Question

Suppose you are designing a sliding window protocol for a 1-Mbps point-to- point link to the moon, which has a one-way latency of \(1.25\) seconds. Assuming that each frame carries \(1 \mathrm{~KB}\) of data, what is the minimum number of bits you need for the sequence number?

Short answer

9 bits

Step-by-step solution

Determine the Bandwidth-Delay Product

Compute the bandwidth-delay product. The bandwidth is given as 1 Mbps (which is equal to 1 million bits per second), and the one-way latency is 1.25 seconds. Hence, the round-trip time (RTT) is twice the one-way latency, which is 2.5 seconds.

Calculate the Required Buffer Size

Multiply the bandwidth by the RTT to get the bandwidth-delay product. This product represents the total number of bits that can be in transit at any given time:\[ \text{Bandwidth-Delay Product} = 1,000,000 \text{ bps} \times 2.5 \text{ s} = 2,500,000 \text{ bits} \text{ or } 2,500 \text{ KB} \text{ (since 1 KB = 8 kb)} \]

Determine the Frame Size

Each frame carries 1 KB of data. So now, calculate the number of frames that can fit within the bandwidth-delay product by dividing the total number of bits in transit by the frame size in bits (1 KB = 8,000 bits):\[ \text{Number of Frames} = \frac{2,500,000 \text{ bits}}{8,000 \text{ bits/frame}} = 312.5 \text{ frames} \]

Calculate the Sequence Number Size

The sequence number should be able to cover the number of frames in transit. Therefore, calculate the minimum number of bits required for the sequence number (N). Since the number of frames is 312.5, and sequence numbers must be whole numbers, round up to the next integer, 313. The minimum number of bits, N, satisfies: \(2^N \ge 313\).\[ N = \text{ceil}(\text{log}_2(313)) = 9 \text{ bits} \]

Key concepts

Bandwidth-Delay Product

The bandwidth-delay product (BDP) is a critical concept when designing a sliding window protocol. It helps determine the capacity of the network path in terms of bits that can be in transit. To calculate BDP, you need to multiply the network bandwidth by the round-trip time (RTT).
In our given problem, the bandwidth is 1 Mbps, and the one-way latency to the moon is 1.25 seconds. Thus, the RTT is 2.5 seconds (since RTT is twice the one-way latency).
Using these values, the bandwidth-delay product is calculated as follows:
\[ \text{BDP} = 1,000,000 \text{ bps} \times 2.5 \text{ s} = 2,500,000 \text{ bits} \text{ or } 2,500 \text{ KB (since 1 KB = 8 kb)} \]
This means that 2,500 KB worth of data can be in transit at any time on this connection.

Sequence Number Calculation

Determining the correct sequence number size is essential for effective protocol design. Sequence numbers are used to keep track of frames in transit and to ensure that data is transmitted and received in the correct order.
Since each frame carries 1 KB of data, the number of frames that can fit within the BDP needs to be calculated. Given that each frame is 1 KB (or 8,000 bits), we find the number of frames as follows:
\[ \text{Number of Frames} = \frac{2,500,000 \text{ bits}}{8,000 \text{ bits/frame}} = 312.5 \text{ frames} \]
To ensure complete coverage, sequence numbers must account for the smallest integer greater than or equal to 312.5, which is 313. Hence, the minimum number of bits (N) required satisfies: \[2^N \text{ bits} \] and the smallest N that satisfies this is 9 bits, because: \(\text{ceil}(\text{log}_2 313) = 9\).
Therefore, we need a 9-bit sequence number to cover all possible frames in transit.

Point-to-Point Link Latency

Point-to-point link latency refers to the time it takes for a signal to travel from the sender to the receiver. In the context of our problem, the one-way latency to the moon is given as 1.25 seconds. Latency is a crucial factor in designing network protocols as it directly influences the round-trip time (RTT).
RTT is essential for understanding how long it will take for a signal to make a round trip from the sender to the receiver and back. The formula is:
\[ \text{RTT} = 2 \times \text{One-way Latency} \ \text{RTT for our example} = 2 \times 1.25 \text{ s} = 2.5 \text{ s} \]
This helps determine the maximum data that can be in transit, as shown in the bandwidth-delay product section. Efficiently managing this with proper buffering and sequencing ensures that the network does not suffer from issues like congestion and frame loss.

Frame Size in Networks

Frame size is a significant aspect of network performance and protocol efficiency. Frames are units of data sent over the network, and in our scenario, each frame carries 1 KB (kilobyte) of data. Understanding and determining the frame size helps in correctly calculating the sequence numbers and managing the sliding window protocol.
From the problem statement:
\[ \text{Frame Size} = 1 \text{ KB} = 8,000 \text{ bits (since 1 KB = 8 kb)} \]
Knowing the frame size allows us to calculate how many frames can be in transit given the BDP. It is vital to align frame sizes with network capabilities and protocol requirements to ensure efficient data transmission and minimize delays or errors.