P10 Unit 7

Pradeep Kumar Chagantipati
Cuw id: F00437246
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 di and the transmission rate of link i, for i = 1, 2, 3. The packet switch delays each packet by d (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*10^8 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

Packet is able to propagate over the link which can be calculated as d1/s1 = 5000*10^3/2.5*10^8
= 5000/2.5*10^5 = 0.02 sec where d1 is the Length of first link
The packet switch adds some delay after receiving the whole packet = 3msec
The first end system requires to transmit the packet to the second link = L/R2
Where L is the packet length,
L/R1= (1500*8)/2*10^6 = 12000/2*10^6 = 0.006sec
Packet is able to propagate over the link which can be calculated as d2/s2 = 4000*10^3/ 2.5*10^8
= 4000/2.5*10^5= 0.016 sec where d2 is the Length of second link
The packet switch adds some delay after receiving the whole packet = 3msec
The first end system requires to transmit the packet to the third link = L/R3
Where L is the packet length,
= L/R1= (1500*8)/2*10^6 = 12000/2*10^6 = 0.006sec
Packet is able to propagate over the link which can be calculated as d3/s3 = 1000*10^3/ 2.5*10^8
= 1000/2.5*10^5= 0.004 sec where d3 is the Length of third link

For end to end delay-
End to end delay = transmission delay of all the 3 links + propagation delays of all the 3 links + switch process delay
= L/R1+ L/R2+ L/R3+ d1/s1+ d2/s2+ d3/s3+0.003+0.003
= 0.006+0.006+0.006+0.02+0.016+0.004+0.003+0.003
= 0.064
Therefore, end to end delay is 0.064