ASSIGNMENT 3(Based on Two-Phase Method, Special cases of L.P.P.)

Q.1. Suppose z∗ represents the auxiliary objective function to be maximized in Phase I. Which of thefollowing statements is/are true?

(a) Max z∗= 0 and all the artificial vectors disappear from the final basis implies we get a basic feasiblesolution.

(b) Max z∗= 0 and one or more artificial vectors appear in the final basis at zero level implies we get abasic feasible solution of the problem but there may be redundancy in the original constraint equations.

(c) Max z∗¡ 0 and one or more artificial vectors appear in the final basis at positive level implies thereis no feasible solution of the original problem.

(d) All of the above.

Q.2. What is the coefficient of artificial variables in the auxiliary objective function z∗ in Phase I of theTwo-Phase Method?

(a) 0 (b) 1 (c) -1 (d) None of the above

Q.3. Solve the following L.P.P. by two-phase method :

Minimize z = 3x1 + 5x2

subject to x1 + 2x2 ≥ 8

3x1 + 2x2 ≥ 12

5x1 + 6x2 ≤ 60

x1, x2 ≥ 0.

(a) zmin = 21 (b) zmin = 17 (c) zmin = 0 (d) None of the above

Q.4. Use two-phase simplex method to solve the L.P.P.

Minimize z = 4x1 + x2

subject to x1 + 2x2 ≤ 3

4x1 + 3x2 ≥ 6

3x1 + x2 = 3

x1, x2 ≥ 0.

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(a) zmin =23

5(b) zmin =

18

5(c) Unbounded solution (d) No feasible solution

Q.5. Solve the following problem using two-phase simplex method.

Maximize z = 5x1 + 3x2

subject to 2x1 + x2 ≤ 1

3x1 + 4x2 ≥ 12

x1, x2 ≥ 0.

(a) zmax = 120 (b) zmax = 78 (c) Unbounded solution (d) No feasible solution

Q.6. Solve the following problem using two-phase simplex method.

Maximize z = 2x1 + 3x2 + x3

subject to − 3x1 + 2x2 + 3x3 = 8

−3x1 + 4x2 + 2x3 = 7

x1, x2, x3 ≥ 0.

(a) zmax = 150 (b) zmax = 174 (c) Unbounded solution (d) No feasible solution

Q.7. Solve the following L.P.P. by simplex method.

Maximize z = 2x1 + 5x2

subject to x1 + 2x2 ≤ 8

x1 ≤ 4

x2 ≤ 3

x2 ≥ 0, x1 is unrestricted in sign .

(a) zmax = 35 (b) zmax = 19 (c) zmax = 188 (d) zmax = 65

Q.8. Solve the following L.P.P. by simplex method.

Maximize z = 2x1 + 3x2

2

subject to − x1 + 2x2 ≤ 4

x1 + x2 ≤ 6

x1 + 3x2 ≤ 9

x1, x2 are unrestricted in sign .

(a) zmax =27

2(b) zmax =

13

2(c) zmax = 83 (d) zmax = 62

Q.9. Solve the following L.P.P. by simplex method.

Maximize z = 3x1 + 2x2 + x3

subject to 2x1 + 5x2 + x3 = 12

3x1 + 4x2 = 11

x1, x2 ≥ 0, x3 is unrestricted in sign .

(a) zmax = 12 (b) zmax = 17 (c) zmax = 32 (d) None of the above

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ANSWERS

Q.1. (d)

Q.2. (c)

Q.3. (a)

Q.4. (b)

Q.5. (d)

Q.6. (c)

Q.7. (b)

Q.8. (a)

Q.9. (d)

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