1. Setting TCP Window Size for Legacy LAN VoIP Niko Färber January 10, 2001

2. Outline LAN simulations Review of previous work Influence of buffer size / number of nodes on “optimal” TCP window size Why is W=4 a good choice for all simulations? Analysis of NxW = BxD Analysis of D Rule of thumb for setting W

3. Review: LAN Simulation Scenario Balanced N-N communication, N=16 hosts, 1 switch File size PDF: Log-Normal Idle time ~ file size / load 100 KB buffer, Drop Tail 10 Mbps Full duplex Evaluation Voice QoS: delay jitter = E{|d i - d i+1 |} Data QoS: goodput =  B i /  T i

4. Review: LAN Simulation Results N=16 nodes, B=100 KB buffer/port goodput [Mbps] delay jitter [ms]  = 0.5 w = 1 2 4 8 16 32 64 0.4 0.3 0.1 0.2 Low load Voice uncritical Data critical High load Voice critical Data uncritical W=4 is good choice for all loads!

5. Extended LAN simulations Change number of nodes N = {4, 8, 16, 32} Change buffer size B = {100, 200, 300} KB Result: W=4 is always good choice

6. LAN Simulation – Example 1 N=4, B=100 goodput [Mbps] delay jitter [ms]

7. LAN Simulation – Example 2 N=16, B=200 goodput [Mbps] delay jitter [ms]

8. Bandwidth-Delay Product To keep pipe full: NxW = BxD W=const. ok for changing load assuming E{N} ~ E{D} ~ Delay includes all delay components, i.e., total round trip time Propagation Packetization Queuing (~  ) Processing N: number of current connections W: TCP window size for each connection [bit] B: bandwidth of link [bit/s] D: round trip time [s]

9. Analysis of Delay Components For example, P = 1480+20 Byte A = 40+20 Byte B = 10 Mbps Dominant component: 1 x d data for each forward link traversed (W=2) To cover any additional queuing and/or processing delay, some additional safety margin is necessary (W=4) d proc, rx d prop = 512/B d data = P/B d Q,fwd d prop = 512/B d data = P/B d proc, tx d Q,bwd d prop = 512/B d ack = A/B d prop = 512/B d ack = A/B tx rx sw d min = 4 d prop + 2 d data + 2 d ack = 3376 Byte/B = 2.3 data packets

10. Verification of Delay Analysis TCP window size (W ) should increase proportional to delay (D ) because of NxW = BxD Dominant contribution to D is packetization time d data D and (hence W ) should be roughly proportional to Number of hops H Packet size P tx rx sw H = {2, 3, 5, 9} P = {128, 256, 512, 1024} [Byte] At what TCP window size there is no further gain in throughput?

11. Number of Hops TCP window size [byte] Throughput [Mbps] H = 2 3 5 9 P = 128 Byte

12. Packet Size TCP window size [byte] Throughput [Mbps] H = 2 P = 128 256 512 1024

13. Conclusions Rule of thumb for setting TCP window size: W [packets] = 2 x H Verified for H=2 (1 switch) by extensive simulative results General relationship of W ~ H and W ~ P can be explained analytically and is verified by simulations Fixed W for different loads can be explained when D ~ and N ~