Industrial Engineering Miscellaneous


Industrial Engineering Miscellaneous

Industrial Engineering

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









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    NA

    Correct Option: C

    NA


  1. In an assembly line for assembling toys, five workers are assigned tasks which take times of 10, 8, 6, 9 and 10 minutes respectively. The balance delay for line is









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    Assuming cycle time = 10 + 8 + 6 + 9 + 10 = 43

    Balance delay = 1 -
    ∑ ti
    = 1-
    43
    = 0.14
    n × te5 × 10

    = 14%

    Correct Option: C

    Assuming cycle time = 10 + 8 + 6 + 9 + 10 = 43

    Balance delay = 1 -
    ∑ ti
    = 1-
    43
    = 0.14
    n × te5 × 10

    = 14%



  1. If the demand for an item is doubled and the ordering cost halved, the economic order quantity









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    EOQ ∝ √DC0

    EOQ'
    =
    EOQ
    D'C0' DC0

    D' = 2D
    C0' =
    1
    C0
    2

    EOQ'
    =
    EOQ
    { 2D × ( C0 / 2 ) } DC0

    ⇒ EOQ' = EOQ

    Correct Option: A

    EOQ ∝ √DC0

    EOQ'
    =
    EOQ
    D'C0' DC0

    D' = 2D
    C0' =
    1
    C0
    2

    EOQ'
    =
    EOQ
    { 2D × ( C0 / 2 ) } DC0

    ⇒ EOQ' = EOQ


  1. 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









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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



  1. In an ideal inventory control system, the economic lot size for a part is 1000. If the annual demand for the part is doubled, the new economic lot size required will be









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    EOQ ∝ √D
    EOQ = 1000

    EOQ1
    =
    EOQ
    D D

    EOQ1 = 1000 ×
    2D
    = 1000√2
    D

    Correct Option: D

    EOQ ∝ √D
    EOQ = 1000

    EOQ1
    =
    EOQ
    D D

    EOQ1 = 1000 ×
    2D
    = 1000√2
    D