Grass Valley NV9000-SE v.5.0 - Manual (Page 468)

450
Tielines
Number of Tieline Paths
(3-hop)
AD = 30 x 30 x 30 / (2 x 5) = 27,000 / 10 = 2700
All Paths
30 + 30 + 30 + 450 + 180 + 2700 = 3420.
Four Connections
(1-hop)
AB = (5 x 2) x 4 = 40
BC = (2 x 5) x 4 = 40
CD = (5 x 2) x 4 = 40
(2-hop)
AC = 40 x 40 / 2 = 1600 / 2 = 800
BD = 40 x 40 / 5 = 1600 / 5 = 320
(3-hop)
AD = 40 x 40 x 40 / (2 x 5) = 64,000 / 10 = 6400
All Paths
40 + 40 + 40 + 800 + 320 + 6400 = 7640.
Comparison
These are the data:
This is the chart for the example, showing the totals for the 4 cases:
Variation in Tieline Count
Thye previous examples used a uniform number of tielines between partitions because it made
the examples easier to understand. When the number of tielines is not uniform, the calculations
are not more difficult.
Connections 1-Hop Paths 2-Hop Paths 3-Hop Paths Total
1 x 3 = 3 30 70 100 200
2 x 3 = 6 60 280 800 1140
3 x 3 = 9 90 630 2700 3420
4 x 3 = 12 120 1120 6400 7640
3 6 9 12
Tieline Cables
1,000
2000
3,000
4,000
5,000
6,000
7,000
8,000
9.000
10,000
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