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 DS0s are in an ODU4?

An ODU4 carries 1,290,240 DS0s, per ITU-T G.709, ITU-T G.707 and ANSI T1.107. This path maps STM-64 into ODU2 as a client, then follows that client's tributaries.

1,290,240

DS0s in an ODU4

Working

ODU4 → 10 × ODU2 → 1 × STM-64 → 64 × STM-1 → 84 × DS1 → 24 × DS0

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.
84 × DS1 in each STM-1 ITU-T G.70784 VC-11 tributaries: 3 TUG-3 × 7 TUG-2 × 4 TU-11.
24 × DS0 in each DS1 ANSI T1.10724 payload timeslots. The framing bit adds 8 kbit/s and brings the line rate to 1.544 Mbit/s.

10 ODU2s
10 × 1 = 10 STM-64s
10 × 1 × 64 = 640 STM-1s
10 × 1 × 64 × 84 = 53,760 DS1s
10 × 1 × 64 × 84 × 24 = 1,290,240 DS0s

Leftover capacity

22.219085815 Gbit/s of each ODU4 line rate (104.794445815 Gbit/s) sits outside 1,290,240 × 64 kbit/s of DS0 line rate (82.575360 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 DS0s by bit rate only. The standard mapping is 1,290,240. Use the mapping.

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