Question:easy
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 5, 2026
- \( 3n \)
- \( n/3 \)
- \( n/6 \)
- \( 3/n \)
Show Solution
The Correct Option is B
Solution and Explanation
Step 1: Recall the general M/M/1 relations:
For any M/M/1 queue, the average number in the system is $L = \dfrac{\rho}{1-\rho}$ where $\rho = \lambda/\mu$ is the server utilization, and the average time in the system is $W = \dfrac{1}{\mu - \lambda}$.
These two expressions are not independent, they are connected through the arrival rate $\lambda$.
Step 2: Connect $L$ and $W$ through $\lambda$:
Dividing the expression for $W$ by the expression for $L$:
\[ \frac{W}{L} = \frac{1/(\mu-\lambda)}{\rho/(1-\rho)} \]
Since $\rho = \lambda/\mu$, we have $1 - \rho = (\mu - \lambda)/\mu$, so $\dfrac{\rho}{1-\rho} = \dfrac{\lambda}{\mu-\lambda}$.
Substituting this back:
\[ \frac{W}{L} = \frac{1/(\mu-\lambda)}{\lambda/(\mu-\lambda)} = \frac{1}{\lambda} \]
Step 3: Interpret the result:
This shows $W = L/\lambda$ for any M/M/1 queue, which is exactly Little's Law recovered from the individual formulas rather than quoted directly.
So regardless of what $\mu$ actually is, as long as the queue is stable ($\rho<1$, which is given), this ratio holds.
Step 4: Apply it to the given data:
We are told $L = n$ and $\lambda = 3$ per minute, so:
\[ W = \frac{L}{\lambda} = \frac{n}{3} \]
This confirms option (B) without needing to know the actual service rate $\mu$.
Final Answer:
Deriving from the underlying M/M/1 formulas gives the same relation as Little's Law. \[ \boxed{W = \dfrac{n}{3}\ \text{minutes}} \]
For any M/M/1 queue, the average number in the system is $L = \dfrac{\rho}{1-\rho}$ where $\rho = \lambda/\mu$ is the server utilization, and the average time in the system is $W = \dfrac{1}{\mu - \lambda}$.
These two expressions are not independent, they are connected through the arrival rate $\lambda$.
Step 2: Connect $L$ and $W$ through $\lambda$:
Dividing the expression for $W$ by the expression for $L$:
\[ \frac{W}{L} = \frac{1/(\mu-\lambda)}{\rho/(1-\rho)} \]
Since $\rho = \lambda/\mu$, we have $1 - \rho = (\mu - \lambda)/\mu$, so $\dfrac{\rho}{1-\rho} = \dfrac{\lambda}{\mu-\lambda}$.
Substituting this back:
\[ \frac{W}{L} = \frac{1/(\mu-\lambda)}{\lambda/(\mu-\lambda)} = \frac{1}{\lambda} \]
Step 3: Interpret the result:
This shows $W = L/\lambda$ for any M/M/1 queue, which is exactly Little's Law recovered from the individual formulas rather than quoted directly.
So regardless of what $\mu$ actually is, as long as the queue is stable ($\rho<1$, which is given), this ratio holds.
Step 4: Apply it to the given data:
We are told $L = n$ and $\lambda = 3$ per minute, so:
\[ W = \frac{L}{\lambda} = \frac{n}{3} \]
This confirms option (B) without needing to know the actual service rate $\mu$.
Final Answer:
Deriving from the underlying M/M/1 formulas gives the same relation as Little's Law. \[ \boxed{W = \dfrac{n}{3}\ \text{minutes}} \]
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