How to calculate the theoretical bandwidth of a communication system

Backplane: backplane or switch fabric; line speed; 10G Ethernet; 1 gethernet

I. calculation formula description
The backboard bandwidth of A vswitch is the maximum amount of data that can be transferred between the vswitch interface processor or interface card and the data bus. The backboard bandwidth indicates the total data exchange capability of a vswitch. The unit is Gbps. It is also called the switching bandwidth. The bandwidth of A vswitch ranges from several Gbps to hundreds of Gbps. The higher the bandwidth of A vswitch, the stronger the ability to process data, but the higher the design cost.
Generally, the calculation method is as follows:
(1) wire speed backboard bandwidth
Measure the test taker's knowledge about the total bandwidth provided by all ports on the vswitch. The formula is as follows: number of ports x corresponding port rate x 2 (full duplex mode). If the total bandwidth is less than or equal to the nominal backboard bandwidth, the bandwidth is linear.
(2) Layer 2 packet forwarding speed
Second-layer packet forwarding rate = number of Gigabit ports x 1.488 mpps + number of 0.1488 MB ports x mpps + number of other types of ports x calculation method, if this rate can be less than or equal to the forwarding rate of the nominal second-layer packet, the switch can achieve the line speed when performing Layer 2 switching.
(3) Layer 3 packet forwarding speed
Layer 3 packet forwarding rate = number of Gigabit ports x 1. 488 mpps + 0.1488 MB port count × mpps + number of other types of ports × calculation method. If this rate can be less than or equal to the forwarding rate of a nominal layer-3 packet, the switch can achieve line speed when performing layer-3 switching.
200 M Ethernet, full duplex is 100/8 m, the unit is bit/s, 12.5 m is converted into byte is 12500000 = M byte/s, the conversion is bytes. Therefore, in an Ethernet packet, the minimum packet size is 64 bytes/s, plus 8 bytes of leading bytes and the gap between 12 bytes of frames, the total is 84 bytes. In this case, 12500000/84 = 148809 is used to obtain the maximum number of packet forwarding requests per second in a 148809 M throughput unidirectional environment, which is 148.8 k pps. Same as above, in two-way 297618 M throughput Ethernet, the number of forwards per second is 297.6. If it is converted to K, the packet forwarding rate is k pps.
This is the packet forwarding rate that can be achieved in layer-2 switching. However, if a router is on a layer-3 route or even nat is enabled, the packet forwarding rate will be greatly reduced, this value is really worth the attention of users, so we can see that many sellers have been emphasizing the number of 148810 packets forwarded by packets. In fact, this is the theoretical limit of layer-2 switching, it is not the value of the real router when the layer-3 is working.
[When we talk about packet forwarding to PPS, we usually only discuss the work order, for example, the 1ge 1.488 mpps on layer-2 .]
[When talking about packet forwarding rate PPS, the router product usually refers to the forwarding rate of layer3 .]
Therefore, if the above three conditions can be met, we will say that this switch is truly linear and non-blocking.
The utilization rate of the backboard Bandwidth Resources is closely related to the internal structure of the vswitch. Currently, the internal structure of a vswitch mainly includes the following types: first, the shared memory structure, which relies on the central switching engine to provide high-performance connections across all ports, the core engine checks each input packet to determine the route. This method requires a lot of memory bandwidth and high management costs. Especially with the increase of switch ports, the price of the central memory will be very high, so the switch kernel becomes a bottleneck for performance implementation; the second is the cross-bus structure. It can establish direct point-to-point connections between ports, which is good for single-point transmission performance, but not suitable for multi-point transmission. The third is the hybrid cross-bus structure, this is a hybrid cross-bus implementation method. Its design idea is to divide an integrated cross-Bus Matrix into small cross matrices and connect them through a high-performance bus in the middle. The advantage is that the number of Cross buses is reduced, the cost is reduced, and the bus contention is reduced. However, the bus connected to the cross matrix becomes a new performance bottleneck.

