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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  ?

 

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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

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