11 Performance Calculations

Performance Calculations

Like other network technologies, 802.11 has both a "headline" rate and the actual number of user bits that can be moved through the network. In many cases, the payload throughput is less than the headline rate. In the 802.11 world, the payload throughput is much lower than the headline rate due to the protocol overhead. Payload throughput of half the headline rate is a good rule of thumb.

It is often helpful to calculate a theoretical maximum throughput to compare real-world results to. 802.11 breaks up data transmission into a series of atomic operations. Each atomic operation has several components which must be accounted for; a basic atomic operation is shown in Figure 25-1. The exchange shown is the simplest, consisting of an 802.11 Data frame and its corresponding acknowledgment. Naturally, exchanges can be much more complicated if they involve fragmentation, or are required to use the 802.11g protection mechanism discussed in Chapter 14.

Figure 25-1. Atomic operation

Each atomic operation is composed of multiple frames separated by interframe spacing. To gain access to the medium, the frame sequence generally starts with a Distributed Inter-frame Space (DIFS), and the frames in the sequence are separated by Short Inter-frame Spaces (SIFS). Values for the DIFS and SIFS are dependent on the particular PHY in use, and can be found in the relevant chapters.

Each frame in the air must be transmitted by the physical layer, and therefore has a physical layer header consisting of a preamble plus additional fields to help decode the frame. Generally speaking, the preamble is the dominant component of the physical header transmission time. The physical layer frame's body carries the MAC frame, which is adds 28 bytes. Within the MAC frame, an 8-byte SNAP header brings the total encapsulation overhead of a data frame to 36 bytes. Encryption headers may add additional length to the MAC header.

The higher-layer protocol packet itself is transmitted according to the rules of the particular PHY in use. Each PHY has a minimum block, or symbol, size, in bits. 802.11b networks running at 11 Mbps use 8-bit symbols. 802.11a and 802.11g use symbols that carry much more data, and the symbol size depends on the data rate. At 54 Mbps, the symbol size is 216 bits. (The eventual 802.11n will have even larger block sizes.) Each block of data requires a fixed amount of time to transmit. Different data rates work by packing more or fewer data bits into a block, not by transmitting blocks faster or slower. For example, a 1,536 byte payload for the physical frame consists of 12,288 bits. In the 802.11a PHY, the frame requires the transmission of 57 symbols.

To find the time required for an atomic exchange, break it apart into its constituent pieces, and add up the time required for each. Start by adding up any inter-frame spacing. For many atomic exchanges, there will only be one DIFS and one SIFS in between frames, though the trend is towards more complex frame structures. With the inter-frame spacing accounted for, divide each frame up into its basic components, based on the physical layer that will be used. Frequently, each frame will be divided in the same way, although protection mechanisms are a notable exception because an older transmission type must be used. Each frame has a physical header, followed by the physical payload. The physical payload is the MAC frame, which consists of the MAC header, a SNAP header on any data frame, followed by the higher-layer protocol packet, and concluding with a trailer.

Example Calculation

As an example, consider the transmission of a single 1,500 byte data frame on an 802.11a network. This analysis is intended to be an example of how to dissect the components of the frame, rather than a detailed model, so the discussion neglects any higher-layer protocol effects.

In 802.11a, the SIFS is 16 microseconds, and the DIFS is 34 microseconds. Exchanging a single data frame requires that the station wait for one DIFS, transmit its data frame, and wait for the acknowledgment. The 802.11 acknowledgment is transmitted after one SIFS. There is a total of 50 microseconds of interframe spacing.

Next, break apart the data frame. The preamble and signal field in 802.11a requires 20 microseconds. The 1,500 byte data packet is encapsulated in a MAC frame with SNAP header, pushing the size up to 1,536 bytes. If the frame is protected by CCMP, an additional 16 bytes of encapsulation is added for the header and integrity check, bringing the total payload in the physical frame to 1,552 bytes, or 12,416 bits. At 54 Mbps, the physical frame payload is broken up into 216-bit symbols, so 58 will be required to transmit the data. Each symbol requires 4 microseconds, for a total of 232 microseconds for the data frame.

The 802.11 acknowledgment is only 14 bytes (112 bits) of payload, and can easily fit into a single 54 Mbps symbol. Some chipsets may use lower data rates. At data rates of 36 Mbps and up, only one symbol is required. Extra symbols are required at lower data rates: two for the 18 and 24 Mbps rates, three at 12 Mbps, four at 9 Mbps, and five at 6 Mbps. Say that the acknowledgment is transmitted at 24 Mbps, which is a reasonable balance between minimizing transmission time and retaining reliable transmission. Three symbols require 12 microseconds, for a total of 32 microseconds including the header.

Adding it all together, the 1,500 byte payload requires 50 microseconds of spacing, 232 microseconds to transmit, and 32 microseconds to acknowledge, for a total of 314 microseconds. If no contention ever occurs, 3,184 of these atomic exchanges can occur per second. Each exchange moves 1,500 bytes of payload, for a total payload rate of 38 Mbps. Under nearly ideal conditions, protocol overhead takes 30% of the headline bit rate.

Other components to a performance model

The previous discussion made several assumptions that are exceedingly generous. Most real-world 802.11 networks are lucky to get user payloads of 55-60% of the headline rate. Any serious network would run a higher-layer protocol, adding further encapsulation overhead. TCP/IP, for example, steals 40 payload bytes for its own headers, and requires protocol acknowledgments in the reverse direction. Depending on exactly how TCP/IP is incorporated into the model, it will further diminish throughput.

The previous example also made the implicit assumption that there was never contention for the medium, which is unrealistic. Even though the 802.11 MAC attempts to avoid collisions, they are a fact of life. Some engineers believe that TCP/IP may even exacerbate collisions because it tries to send session-layer acknowledgments back to the sender while the sender may continue to stream data.

Real-world networks also have many other sources of protocol overhead. The previous calculation did not attempt to estimate the overhead of sending Beacon frames, or how typical frame loss rates would diminish throughput.

Block acknowledgments

It is relatively straightforward to build on the previous example to show why block acknowledgments are such a powerful concept. Over 10% of the transmission time for the atomic exchange is required for inter-frame spacing and 802.11 acknowledgment. By changing the 1:1 ratio of data frames to acknowledgments, the proportion of airtime used for overhead operations can be shrunk much farther.