Question
In this problem, we explore the use of small packets for Voice-over-IP appli- cations. One of the drawbacks of a small packet size is that a large fraction of link bandwidth is consumed by overhead bytes. To this end, suppose that the packet consists of P bytes and 5 bytes of header.
a. Consider sending a digitally encoded voice source directly. Suppose the source is encoded at a constant rate of 128 kbps. Assume each packet is entirely filled before the source sends the packet into the network. The time required to fill a packet is the packetization delay. In terms of L, determine the packetization delay in milliseconds.
b. Packetization delays greater than 20 msec can cause a noticeable and unpleasant echo. Determine the packetization delay for L = 1,500 bytes (roughly corresponding to a maximum-sized Ethernet packet) and for L = 50 (corresponding to an ATM packet).
c. Calculate the store-and-forward delay at a single switch for a link rate of R = 622 Mbps for L = 1,500 bytes, and for L = 50 bytes.
d. Comment on the advantages of using a small packet size.
Answer
a)
To determine the packetization delay in milliseconds, we need to calculate the time required to fill a packet.
Given: Packet size (including header) = P + 5 bytes
Source encoding rate = 128 kbps
Convert the encoding rate to bytes per second:
128 kbps = 128,000 bits per second
1 byte = 8 bits
128,000 bits per second / 8 = 16,000 bytes per second
The time required to fill a packet is the packetization delay, which can be calculated is as follows:
Packetization delay = (P + 5) / (16,000 bytes per second) * 1000 milliseconds
b)
For L = 1,500 bytes (Ethernet packet):
Packetization delay = (1,500 + 5) / (16,000 bytes per second) * 1000 milliseconds
For L = 50 bytes (ATM packet):
Packetization delay = (50 + 5) / (16,000 bytes per second) * 1000 milliseconds
c)
The store-and-forward delay at a single switch can be calculated using the formula:
Store-and-forward delay = Packet length / Link rate
For L = 1,500 bytes and R = 622 Mbps:
Store-and-forward delay = 1,500 bytes / (622 Mbps * (1,000,000 bits per second / 8)) seconds
For L = 50 bytes and R = 622 Mbps: Store-and-forward delay = 50 bytes / (622 Mbps * (1,000,000 bits per second / 8)) seconds
d)
Advantages of using a small packet size:
- Reduced overhead: Small packets have fewer header bytes compared to larger packets, resulting in less overhead and more efficient utilization of the link bandwidth.
- Lower latency: Smaller packets have shorter transmission times, leading to lower end-to-end latency in real-time applications like Voice-over-IP (VoIP).
- Improved network responsiveness: Small packets can be processed and transmitted more quickly, allowing for faster delivery of data and better responsiveness in interactive applications.
- Better error recovery: If a small packet encounters an error during transmission, only a small portion of data needs to be retransmitted, reducing the impact of errors on overall throughput.
- Improved congestion control: Smaller packets can provide finer granularity for congestion control mechanisms, enabling more precise adjustment of transmission rates based on network conditions.
However, using small packet sizes can also have some disadvantages, such as increased overhead due to a larger number of packets and potentially higher processing overhead at network devices. It is important to strike a balance between packet size, overhead, and application requirements to optimize network performance.