One objective of a data communication network is to move the highest possible volume of accurate information through the network. The higher the volume, the greater the resulting network’s efficiency and the lower the cost. Network efficiency is affected by characteristics of the circuits such as error rates and maximum transmission speed, as well as by the speed of transmitting and receiving equipment, the error-detection and control methodology, and the protocol used by the data link layer.
Each protocol we discussed uses some bits or bytes to delineate the start and end of each message and to control error. These bits and bytes are necessary for the transmission to occur, but they are not part of the message. They add no value to the user, but they count against the total number of bits that can be transmitted.
Each communication protocol has both information bits and overhead bits. Information bits are those used to convey the user’s meaning. Overhead bits are used for purposes such as error checking and marking the start and end of characters and packets. A parity bit used for error checking is an overhead bit because it is not used to send the user’s data; if you did not care about errors, the overhead error checking bit could be omitted and the users could still understand the message.
Transmission efficiency is defined as the total number of information bits (i.e., bits in the message sent by the user) divided by the total bits in transmission (i.e., information bits plus overhead bits). For example, let’s calculate the transmission efficiency of asynchronous transmission. Assume we are using 7-bit ASCII. We have 1 bit for parity, plus 1 start bit and 1 stop bit. Therefore, there are 7 bits of information in each letter, but the total bits per letter is 10 (7 + 3). The efficiency of the asynchronous transmission system is 7 bits of information divided by 10 total bits, or 70 percent.
In other words, with asynchronous transmission, only 70 percent of the data rate is available for the user; 30 percent is used by the transmission protocol. If we have a communication circuit using a dial-up modem receiving 56 Kbps, the user sees an effective data rate (or throughput) of 39.2 Kbps. This is very inefficient.
We can improve efficiency by reducing the number of overhead bits in each message or by increasing the number of information bits. For example, if we remove the stop bits from asynchronous transmission, efficiency increases to 7/9, or 77.8 percent. The throughput of a dial-up modem at 56 Kbps would increase 43.6 Kbps, which is not great but is at least a little better.
The same basic formula can be used to calculate the efficiency of synchronous transmission. For example, suppose we are using SDLC. The number of information bits is calculated by determining how many information characters are in the message. If the message portion of the frame contains 100 information characters and we are using an 8-bit code, then there are 100 x 8 = 800 bits of information. The total number of bits is the 800 information bits plus the overhead bits that are inserted for delineation and error control. Figure 4.9 shows that SDLC has a beginning flag (8 bits), an address (8 bits), a control field (8 bits), a frame check sequence (assume we use a CRC-32 with 32 bits), and an ending flag (8 bits). This is a total of 64 overhead bits; thus, efficiency is 800/(800 + 64) = 92.6 percent. If the circuit provides a data rate of 56 Kbps, then the effective data rate available to the user is about 51.9 Kbps.
This example shows that synchronous networks usually are more efficient than asynchronous networks and some protocols are more efficient than others. The longer the message (1,000 characters as opposed to 100), the more efficient the protocol. For example, suppose the message in the SDLC example were 1,000 bytes. The efficiency here would be 99.2 percent, or 8,000/(8000 + 64), giving an effective data rate of about 55.6 Kbps.
The general rule is that the larger the message field, the more efficient the protocol. So why not have 10,000-byte or even 100,000-byte packets to really increase efficiency? The answer is that anytime a frame is received containing an error, the entire frame must be retransmitted. Thus, if an entire file is sent as one large packet (e.g., 100K) and 1 bit is received in error, all 100,000 bytes must be sent again. Clearly, this is a waste of capacity. Furthermore, the probability that a frame contains an error increases with the size of the frame; larger frames are more likely to contain errors than are smaller ones, simply due to the laws of probability.
Figure 4.12 Frame size effects on throughput
Thus, in designing a protocol, there is a trade-off between large and small frames. Small frames are less efficient but are less likely to contain errors and cost less (in terms of circuit capacity) to retransmit if there is an error (Figure 4.12).
Throughput is the total number of information bits received per second, after taking into account the overhead bits and the need to retransmit frames containing errors. Generally speaking, small frames provide better throughput for circuits with more errors, whereas larger frames provide better throughput in less-error-prone networks. Fortunately, in most real networks, the curve shown in Figure 4.12 is very flat on top, meaning that there is a range of frame sizes that provide almost optimum performance. Frame sizes vary greatly among different networks, but the ideal frame size tends to be between 2,000 and 10,000 bytes.
Sleuthing for the Right Frame Size
MANAGEMENT FOCUS
Optimizing performance in a network, particularly a client-server network, can be difficult because few network managers realize the importance of the frame size. Selecting the right — or the wrong —frame size can have greater effects on performance than anything you might do to the server.
Standard Commercial, a multinational tobacco and agricultural company, noticed a decrease in network performance when they upgraded to a new server. They tested the effects of using frame sizes between 500 bytes to 32,000 bytes. In their tests, a frame size of 512 bytes required a total of 455,000 bytes transmitted over their network to transfer the test messages. In contrast, the 32,000-byte frames were far more efficient, cutting the total data by 44 percent to 257,000 bytes.
However, the problem with 32,000-byte frames was a noticeable response time delay because messages were saved until the 32,000-byte frames were full before transmitting.
The ideal frame size depends on the specific application and the pattern of messages it generates. For Standard Commercial, the ideal frame size appeared to be between 4,000 and 8,000. Unfortunately, not all network software packages enable network managers to fine-tune frame sizes in this way.
So why are the standard sizes of Ethernet frames about 1,500 bytes? Because Ethernet was standardized many years ago when errors were more common. Jumbo and super jumbo frame sizes emerged from higher speed, highly error free fiber-optic networks.
Calculating the actual throughput of a data communications network is complex because it depends not only on the efficiency of the data link protocol but also on the error rate and number of retransmissions that occur. Transmission rate of information bits (TRIB) is a measure of the effective number of information bits that is transmitted over a communication circuit per unit of time. The basic TRIB equation from ANSI is shown in Figure 4.13, along with an example.
figure 4.13 Calculating TRIB (transmission rate of information bits)
