###### Question 8266 – Computer-Networks

May 3, 2024###### Question 10717 – GATE 2007-IT

May 3, 2024# Question 8427 – Computer-Networks

Consider a network connected two systems located 8000 kilometers apart. The bandwidth of the network is 500 × 10^{6} bits per second. The propagation speed of the media is 4 × 10^{6} meters per second. It is needed to design a Go-Back-N sliding window protocol for this network. The average packet size is 10^{7} bits. The network is to be used to its full capacity. Assume that processing delays at nodes are negligible. Then the minimum size in bits of the sequence number field has to be ___________.

Correct Answer: A

Question 54 Explanation:

<img src="https://solutionsadda.in/wp-content/uploads/2020/01/g3.jpg"

η=100%

n = ?

∴a=T

Given Protocol, Go-back-N protocol, So η = w/(1+2a) where w = 2

100/100 = w/(1+2a) ⇒ w = 1+2a

⇒ 2

⇒ 2

⇒ 2

⇒ n = 8

η=100%

n = ?

∴a=T

_{p}/T_{n}=2/0.02=100Given Protocol, Go-back-N protocol, So η = w/(1+2a) where w = 2

^{n}-1100/100 = w/(1+2a) ⇒ w = 1+2a

⇒ 2

^{(n-1)}= 1+2(100)⇒ 2

^{n}– 1 = 201⇒ 2

^{n}= 202 ⇒ 2^{n}= 2^{8}⇒ n = 8

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