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Basic negociation query

Queen245_goa
Level 1
Level 1

Dear All,

I had some queries running in my mind which iam sure could be cleared in this forum

My scenario is like this...

Client-1(b mode)------ AP(auto or mixed mode) ------ Client-2(g mode)

Now suppose Client-1 wants to transfer data to client-2 via AP, what would be the rate of transfer of data

a) 11 Mbps (b mode) - lowest of the two rates

b) 11 Mbps (b mode) from Client-1 to AP and 54Mbps from AP to Client-2 and vice versa.

can someone pl throw some light here.

looing forward for your replies.

- Queenie

1 Accepted Solution

Accepted Solutions

scottmac
Level 10
Level 10

It would be at the rate of the lowest throughout link in terms of overall throughput.

Each link will transmit the data at it's native rate. For example, if you had something like this:

PC--10BaseT--(10/100 switch)--100BaseT--Server

The segment between the PC and switch is talking at 10 Meg, the segment between the switch and server is talking at 100 meg. The most you could push from PC --> Server is 10Meg, because that is the "slowest" link. The switch is catching at 10 and throwing at 100.

For data going the other way (100-->10) then the switch must buffer the data arriving at the faster rate until either the 10meg link transmits it, or the data expires in the buffer.

It's not an exact analogy because of the way the wireless protocols schedule the transmission and reception of the traffic (wired switches don't), but the general concept is the same.

Good Luck

Scott

View solution in original post

3 Replies 3

didyap
Level 6
Level 6

When you configure the 802.11g access point radio for best throughput, the access point sets all data rates to basic (required). This setting blocks association from 802.11b client devices. The best throughput option appears on the web-browser interface Express Setup and Radio Settings pages and in the speed CLI configuration interface command.

The wireless device always attempts to transmit at the highest data rate that is set to Basic. If there are obstacles or interference, the wireless device steps down to the highest rate that allows data transmission.

The data rates supported on an IEEE 802.11b, 2.4 GHz radio are 1 Mbps, 2 Mbps, 5.5 Mbps and 11 Mbps.

The data rates supported on an IEEE 802.11g, 2.4 GHz radio are 1 Mbps, 2 Mbps, 5.5 Mbps, 6 Mbps, 9 Mbps, 11 Mbps, 12 Mbps, 18 Mbps, 24 Mbps, 36 Mbps, 48 Mbps and 54 Mbps.

The data rates supported on an IEEE 802.11a, 5 GHz radio are 6 Mbps, 9 Mbps, 12 Mbps, 18 Mbps, 24 Mbps, 36 Mbps, 48 Mbps and 54 Mbps.

scottmac
Level 10
Level 10

It would be at the rate of the lowest throughout link in terms of overall throughput.

Each link will transmit the data at it's native rate. For example, if you had something like this:

PC--10BaseT--(10/100 switch)--100BaseT--Server

The segment between the PC and switch is talking at 10 Meg, the segment between the switch and server is talking at 100 meg. The most you could push from PC --> Server is 10Meg, because that is the "slowest" link. The switch is catching at 10 and throwing at 100.

For data going the other way (100-->10) then the switch must buffer the data arriving at the faster rate until either the 10meg link transmits it, or the data expires in the buffer.

It's not an exact analogy because of the way the wireless protocols schedule the transmission and reception of the traffic (wired switches don't), but the general concept is the same.

Good Luck

Scott

Thanks Scott for the wonderful explaination.

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