 

Ii. Port Rate Calculation
The minimum packet length for Ethernet transmission is 64 bytes, And the POs port is 40 bytes. The packet forwarding speed is measured by the number of 64 bytes of data packets (minimum packet) sent per unit time. For Gigabit Ethernet, the calculation method is as follows: 1,000,000,000 bps/8bit/(64 + 8 + 12) byte = 1,488,095 PPS Description: When the Ethernet frame is 64 byte, the fixed overhead of the frame gap between 8-byte frames and 12-byte frames must be considered. Therefore, the packet forwarding rate of A 1-gigabit Ethernet port when forwarding a 64-byte packet is 1.488 mpps. The line-rate port forwarding rate of Fast Ethernet is exactly 148.8 of that of Gigabit Ethernet, which is KPPS.
Assume that a vswitch has one A, B, and C interfaces, and their packet forwarding rate is X, Y, and Z respectively.
64 + 8 + 12: Based on 64-byte grouping test (minimum packet length of Ethernet transmission is 64 bytes); 8 Ethernet, each frame header must contain an 8-byte leading character. The minimum frame gap is 12 bytes. Then multiply by 8 is converted to bit.
Forwarding bandwidth = packet forwarding rate * 8*(64 + 8 + 12) = 1344 * packet forwarding Rate
Therefore:
Vswitch forwarding bandwidth = x * 8*(64 + 8 + 12) + y * 8*(64 + 8 + 12) + z * 8*(64 + 8 + 12)
= (X + y + Z) * 1344;
= Switch packet forwarding rate * 1344
10 Gigabit Ethernet 14.88 mpps 2 Gigabit Ethernet 1.488 mpps 0.1488 Gigabit Ethernet mpps
OC-3 POS 0.29 mpps OC-12 POS 1.17 mpps OC-48 POS 468 mpps

Iii. Total port Rate
In Ethernet, each frame header must contain an 8-byte leading character, which is used to tell the listener that data is coming. Then, there must be a frame gap between each frame in the Ethernet, that is, after each frame is sent, wait for a while before another frame is sent. In the Ethernet standard, the minimum value is 12 bytes, however, the frame gap may be larger than 12 bytes in practice. Here I use the minimum value. Each frame requires a fixed overhead of 20 bytes. Now let's calculate the actual throughput of a single port of the switch: 148,809 × (64 + 8 + 12) × 8 ≈ 100 Mbps, through this formula, we can see that the actual data exchange volume accounts for 64/84 = 76%. The "line rate" data throughput of the switch port link is actually only 76 Mbps, and the other part is used to handle additional overhead, both of them are standard Mbit/s or Gbit/s.
Vswitch board bandwidth calculation method
The backboard bandwidth is the maximum amount of data that can be transferred between the vswitch interface processor or interface card and the data bus. The higher the bandwidth of the backboard of A vswitch, the stronger the ability to process data, but the design cost will also rise.
However, how can we check whether the bandwidth of A vswitch is sufficient? Obviously, the estimation method is useless. I think we should consider the following two aspects:
1) The sum of the total port capacity X ports should be 2 times smaller than the backboard bandwidth, enabling full-duplex non-blocking switching, proving that the switch has the conditions to maximize the data exchange performance.
2) Full configuration throughput (mpps) = number of fully configured GE ports x 1. 488mpps the theoretical throughput of One gigabit port when the packet length is 64 bytes is 1.488 mpps. For example, A vswitch that can provide up to 64 Gigabit ports must have a full configuration throughput of 64 × 1.488 mpps = 95.2 mpps to ensure that all ports are working at the same speed, provides non-blocking packet switching. If a vswitch can provide a maximum of 176 Gigabit ports and the declared throughput is less than 261.8 mpps (176x1.488 mpps = 261.8 ), the user has reason to think that the switch adopts a blocking structure design.
Generally, the switches that both meet the requirements are qualified switches.
When the backplane is relatively large and the throughput is relatively small, what is the software effect besides retaining the ability to upgrade and expand? ? There is a problem with the design of the dedicated chip circuit; the backboard is relatively small. The overall performance of vswitches with relatively high throughput is relatively high. However, the Board bandwidth can be believed by the manufacturer, but the throughput cannot be believed by the manufacturer. because the latter is a design value, the test is very difficult and the significance is not great.
The switch's back-board speed is generally Mbps, which refers to the second layer,
Mpps is used for switching between three or more layers.