D-Link DGS-3324SR Product Manual - Page 36
Number of Switches * 2, 2 * 4 + 3 * 2 + Number of Switches * 2
UPC - 790069262067
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xStack DGS/DXS-3300 Series Layer 3 Stackable Gigabit Ethernet Switch User Manual 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-3350SR (Token Cost = 4), and three DXS-3326GSR (Token Cost = 2). In this case the total Token Cost would be: (1 * 2) + (2 * 4) + (3 * 2) = 16 If the user then wanted to add the maximum number of DGS-3324SR Switches (Token Cost = 2) to this stack: (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 Therefore, in this case the user could add extra eight DGS-3324SR switches to this ring stack. The entire stack would then consist of nine 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: 9 * 2 + 2 * 4 + 3 * 2 ≤ 32 Although the Token Cost is less than 32, the number of switch boxes is 14, which exceeds the maximum number of 12. Thus, only extra six DGS-3324SRs 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 DXS3326GSRs (no modules). Our switch count would equal ten and our token cost would equal thirty (18 + 6 + 6 = 30 ≤ 32). Success! • Make a ring stack consisting of four DGS-3324SRs, five DXS-3326GSRs (no modules), three DXS-3350SRs (no modules). Our switch count would equal twelve and our token cost would equal thirty (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 DGS3324SRs and six DXS-3326GSRs (12 + 20 = 32). This is the maximum number of switch boxes allowed in a ring stack. Our switch count stays at twelve and our token cost becomes thirty-two (2 + 2 + 2 + 2 + 24 = 32 ≤ 32). Success! 21