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 E2s are in an OTU4?
An OTU4 carries 7,680 E2s, per ITU-T G.709, ITU-T G.707 and ITU-T G.702. This path maps STM-64 into ODU2 as a client, then follows that client's tributaries.
7,680
E2s in an OTU4
Working
OTU4 → 1 × ODU4 → 10 × ODU2 → 1 × STM-64 → 64 × STM-1 → 3 × E3 → 4 × E2
- 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.
- 3 × E3 in each STM-1 ITU-T G.707Three VC-3s, each carrying an E3.
- 4 × E2 in each E3 ITU-T G.702Four E2s.
- 1 ODU4
- 1 × 10 = 10 ODU2s
- 1 × 10 × 1 = 10 STM-64s
- 1 × 10 × 1 × 64 = 640 STM-1s
- 1 × 10 × 1 × 64 × 3 = 1,920 E3s
- 1 × 10 × 1 × 64 × 3 × 4 = 7,680 E2s
Leftover capacity
46.929333568 Gbit/s of each OTU4 line rate (111.809973568 Gbit/s) sits outside 7,680 × 8.448 Mbit/s of E2 line rate (64.880640 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 13,235.08, which floors to 13,235 E2s by bit rate only. The standard mapping is 7,680. Use the mapping.
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