Chapter 2: Problem 25
Suppose you are designing a sliding window protocol for a 1-Mbps point-to- point link to a stationary satellite revolving around the earth at \(3 \times 10^{4} \mathrm{~km}\) altitude. Assuming that each frame carries \(1 \mathrm{~KB}\) of data, what is the minimum number of bits you need for the sequence number in the following cases? Assume the speed of light is \(3 \times 10^{8} \mathrm{~m} / \mathrm{s}\). (a) \(\mathrm{RWS}=1\) (b) RWS = SWS
Short Answer
Expert verified
(a) 5 bits (b) 5 bits
Step by step solution
01
Calculate the propagation delay
The first step is to calculate the time it takes for a signal to travel from the Earth to the satellite. This distance is given as 30,000 km or 30,000,000 m. The speed of light is given as 3 \times 10^8 m/s. Using the formula for time, we get: \[\text{Propagation delay} = \frac{ \text{distance}}{ \text{speed}} = \frac{30,000,000 \text{ m}}{3 \times 10^8 \text{ m/s}} = 0.1 \text{ s} \]This is the one-way delay. Since communication involves sending a frame and receiving an acknowledgment, the round-trip delay (RTT) is twice the propagation delay:\[ \text{RTT} = 0.1 \text{ s} \times 2 = 0.2 \text{ s} \]
02
Calculate the transmission delay
Next, calculate how long it takes to send a frame. The bandwidth of the link is 1 Mbps, and each frame carries 1 KB of data. Converting 1 KB (kilobyte) to bits, we have 1 KB = 8 \times 10^3 bits. Using the formula for transmission time, we get: \[\text{Transmission delay} = \frac{ \text{Frame size}}{ \text{Bandwidth}} = \frac{8 \times 10^3 \text{ bits}}{1 \times 10^6 \text{ bits/s}} = 0.008 \text{ s} \]
03
Determine the size of the sequence number for (a) RWS = 1
For (a), where the Receiver Window Size (RWS) = 1, we only need to ensure that the receiver accepts one frame at a time. The window size must be at least the Bandwidth-Delay Product (BDP), which is the product of the bandwidth and the RTT:\[ \text{BDP} = \text{Bandwidth} \times \text{RTT} = 1 \times 10^6 \text{ bits/s} \times 0.2 \text{ s} = 200,000 \text{ bits} \]Since each frame is 8,000 bits, we need enough sequence numbers to cover this bandwidth-delay product:\[\text{Number of frames in transit} = \frac{\text{BDP}}{\text{Frame size}} = \frac{200,000 \text{ bits}}{8,000 \text{ bits/frame}} = 25 \text{ frames} \]Thus, the sequence number must be able to represent 25 frames. In binary, the minimum number of bits required to represent 25 is 5 bits (as 2^5 = 32).
04
Determine the size of the sequence number for (b) RWS = SWS
For (b), where RWS = SWS (Sender Window Size), the window size must also cover the RTT. The sliding window protocol will need as many sequence numbers as there are frames in transit, which we calculated earlier:\[\text{Number of frames in transit} = 25 \text{ frames} \]Again, the sequence number must be able to represent these 25 frames. Thus, the same minimum number of bits is required, which is 5 bits.
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Propagation Delay
Propagation delay refers to the time it takes for a signal to travel from the sender to the receiver. In this exercise, the signal travels from Earth to a satellite at an altitude of 30,000 km.
Since the speed of light is constant at 3 × 10^8 m/s, we can calculate the delay using:\[\text{Propagation delay} = \frac{\text{distance}}{\text{speed}} = \frac{30,000,000 \text{ m}}{3 \times 10^8 \text{ m/s}} = 0.1 \text{ s}\]
This calculation gives us the one-way delay.
To find the Round-Trip Time (RTT), we need to consider the time for the signal to travel to the satellite and back:\[\text{RTT} = 0.1 \text{ s} \times 2 = 0.2 \text{ s}\]
Since the speed of light is constant at 3 × 10^8 m/s, we can calculate the delay using:\[\text{Propagation delay} = \frac{\text{distance}}{\text{speed}} = \frac{30,000,000 \text{ m}}{3 \times 10^8 \text{ m/s}} = 0.1 \text{ s}\]
This calculation gives us the one-way delay.
To find the Round-Trip Time (RTT), we need to consider the time for the signal to travel to the satellite and back:\[\text{RTT} = 0.1 \text{ s} \times 2 = 0.2 \text{ s}\]
Transmission Delay
Transmission delay is the time it takes to push all the frame's bits onto the wire.
It depends on the size of the data being sent and the bandwidth of the link.
