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 OTU4?
An OTU4 carries 1,209,600 E0s, per ITU-T G.709, ITU-T G.707 and ITU-T G.704. This path maps STM-64 into ODU2 as a client, then follows that client's tributaries.
1,209,600
E0s in an OTU4
Working
OTU4 → 1 × ODU4 → 10 × ODU2 → 1 × STM-64 → 64 × STM-1 → 63 × E1 → 30 × E0
- 1 × ODU4 in each OTU4 ITU-T G.709The OTU4 frame carries exactly one ODU4.
- 10 × ODU2 in each ODU4 ITU-T G.709Ten ODU2s use all 80 tributary slots, eight slots each.
- 1 × STM-64 in each ODU2 ITU-T G.709Client mappingCBR STM-64 mapped into OPU2.
- 64 × STM-1 in each STM-64 ITU-T G.70764 × 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 ODU4
- 1 × 10 = 10 ODU2s
- 1 × 10 × 1 = 10 STM-64s
- 1 × 10 × 1 × 64 = 640 STM-1s
- 1 × 10 × 1 × 64 × 63 = 40,320 E1s
- 1 × 10 × 1 × 64 × 63 × 30 = 1,209,600 E0s
Leftover capacity
34.395573568 Gbit/s of each OTU4 line rate (111.809973568 Gbit/s) sits outside 1,209,600 × 64 kbit/s of E0 line rate (77.414400 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 1,747,030.83, which floors to 1,747,030 E0s by bit rate only. The standard mapping is 1,209,600. Use the mapping.
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