- Posts: 27
- Thank you received: 0
Frame-Relay Traffic shaping
16 years 11 months ago #24134
by tsunami
Frame-Relay Traffic shaping was created by tsunami
Hi All,
I need your suggestions on the Frame-relay and shaping concepts.
I wanted to know the process flow on the frame relay interface when shaping is enabled.
Let us assume two Routers R1 and R2 connected to a frame relay cloud.
Both R1 and R2 has a line access rate of 128kbps. The CIR is set to 64kbps on both in and out directions.
The Committed burst is 8000 bits per timing interval. The Excess burst is zero.
To summarize:
1. Line Access rate on each routers = 128kbps.
2. Comitted Information rate = 64kbps.
3. Comitted Burst = 8000 bits/Tc.
4. Timing inteval = 125ms.
5. Excess burst = 0.
Now the theory:
one ICMP packet of 1500 bytes is sent from R1 to R2.
R1 places this packet in the TX-ring for framing.
So the total frame size would be 1506 bytes with 6 bytes of header information.
2bytes flags, 2 bytes address, 2bytes of FCS.
So the frame size would be 12048 bits.
Router R1 has a Tc=125ms and Bc=8000 bits.
So, in the first Tc (between 0 and 125ms), it transmits 8000 bits which takes 62.5ms.
The router sits idle for the next 62.5ms. The router transmits the next 4048 bits (between 125 and 250ms).
This time the router just takes 31.625ms because there are only 4048 bits.
The one-way delay for the ICMP Packet to reach from R1 to R2 is 125ms + 31.625ms = 156.625ms.
So the round trip time would be twice (because R2 has identical settings) = 313.25ms.
But the result deviated from the theory:
The clock rate on both routers were set to 128000. The Router R1 used DLCI 102 and R2 used 201.
R1's map-class
map-class frame-relay DCLASS
frame-relay cir in 64000
frame-relay cir out 64000
frame-relay bc in 8000
frame-relay bc out 8000
frame-relay be in 0
frame-relay be out 0
R2's map-class
map-class frame-relay DCLASS
frame-relay cir in 64000
frame-relay cir out 64000
frame-relay bc in 8000
frame-relay bc out 8000
frame-relay be in 0
frame-relay be out 0
R1's show traffic-shape
R1#sh traffic-shape
Interface Se1/0
Access Target Byte Sustain Excess Interval Increment Adapt
VC List Rate Limit bits/int bits/int (ms) (bytes) Active
102 64000 1000 8000 0 125 1000 -
R2's show traffic-shape
R1#sh traffic-shape
Interface Se1/0
Access Target Byte Sustain Excess Interval Increment Adapt
VC List Rate Limit bits/int bits/int (ms) (bytes) Active
102 64000 1000 8000 0 125 1000 -
However, the ping from R1 to R2 gives an RTT of 140ms.
R1#ping
Protocol [ip]:
Target IP address: 10.10.1.2
Repeat count [5]: 1
Datagram size [100]: 1500
Timeout in seconds [2]:
Extended commands [n]:
Sweep range of sizes [n]:
Type escape sequence to abort.
Sending 1, 1500-byte ICMP Echos to 10.10.1.2, timeout is 2 seconds:
!
Success rate is 100 percent (1/1), round-trip min/avg/max = 140/140/140 msCan anybody put my thoughts in correct direction ?
netsharath
I need your suggestions on the Frame-relay and shaping concepts.
I wanted to know the process flow on the frame relay interface when shaping is enabled.
Let us assume two Routers R1 and R2 connected to a frame relay cloud.
Both R1 and R2 has a line access rate of 128kbps. The CIR is set to 64kbps on both in and out directions.
The Committed burst is 8000 bits per timing interval. The Excess burst is zero.
To summarize:
1. Line Access rate on each routers = 128kbps.
2. Comitted Information rate = 64kbps.
3. Comitted Burst = 8000 bits/Tc.
4. Timing inteval = 125ms.
5. Excess burst = 0.
Now the theory:
one ICMP packet of 1500 bytes is sent from R1 to R2.
R1 places this packet in the TX-ring for framing.
So the total frame size would be 1506 bytes with 6 bytes of header information.
2bytes flags, 2 bytes address, 2bytes of FCS.
So the frame size would be 12048 bits.
Router R1 has a Tc=125ms and Bc=8000 bits.
So, in the first Tc (between 0 and 125ms), it transmits 8000 bits which takes 62.5ms.
The router sits idle for the next 62.5ms. The router transmits the next 4048 bits (between 125 and 250ms).
This time the router just takes 31.625ms because there are only 4048 bits.
The one-way delay for the ICMP Packet to reach from R1 to R2 is 125ms + 31.625ms = 156.625ms.
So the round trip time would be twice (because R2 has identical settings) = 313.25ms.
But the result deviated from the theory:
The clock rate on both routers were set to 128000. The Router R1 used DLCI 102 and R2 used 201.
R1's map-class
map-class frame-relay DCLASS
frame-relay cir in 64000
frame-relay cir out 64000
frame-relay bc in 8000
frame-relay bc out 8000
frame-relay be in 0
frame-relay be out 0
R2's map-class
map-class frame-relay DCLASS
frame-relay cir in 64000
frame-relay cir out 64000
frame-relay bc in 8000
frame-relay bc out 8000
frame-relay be in 0
frame-relay be out 0
R1's show traffic-shape
R1#sh traffic-shape
Interface Se1/0
Access Target Byte Sustain Excess Interval Increment Adapt
VC List Rate Limit bits/int bits/int (ms) (bytes) Active
102 64000 1000 8000 0 125 1000 -
R2's show traffic-shape
R1#sh traffic-shape
Interface Se1/0
Access Target Byte Sustain Excess Interval Increment Adapt
VC List Rate Limit bits/int bits/int (ms) (bytes) Active
102 64000 1000 8000 0 125 1000 -
However, the ping from R1 to R2 gives an RTT of 140ms.
R1#ping
Protocol [ip]:
Target IP address: 10.10.1.2
Repeat count [5]: 1
Datagram size [100]: 1500
Timeout in seconds [2]:
Extended commands [n]:
Sweep range of sizes [n]:
Type escape sequence to abort.
Sending 1, 1500-byte ICMP Echos to 10.10.1.2, timeout is 2 seconds:
!
Success rate is 100 percent (1/1), round-trip min/avg/max = 140/140/140 msCan anybody put my thoughts in correct direction ?
netsharath
Time to create page: 0.117 seconds