Industrial Engineering Miscellaneous
- The jobs arrive at a facility, for service, in a random manner. The probability distribution of number of arrivals of jobs in a fixed time interval is
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Since arrival rates depends upon the time factor, so accordingly graph can be chosen from Poisson distribution, but normal distribution expresses same result throughout.Correct Option: B
Since arrival rates depends upon the time factor, so accordingly graph can be chosen from Poisson distribution, but normal distribution expresses same result throughout.
- Demand during lead time with associated probabilities is shown below:
Expected demand during lead time is
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Probability
Expected demand during lead time
⇒ 7.5 + 9.8 + 15.75 + 16 + 25.5 = 74.55Correct Option: C
Probability
Expected demand during lead time
⇒ 7.5 + 9.8 + 15.75 + 16 + 25.5 = 74.55
- Two machines of the same production rate are available for use. On machine 1, the fixed cost is Rs. 100 and the variable cost is Rs. 2 per piece produced. The corresponding numbers for the machine are Rs. 200 and Rs. 1 respectively. For certain strategic reasons both the machines are to be used concurrently. The sale price of the first 800 units is Rs. 3.50 per unit & and subsequently it is only Rs. 3.00. The breakeven production rate for each machine is
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For Machine M1:
Fixed cost = Rs 100
Variable cost = Rs 2 per piece
For Machine M2:
Fixed cost = Rs 200
Variable cost = 1 Rs per piece
Total cost = Fixed cost + Variable cost × Number of units.
Total cost of production on machine M1 & M2 is
⇒ 100 + 2n + 200 + n = 300 + 2n
For the first 300 units, selling price is 3.50 Rs per unit.
So the break even point
300 + 3n = 3.50 (n+n)
300 + 3n = 3.50 (2 n)
7n = 300 + 3n = n = 75Correct Option: A
For Machine M1:
Fixed cost = Rs 100
Variable cost = Rs 2 per piece
For Machine M2:
Fixed cost = Rs 200
Variable cost = 1 Rs per piece
Total cost = Fixed cost + Variable cost × Number of units.
Total cost of production on machine M1 & M2 is
⇒ 100 + 2n + 200 + n = 300 + 2n
For the first 300 units, selling price is 3.50 Rs per unit.
So the break even point
300 + 3n = 3.50 (n+n)
300 + 3n = 3.50 (2 n)
7n = 300 + 3n = n = 75
- An electronic equipment manufacturer has decided to add a component subassembly operation that can produce 80 units during a regular 8-hour shift. This operation consists of three activities as below
For line balancing the number of work stations required for the activities M, E and T would respectively be
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Number of units produce in a day = so units working hour in a day = 8 hours
Time taken to produce 1 unit = 8 × 60 = 6 min 80
T = 6 min
Number of work station are the whole number more the function.
So no of work station required for the activities, M, E & T would be 2, 3, 4, 1Correct Option: A
Number of units produce in a day = so units working hour in a day = 8 hours
Time taken to produce 1 unit = 8 × 60 = 6 min 80
T = 6 min
Number of work station are the whole number more the function.
So no of work station required for the activities, M, E & T would be 2, 3, 4, 1
- A maintenance service facility has Poisson arrival rates, negative exponential service time and operates on a 'first come first served' queue discipline. Break-downs occur on an average of 3 per day with a range of zero to eight. The maintenance crew can service an average of 6 machines per day with a range of zero to seven. The mean waiting time for an item to be serviced would be
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Given: λ = 3 per day (arrival rate)
μ = 6 per day (service rate)
Mean waiting time for an item to serviced= λ = 3 = 1 × 1 = 1 day. μ(μ − λ) 6(6 − 3) 2 3 6 Correct Option: A
Given: λ = 3 per day (arrival rate)
μ = 6 per day (service rate)
Mean waiting time for an item to serviced= λ = 3 = 1 × 1 = 1 day. μ(μ − λ) 6(6 − 3) 2 3 6
