Consider channel capacity is 100 Mbps, mean frame length is 10000 bits and average frame arrival rate is 5000 frames/sec. Applying simple queuing theory, the mean frame delay to send a frame onto the channel is:

Question

Accepted Answer

The correct option is (1):
This is a classic problem in network performance analysis that applies the formula for mean delay in an M/M/1 or M/G/1 queuing model. The mean frame delay ((W)) is the total average time a frame spends in the system, which includes the time it waits in the queue plus the time it takes to be transmitted. First, we must calculate the Transmission Time ((T\_x)) for a single frame: (T\_x = \text{Length} / \text{Capacity} = 10000 \text{ bits} / 100 \times 10^6 \text{ bits/sec} = 0.0001 \text{ seconds}), or 100 microseconds. Next, we determine the Channel Utilization or Traffic Intensity ((\rho)), which is the fraction of time the channel is busy: (\rho = \text{Arrival Rate} \times T\_x = 5000 \text{ frames/sec} \times 0.0001 \text{ sec/frame} = 0.5). With the system stable ((\rho \< 1)), the mean delay formula is (W = \frac{T\_x}{1 - \rho}). Substituting the calculated values gives (W = \frac{0.0001 \text{ sec}}{1 - 0.5} = \frac{0.0001}{0.5} = 0.0002 \text{ seconds}), which correctly converts to 200 micro seconds.

The incorrect option is (2):
The value 100 micro seconds represents only the **Transmission Time ((T\_x))** for a single frame, which is the service time of the channel. The Transmission Time is correctly calculated as $10000 \text{ bits} / 100 \text{ Mbps} = 10^{-4} \text{ seconds}$, or 100 microseconds. However, the problem asks for the *mean frame delay* in the system, which includes both the queuing delay and the transmission time. Since the traffic intensity ((\rho)) is 0.5, meaning the channel is busy half the time, there will be a substantial average waiting time in the queue. Therefore, the total mean delay must be greater than the transmission time, invalidating this option.

The incorrect option is (3):
A delay of 0.5 seconds is grossly incorrect and represents a significant error in calculation, potentially by thousands of times the actual delay. This figure suggests the system is either extremely overloaded or that a fundamental unit or formula error was made. In a stable system with 50% utilization, the total average delay is only twice the transmission time, which is 200 microseconds, not 500,000 microseconds. Choosing this option indicates a lack of understanding of the expected order of magnitude for delays in a high-speed networking environment.

The incorrect option is (4):
The value 2 mili seconds, or 2000 microseconds, is ten times greater than the correct delay. This result would be obtained if a student incorrectly used a traffic intensity of $\rho = 0.9$ in the queuing formula, as $\frac{1}{1-0.9} = 10$, leading to a total delay ten times the transmission time. Since the correct utilization is $\rho = 0.5$, which only leads to a factor of two increase, this option is based on an incorrect calculation of the channel utilization or a misapplication of the queuing delay formula. It does not reflect the given input parameters of a 100 Mbps channel and a 5000 frames/sec arrival rate.