Applied Test Series

Question

How big would the window size have to be for the channel utilization to be greater than 98 percent and the bandwidth is 1 Gbps? Suppose that the size of a packet is 1,500 bytes, including both header fields and data. The one way propagation delay is 15 milliseconds for the channel  ?

 

Answer 1

Tt= (1500 bytes * 8) / $10^{9}$ = 0.00001.2 seconds = 0.012 milliseconds

utilization = 0.98 = $\frac{W_{s}}{1+2a}$

$a = \frac{t_{p}}{t_{t}}$

(0.012x) / 30.012 = 0.98

0.012x = 29.41176

x = 2450.98

The window size would have to be approximately 2451 packets

Comments

Hii,

In this Question Window size should be 2450 not 2451. For calculating window size we should consider floor value not ceil value.please correct me if i’m wrong.

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Yes you are right , we consider floor value for calculating window size, but see in this question it is asked to calculate value of window size to get utilization more that 98% . Here equation we are getting is x > 2450.98 hence answer is 2451