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 ODU0?
An ODU0 carries 48 E2s, per ITU-T G.709, ITU-T G.707 and ITU-T G.702. This path maps STM-4 into ODU0 as a client, then follows that client's tributaries.
48
E2s in an ODU0
Working
ODU0 → 1 × STM-4 → 4 × STM-1 → 3 × E3 → 4 × E2
- 1 × STM-4 in each ODU0 ITU-T G.709Client mappingCBR STM-4 mapped into OPU0.
- 4 × STM-1 in each STM-4 ITU-T G.7074 × 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 STM-4
- 1 × 4 = 4 STM-1s
- 1 × 4 × 3 = 12 E3s
- 1 × 4 × 3 × 4 = 48 E2s
Leftover capacity
838.656 Mbit/s of each ODU0 line rate (1.244160 Gbit/s) sits outside 48 × 8.448 Mbit/s of E2 line rate (405.504 Mbit/s). That remainder is overhead, fixed stuff or justification. It is not another whole tributary.
By bit rate only
Dividing the line rates gives 147.27, which floors to 147 E2s by bit rate only. The standard mapping is 48. Use the mapping.
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