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 OTU4?
An OTU4 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 OTU4
Working
OTU4 → 1 × ODU4 → 10 × ODU2 → 1 × STM-64 → 64 × STM-1 → 84 × DS1 → 24 × DS0
- 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.
- 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 ODU4
- 1 × 10 = 10 ODU2s
- 1 × 10 × 1 = 10 STM-64s
- 1 × 10 × 1 × 64 = 640 STM-1s
- 1 × 10 × 1 × 64 × 84 = 53,760 DS1s
- 1 × 10 × 1 × 64 × 84 × 24 = 1,290,240 DS0s
Leftover capacity
29.234613568 Gbit/s of each OTU4 line rate (111.809973568 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 1,747,030.83, which floors to 1,747,030 DS0s by bit rate only. The standard mapping is 1,290,240. Use the mapping.
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