Industrial Engineering Miscellaneous Easy Questions and Answers | Page - 8

Industrial Engineering Miscellaneous

If at the optimum in a linear programming problem, a dual variable corresponding to a particular primal constraint is zero, then it means that

1. Right hand side of the primal constraint can altered without affecting the optimum solution
1. Changing the right hand side of the primal constraint will disturb the optimum solution
1. The objectives function is unbounded
1. The problem is degenerate

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NA

Correct Option: C

NA

In PERT, the distribution of activity times is assumed to be

1. Normal
1. Gamma
1. Beta
1. Exponential

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NA

Correct Option: C

NA

A dummy activity is used in PERT network to describe

1. Precedence relationship
1. Necessary time delay
1. Resource restriction
1. Resource idleness

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Dummy activities often have a zero completion time & are used to represents precendence relationship.

Correct Option: A

Dummy activities often have a zero completion time & are used to represents precendence relationship.

When the annual demand of a product is 24000 units, the EOQ (Economic Order Quantity) is 2000 units. If the annual demand is 48000 units the most appropriate EOQ will be

1. 1000 units
1. 2000 units
1. 2800 units
1. 4000 units

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EOQ ∝ √D
EOQ₂ = EOQ₁
√D₂ √D₁

EOQ₂ = EOQ₁ × √D₂
√D₁

EOQ₂ = 2000 × √4800 = 2828
√2400

Correct Option: C

EOQ ∝ √D
EOQ₂ = EOQ₁
√D₂ √D₁

EOQ₂ = EOQ₁ × √D₂
√D₁

EOQ₂ = 2000 × √4800 = 2828
√2400

A company produces two types of toys: P and Q. Production time of Q is twice that of P and the company has a maximum of 2000 time units per day. The supply of raw material is just sufficient to produce 1500 toys (of any type) per day. Toy type Q requires an electric switch which is available @ 600 pieces per day only. The company makes a profit of Rs. 3 and Rs. 5 on type P and Q respectively. For maximization of profits, the daily production quantities of P and Q toys should respectively be

1. 1000, 500
1. 800, 600
1. 500, 1000
1. 1000, 1000

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Z_max = 3P + 5Q
subject to P + 2Q ≤ 2000
P + Q ≤ 1500
Q ≤ 600
P, Q ≥ 0

Feasible solution (O A B C D)
Since At point A (1500,0)
Z = 3 × 1500 + 5 × 0 = 4500
At Point B (1000, 500)
Z = 3 × 1000 + 5 × 500 = 5500
At point C (800, 600)
Z = 3 × 800 + 5 × 600 = 5400
At Point O (0, 600)
Z = 3 × 0 + 5 × 600 = 3000
Hence Z in maximum at B (1000,500)
P = 1000 units and Q = 500 units

Correct Option: A

Z_max = 3P + 5Q
subject to P + 2Q ≤ 2000
P + Q ≤ 1500
Q ≤ 600
P, Q ≥ 0

Feasible solution (O A B C D)
Since At point A (1500,0)
Z = 3 × 1500 + 5 × 0 = 4500
At Point B (1000, 500)
Z = 3 × 1000 + 5 × 500 = 5500
At point C (800, 600)
Z = 3 × 800 + 5 × 600 = 5400
At Point O (0, 600)
Z = 3 × 0 + 5 × 600 = 3000
Hence Z in maximum at B (1000,500)
P = 1000 units and Q = 500 units