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 ODU4?

An ODU4 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 ODU4

Working

ODU4 → 10 × ODU2 → 1 × STM-64 → 64 × STM-1 → 63 × E1 → 30 × E0

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.

10 ODU2s
10 × 1 = 10 STM-64s
10 × 1 × 64 = 640 STM-1s
10 × 1 × 64 × 63 = 40,320 E1s
10 × 1 × 64 × 63 × 30 = 1,209,600 E0s

Leftover capacity

27.380045815 Gbit/s of each ODU4 line rate (104.794445815 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 1,637,413.21, which floors to 1,637,413 E0s by bit rate only. The standard mapping is 1,209,600. Use the mapping.

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