Question

Consider the three hosts A, B and C in the network. The host A transmits the packet of size 700 bytes to host C through the host B. The propagation speed from A to B is
2x10 m/sec and the propagation speed from B to C is 3x10 m/sec. The dista ce
between the hosts A and B is 1000 km and the distance between B and C is 1500 km.
The link bandwidths of A-B and B-C are same that is 5 Mbps. What is the end-to-end
delay for the packet?
a) 10.02 ms
b) 12.24 ms
c) 16.12 ms
d) 15.16 ms​

Answer 1

End-to-end  delay for the packet is 12.24ms

Given packet size  = 700 bytes.

Propogation speed from A to B is 2x10^8m/s , (In the question its given  wrong)

• Vab = 2x10^8m/s

Propogation speed from B to C is 3x10^8m/s

• Vbc = 3x10^8m/s

distance  between the hosts A and B is

• Dab = 1000 km and

the distance between B and C is

• Dbc = 1500 km.

The link bandwidths of A-B and B-C are same that is

• B = 5 Mbps

Total Delay = Total Propagation delay + Total transmission delay

Propagation delay:

• Pd = link length/ signal speed
• Pd( for AB) = Dab/Vab = 1000km/Vab = 10^6/2x10^8 = 5ms
• Pd( for BC) = Dbc/Vbc = 1500km/Vbc = 1.5x10^6/3x10^8 = 5ms

Transmission Delay:

• td = data length/signal bandwith
• td(AB) = 700 bytes/5x10^6bits = 700x8/5 = 1.12 ms

• td(BC) = 700 bytes/5x10^6bits = 700x8/5 = 1.12 ms

Hence total end to end delay = 5 + 5 + 1.12 + 1.12 = 12.24ms

Answer 2

Answer:

12.24

Step-by-step explanation: