Chapter 5: Problem 9
You are hired to design a reliable byte-stream protocol that uses a sliding window (like TCP). This protocol will run over a 100-Mbps network. The RTT of the network is \(100 \mathrm{~ms}\), and the maximum segment lifetime is 60 seconds. (a) How many bits would you include in the AdvertisedWindow and SequenceNum fields of your protocol header? (b) How would you determine the numbers given above, and which values might be less certain?
Short Answer
Expert verified
(a) AdvertisedWindow field: 21 bits, SequenceNum field: 30 bits. (b) Calculations are based on given network bandwidth and RTT; values may vary due to variable factors.
Step by step solution
01
Calculate the Bandwidth-Delay Product
The Bandwidth-Delay Product (BDP) determines the amount of data that can be in transit in the network. First, convert the network bandwidth into bits per second: \(100 \text{ Mbps} = 100 \times 10^6 \text{ bps}\)Then, calculate the BDP using the given RTT: \[ \text{BDP} = \text{Bandwidth} \times \text{RTT} \ \text{BDP} = 100 \times 10^6 \text{ bps} \times 0.1 \text{ s} = 10 \times 10^6 \text{ bits} = 10 \text{ Mb} \ \text{BDP} = 1.25 \text{ MB} \ \text{BDP} = 1.25 \times 10^6 \text{ bytes} \]
02
Determine the AdvertisedWindow Size
The AdvertisedWindow size should be at least equal to the BDP to fully utilize the network. Therefore, \[ \text{AdvertisedWindow} \text{Size} = 1.25 \times 10^6 \text{ bytes} \]
03
Calculate the Number of Bits for AdvertisedWindow field
To determine the number of bits required for the AdvertisedWindow field, find the smallest power of 2 that can represent the AdvertisedWindow size. Using the formula \( n = \text{ceil}(\text{log}_2(\text{AdvertisedWindow Size})) \) : \[ \text{bits} = \text{ceil}(\text{log}_2(1.25 \times 10^6)) \ \text{bits} = \text{ceil}(20.93) = 21 \]
04
Determine the Sequence Number Range
The Sequence Number must accommodate the maximum segment lifetime, which is the maximum time any segment can be in the network.Using the formula: \[ \text{Sequence Number Range} = \text{Bandwidth} \times \text{Maximum Segment Lifetime} \ \text{Sequence Number Range} = 100 \times 10^6 \text{ bps} \times 60 \text{ s} = 6 \times 10^9 \text{ bits} = 750 \times 10^6 \text{ bytes} \]
05
Calculate the Number of Bits for SequenceNum field
Like in the AdvertisedWindow calculation, determine the number of bits required to represent the Sequence Number Range.Using the formula \( n = \text{ceil}(\text{log}_2(\text{Sequence Number Range})) \) : \[ \text{bits} = \text{ceil}(\text{log}_2(750 \times 10^6)) \ \text{bits} = \text{ceil}(29.49) = 30 \]
06
Analyze Uncertain Values
Values determined from the network's given parameters (RTT, bandwidth) and protocol constraints (maximum segment lifetime). Uncertain values often depend on variable factors such as changing RTTs, bandwidth fluctuations, and maximum segment lifetime assumptions based on the specific network configuration.
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Bandwidth-Delay Product (BDP)
The Bandwidth-Delay Product (BDP) tells us how much data can be in transit in the network at any given time. Imagine a pipe where water flows through; the length of the pipe and the speed at which water flows dictate how much water is within it. Here, bandwidth is the flow rate, and delay is the length of the pipe.
To compute BDP: Convert the network bandwidth from Mbps to bps. Multiply the bandwidth by the Round Trip Time (RTT). For instance, with bandwidth of 100 Mbps and RTT of 100 ms:
\[ \text{BDP} = 100 \times 10^6 \text{ bps} \times 0.1 \text{ s} = 10 \times 10^6 \text{ bits} = 1.25 \times 10^6 \text{ bytes} \]
The BDP indicates the storage needed to handle the data in transit, which is crucial for optimal network performance.
To compute BDP: Convert the network bandwidth from Mbps to bps. Multiply the bandwidth by the Round Trip Time (RTT). For instance, with bandwidth of 100 Mbps and RTT of 100 ms:
\[ \text{BDP} = 100 \times 10^6 \text{ bps} \times 0.1 \text{ s} = 10 \times 10^6 \text{ bits} = 1.25 \times 10^6 \text{ bytes} \]
The BDP indicates the storage needed to handle the data in transit, which is crucial for optimal network performance.
Advertised Window
The Advertised Window size is essential for controlling the flow of data, ensuring the sender doesn’t overwhelm the receiver. The size of the Advertised Window should be at least equal to the BDP.
To determine the size, use the BDP calculated earlier. In our example:
\[ \text{AdvertisedWindow Size} = 1.25 \times 10^6 \text{ bytes} \]
This means our Advertised Window size should be around 1.25 MB. To represent this in bits, we use the smallest power of 2 that can accommodate the AdvertisedWindow size:
\[ \text{bits} = \text{ceil}(\text{log}_2(1.25 \times 10^6)) = 21 \]
To determine the size, use the BDP calculated earlier. In our example:
\[ \text{AdvertisedWindow Size} = 1.25 \times 10^6 \text{ bytes} \]
This means our Advertised Window size should be around 1.25 MB. To represent this in bits, we use the smallest power of 2 that can accommodate the AdvertisedWindow size:
\[ \text{bits} = \text{ceil}(\text{log}_2(1.25 \times 10^6)) = 21 \]
Sequence Number Range
The Sequence Number Range must cover all possible bytes sent in the network. This range ensures that every byte gets a unique identifier. We calculate it based on the network bandwidth and the maximum segment lifetime.
Given a bandwidth of 100 Mbps and a segment lifetime of 60 seconds:
\[ \text{Sequence Number Range} = 100 \times 10^6 \text{ bps} \times 60 \text{ s} = 6 \times 10^9 \text{ bits} = 750 \times 10^6 \text{ bytes} \]
To represent this in bits, we use:
\[ \text{bits} = \text{ceil}(\text{log}_2(750 \times 10^6)) = 30 \]
Given a bandwidth of 100 Mbps and a segment lifetime of 60 seconds:
\[ \text{Sequence Number Range} = 100 \times 10^6 \text{ bps} \times 60 \text{ s} = 6 \times 10^9 \text{ bits} = 750 \times 10^6 \text{ bytes} \]
To represent this in bits, we use:
\[ \text{bits} = \text{ceil}(\text{log}_2(750 \times 10^6)) = 30 \]
Sliding Window Protocol
The sliding window protocol is a key method for controlling data flow in a network. Think of it as a window moving over the data stream, which lets a specific range of data be sent and acknowledged at given times.
The sender transmits several frames before needing an acknowledgment for the first one, enhancing throughput and efficiency. As each frame is acknowledged, the window slides, allowing the next set of frames to be sent. The size of this window is typically determined by the BDP and Advertised Window values.
Advantages include better utilization of bandwidth and decreased latency owing to continuous data flow rather than waiting for individual acknowledgments.
The sender transmits several frames before needing an acknowledgment for the first one, enhancing throughput and efficiency. As each frame is acknowledged, the window slides, allowing the next set of frames to be sent. The size of this window is typically determined by the BDP and Advertised Window values.
Advantages include better utilization of bandwidth and decreased latency owing to continuous data flow rather than waiting for individual acknowledgments.
TCP
Transmission Control Protocol (TCP) is a reliable byte-stream protocol widely used in Internet communications. It ensures data is delivered correctly and in order, managing both flow and congestion control.
Key Features of TCP:
Key Features of TCP:
- Reliability: It checks data integrity and resends lost packets.
- Flow Control: Uses the Advertised Window to prevent sender overload.
- How it works: It sends data in segments, and each byte has a sequence number. As the receiver acknowledges segments, the sender slides its window and sends more.
- Congestion Control: Changes its transmission rate based on network congestion.
Network RTT
Round Trip Time (RTT) measures the time it takes for a signal to travel from the sender to the receiver and back. It is a crucial factor for calculating the BDP and affects the overall efficiency of data transmission.
A higher RTT can result in longer delays but allows more data to be in transit. For the RTT given in our example (100 ms), it influences the size of the sliding window and Advertised Window.
Effective managing of RTT helps optimize network performance, ensuring smoother and faster communication between devices.
A higher RTT can result in longer delays but allows more data to be in transit. For the RTT given in our example (100 ms), it influences the size of the sliding window and Advertised Window.
Effective managing of RTT helps optimize network performance, ensuring smoother and faster communication between devices.
