Question

Consider a packet of length L which begins at end system A and travels over three links to a destination end system. These three links are connected by two packet switches. Let di, si, and Ri denote the length, propagation speed, and the transmission rate of link i, for i The packet switch delays each packet by dproc. Assuming no queuing delays, in terms of di, si, and Ri, (i=1,2,3), and L, what is the total end-to-end delay for the packet? Suppose now the packet is 1,500 bytes, the propagation speed on both links is 2.5 . 108 m/s, the transmission rates of all three links are 2Mbps, the packet switch processing delay is 3 msec, the length of the first link is 5,000 km, the length of the second link is 4,000 km, and the length of the last link is 1,000 km. For these values, what is the end-to-end delay? Now suppose R1 = R2 = R3 = R and dproc = 0. Further suppose the packet switch does not store-and-forward packets but instead immediately transmits each bit it receives before waiting for the packet to arrive. What is the end-to-end delay?

Answer 1

Answer:

Answer explained below

Explanation:

packet length = L

link i Length = di

propagation speed = si

transmission rate = Ri

the first end system requires to transmit the packet onto the first link = L/R1

the first end system requires to transmit the packet onto the second link = L/R2

the first end system requires to transmit the packet onto the third link = L/R3

the packet propagates over the first link = d1/s1

the packet propagates over the second link = d2/s2

the packet propagates over the third link = d3/s3

end to end delay = L/R1+L/R2+L/R3+d1/s1+d2/s2+d3/s3+d(proc)+ d(pro)

packet size = 1500 bytes

propagation speed on both links = 2.5 x 10^8

transmission rate of all three links = 2 Mbps

packet switch processing delay = 3 msec

length of the first link = 5000 km

length of the second link = 4000km

length of the last link = 1000 km

the first end system requires to transmit the packet onto the first link = L/R1 = .006 sec

packet propagates over the first link = .02 sec

the packet switch requires to transmit the packet onto the first link = L/R1 = .006 sec

packet propagates over the second link = .016 sec

the packet switch requires to transmit the packet onto the third link = L/R1 = .006 sec

packet propagates over the third link = .004 sec

end to end delay = .006+.006+.006+.02+.016+.004+.003+.003

= 0.064 sec