). x3 ratio is least. The pivot for x6 is x1 as among(60/1,30/0,-50/3) x1 ratio is least.The pivot for x7 is x3 as among(60/8,-30/6,50/7) x3 ratio is least.
f)Rule-1 is satisfied hence optimal solution. with x5=39, x3=0 and x1=88. and z=75
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