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Understanding multipliers and divisors value in calculating OTN frame rates (255,239,238,237 etc) for OPUk,ODUk and OTUk

June 19, 2016

**Multiplicative factor is just a simple math :eg. for ODU1/OPU1=3824/3808={(239*16)/(238*16)}

Here value of multiplication factor will give the number of times for rise in the frame size after adding header/overhead.

Example:let consider y=(x+delta[x])/x; In terms of OTN frame here delta[x] is increment of Overhead.

As we are using Reed Soloman(255,239) i.e we are dividing 4080bytes in sixteen frames (The forward error correction for the OTU-k uses 16-byte interleaved codecs using a Reed- Solomon S(255,239) code. The RS(255,239) code operates on byte symbols.).Hence 4080/16=255.

Try to understand using OTN frames now. I have tried to make it legible.

As we know that OPU1 payload rate= 2.488 Gbps (OC48/STM16) and is  frame size is 4*3808 as below.

*After adding OPU1 and ODU1 16 bytes overhead: Frames could be fragmented into following number of chunks.

 3808/16 = 238, (3808+16)/16 = 239

So, ODU1 rate: 2.488 x 239/238** ~ 2.499Gbps

*Now after adding  FEC bytes

  OTU1 rate: ODU1 x 255/239 = 2.488 x 239/238 x 255/239

  =2.488 x 255/238 ~2.667Gbps 

 

Now let’s have a small discussion over different multiplier and divisor scenarios that will make it clearer to understand.

We know that an OTU frame 4 * 4080 bytes (= 255 * 16 * 4)

 OPU representing the Payload (3824-16) * 4 * 4 = 3808 bytes (= 238 * 16 * 4) .

OPU1 is exactly the rate of STM-16.

Now,

ODU1 = (3824/3808) * OPU1 = ((16 * 239) / (238 16 *)) * OPU1 = (239/238) * STM-16 

OTU1 = (4080/3808) * OPU1 = ((255 * 16) / (238 * 16)) * OPU1 = (255/238) * STM-16 

 

OPU2 contains 16 * 4 = 64 bytes of fixed stuff (FS) added to the 1905 to 1920 .

OPU2 * ((238 * 16 * 4-16 * 4) / (238 * 16 * 4)) = STM-64 rate

OPU2 = 238 / (238-1) * STM-64 = 238/237* STM-64 rate 

ODU2 = (239/237) * STM-64 rate ,

similarly

 OTU2 = ( 255/237) * STM-64 rate 

 OPU3 Including 2 * 16 * 4 = 128 fixed stuff (FS) bytes added to the 1265 ~ 1280 and 2545 ~ 2560

 OPU3 * ((238 * 16 * 4-2 * 16 * 4) / (238 * 16 * 4)) = rate of STM-256

OPU3 = 238 / (238-2) * STM-256 = 238/236 * STM-256

ODU3 = (239 / 236) * STM-256

OTU3 = (255/236) * STM-256

The OTU4 was required to transport ten ODU2e signals, which have a non-SDH based clock frequency as basis. The OTU4 clock should be based on the same SDH clock as the OTU1, OTU2 and OTU3 and not on the 10GBASE-R clock, which determines the ODU2e frequency. An exercise was performed to determine the necessary divider in the factor 255/divider, and the value 227 was found to meet the requirements (factor 255/227). Note that this first analysis has indicated that a future 400 Gbit/s OTU5 could be created using a factor 255/226 and a 1 Tbit/s OTU6 using a factor 255/225.

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Can you share the reference where this really good explanation was taken or based on.

It was driven me crazy. And now I can sleep.

Thanks a ton for such a detail explanation..!!
though i have one doubt what is the need of Fix stuffing byte in OPU2 and OPU3 frame.

Hi Akhilesh,

Hope this will answer your query.

we are adding FS bytes to equalize the payload size with the standard available size..example

we know that OPU2 is 9.953Gbps which is the rate of standard STM64 signal........(1)

now lets consider mapping 4xODU1 into OPU2
=4xODU1=4X2.498=9.9951Gbps................(2)

Now if you compare both (1) and (2),you will find a difference in the rate and it will be around

9.953-9.995=41.82Mbps or 4182ppm.........(3) but we know that through available mapping procedures we cannot compensate difference of 4182ppm.

We can compensate for +/-20ppm as per standard.So we definitely need to add some stuffing byte so that justification could be achieved with fixed size JC bytes as we do stuffing in SDH frames.

so we add as much number of standard bytes so that we can atleast reach justification of +/-20ppm so if you add 16x4 FS then the
rate for OPU2 will be 9.9952Gbps...............(4)

now after adding FS bytes ;rated difference will be 9.9952-9.9951 ==177Kbps=18ppm.....................(5)

value of (5) i.e 18ppm is under limit and ofcourse we can compensate such difference with standarized +/-20ppm margin.

normally placing of FS bytes is in multiple of 16byte due to OTN frame reed soloman FEC capability.

Nice One



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