Question

In this problem, we explore the use of small packets for Voice-over-IP applications. 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 \(128 \mathrm{kbps}\). 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 \(R=622 \mathrm{Mbps}\) for \(L=1,500\) bytes, and for \(L=50\) bytes. d. Comment on the advantages of using a small packet size.

Short answer

The packetization delay is 93.44 ms for L=1,500 and 2.81 ms for L=50; store-and-forward delays are 0.019 ms and 0.00064 ms, respectively. Small packets reduce delays and echo, aiding prompt network transmission.

Step-by-step solution

Define Packetization Delay Formula

The packetization delay is the time it takes for one packet to fill up before being sent. The rate at which data is encoded is given as 128 kbps. The formula for packetization delay, given in seconds, is \( \text{Packetization Delay} = \frac{P}{ ext{Transmission Rate}} \). Since 128 kbps is equivalent to 128,000 bits per second and \(P\) is the payload size in bits, first we convert \(P\) from bytes to bits by multiplying by 8.

Compute Packetization Delay in General Terms

Given the packet size consists of \(P\) bytes payload and 5 bytes header, the total packet size in bits is \((P + 5) \times 8\). However, for packetization delay we're concerned only about the payload, \(P \times 8\) bits. Thus, the packetization delay \(T_p\) in milliseconds is calculated as: \[ T_p (\text{ms}) = \frac{P \times 8}{128,000} \times 1000 \] which simplifies to \[ T_p = \frac{P \times 8}{128} \].

Calculate Packetization Delay for L=1,500 bytes

Substitute \(L=1,500\) bytes into the formula. Payload \(P = L - 5 = 1,495\) bytes. Thus: \[ T_p = \frac{1,495 \times 8}{128} = 93.4375 \text{ ms} \].

Calculate Packetization Delay for L=50 bytes

For \(L=50\) bytes, payload \(P = 50 - 5 = 45\) bytes. Substitute into formula: \[ T_p = \frac{45 \times 8}{128} = 2.8125 \text{ ms} \].

Define Store-and-Forward Delay Formula

Store-and-forward delay is the time taken to transmit a packet across a network link. Given the link rate \(R = 622\) Mbps, the total packet size in bits is \((P+5) \times 8\). The store-and-forward delay \(T_s\) is \[ T_s = \frac{(P + 5) \times 8}{R} \].

Calculate Store-and-Forward Delay for L=1,500 bytes

Substitute \(P = 1,495\) bytes into the formula, \((P+5) = 1,500\) bytes: \[ T_s = \frac{1,500 \times 8}{622,000,000} \times 1,000 = 0.019 \text{ ms} \].

Calculate Store-and-Forward Delay for L=50 bytes

Substitute \(P = 45\) bytes into the formula, \((P+5) = 50\) bytes: \[ T_s = \frac{50 \times 8}{622,000,000} \times 1,000 = 0.00064 \text{ ms} \].

Comment on Small Packet Size Advantages

Using smaller packets, such as with ATM (L=50), results in lower packetization delay and less noticeable echo. Additionally, they facilitate faster transmission through the network since smaller packets require shorter store-and-forward delays.

Key concepts

Packetization Delay

In Voice-over-IP (VoIP) applications, packetization delay is crucial because it dictates how quickly packets of data can be sent over the network. This delay is the time taken to accumulate one full packet of voice data before it can be transmitted. It depends on the rate at which data is encoded and the packet's payload size. For example, a voice source encoded at 128 kbps implies that data can be encoded at 128,000 bits per second. If the packet size is defined in bytes, we first need to convert these bytes to bits by multiplying by 8. The packetization delay can be expressed as:\[T_p = \frac{P \times 8}{128,000} \times 1000\]where \(P\) is the number of bytes in the payload. When you have a large packet size like 1,500 bytes (the usual size for Ethernet packets), the delay can be significant. For example, here the delay is approximately 93.44 ms, which might cause perceptible echoes during a call. On the other hand, smaller packet sizes, like 50 bytes for ATM packets, have a smaller delay of around 2.81 ms. This results in less echo, providing a smoother VoIP experience.

Store-and-Forward Delay

Store-and-forward delay comes into play when a network switch or router receives a packet and decides to which link it should forward this packet next. This delay includes the time taken for the packet to be entirely received, stored, and then forwarded. This is crucial because in a busy network, efficient forwarding can make a huge difference in overall performance. Given a link rate \(R\), the formula for calculating store-and-forward delay is:\[T_s = \frac{(P + 5) \times 8}{R} \times 1000\]where \(P\) is the payload size in bytes, the 5 bytes represent the header, and \(R\) is the link rate in bits per second.For a link rate of 622 Mbps, the store-and-forward delay for a 1,500-byte packet is about 0.019 ms, whereas for a 50-byte packet, it's incredibly brief at about 0.00064 ms. This small delay illustrates how different packet sizes influence the speed at which data travels through the network.

Link Bandwidth Overhead

Link bandwidth overhead becomes apparent when dealing with small packet sizes in Voice-over-IP (VoIP) networks. In the context of packet data, every packet requires a header, which doesn't carry payload data but is necessary for directing the packet to its destination. Even though headers are usually small (5 bytes in this exercise), they represent a significant portion of the data when packets themselves are tiny. For instance, with a 50-byte packet, the 5-byte header forms a substantial part of the packet, affecting the percentage of actual usable data being transmitted. This means more bandwidth is consumed by overhead, staying proportional to the packet's size. This reveals a trade-off: while smaller packets minimize delays and reduce echo, they inadvertently increase the overhead per byte delivered, leading to a potential decrease in efficiency. Understanding this balance is key in designing networks that rely heavily on small data packets, as often seen in VoIP scenarios.