96. When using a simple moving average to forecast demand, one would

1. 1. given equal weight to all demand data

   
   
   

   2. assign more weight to the recent demand data

   
   
   

   3. include new demand data in the average without discarding the earlier data

   
   
   

   4. include new demand data in the average after discarding some of the earlier demand data
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: D

   NA

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97. In a time series forecasting model, the demand for five time periods was 10,13,15,18 and 22. A linear regression fit resulted in an equation F = 6.9 + 2.9 where F is the forecast for period f. The sum of absolute deviations for the five data is

1. 1. 2.2

   
   
   

   2. 0.2

   
   
   

   3. –1.2

   
   
   

   4. 24.3

   
   
   

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1. View Hint View Answer Discuss in Forum

   F = 6.9 + 2.9 t
   F₁ = 6.9 + 2.9 × 1 ⇒ 9.8
   F₂ = 6.9 + 2.9 × 2 ⇒ 12.7
   F₃ = 6.9 + 2.9 × 3 ⇒ 15.6
   F₄ = 6.9 + 2.9 × 4 ⇒ 18.5
   F₅ = 6.9 + 2.9 × 5 = 21.4
   D₁ = 10 D₂ = 13 0₃ = 15
   D₄ = 16 D₅ = 22
   SAD = ∑ |e_i|
   SAD = |10 − 9.8| + |13 − 12.7| + |15 − 15.6| + |18 − 18.5| + |22 − 21.4 |
   Absolute Deviation = 2.2

   Correct Option: A

   F = 6.9 + 2.9 t
   F₁ = 6.9 + 2.9 × 1 ⇒ 9.8
   F₂ = 6.9 + 2.9 × 2 ⇒ 12.7
   F₃ = 6.9 + 2.9 × 3 ⇒ 15.6
   F₄ = 6.9 + 2.9 × 4 ⇒ 18.5
   F₅ = 6.9 + 2.9 × 5 = 21.4
   D₁ = 10 D₂ = 13 0₃ = 15
   D₄ = 16 D₅ = 22
   SAD = ∑ |e_i|
   SAD = |10 − 9.8| + |13 − 12.7| + |15 − 15.6| + |18 − 18.5| + |22 − 21.4 |
   Absolute Deviation = 2.2

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98. In a single server infinite population queuing model. Arrivals follow a Poisson distribution with mean λ = 4 per hour. The service times are exponential with mean service time equal to 12 minutes. The expected length of the queue will be
    

1. 1. 4
      

   
   
   

   2. 3.2
      

   
   
   

   3. 1.25
      

   
   
   

   4. 24.3
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   λ =	4	, μ =	1
   	60		12

   
   

   ∴ ρ =	λ	=	48	= 0.8
   	μ		60

   
   

   Wq = queue length =	ρ2	=	(0.8)2	= 3.2
   	(1 – ρ)		(1 – 0.8)

   
   

   Correct Option: B

   λ =	4	, μ =	1
   	60		12

   
   

   ∴ ρ =	λ	=	48	= 0.8
   	μ		60

   
   

   Wq = queue length =	ρ2	=	(0.8)2	= 3.2
   	(1 – ρ)		(1 – 0.8)

   
   

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99. At a production machine, parts arrive according to a Poisson process at the rate of 0.35 parts per minute. Processing time for parts have exponential distribution with mean of 2 minutes. What is the probability that a random part arrival finds that there are already 8 parts in the system (in machine + in queue)?
    

1. 1. 0.0247
      

   
   
   

   2. 0.0576
      

   
   
   

   3. 0.0173
      

   
   
   

   4. 0.082

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given, λ = 0.35/min.,
   

   μ =	1	/ min = 0.5 min
   	2

   
   

   ρ =	λ	=	0.35	= 0.7
   	μ		0.5

   
   Probability of 'n' numbers in the system,
   p_n (t) = ρ_n (1 – ρ)
   = (0.7)₈ (1 – 0.7) = 0.0173.
   

   Correct Option: C

   Given, λ = 0.35/min.,
   

   μ =	1	/ min = 0.5 min
   	2

   
   

   ρ =	λ	=	0.35	= 0.7
   	μ		0.5

   
   Probability of 'n' numbers in the system,
   p_n (t) = ρ_n (1 – ρ)
   = (0.7)₈ (1 – 0.7) = 0.0173.
   

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100. The cost of providing service in a queuing system increases with
     

1. 1. Increased mean time in the queue
      

   
   
   

   2. Increased arrival rate
      

   
   
   

   3. Decreased mean time in the queue
      

   
   
   

   4. Decreased arrival rate

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: C

   NA

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