Industrial Engineering Miscellaneous Easy Questions and Answers | Page - 5

Industrial Engineering Miscellaneous

The sales of a product during the last four years were 860, 880,870 and 890 units. The forecast for the fourth year was 876 units. If the forecast for the fifth year, using simple exponential smoothing, is equal to the forecast using a three period moving average the value of the exponential smoothing constant α is

1. 1/7
1. 1.5
1. 2/7
1. 2/5

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F₄ = 876 units

F₅ = F₃
F₃ = 890 + 870 + 880
2

F₅ = F₄ + α (D₄ – F₄)
880 = 876 + α (890 – 876)
α = 2
7

Correct Option: C

F₄ = 876 units

F₅ = F₃
F₃ = 890 + 870 + 880
2

F₅ = F₄ + α (D₄ – F₄)
880 = 876 + α (890 – 876)
α = 2
7

For a product, the forecast and the actual sales for December 2002 were 25 and 20 respectively. If the exponential smoothing constant (α) is taken as 0.2, the forecast sales for January 2003 would be

1. 21
1. 23
1. 24
1. 27

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F_t+1 = F_t + α (D_t – F_t)
F_{Jan (2003)} = F_Dec + α (D_Dec – F_Dec )
D_Dec ⇒ 20 units
F_Dec = 25 units
F_Jan = 25 + 0.2 (20 – 25)
= 25 + 0.2 (–5) = 25 –1
F_Jan ⇒ 24 units

Correct Option: C

F_t+1 = F_t + α (D_t – F_t)
F_{Jan (2003)} = F_Dec + α (D_Dec – F_Dec )
D_Dec ⇒ 20 units
F_Dec = 25 units
F_Jan = 25 + 0.2 (20 – 25)
= 25 + 0.2 (–5) = 25 –1
F_Jan ⇒ 24 units

In an M/M/1 queuing system, the number of arrivals in an interval of length T is a Poisson random variable i.e. the probability of there being n arrivals in an interval of length T is
e^λT(λT)ⁿ The probability density function f(t) of the inter- arrival time is given by
n!

1. λ²(e^−λ2T)
1. e^−λ²T
  λ²
1. λ²e^−λt
1. e^−λT
  λ

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For any Poisson distribution, the probability density function f(t) is given by λe^{– λt} .

Correct Option: C

For any Poisson distribution, the probability density function f(t) is given by λe^{– λt} .

The number of customers arriving at a railway reservation counter is Poisson distributed with an arrival rate of eight customers per hour. The reservation clerk at this counter takes six minutes per customer on an average with an exponentially distributed service time. The average number of the customers in the queue will be

1. 3
1. 3.2
1. 4
1. 4.2

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Average number of customer in queue
= λ² = 8 × 8 = 3.2
μ(μ − A) 10 × (10 − 8)

Correct Option: B

Average number of customer in queue
= λ² = 8 × 8 = 3.2
μ(μ − A) 10 × (10 − 8)

Demand during lead time with associated probabilities is shown below:

Expected demand during lead time is

1. 1789
1. 12.4
1. 74.55
1. 11,.46

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Probability

Expected demand during lead time
⇒ 7.5 + 9.8 + 15.75 + 16 + 25.5 = 74.55

Correct Option: C

Probability

Expected demand during lead time
⇒ 7.5 + 9.8 + 15.75 + 16 + 25.5 = 74.55