Measure actual throughput on a FE or Gigabit LAN connection
I have been looking into an issue with a customers LAN investigating slow throughput.
A couple of things I have learned and have confused me on how to determine the actual speed of a connection between a PC and its switch/router...
Why am i looking at this?
I have always been curious about the impact poorly deployed cabling has on actual LAN performance. If for example a cable ran past a device that induced some sort of interference that slowed traffic across the cable, then Windows may be oblivious to it, and all the testing at switch and device level may not highlight the fact that the cable run from the outlet to the patch panel is the problem.
Your collective thoughts/knowledge and understanding apprecaited
A couple of things I have learned and have confused me on how to determine the actual speed of a connection between a PC and its switch/router...
1.
When I throttle the port that a PC is connected, to 512Kb/s using QOS, the PC still says the link speed is 1Gb/s2.
When I set the the port speed to say 10Mb/s Windows says it is connected to a 10Mb/s link.3.
The difference suggests Windows or the NIC, asks the Switch for the link speed and doesn't actually measure the line speed.Why am i looking at this?
I have always been curious about the impact poorly deployed cabling has on actual LAN performance. If for example a cable ran past a device that induced some sort of interference that slowed traffic across the cable, then Windows may be oblivious to it, and all the testing at switch and device level may not highlight the fact that the cable run from the outlet to the patch panel is the problem.
Your collective thoughts/knowledge and understanding apprecaited
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Dropped packets usually occur due to network congestion and queues that are full. If there is not sufficient space to queue new packets, since the queue is FIFO, the newest packets are discarded. Once the queue has room, it will accept packets again.
Induced current can bring the signal voltage out of tolerance. On a twisted pair, interference should be near identical on both wires, since the close twist will equally expose each wite to the same source if interference.
For example, an NRZI-encoded signal expects a swing from +1VDC to -1V to indicate a binary unit of '1'. The next timing cycle, the voltage stays at -1V. This indicates binary '0'. (NRZI means non-return to zero, inverted. The signal should never be at zero.)
+1 = 1
+1 = 0 (no change)
-1 = 1 (change)
In MLT-3 encoding, there are theree states: +1, 0, and -1.
If the induced current adds 50mV, the twisted pair gives a differential that cancels out the added voltage:
EXAMPLE 1 - no induced voltage
D1 = +1.0V
D2 = -1.0V
D1 - D2 = +2.0V
EXAMPLE 2 - induced voltage = +100mV
D1 = +1 +0.10V = +1.1V
D2 = -1V +0.10V = -0.9 V
D1 - D2 = +2.0V
EXAMPLE 3 - induced voltage = +700mV
D1 = +1 +0.70V = +1.7V
D2 = -1V +0.70V = -0.3 V
D1 - D2 = +2.0V
Looks all the same, right? But, in Example 3, the D2 arrives at -0.3V. That's pretty close to zero. How does the receiver know that that is actually a signal, especially if the tolerence might be +/- 0.4V ?
These are made-up numbers, but you get the idea. Induced current is additive on the length of exposed wire. Crossed wires at 90-degrees are less voltage change than the same cables run in parallel for 5-10 feet.
Induced current can bring the signal voltage out of tolerance. On a twisted pair, interference should be near identical on both wires, since the close twist will equally expose each wite to the same source if interference.
For example, an NRZI-encoded signal expects a swing from +1VDC to -1V to indicate a binary unit of '1'. The next timing cycle, the voltage stays at -1V. This indicates binary '0'. (NRZI means non-return to zero, inverted. The signal should never be at zero.)
+1 = 1
+1 = 0 (no change)
-1 = 1 (change)
In MLT-3 encoding, there are theree states: +1, 0, and -1.
If the induced current adds 50mV, the twisted pair gives a differential that cancels out the added voltage:
EXAMPLE 1 - no induced voltage
D1 = +1.0V
D2 = -1.0V
D1 - D2 = +2.0V
EXAMPLE 2 - induced voltage = +100mV
D1 = +1 +0.10V = +1.1V
D2 = -1V +0.10V = -0.9 V
D1 - D2 = +2.0V
EXAMPLE 3 - induced voltage = +700mV
D1 = +1 +0.70V = +1.7V
D2 = -1V +0.70V = -0.3 V
D1 - D2 = +2.0V
Looks all the same, right? But, in Example 3, the D2 arrives at -0.3V. That's pretty close to zero. How does the receiver know that that is actually a signal, especially if the tolerence might be +/- 0.4V ?
These are made-up numbers, but you get the idea. Induced current is additive on the length of exposed wire. Crossed wires at 90-degrees are less voltage change than the same cables run in parallel for 5-10 feet.
ASKER
In my early IT career i worked in a 2000MW Power Station. At that era my first project was to project manage the deployment of Cat3 Shielded, fiber and phone lines around the site. During testing by the contractors who deployed the various mediums, I got to see the effect some interference had on the un-shielded Cat 3 that was being replaced (the test tool looked like an oscilloscope).
Does this constitute what is termed dropped packets?