Question

Multiple choice questions: Circle the correct answer

1) Suppose two hosts, A and B are separated by 1500 kilometers and are connected by a link of R = 2 Mbps. Suppose the propagation speed over the link is 3 x 108 meters/s. Considering sending a file of 100,000 bits from Host A to Host B. Suppose the file is sent continuously as one large message. The maximum number of bits that will be in the link at any time is

(a) 1 bit; (b) 1000 bits; (c) 10,000 bits; (d) 100,000 bits; (e) None of the above.

1.A) Referring to Problem 1), assuming the file is sent continuously as a large message, the total time required to transfer the file to B is

(a) 40 ms; (b) 45 ms; (c) 50 ms; (d) 55 ms, (e) None of the above.

1.B) Referring to Problem 1), suppose now the file is broken up into 10 packets with each packet containing 10,000 bits. Suppose that each packet is acknowledged by the receiver and the transmission time of an acknowledgement packet is negligible. Finally, assume that the sender cannot send a packet until the preceding one is acknowledged. The total time required for the B to receive the whole file is

(a) 140 ms; (b) 145 ms; (c) 290 ms; (d) 300 ms; (e) None of the above

Answer 1

Step 1

Transmission Time = File Size / Bandwidth = 10^5 bits/ 2Mbps = 50ms

Propagation Drlay = Distance / speed = 1500 Km / 3* 10^8m/sec = 5ms

In 1 sec = 2Mb

5msec = 2* 10^6 * 5 * 10^^-3 = 10 * 1000 bits

Optioc is correct

Step 2

1.A

the total time required to transfer the file to B is =

Transmission Time + Propagation Delay

= 50ms + 5ms = 55ms

Option d is correct

1.B

Transmission Time = File Size / Bandwidth = 10^4 bits/ 2Mbps = 5ms

The total time required for the B to receive the whole file = 9 Packets first * ( Transmission time + 2*Propagation Delay ) + last packettime ( Transmission Time + Propagation Delay)

= 9* ( 5 + 2*5) + ( 5+5)

= 145ms

Option B is correct