Question

Note: b denotes bits and B denotes Bytes (1 Byte = 8 bits). When a node transmits a packet, it doesn’t need to wait for the last packet to be propagated along the link. As soon as one packet is transmitted into the link, the node can start transmitting the next packet

Part1:

For each of the following cases, calculate the latency (from first bit sent to the last bit received). Assume the bandwidth of each link is 2Gbps, and the size of each packet is 120 Kb (Kilo-bits). Note that little b is bit and big B is Byte.Assume the length of each link is 40meters, and propagation speed is 2*10⁶meters per second. There is no queuing or processing delay.

a. If the source sends a message containing 1packet and there are 2routers on the path.

b. If the source sends a message containing 1packet and there are 4routers on the path.

c. If the source sends a message containing 10packets and there are 2routers on the path.

d. If the source sends a message containing 10packets and there are 4routers on the path.

Part2:

Now keep all assumptions from part1 except processing delay. Assume each router begins retransmitting a packet after it has finished receiving the packet and processed it for 4microseconds (d-proc = 4μs). Processing delay is zero at source. Solve the problem for each of the 4 above-mentioned cases with this assumption.

Answer 1

We first do some calculations, and derive a generic formula.

When any link is being used for the first time, the first-bit-propogation-delay, d-prop

= (Link-Length in metres) / (Link-Propogation-Speed in metres/second)

= 40 / (2 * 10^6)

= 20 * 10^-6 seconds

= 20 us

Next, the time for packet-propogation (link-time) , once first-bit-propogation-delay (d-prop) has been overcome

= (Packet-Size in bits) / (Link-Bandwidth in bps)

= (120 * 10^3) / (2 * 10^9)

= 60 * 10^-6 seconds

= 60 us

Now, let,

• p = Number of packets
• r = Number of routers
• d-proc = Processing-delay at router

For first packet, there will be propogation-delay involved at every link, as the links are just starting up.

So,

• Time taken to travel from source to first router = d-prop + link-time = (20 + 60) us = 80 us
• Time taken to travel from router to next destination = d-prop + d-proc + link-time = (d-proc + 80) us

So, total time for first packet to travel from source to destination = (80 + r(d-proc + 80)) us

For second packet onwards, there will be no propogation-delay involved.

So,

• Time taken to travel from source to first router = link-time = 60 us
• Time taken to travel from router to next destination = d-proc + link-time = (d-proc + 60) us

So, total time for second-packet onwards, to travel from source to destination = (60 + r(d-proc + 60)) us

Thus, total time for "p" packets, to travel across "r" routers, with "d-proc" delay at each router

= (Time for first-packet) + (Time for subsequent p-1 packets)

= [80 + r(d-proc + 80) + (p - 1)(60 + r(d-proc + 60))] us

Part 1, a)

Here, p = 1, r = 2, d-proc = 0

So, total time = [80 + 2(80)]

= 240 us

Part 1, b)

Here, p = 1, r = 4, d-proc = 0

So, total time = [80 + 4(80)]

= 400 us

Part 1, c)

Here, p = 10, r = 2, d-proc = 0

So, total time = [80 + 2(80) + 9(60 + 2(60))]

= [240 + 9(180)]

= 1860 us

Part 1, d)

Here, p = 10, r = 4, d-proc = 0

So, total time = [80 + 4(80) + 9(60 + 4(60))]

= [400 + 9(300)]

= 3100 us

Part 2, a)

Here, p = 1, r = 2, d-proc = 4

So, total time = [80 + 2(4 + 80)]

= 248 us

Part 2, b)

Here, p = 1, r = 4, d-proc = 4

So, total time = [80 + 4(4 + 80)]

= 416 us

Part 2, c)

Here, p = 10, r = 2, d-proc = 4

So, total time = [80 + 2(4 + 80) + 9(60 + 2(4 + 60))]

= [248 + 9(188)]

= 1940 us

Part 2, d)

Here, p = 10, r = 4, d-proc = 4

So, total time = [80 + 4(4 + 80) + 9(60 + 4(4 + 60))]

= [416 + 9(316)]

= 3260 us