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# Computer Networking

## Questions Answers

Station A uses 32 byte packets to transmit messages to station B using sliding window protocol. The round trip delay between A and B is 80 milliseconds and the bottleneck bandwidth on the path between A and B is 128 kbps. What is the optimal window size that A should use ?

Solution

Band width delay product

=round trip delay x bottle neck band width

= 80 x 128 x 10-3 x 1024 bits

= (80 x 128 x 10-3 x 1024)/8 byte

Optimal size window = (80 x 128 x 10-3 x 1024) / (8x32)

= 40

In a token ring network the transmission speed is 107 bps and propogation speed is 200 meters ---s. The 1 bit delay in this network is equivalent to

Solution

107 bits need 1 second

∴ 1 bit needs (1/107) seconds = 1/10  μs

Since 1  μs corresponds to 20 meter cable
∴ 1/10   μs corresponds to 20 meter cable

The distance between two stations M and N is L kilometres . All frames are K bits long. The propagation delay per kilometre is t seconds. Let Rbits/second be the channel capapcity. Assuming that processing delay is negligible ,what is  the minimum number of bits for the sequence number field in a frame for maximum utilization, when sliding window protocol is used

Solution

Total propogation delay = Lt sec

Round trip time = 2 x Lt = 2 Lt sec
No of bits transmitted in round trip= 2LtR bits

Number of frames = 2LtR/K

Let number of bits in sequent number be b

2b={2LtR / K}

B= {log2 (2LtR / K)}

The address of class B host is to be split into subnets with a 6 bit subnet number. What is the maximum number of subnets and the maximum number of hosts in each subnet ?

Solution

Maximum number of subnets

=26-2=62

Maximum number of host in each subnet = 210-2=1022

What signal to noise ratio is needed to put a T carrier on a 10-Khz line ?
Solution
To send a T1 signal we need
Hlog2(1+(s/n))=1.544 x 106 with H 50,000
S / N =230-1.48
93 db

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