Made Easy Workbook - Computer Networks

Question

Consider two machines, A and B, connected by a 100 Mbps Ethernet with three store and forward relay switches in path between them. Suppose that no other machines are using the Ethernet, that each link introduces a propagation delay 12 microsecond, and that switch being transmitting a packet immediately after receiving the last bit of packet. What is the total transfer time for a 1500 bytes packet, as measured from transmission of first bit from A to receipt of last bit at B?

Comments

Got it, Answer is 528. But do let me know you approach, might be that is better than mine.

Answer 1

TRANSMISSION TIME = 1500*8/100*10^6 = 120 MICROSEC

PROPAGATION TIME FOR EACH  = 12 MICROSEC

SINCE THREE STORE AND FORWARD SWITCHES , SO TOTAL TIME = 4*TT + 4*PT = 4*120+4*12 = 528.

Comments

Why you are taking 4*TT amd 4*PT?

Can you please elaborate.

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1*TT for A and 1*TT for each relay switch this makes 4*TT

4*PT as there are 4 links between A and B

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What about the acknowledgement?? Since it's told for the last bit receipt so would not we have to take 4[T.T+ 2xP.T] ??