Quantity is how many containers you have. The result is how many tributaries fit, the path, and the leftover line rate. The choices stay in the URL, so you can share them.

Result: standard mapping

How many E0s are in an OTU1?

An OTU1 carries 30,240 E0s, per ITU-T G.709, ITU-T G.707 and ITU-T G.704. This path maps STM-16 into ODU1 as a client, then follows that client's tributaries.

30,240

E0s in an OTU1

Working

OTU1 → 1 × ODU1 → 1 × STM-16 → 16 × STM-1 → 63 × E1 → 30 × E0

1 × ODU1 in each OTU1 ITU-T G.709The OTU1 frame carries exactly one ODU1. The line-rate gap is OTU overhead and FEC.
1 × STM-16 in each ODU1 ITU-T G.709Client mappingCBR STM-16 mapped into OPU1.
16 × STM-1 in each STM-16 ITU-T G.70716 × STM-1.
63 × E1 in each STM-1 ITU-T G.70763 VC-12 tributaries: 3 TUG-3 × 7 TUG-2 × 3 TU-12.
30 × E0 in each E1 ITU-T G.70430 voice channels in 32 timeslots. TS0 is framing and TS16 is channel-associated signalling.

1 ODU1
1 × 1 = 1 STM-16
1 × 1 × 16 = 16 STM-1s
1 × 1 × 16 × 63 = 1,008 E1s
1 × 1 × 16 × 63 × 30 = 30,240 E0s

Leftover capacity

730.697143 Mbit/s of each OTU1 line rate (2.666057143 Gbit/s) sits outside 30,240 × 64 kbit/s of E0 line rate (1.935360 Gbit/s). That remainder is overhead, fixed stuff or justification. It is not another whole tributary.

By bit rate only

Dividing the line rates gives 41,657.14, which floors to 41,657 E0s by bit rate only. The standard mapping is 30,240. Use the mapping.

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