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(Common data question P10, P11)

P10. 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 $d _i, s _i, and R _i$ denote the length, propagation speed, and the transmission rate of link i, for i = 1, 2, 3. The packet switch delays each packet by $d _{proc}$. Assuming no queuing delays, in terms of $d _i, s _i, R _i, (i = 1,2,3), and L$, what is the total end-to-end delay for the packet? Suppose now the packet is 1,500 bytes, the propagation speed on all three links is $2.5 \times 108$m/s, the transmission rates of all three links are 2 Mbps, 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?

P11.  In the above problem, suppose R1 = R2 = R3 = R and $d _{proc} = 0$. Further, suppose the packet switch does not store-and-forward packets but instead immediately transmits each bit it receives before waiting for the entire packet to arrive. What is the end-to-end delay?

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three links are connected by two packet switches 

2 times dpoc will be added i.e 3msec*2 = 0.006 sec

Tt = L/R = 1500 bytes/ 2Mbps = 0.006 sec

for three links 6ms*3 = 18ms

and now Tp is individually calculated as they are different 

Tp= D/S 

Tp1 = 5000km/(2.5X10^8 m/sec) = 0.020 sec

Tp2 = 4000km/(2.5X10^8 m/sec) = 0.016 sec

Tp3 = 1000km/(2.5X10^8 m/sec) = 0.004 sec

adding all we get

0.018+0.006+0.020+0.016+0.004= 0.064 sec = 64msec 

 

now in 2nd case, we are given R1+R2+R3=R

and dpoc =0 

so only a one-time transmission delay will be used  for loading the packet and with continuous transmission 

now add Tt+Tp

 Tt = L/R = 1500 bytes/ 2Mbps = 0.006 sec 

and

Tp= D/S 

Tp1 = 5000km/(2.5X10^8 m/sec) = 0.020 sec

Tp2 = 4000km/(2.5X10^8 m/sec) = 0.016 sec

Tp3 = 1000km/(2.5X10^8 m/sec) = 0.004 sec

==> 0.006+0.020+0.016+0.004= 0.046 sec = 46msec

 

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