Industrial Engineering Miscellaneous

Industrial Engineering Miscellaneous

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Industrial Engineering

Industrial Engineering Miscellaneous
  1. A standard machine tool and an automatic machine tool are being compared for the production of a component.

    For line balancing the number of work stations required for the activities M, E and T would respectively be








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    Given data

    Total cost of z1 component by using standard machine tool,

    (TC)1 =
    30
    		 +
    22 × z1
    		× 200 = 100 +
    2200
    		z1
    606030

    Total cost of z2 component by using Automatic Machine tool,
    (TC)2 = 2 +
    5
    		× z2 × 800 = 1600 +
    2000
    		z2
    6030

    Let break even point be z number of components
    ∴  100 +
    2200
    		z = 1600 +
    2000z
    3030

    or  
    200
    		z = 1500
    30

    or   z =
    1500 × 30
    		 = 225
    200


    Alternately:
    Let N be the Break even number
    At Break even point
    1
    		 +
    22
    		.N = 2 +
    5N
    		 ×
    800
    26060200

    ⇒ 
    1
    		 +
    22
    		N = 2 +
    5N
    		 ×
    800
    26060200

    ⇒ 
    1
    		 +
    22
    		N = 8 +
    20N
    26060

    ⇒ 
    22
    		N −
    20
    		N = −
    1
    60602

    ⇒ 
    22 − 20
    		N =
    16 − 1
    602

    ⇒  N =
    15 × 60
    		 = 225
    2 × 2

    Correct Option: D

    Given data

    Total cost of z1 component by using standard machine tool,

    (TC)1 =
    30
    		 +
    22 × z1
    		× 200 = 100 +
    2200
    		z1
    606030

    Total cost of z2 component by using Automatic Machine tool,
    (TC)2 = 2 +
    5
    		× z2 × 800 = 1600 +
    2000
    		z2
    6030

    Let break even point be z number of components
    ∴  100 +
    2200
    		z = 1600 +
    2000z
    3030

    or  
    200
    		z = 1500
    30

    or   z =
    1500 × 30
    		 = 225
    200


    Alternately:
    Let N be the Break even number
    At Break even point
    1
    		 +
    22
    		.N = 2 +
    5N
    		 ×
    800
    26060200

    ⇒ 
    1
    		 +
    22
    		N = 2 +
    5N
    		 ×
    800
    26060200

    ⇒ 
    1
    		 +
    22
    		N = 8 +
    20N
    26060

    ⇒ 
    22
    		N −
    20
    		N = −
    1
    60602

    ⇒ 
    22 − 20
    		N =
    16 − 1
    602

    ⇒  N =
    15 × 60
    		 = 225
    2 × 2

  1. 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 = 75

    Correct 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

  1. 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 = 74.55

    Correct Option: C

    Probability

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

  1. 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.

  1. The symbol used for Transport in work study is








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    NA

    Correct Option: A

    NA