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 ODU0?

An ODU0 carries 8,064 DS0s, per ITU-T G.709, ITU-T G.707 and ANSI T1.107. This path maps STM-4 into ODU0 as a client, then follows that client's tributaries.

8,064

DS0s in an ODU0

Working

ODU0 → 1 × STM-4 → 4 × STM-1 → 84 × DS1 → 24 × DS0

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.
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.

1 STM-4
1 × 4 = 4 STM-1s
1 × 4 × 84 = 336 DS1s
1 × 4 × 84 × 24 = 8,064 DS0s

Leftover capacity

728.064 Mbit/s of each ODU0 line rate (1.244160 Gbit/s) sits outside 8,064 × 64 kbit/s of DS0 line rate (516.096 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 19,440, which floors to 19,440 DS0s by bit rate only. The standard mapping is 8,064. Use the mapping.

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