Question

Calculate the latency (from first bit sent to last bit received) for the following: (a) 1-Gbps 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 (b) but assume the switch implements "cut-through" switching: It is able to begin retransmitting the packet after the first 128 bits have been received.

Short answer

Part (a): 30 microseconds. Part (b): 60 microseconds. Part (c): 30.384 microseconds.

Step-by-step solution

Calculate Transmission Delay for Part (a)

Calculate the transmission delay, which is the time it takes to push all 5000 bits onto the network. With a 1-Gbps Ethernet link, the transmission delay is \[ \text{Transmission Delay} = \frac{\text{Packet Size}}{\text{Link Speed}} = \frac{5000 \text{ bits}}{1 \times 10^9 \text{ bits/second}} = 5 \text{ microseconds}. \]

Calculate Total Latency with Single Switch

Include the propagation delay for the link and the store-and-forward operation by the switch. Since the switch adds another transmission delay after receiving the packet, we have \[ \text{Total Latency} = \text{Transmission Delay} + \text{Propagation Delay} + \text{Transmission Delay} + \text{Propagation Delay} \] substituting the values, \[ \text{Total Latency} = 5 \text{ microseconds} + 10 \text{ microseconds} + 5 \text{ microseconds} + 10 \text{ microseconds} = 30 \text{ microseconds}. \]

Calculate Total Latency with Three Switches for Part (b)

Each link still adds a propagation delay of 10 microseconds. As there are now three switches, the latency calculation includes additional transmission and propagation delays: \[ \text{Total Latency} = 4 \times \text{Transmission Delay} + 4 \times \text{Propagation Delay} \] substituting the values, \[ \text{Total Latency} = 4 \times 5 \text{ microseconds} + 4 \times 10 \text{ microseconds} = 20 \text{ microseconds} + 40 \text{ microseconds} = 60 \text{ microseconds}. \]

Calculate Transmission Delay for First 128 Bits for Part (c)

For cut-through switching, the switch starts retransmitting after receiving the first 128 bits. The transmission time for first 128 bits is \[ \frac{128 \text{ bits}}{1 \times 10^9 \text{ bits/second}} = 0.128 \text{ microseconds}. \]

Calculate Total Latency with Cut-Through Switching

Now include initial transmission delay for first 128 bits, and the entire latency calculation for three switches: \[ \text{Total Latency} = 3 \times (\text{Initial Transmission Delay} + \text{Propagation Delay}) + \text{Propagation Delay} + \text{Final Transmission Delay} \] substituting the values, \[ \text{Total Latency} = 3 \times (0.128 \text{ microseconds} + 10 \text{ microseconds}) + 10 \text{ microseconds} + 4.872 \text{ microseconds} = 30.384 \text{ microseconds}. \]

Key concepts

Transmission Delay

Transmission delay is the time it takes for all the bits of the packet to be pushed onto the network. This can be calculated using the formula: \ \( \text{Transmission Delay} = \frac{\text{Packet Size}}{\text{Link Speed}} \).

For example, with a 1-Gbps Ethernet connection and a packet size of 5000 bits, the transmission delay is calculated as: \
\[ \text{Transmission Delay} = \frac{5000 \text{ bits}}{1 \times 10^9 \text{ bits/second}} = 5 \text{ microseconds} \].

This means it takes 5 microseconds to push the entire packet onto the network. Note that the transmission delay depends on both the packet size and the speed of the network link. Faster links and smaller packets will have shorter transmission delays.

Propagation Delay

Propagation delay is the time it takes for a signal to travel from the sender to the receiver across the network medium. Unlike transmission delay, which depends on how much data is being sent, propagation delay depends on the distance between the sender and receiver and the speed at which the signal travels through the medium.

Let's consider the following fixed value: a link introduces a propagation delay of 10 microseconds. If a signal needs to travel through multiple links, the propagation delay adds up. For example, if it passes through 4 links:

• Each link: 10 microseconds
• 4 links: 4 x 10 microseconds = 40 microseconds

Therefore, the farther the distance or the more links in between, the higher the total propagation delay.

Cut-Through Switching

Cut-through switching is a method used by network switches to start forwarding a packet after only a part of it (usually the header) has been received, rather than waiting for the entire packet to be received.

In the context of our problem, let's assume a switch that can start retransmitting a packet after receiving the first 128 bits. The transmission time for these first 128 bits is: \ \[ \frac{128 \text{ bits}}{1 \times 10^9 \text{ bits/second}} = 0.128 \text{ microseconds} \].

This method significantly reduces the overall latency because the switch does not wait for the entire packet before starting to forward it. For a network with multiple switches, cut-through switching can greatly reduce the cumulative delay, making networks faster and more responsive.

Store-and-Forward Switching

Store-and-forward switching is another method used by network switches. In this method, the switch receives the entire packet before it starts forwarding it. This can introduce additional delay because the switch must wait until the complete packet has been received.

For instance, if each switch has to receive the entire 5000-bit packet before forwarding it, this adds a transmission delay for every switch in the path.

• If there are three switches, with each adding 5 microseconds of transmission delay:
• Total Delay due to switches: 3 switches x 5 microseconds = 15 microseconds

Store-and-forward switching can provide better error checking (since the whole packet is received before forwarding), but at the cost of added latency compared to cut-through switching.