Industrial Engineering Miscellaneous

Industrial Engineering Miscellaneous

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Industrial Engineering

Industrial Engineering Miscellaneous
  1. The distribution of lead time demand for an item is as follows :

    The reorder level is 1.25 times the expected value of the lead time demand. The service level is








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    The expected value of lead time demand
    ⇒ 80 × 0.20 + 100 × 0.25 + 120 × 0.30 + 140 × 0.25 = 112
    Reorder level is 1.25 times the lead time demand.
    So, Reorder value = 1.25 × 112 = 140
    ∴ Here, we can see that both the maximum demand and reorder value are equal, service level = 100%

    Correct Option: D

    The expected value of lead time demand
    ⇒ 80 × 0.20 + 100 × 0.25 + 120 × 0.30 + 140 × 0.25 = 112
    Reorder level is 1.25 times the lead time demand.
    So, Reorder value = 1.25 × 112 = 140
    ∴ Here, we can see that both the maximum demand and reorder value are equal, service level = 100%

  1. A residential school stipulates the study hours as 8 : 00 pm to 10 : 30 pm. Warden makes random checks on a certain student 11 occasions a day during the study hours over a period of 10 days and observes that he is studying on 71 occasions. Using 95% confidence interval, the estimated minimum hours of his study during that 10 day period is








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    Number of total observation in 10 days = 11 × 10 = 110
    No. of observation when studying = 71

    ∴ P =
    71
    		 = 0.6455
    110

    (Probability of studying)
    Total studying hour in 10 days = (2.5 hr) × 10
    Hence, Minimum no. of hours of studying in 10 days = (25 hr) × p = 25 × 0.6455 = 16.13 hrs.

    Correct Option: C

    Number of total observation in 10 days = 11 × 10 = 110
    No. of observation when studying = 71

    ∴ P =
    71
    		 = 0.6455
    110

    (Probability of studying)
    Total studying hour in 10 days = (2.5 hr) × 10
    Hence, Minimum no. of hours of studying in 10 days = (25 hr) × p = 25 × 0.6455 = 16.13 hrs.

  1. The expected time (te) of a PERT activity in terms of optimistic time (to), pessimistic time (tp) and most likely time (tl) is given by








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    NA

    Correct Option: A

    NA

Direction: Consider a PERT network for a project involving six tasks (a to f)

  1. The standard deviation of the critical path of the project is








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    √25 + 81 + 36 + 9 = √151 days.

    Correct Option: A

    √25 + 81 + 36 + 9 = √151 days.

  1. The expected completion time of the project is








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    ⇒ a – c – e – f
    ⇒ 30 + 60 + 45 + 20 = 155 days

    Correct Option: D


    ⇒ a – c – e – f
    ⇒ 30 + 60 + 45 + 20 = 155 days