DXS-3326GSR Gigabit Layer 3 Switch
4 * Number of Switches
≤
32
Number of Switches
≤
32/4
Number of Switches
≤
8
In this case, a maximum of eight DXS-3326GSR switches (where all of these switches have a 10G uplink) can be ring
stacked.
Adding a different switch type to an existing stack
In this example, there are three different switch types, each with different token costs. There is one DGS-3324SR (Token
Cost = 2), two DXS-3350SRs (Token Cost = 4), and three
DXS-3326GSRs (Token Cost = 2). In this case the total Token
Cost would be:
(1 * 2) + (2 * 4) + (3 * 2) = 16
If you then wanted to add the maximum number of DXS-3326GSR Switches (Token Cost = 2) to this stack:
(1*2 + 2 * 4 + 3 * 2) + Number of Switches * 2
≤
32
16 + Number of Switches * 2
≤
32
Number of Switches * 2
≤
32 – 16 = 16
Number of Switches
≤
16/2 =8
So, in this case, you could add an extra eight DXS-3326GSR switches to this ring stack. The entire stack would then
consist of eight DGS-3326GSRs, 1 DGS-3324SRs (Token Cost = 2), two DXS-3350SRs (Token Cost = 4), and three DXS-
3326GSRs (Token Cost = 2). This gives a total Token Cost for the stack of:
1 * 2+ 8 * 2 + 2 * 4 + 3 * 2
≤
32
Although the Token Cost is 32, the number of switch boxes is 14, which exceeds the maximum number of 12. Thus, only
seven extra DXS-3326GSRs can be added to the ring stack.
For further examples, we can:
Make a ring stack consisting of four DXS-3350SRs (one with module), three DGS-3324SRs, and three
DXS-3326GSRs (no modules). Our switch count would equal 10 and our token cost would equal 32 (18 + 8+ 6
= 32
≤
32). Success!
•
•
•
Make a ring stack consisting of four DGS-3324SRs, five DXS-3326GSRs (no modules), and three
DXS-3350SRs (no modules). Our switch count would equal 12 and our token cost would
equal 30 (8 + 10 + 12
= 30
≤
32). Success!
Add four 10G modules to an existing ring stack (2+2+2+2=8). Using a stack consisting of six DGS-3324SRs
and six DXS-3326GSRs (12 + 20 = 32). This is the maximum number of switch boxes allowed in a ring stack.
Our switch count stay at 12 and our token cost becomes 32 (2 +2 + 2 +2+ 24 = 32
≤
32). Success!
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