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
