Industrial Engineering Miscellaneous Difficult Questions and Answers | Page - 1

Industrial Engineering Miscellaneous

A component can be produced by any of the four processes I, II, III and IV. The fixed cost and the variable cost for each of the processes are listed below. The most economical process for producing a batch of 100 pieces is

1. I
1. II
1. III
1. IV

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TC = Fixed Cost + Variable cost
TC_I = 20 + 100 × 3 = Rs.320
TC_II = 50 + 100 × 1 = Rs.150
TC_III = 40 + 100 × 2 = Rs.240
TC_IV = 10 + 100 × 4 = Rs.410
Since total cost for the Process II is minimum
So II is cost for economical point of view

Correct Option: B

TC = Fixed Cost + Variable cost
TC_I = 20 + 100 × 3 = Rs.320
TC_II = 50 + 100 × 1 = Rs.150
TC_III = 40 + 100 × 2 = Rs.240
TC_IV = 10 + 100 × 4 = Rs.410
Since total cost for the Process II is minimum
So II is cost for economical point of view

A linear programming problem is shown below:
Maximise 3x + 7y
Subjeot to 3x + 7y ≤ 10
4x + 6y ≤ 8
x, y ≥ 0

1. an unbounded objective function
1. exactly one optimal solution
1. exactly two optimal solution
1. infinitely many optimal solution

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z = 3x + 7y
Constraints 3x + 7y ≤ 10
4x + 6y < 8; x, y > 0
Corresponding equations
3x + 7y = 10; 4x + 6y = 8

A (0, 4/3) z = 9.23
B (2, 0) z = 6
Thus, exactly one optimal solution.
Hence, the correct option is (b).

Correct Option: B

z = 3x + 7y
Constraints 3x + 7y ≤ 10
4x + 6y < 8; x, y > 0
Corresponding equations
3x + 7y = 10; 4x + 6y = 8

A (0, 4/3) z = 9.23
B (2, 0) z = 6
Thus, exactly one optimal solution.
Hence, the correct option is (b).

A manufacturer can produce 12000 bearings per day. The manufacturer received an order of 8000 bearings per day from a customer. The cost of holding a bearing in stock is Rs. 0.20 per month. Setup cost per production run is Rs. 500. Assuming 300 working days in a year, the frequency of production run should be

1. 4.5 days
1. 4.5 months
1. 6.8 days
1. 6.8 months

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D = 8000 × 300 bearings/year
P = 12000 bearings/day
C = 8000 bearings/day
C_h = 0.12 × 12/year
C_O = 500/Production
Q* = √ 2 × D × C_O × P
C_h P - C

⇒ √ 2 × 240000 × 500 × 12000
0.20 × 12 12000 - 8000

⇒ 54772.526
T = Q*
C

T ⇒ 6.84 days

Correct Option: C

D = 8000 × 300 bearings/year
P = 12000 bearings/day
C = 8000 bearings/day
C_h = 0.12 × 12/year
C_O = 500/Production
Q* = √ 2 × D × C_O × P
C_h P - C

⇒ √ 2 × 240000 × 500 × 12000
0.20 × 12 12000 - 8000

⇒ 54772.526
T = Q*
C

T ⇒ 6.84 days

Consider the following data with reference to elementary deterministic economic order quantity model

The total number of economic orders per year to meet the annual demand is _______.

1. 54
1. 50
1. 56
1. 48

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D = 100000
C = 10 Rs
C_h = 1.5
C_O = 30/order
EOQ(Q*) = √ 2 × D × C_O
C_h

= √ 2 × 100000 × 30 = 2000 units
1.5

No of orders N* = D
Q*

N* = 100000 ⇒ 50
2000

Correct Option: B

D = 100000
C = 10 Rs
C_h = 1.5
C_O = 30/order
EOQ(Q*) = √ 2 × D × C_O
C_h

= √ 2 × 100000 × 30 = 2000 units
1.5

No of orders N* = D
Q*

N* = 100000 ⇒ 50
2000

A company uses 2555 units of an item annually. Delivery lead time is 8 days. The reorder point (in number of units) to achieve optimum inventory is

1. 7
1. 8
1. 56
1. 60

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Reorder level = L_T × d + SS ↑_{If guessed}
Q ⇒ 2555 units
L_s = 8 days
Cycle Time = 365 days

Q = n
T t

t = Q t
T

t = 2555 × 8 = 56 units
365

Correct Option: C

Reorder level = L_T × d + SS ↑_{If guessed}
Q ⇒ 2555 units
L_s = 8 days
Cycle Time = 365 days

Q = n
T t

t = Q t
T

t = 2555 × 8 = 56 units
365