Problem 2 - Computer Networks: Global Edition

Question

A group of \(\mathrm{N}\) stations share a \(56-\mathrm{kbp}\) s pure ALOHA channel. Each station outputs a 1000 -bit frame on average ance every \(100 \mathrm{sec}\), even if the previous one has not yet been sent (e.g., the stations can buffer outgoing frames). What is the maximum value of N?

Step-by-Step Solution

Answer

The maximum value of \( N \) is 10.

Step 1: Understand Pure ALOHA

The pure ALOHA protocol allows the stations to send frames at any time. If two frames overlap in any way, they collide and must be resent. The theoretical maximum efficiency of pure ALOHA is 18.4%, meaning that the maximum utilization of the channel is about 18.4% before collisions start affecting the transmission.

Step 2: Calculate Frame Time

First, we need to calculate how long it takes to send a single frame. Given a 1000-bit frame and a 56 kbps channel, the frame time is the frame size divided by channel bandwidth: \( \text{Frame Time} = \frac{1000}{56000} \) seconds.

Step 3: Calculate Frame Rate

Each station sends a frame every 100 seconds. So, the rate at which a single station sends frames is \( \frac{1}{100} \) frames per second.

Step 4: Find Total Channel Utilization

Multiply the frame time by the number of stations \( N \) and the frame rate to find the total channel utilization, which must not exceed the maximum efficiency of pure ALOHA (18.4%). This gives \( N \times \frac{1000}{56000} \times \frac{1}{100} \leq 0.184 \).

Step 5: Solve for Maximum N

To solve for the maximum number of stations, rearrange the equation: \( N \leq \frac{0.184 imes 56000 imes 100}{1000} \). Doing this calculation results in \( N \leq 10.304 \). Since \( N \) must be a whole number, the largest possible value for \( N \) is 10.

Key Concepts

Channel Utilization

Channel utilization refers to how effectively the capacity of a network channel is used by data transmissions. In the context of the pure ALOHA protocol, utilization is crucial because this protocol allows devices to transmit data at random, which may lead to collisions. If two frames overlap during transmission, they will collide, and both frames must be resent. This can lower the channel's overall efficiency.

• Pure ALOHA has a theoretical maximum channel utilization of 18.4%.
• This means that only 18.4% of the time is utilized effectively for successful data transmission.
• The rest of the time—nearly 80%—can be affected by collisions, where data must be retransmitted.

Understanding channel utilization helps us determine how many devices can efficiently share the channel without exceeding this threshold, thereby minimizing transmissions that potentially collide.

Frame Time Calculation

Frame time calculation is essential to understand how long it takes for a single frame of data to be completely transmitted. This is calculated by dividing the size of the frame by the channel's bandwidth.
For instance, in the given problem:

• Frame Size: 1000 bits
• Channel Bandwidth: 56 kbps (kilobits per second)

To find the frame time, do the following calculation: \[\text{Frame Time} = \frac{1000}{56000} \approx 0.017857 \text{ seconds}\]This calculation tells us that each frame takes approximately 0.018 seconds to be sent. Knowing the frame time is critical for managing how data is scheduled and transmitted, especially in systems like pure ALOHA where you want to prevent too many data overlaps that cause collisions.

Maximum Efficiency

Achieving maximum efficiency in a network protocol like pure ALOHA involves managing the number of devices and their data transmissions effectively. To ensure that the network functions optimally, we must stay within the bounds of the protocol's efficiency limit of 18.4%.
In practice:

• The equation used for calculation ensures that channel utilization does not exceed this efficiency threshold.
• By rearranging and solving the equation \( N \times \frac{1000}{56000} \times \frac{1}{100} \leq 0.184 \), we can find the maximum number of devices \( N \) that can use the channel efficiently.
• The calculation results in a maximum of 10 devices, considering that \( N \) must be a whole number.

This approach ensures that the data from multiple stations is transmitted without excessive collisions, thereby allowing the network to operate at its highest possible efficiency under the constraints of the pure ALOHA protocol.