For this problem, use a formula from this chapter, but first state the formula. Frames arrive randomly at a 100-Mbps channel for transmission. If the channel is busy when a frame arrives, it waits its turn in a queue. Frame length is exponentially distributed with a mean of 10,000 bits/frame. For each of the following frame arrival rates, give the delay experienced by the average frame, including both queueing time and transmission time.
a. 90 frames/ sec.
b. 900 frames/ sec.
c. 9000 frames/ sec

(a). 0.1 × 10^-3 = 0.1 msec.

(b). 0.11 × 10^-3 = 0.11 msec.

(c). 1 × 10^-3. = 1 msec.

Explanation:

So, in order to solve this problem or question there is the need to use the Markov queuing formula which is represented mathematically as below;

Mean time delay, t = /(mean frame length,L) × channel capacity, C - frame arrival rate, F. (1).

(a). Using equation (1) above, the delay experience by 90 frames/sec.

=> 1/(10^-4× 10^8 - 90). = 0.1 × 10^-3= 0.1 msec.

(b). Using equation (1) above, the delay experience by 900 frames/sec.

=> 1/(10^-4× 10^8 - 900). = 0.11 × 10^-3= 0.11 msec.

(c). Using equation (1) above, the delay experience by 90 frames/sec.

=> 1/(10^-4× 10^8 - 9000). = 1 × 10^-3= 1 msec.

Hence, the operating queuing system is; 900/10^-4 × 10^8 = 0.9.

answer; i believe that the above statement is (true);

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