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Calculate the latency (from first bit sent to last bit received) for the following: (a) 10-Mbps Ethernet with a single store-and-forward switch in the path, and a packet size of 5000 bits. Assume that each link introduces a propagation delay of \(10 \mu \mathrm{s}\) and that the switch begins retransmitting immediately after it has finished receiving the packet. (b) Same as (a) but with three switches. (c) Same as (a) but assume the switch implements "cut-through" switching: It is able to begin retransmitting the packet after the first 200 bits have been received.

Short Answer

Expert verified
Part (a): 520 μs, part (b): 540 μs, part (c): 520 μs.

Step by step solution

01

Calculate transmission time

The transmission time is the time it takes to send the entire packet over the link. It is given by the formula: \[ \text{Transmission Time} = \frac{\text{Packet Size}}{\text{Bandwidth}} \] For a 10-Mbps Ethernet and a packet size of 5000 bits: \[ \text{Transmission Time} = \frac{5000 \text{ bits}}{10 \times 10^6 \text{ bits/s}} = 0.0005 \text{ s} = 500 \text{ μs} \]
02

Calculate propagation delay

The propagation delay is the time it takes for a bit to travel from the sender to the receiver. This is given as 10 μs per link. For part (a) with one link: \[ \text{Propagation Delay} = 10 \text{ μs} \]
03

Calculate total latency for part (a)

For part (a), with one store-and-forward switch: \[ \text{Total Latency} = \text{Transmission Time} + 2 \times \text{Propagation Delay} \] Because the switch introduces its own propagation delay: \[ \text{Total Latency} = 500 \text{ μs} + 2 \times 10 \text{ μs} = 520 \text{ μs} \]
04

Calculate total latency for part (b)

For part (b), with three store-and-forward switches, we have four links: \[ \text{Total Latency} = \text{Transmission Time} + 4 \times 10 \text{ μs} \] \[ \text{Total Latency} = 500 \text{ μs} + 40 \text{ μs} = 540 \text{ μs} \]
05

Calculate transmission time for cut-through switching in part (c)

For cut-through switching, the switch starts retransmitting after receiving 200 bits. \[ \text{Transmission Time}_{\text{initial}} = \frac{200 \text{ bits}}{10 \times 10^6 \text{ bits/s}} = 20 \text{ μs} \]
06

Calculate remaining transmission time for each link in part (c)

After the first 200 bits, the remaining 4800 bits are retransmitted immediately through the switch: \[ \text{Transmission Time}_{\text{remaining}} = \frac{4800 \text{ bits}}{10 \times 10^6 \text{ bits/s}} = 480 \text{ μs} \]
07

Calculate total latency for part (c)

Combining the initial and remaining transmissions with the propagation delays: \[ \text{Total Latency} = \text{Transmission Time}_{\text{initial}} + \text{Transmission Time}_{\text{remaining}} + 2 \times 10 \text{ μs} \] \[ \text{Total Latency} = 20 \text{ μs} + 480 \text{ μs} + 20 \text{ μs} = 520 \text{ μs} \]

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Transmission Time
Transmission time is a crucial concept in network latency calculations. It refers to the time required to send the entire packet from one node to another over a network link.
This can be calculated using the formula: \ \[ \text{Transmission Time} = \frac{\text{Packet Size}}{\text{Bandwidth}} \]
Where:
  • Packet Size is the size of the packet in bits
  • Bandwidth is the capacity of the link in bits per second (bps)
For example, if you have a 10-Mbps (Megabits per second) Ethernet link and you are sending a packet of 5000 bits, the calculation is:
\ \[ \text{Transmission Time} = \frac{5000 \text{ bits}}{10 \times 10^6 \text{ bits/s}} = 0.0005 \text{ s} = 500 \text{ μs} \] This means it takes 500 microseconds to transmit the packet over this link.
Propagation Delay
Propagation delay is another critical factor in network latency. It is the time a single bit takes to travel from the sender to the receiver over the network link.
It is typically influenced by the speed of light in the transmission medium (which can be fiber, copper, etc.). For example, in the given problem, the propagation delay for each link is given as 10 microseconds (\text{ μs}).
So, if you have one link, the propagation delay is simply 10 μs. But, if you have multiple links or switches in between, you need to account for each segment.
For example:
  • With one link, the propagation delay is 10 μs
  • With three switches (four links), the total propagation delay is \[4 \times 10 \mu \text{s} = 40 \mu \text{s}\]
Store-and-Forward Switch
A store-and-forward switch is a type of network switch that receives the entire packet before retransmitting it to the next destination.
This introduces additional latency because the switch has to wait until the entire packet arrives. Only then it begins forwarding it.
For instance, if the transmission time for a 5000-bit packet over a 10-Mbps link is 500 microseconds (\text{ μs}), and the propagation delay per link is 10 μs, the total latency for one store-and-forward switch is: \ \[ \text{Total Latency} = \text{Transmission Time} + 2 \times \text{Propagation Delay} = 500\mu \text{s} + 2 \times 10 \mu \text{s} = 520 \mu \text{s} \] If you have more store-and-forward switches, you need to add the propagation delays for each additional link.
For example, with three store-and-forward switches (having four links in total): \ \[ \text{Total Latency} = 500 \mu \text{s} + 4 \times 10 \mu \text{s} = 540 \mu \text{s} \]
Cut-Through Switching
Cut-through switching is a technique used by some network switches to reduce latency. Unlike store-and-forward switches, a cut-through switch begins forwarding the packet as soon as it receives the first part of the packet.
This can significantly reduce the overall latency as it does not wait for the entire packet to arrive.
For example, if a switch starts forwarding after the first 200 bits of a 5000-bit packet, the initial transmission time is: \ \[ \text{Transmission Time}_{\text{initial}} = \frac{200 \text{ bits}}{10 \times 10^6 \text{ bits/s}} = 20 \text{ μs} \] The remaining 4800 bits are transmitted while the initial 200 bits are already being forwarded, so the remaining transmission time is: \ \[ \text{Transmission Time}_{\text{remaining}} = \frac{4800 \text{ bits}}{10 \times 10^6 \text{ bits/s}} = 480 \text{ μs} \] Combining this with the propagation delay for paths with multiple links and switches, the total latency can be calculated. For example, with one cut-through switch and two links:
\ \[ \text{Total Latency} = 20 \text{ μs} + 480 \text{ μs} + 2 \times 10 \text{ μs} = 520 \text{ μs} \]

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