# GATE | GATE-CS-2015 (Set 3) | Question 46

Since it is a network that uses switch, every packet goes through two links, one from source to switch and other from switch to destination.

Since there are 10000 bits and packet size is 5000, two packets are sent. Transmission time for each packet is 5000 / 107</sup microseconds.

Two hosts are connected via a packet switch with 107 bits per second links. Each link has a propagation delay of 20 microseconds. The switch begins forwarding a packet 35 microseconds after it receives the same. If 10000 bits of data are to be transmitted between the two hosts using a packet size of 5000 bits, the time elapsed between the transmission of the first bit of data and the reception of the last bit of the data in microseconds is _________.
(A) 1075
(B) 1575
(C) 2220
(D) 2200

Explanation: Sender host transmits first packet to switch, the transmission time is 5000/107 which is 500 microseconds. After 500 microseconds, the second packet is transmitted. The first packet reaches destination in 500 + 35 + 20 + 20 + 500 = 1075 microseconds. While the first packet is traveling to destination, the second packet starts its journey after 500 microseconds and rest of the time taken by second packet overlaps with first packet. So overall time is 1075 + 500 = 1575.

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