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Written by fallleaf
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Friday, 03 March 2006 |
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从资料上找了一个SDH系统端到端计算时延的例子，是全英文的，试着翻译了几句。 End-to-end delay has three components. Delay can be introduced in the network as serialization delay, propagation delay, or NE processing delay. Serialization delay has the most significant impact on end-to-end delay.
端到端的时延包含三各部分，它们是串行时延，传播时延，网元处理时延。串行时延对端到端时延的影响最大。
Serialization delay— The frame sizes or maximum transmission unit (MTU) used by customers plays an important part in the serialization delay. A Layer 2 frame cannot be processed by a receiver until all bits have been received by the receiver and a cyclic redundancy check (CRC) has been performed. Let us examine the serialization delay introduced by SONET/SDH NEs.
串行时延－用户数据的帧长度或者MTU（最大传输单位）在串行时延中起着重要作用。二层的帧必须等到所有比特全部接收通过CRC（循环冗余校验）效验后才能被处理。现在来计算一下SDH系统的串行时延。
For SONET STS-1:
对于SONET STS-1:
(810 bytes/frame * 8 bits/byte)/(51.84 * 10^6 bits/sec) = 125 microseconds
= 0.125ms
For SONET STS-3/SDH STM-1:
对于SONET STS-3/SDH STM-1
(810 bytes/frame * 8 bits/byte)/(155.52 * 10^6 bits/sec) = 41.66 microseconds
For SONET/SDH STS-12/STM-4:
(810 bytes/frame * 8 bits/byte)/(622.08 * 10^6 bits/sec) = 10.41 microseconds
For SONET/SDH STS-48/STM-16:
(810 bytes/frame * 8 bits/byte)/(2488.32 * 10^6 bits/sec) = 2.60 microseconds
For SONET/SDH STS-192/STM-64:
(810 bytes/frame * 8 bits/byte)/(9953.28 * 10^6 bits/sec) = 0.65 microseconds
The preceding calculations reveal that the largest delay of 125µs is introduced at the STS-1 rate. However, this is negligible compared to the delay introduced by the CPE device at a lower-bandwidth service interface. If a customer decides to use a 9000-byte frame size over a DS1 interface, for example, the serialization delay can be calculated as follows:
(9000 bytes/frame * 8 bits/byte)/1,536,000 bps = 0.0468 seconds or 46.875 ms
Using such a large MTU has just rendered the SLA requirement of 25 ms impossible to achieve. It is precisely for this reason that SLA agreements must guarantee delay time(s) at certain fixed MTU size(s).
For MPLS-compatible Ethernet frame sizes of 1548 bytes MTU operating at 10-Mbps link speed, the serialization delay can be calculated as follows:
(1548 bytes/frame * 8 bits/byte)/10,000,000 bps = 0.001238 seconds or 1.238 ms
For 1500-byte MTU serial High-level Data Link Control (HDLC) or PPP encapsulation operating over DS1 links, the serialization delay can be calculated as follows:
(1500 bytes/frame * 8 bits/byte)/1,536,000 bps = 0.007812 seconds or 7.8 ms
For 1500-byte MTU serial HDLC or PPP encapsulation operating over E1 links, the serialization delay can be calculated as follows:
(1500 bytes/frame * 8 bits/byte)/2,048,000 bps = 0.00585 seconds or 5.8 ms
Propagation delay— Propagation delay is introduced due to the finite speed of light and the laws of physics. Propagation delay introduces asymmetrical delay in large UPSR/SNCP rings and must be taken into consideration during optical network design. The nominal velocity of propagation (NVP) is defined as the ratio of the speed of light in a vacuum (c) to the refractive index of the material. The value of c is a constant fixed at 300,000 km/second. The refractive index for SMF is 1.5.
NVP = (c)/(Refractive index of fiber)
The NVP for SMF can be calculated as follows:
NVP = (300,000 km/second)/(1.5) = 200,000 km/sec
The inverse of this equation results in the propagation delay of SMF at 0.005 ms/km. This means the propagation delay for SMF in milliseconds can be calculated as follows:
Propagation delay = Fiber length (km) * 0.005 ms/km
NE processing delay— This is the delay measured in subunits of seconds that results from a signal being processed by OEO NEs, such as add/drop multiplexers (ADMs) and 3R regenerators, on the way to its destination. According to ITU-T specifications, the signal must ingress and egress an optical NE within 450 microseconds (0.45 ms). This means the end-to-end NE processing delay in milliseconds can be calculated as follows:
NE processing delay = Total number of NEs on the path * 0.45 ms
The calculation for total delay is as follows:
Total delay = Serialization delay + Propagation delay + NE processing delay
It is our intent to limit the number of end-to-end rings to three. This means that traffic sourced from a collector ring node could traverse over a core ring and then terminate at another collector node. We can also consider a design maximum of 16 NEs (nodes) per ring. This means a signal would traverse over 8 * 3 (24) nodes at worst case. We also have the design premise that the geographic circumference (perimeter) of Washington, D.C. does not exceed 30 km. This means that the signal would not traverse over 30 km/2 (15 km) over a single ring. The maximum distance a signal would traverse is (source collector circumference/2) + (core ring circumference/2) + (destination collector circumference/2) or 15 km + 15 km + 15 km (45 km). The slowest-speed CPE interface the service provider intends to provide customers is the DS1 interface with an MTU of 1500 bytes.
From a design perspective, the equation that needs to be solved is shown here:
Total delay < Serialization delay + Propagation delay + NE processing delay
Total delay < [(MTU * 8 bits/byte)/(Bandwidth of lowest-speed service interface)] + [Fiber length (km) * 0.005 km/s] + [Total number of NEs on the path * 0.45 ms]
Substituting values into the design equation, we get the following:
Total delay < [(1500 * 8)/(1,536,000)] + [45 * 0.005] + [24 * 0.45]
Total delay < [7.8 ms] + [0.225 ms] + [10.8 ms]
Total delay < 18.825 ms
This satisfies our SLA requirement of 25 ms as the end-to-end delay that will not be exceeded.
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