Understanding the network.pdf - Back to Home

10Mbps bt = 100 nsec
16Mbps bt = 63.3 nsec
100Mbps bt = 10 nsec
1Gbps bt = 1 nsec
To calculate the frame transmission time, use the following
calculation:
FTt = TFSb × bt = 672 bits × 100 ns = 67.200 nsec = 67.2 usec
Step 3. Now that we know how much time it takes to transmit a single
frame (FTt), we can calculate how many frames can be transmitted in
one second (FPS) by using this formula:
FPS = 1 sec ÷ FTt = 1 sec ÷ 67.2 usec = 14,880 FPS
The goal of any network transmission device is to be capable of
transmitting at wire rate. Wire rate is the transmission of the smallest
legal-sized data frame at the fast constant rate of transmission
supported by the medium and transmission protocol, without adding
any additional processing latency. Of course, the legal frame sizes
vary depending on the Layer 2 protocol being used. For Ethernet, the
minimum frame size is 72 bytes (64 bytes + 8-byte preamble), with a
maximum of 1,526 (1,518 + 8-byte preamble) or 1,530 with 802.3ac
due to the constraints of CSMA/CD. With Token Ring (which does not
require collision detection), no minimum or maximum frame size is
defined. Instead, the minimum frame size is calculated using the
frame format data and LLC information, which is 43 bytes. The
maximum frame size is dictated by the maximum amount of time a
station can hold the token. So, on a 4Mbps ring, the maximum packet
size is 4,472 bytes; on a 16Mbps ring, it is 17,800 bytes. FDDI, like
Ethernet, supports a minimum and maximum frame size: 32-byte
minimum (28 frame + 4 byte LLC-PDU, no data) and 4,500-byte
maximum.
The reason the smallest legal frame size (64 bytes, in the case of
Ethernet) expresses the FPS rating for port forwarding is because
smaller frames are harder to process than larger ones. When looking
at throughput utilization, the FPS rating is inversely proportional to
the frame length. Mathematically speaking, you can move more small
frames than large ones in a finite period.