UGC NET PAPER 2 June 2019 Question 10 (Shift-I)

Question 10

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10. Consider an LPP given as

Max Z = 2x₁ - x₂ + 2x₃

Subject to the constraints
2x₁ + x₂ ≤ 10
x₁ + 2x₂ - 2x₃ ≤ 20
x₁ + 2x₃ ≤ 5
x₁, x₂, x₃ ≥ 0

What shall be the solution of the LPP after applying first iteration of the Simplex Method?

Options:

1. 1.   x₁=5/2, x₂=0, x₃=0, Z=5
2. 2.   x₁=0, x₂=0, x₃=5/2, Z=5
3. 3.   x₁=0, x₂=5/2, x₃=0, Z= -5/2
4. 4.   x₁=0, x₂=0, x₃=10, Z=20

Explanation Question 10

Initial simplex table:

Cj (Coef. in Z)		2	-1	2	0	0	0	b	θ = b / key column value (Choose minimum positive value)
Cb	Basic. Var.	X1	X2	X3	S1	S2	S3
0	S1	2	1	0	1	0	0	10	10/0 = ∞
0	S2	1	2	-2	0	1	0	20	20/(-2) = -10
0	S3	1	0	2	0	0	1	5	5/2
Zj = Σ Cb aij		0	0	0	0	0	0
Ci - Zj		2	-1	2	0	0	0

Here key column is X₃ column
and key row is S₃ row because it has minimum positive value for θ
So, Key element is 2 that is intersection of X₃ column and S₃ column

Here, minimum positive value of basic variable is for S₃ that is 5/2, hence S₃ is the variable that leaves the set of basic variables.

Now, choose maximun value from C_i - Z_j row, column X₁ and X₃ having larger value which is 2 and same in both column. we choose column X₃ because it leads to the correct expected solution after first iteration. hence S₃ is the variable that enter the set of basic variables.

Simplex table after first iteration :
Here, we perform the row operation : R₃ → R₃/2 on pivot element
Also, Perform the column operation on R₂ : R₂ → R₂ + 2*R₃

Cj (Coef. in Z)		2	-1	2	0	0	0	b
Cb	Basic. Var.	X1	X2	X3	S1	S2
0	S1	2	1	0	1	0	0	10
0	S2	2	2	0	0	1	0	25
2	X3	1/2	0	1	0	0	1/2	5/2

So, now value of the X₁ = 0,  X₂ = 0,  X₃ = 5/2,  S₁ = 10,  S₂ = 25,  S₃ = 0

After substituting above value in Z = 2x₁ - x₂ + 2x₃

Z = 2x₁ - x₂ + 2x₃
= 2 x 0 - 0 + 2 x (5/2)
= 5

So, Option 2 is the right answer.

Reference Example : THE STEPS OF THE SIMPLEX ALGORITHM

Clear Your Concept using below video reference example of LPP :

Solution of LPP using Simplex Method (maximization problem)

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