# GATE | GATE-CS-2014-(Set-2) | Question 65

Consider the store and forward packet switched network given below. Assume that the bandwidth of each link is 106 bytes / sec. A user on host A sends a file of size 103 bytes to host B through routers R1 and R2 in three different ways. In the first case a single packet containing the complete file is transmitted from A to B. In the second case, the file is split into 10 equal parts, and these packets are transmitted from A to B. In the third case, the file is split into 20 equal parts and these packets are sent from A to B. Each packet contains 100 bytes of header information along with the user data. Consider only transmission time and ignore processing, queuing and propagation delays. Also assume that there are no errors during transmission. Let T1, T2 and T3 be the times taken to transmit the file in the first, second and third case respectively. Which one of the following is CORRECT?

(A) T1 < T2 < T3

(B) T1 > T2 > T3
(C) T2 = T3, T3 < T1
(D) T1 = T3, T3 > T2

Explanation: The important thing to note here is in first case, the whole packet is being transmitted, so no piplelining of packet happens.

In second and third case, we have advantage of pipelining (While packet ‘i’ is being transmitted from R1 to R2,
packet ‘i-1’ is being transmitted from A to R1 at the same time).

Following are complete calculations.

```File Size = 1000 bytes
Transmission Speed of all  links = 10^6 bytes/sec

Ist Case:
= packetsize/bandwidth
= (1000 + 100)/10^6
= 1100 micros
Total time = 3*1100
= 3300 microsec.

Second case:
Transmission time for one link and one part
= (100 + 100)/10^6
=  200 microsec

[Note the pipe-lining in packets.  While
packet 'i' is being transmitted from R1 to R2,
packet 'i-1' is being transmitted from A to R1
at the same time]
Total time = 3*200 + 9*200
= 2400 micro sec

Third Case:
Transmission time for one link and one part
= (50+100)/10^6
= 150microsec
Total time = 3*150+19*150
= 3300 microsec ```

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