Given that each frame is 1 KB (kilobyte) and the link's bandwidth is 1 Mbps (megabit per second), we convert 1 KB to bits: \[1 \text{ KB} = 8 \times 10^3 \text{ bits}\]
Then, the transmission delay is calculated by:\[\text{Transmission delay} = \frac{\text{Frame size}}{\text{Bandwidth}} = \frac{8 \times 10^3 \text{ bits}}{1 \times 10^6 \text{ bits/s}} = 0.008 \text{ s}\]
This tells us how long it takes to fully transmit a single frame over the link.
It depends on the size of the data being sent and the bandwidth of the link.
Given that each frame is 1 KB (kilobyte) and the link's bandwidth is 1 Mbps (megabit per second), we convert 1 KB to bits: \[1 \text{ KB} = 8 \times 10^3 \text{ bits}\]
Then, the transmission delay is calculated by:\[\text{Transmission delay} = \frac{\text{Frame size}}{\text{Bandwidth}} = \frac{8 \times 10^3 \text{ bits}}{1 \times 10^6 \text{ bits/s}} = 0.008 \text{ s}\]
This tells us how long it takes to fully transmit a single frame over the link.
Sequence Number
Sequence numbers are crucial for the sliding window protocol.
They help keep track of the frames sent and those acknowledged.
In this scenario, we need sequence numbers to accommodate the bandwidth-delay product (BDP), which is the product of the link's bandwidth and the RTT. We calculated the BDP as:\[\text{BDP} = \text{Bandwidth} \times \text{RTT} = 1 \times 10^6 \text{ bits/s} \times 0.2 \text{ s} = 200,000 \text{ bits}\]
Since each frame is 8,000 bits, the number of frames in transit is:\[\text{Number of frames in transit} = \frac{\text{BDP}}{\text{Frame size}} = 25 \text{ frames}\]
Therefore, the sequence number must be able to represent these 25 frames. In binary, this can be done with 5 bits.
They help keep track of the frames sent and those acknowledged.
In this scenario, we need sequence numbers to accommodate the bandwidth-delay product (BDP), which is the product of the link's bandwidth and the RTT. We calculated the BDP as:\[\text{BDP} = \text{Bandwidth} \times \text{RTT} = 1 \times 10^6 \text{ bits/s} \times 0.2 \text{ s} = 200,000 \text{ bits}\]
Since each frame is 8,000 bits, the number of frames in transit is:\[\text{Number of frames in transit} = \frac{\text{BDP}}{\text{Frame size}} = 25 \text{ frames}\]
Therefore, the sequence number must be able to represent these 25 frames. In binary, this can be done with 5 bits.
RTT (Round-Trip Time)
RTT or Round-Trip Time measures the time it takes for a signal to travel to the destination and back.
In our situation, RTT is crucial for determining the time window available for acknowledgments and retransmissions.
We previously calculated the RTT from the propagation delay as:\[\text{RTT} = 0.1 \text{ s} \times 2 = 0.2 \text{ s}\]This translates to the total time it takes for a signal to reach the satellite and for the acknowledgment to return.
The RTT determines how fast acknowledgments come back, influencing the efficiency of the sliding window protocol.
In our situation, RTT is crucial for determining the time window available for acknowledgments and retransmissions.
We previously calculated the RTT from the propagation delay as:\[\text{RTT} = 0.1 \text{ s} \times 2 = 0.2 \text{ s}\]This translates to the total time it takes for a signal to reach the satellite and for the acknowledgment to return.
The RTT determines how fast acknowledgments come back, influencing the efficiency of the sliding window protocol.
Bandwidth-Delay Product (BDP)
The Bandwidth-Delay Product (BDP) is a key concept in networking.
It represents the amount of data that can be in transit within the network.
Calculated as the product of bandwidth and RTT, it tells us the pipeline capacity of the link.
For our 1 Mbps link and 0.2 s RTT, the BDP is:\[\text{BDP} = \text{Bandwidth} \times \text{RTT} = 1 \times 10^6 \text{ bits/s} \times 0.2 \text{ s} = 200,000 \text{ bits}\]This value helps determine the sender's window size.
If the window size is less than the BDP, the link won't be fully utilized, causing inefficiencies.Thus, the window size and sequence numbers are set according to the BDP to ensure optimal performance.
It represents the amount of data that can be in transit within the network.
Calculated as the product of bandwidth and RTT, it tells us the pipeline capacity of the link.
For our 1 Mbps link and 0.2 s RTT, the BDP is:\[\text{BDP} = \text{Bandwidth} \times \text{RTT} = 1 \times 10^6 \text{ bits/s} \times 0.2 \text{ s} = 200,000 \text{ bits}\]This value helps determine the sender's window size.
If the window size is less than the BDP, the link won't be fully utilized, causing inefficiencies.Thus, the window size and sequence numbers are set according to the BDP to ensure optimal performance.
