Port Trunking and Load Balancing in Blade Switches
For example: There is a port trunk group with eight member ports, and there is a frame with SA=0x0080C800000B DA=0x0080C800000C.
•SA last three bits is 0b011
•DA last three bits is 0b100
•SA XOR DA is 0b111
In the following example, four trunk ports are link up in the port trunk.
•For algorithm 1, N = 0b011 MOD 4 = 3
The outgoing port would be the fourth link up in the port trunk.
•For algorithm 2, N = 0b100 MOD 4 = 0
The outgoing port would be the first link up in the port trunk.
•For algorithm 3, N = 0b111 MOD 4 = 3
The outgoing port would be the fourth link up in the port trunk.
The following table shows the relationship between the remainder and outgoing ports
Link up port | 1st | 2nd | 3rd | 4th | 5th | 6th | 7th | 8th |
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Last three LSB | 0b000 | 0b001 | 0b010 | 0b011 | X | X | X | X |
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Last three LSB | 0b100 | 0b101 | 0b110 | 0b111 | X | X | X | X |
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In the following example, 6 trunk ports are link up in the port trunk.
•For algorithm 1, N = 0b011 MOD 6 = 3
The outgoing port would be the fourth link up in the port trunk.
•For algorithm 2, N = 0b100 MOD 6 = 4
The outgoing port would be the fifth link up in the port trunk.
•For algorithm 3, N = 0b111 MOD 6 = 1
The outgoing port would be the second link up in the port trunk.
The following table shows the relationship between the remainder and outgoing ports.
Link up port | 1st | 2nd | 3rd | 4th | 5th | 6th | 7th | 8th |
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Last three LSB | 0b000 | 0b001 | 0b010 | 0b011 | 0b100 | 0b101 | X | X |
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Last three LSB | 0b110 | 0b111 | X | X | X | X | X | X |
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HP ProLiant BL |