propagation delay

Question

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 100 bits of data are to be transmitted between the two hosts using a packet size of 500 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 _____.

Answer 1

Since packet size is 500 bits even though we need only 100 bits, we have to send 500 bits. 

Transmission time for 500 bits = Time for 500 bits to travel

= 500/10⁷ seconds = 50 microseconds

So, after 50 microseconds he packet is completely transmitted. But the packet must reach the other end and this happens exactly after propagation delay which here is 20 microseconds. So, the switch receives the packet after 50 + 20 = 70 microseconds.

Given in question, switch forwards after 35 microseconds. So, after 70 + 35 = 105 microseconds, the switch forwards the packet and at 105 + 70 = 175 microseconds the receiver receives the last bit of data. (70 is same as the time to reach the switch as propagation delay is same for the two links as given in question). 

Comments

I believe this is the actual question:
https://gateoverflow.in/8495/gate2015-3_36

---

@arjun sir...how to solve is difference is asked between first and last bit?

---

Sir but in question it is given that switch forwards the packet 35microsec after it receives the same packet.So the switch waits 2nd packet to arrive and then forwards the first packet  35microsec after its arrival.

So first packet total time to arrive= 70microsec(Lets say on timer) upon second packet arrival to switch timer shows (70 +50) microseconds.Total time equals to 70+50+35+70=225microsec.........Sir correct me if i am wrong.......(Here i have taken that 2nd packet transmission starts before first packet reaches the switch)