Question:
In an M/M/1 queuing system, the customers arrive at a rate of 3 per minute, and the average number of customers in the system is \( n \) at steady state. Assume that the server utilization is less than 1.
Which ONE of the following is the average time a customer spends in the system (in minutes) at steady state?
In an M/M/1 queuing system, the customers arrive at a rate of 3 per minute, and the average number of customers in the system is \( n \) at steady state. Assume that the server utilization is less than 1.
Which ONE of the following is the average time a customer spends in the system (in minutes) at steady state?
Which ONE of the following is the average time a customer spends in the system (in minutes) at steady state?
Show Hint
Apply Little's Law, \( L = \lambda W \), with \( L = n \) and \( \lambda = 3 \).
Updated On: Aug 3, 2026
- \( 3n \)
- \( n/3 \)
- \( n/6 \)
- \( 3/n \)
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The Correct Option is B
Solution and Explanation
Step 1: Identify the relevant law:
This problem gives us the average number of customers in a queuing system and the arrival rate, and asks for the average time spent in the system.
This exact relationship is described by Little's Law, a fundamental result that holds for any stable queuing system regardless of the arrival or service distribution.
Step 2: State Little's Law:
Little's Law relates the average number in the system \( L \), the average arrival rate \( \lambda \), and the average time in the system \( W \) as:
\[ L = \lambda \times W \]
Since the system is stated to be in steady state with utilization less than 1, the system is stable and this relation applies directly.
Step 3: Substitute the given values:
Here the arrival rate is \( \lambda = 3 \) customers per minute, and the average number of customers in the system is \( L = n \).
\[ n = 3 \times W \]
Solving for \( W \):
\[ W = \frac{n}{3} \]
Step 4: Rule out the other options:
Option (A) inverts the relation (\( W = 3n \) would apply if \( \lambda = 1/3 \)), option (C) divides by 6 which has no basis here (there is no service rate of 6 given), and option (D) inverts \( n \) itself, which is dimensionally wrong for a time.
Only \( n/3 \) is consistent with Little's Law using the given arrival rate.
Final Answer:
Little's Law applies directly with the given arrival rate. \[ \boxed{W = \dfrac{n}{3}\ \text{minutes}} \]
This problem gives us the average number of customers in a queuing system and the arrival rate, and asks for the average time spent in the system.
This exact relationship is described by Little's Law, a fundamental result that holds for any stable queuing system regardless of the arrival or service distribution.
Step 2: State Little's Law:
Little's Law relates the average number in the system \( L \), the average arrival rate \( \lambda \), and the average time in the system \( W \) as:
\[ L = \lambda \times W \]
Since the system is stated to be in steady state with utilization less than 1, the system is stable and this relation applies directly.
Step 3: Substitute the given values:
Here the arrival rate is \( \lambda = 3 \) customers per minute, and the average number of customers in the system is \( L = n \).
\[ n = 3 \times W \]
Solving for \( W \):
\[ W = \frac{n}{3} \]
Step 4: Rule out the other options:
Option (A) inverts the relation (\( W = 3n \) would apply if \( \lambda = 1/3 \)), option (C) divides by 6 which has no basis here (there is no service rate of 6 given), and option (D) inverts \( n \) itself, which is dimensionally wrong for a time.
Only \( n/3 \) is consistent with Little's Law using the given arrival rate.
Final Answer:
Little's Law applies directly with the given arrival rate. \[ \boxed{W = \dfrac{n}{3}\ \text{minutes}} \]
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