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Measurement and Modelling I:End-to-End Bandwidth Measurements

6.829 Lecture 11

October 10, 2002

Jacob Strauss

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End-to-End Measurements

Goal: Bandwidth-Aware Applications

Is a path fast enough? Which path is faster?

What is bandwidth?

Link Capacity: physical link speed (e.g. 100 Mbps Ethernet)

for link i, link capacity ci.

Path Capacity: minimum capacity link along a path

C= mini=0...n(ci)

Link Utilization: fraction of link capacity used over some time interval

0 ui 1

Available Bandwidth: minimum spare capacity along a path over some interval

A = mini=0...n(1 ui)ci

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Example Topology

C=10 Mbps, u=0.2C=100 Mbps, u=0.95

C=20 Mbps, u=0.5X

Y

Capacity from X to Y?


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Example Topology

C=10 Mbps, u=0.2C=100 Mbps, u=0.95

C=20 Mbps, u=0.5X

Y

Capacity from X to Y? C = 10Mbps

Available bandwidth from X to Y?


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Example Topology

C=10 Mbps, u=0.2C=100 Mbps, u=0.95

C=20 Mbps, u=0.5X

Y

Capacity from X to Y? C = 10Mbps

Available bandwidth from X to Y? A = 5Mbps

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Problem: Capacity Estimation

X Y

C1

C2

C3

How can we measure the path capacity between X and Y?

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Packet Pair on an unloaded path

X Y

tt

C1

C2

C3

Send two packets, of P bits each, back-to-back

Record the difference between arrival times, t.

Path Capacity is Pt

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Packet Pair Complications: Cross Traffic

X Y

t

C1

C2

C3

t

other packet

probe packet

t >P

C

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Packet Pair Complications: Multiple Queues

X Y

C1

C2

C3

t

other packet

probe packet

t

t nyu.ron.lcs.mit.edu

1

500B

500B

sen

tgap

b2b

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New Problem: Available Bandwidth Estimation

Can we use the same packet train to estimate A?


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No, but we can determine whether P/ > A or P/ < A

Later well build on this test to estimate A


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No, but we can determine whether P/ > A or P/ < A

Later well build on this test to estimate A

Basic idea: instead of just comparing wn+1 to wn, compare wn to

wn+1, wn+2, wn+3, . . .

Method: Plot evolution of wn versus n

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Pathload Delay Model

The packet train consists of K packets of size L sent at a constant

rate R.

Path consists ofH links, each with capacity Ci, available bandwidth

Ai, and queue length qki when the kth probe packet arrives


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rate R.



The One-Way-Delay for probe k is:

Dk =Hi=1

L

Ci+

qkiCi


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rate R.



The One-Way-Delay for probe k is:

Dk =Hi=1

L

Ci+

qkiCi

Dk Dk+1Dk =Hi=1

qkiCi

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Increasing Trends

If R > A, then Dk > 0 for k 1

If R A, then Dk = 0 for k 1


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Increasing Trends



Intuition for a single link i with Ai < R:

Queue grows longer with each probe packet that arrives

qki = (L + uiCiT)C1T = (RAi)T > 0

For a single link with Ai R:

qki = 0

By induction, show that Dk > 0 when R > A


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Increasing Trends



Intuition for a single link i with Ai < R:

Queue grows longer with each probe packet that arrives

qki = (L + uiCiT)C1T = (RAi)T > 0

For a single link with Ai R:

qki = 0

By induction, show that Dk > 0 when R > A

Plot Dk if Dk > 0 then R > A

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Complications

Ai not a constant

varies on both short and long time scales

Need to choose K and L carefully

Too short cant tell if Dk > 0

Too long flood link

Compromise use multiple trains

How did we choose R initially?

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Iterative Search for A

maintain two state variables: Rmax and Rmin.

always have Rmin A Rmax .

Pick a new R halfway between Rmax and Rmin

Test whether R A, update either Rmax or Rmin

Stop when Rmax and Rmin are close enough.

In practice the search is more complicated because the outcome of

testing R A is sometimes unsure

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Pathload Block Diagram

Timestamps

Filters

Estimates

Network

Packet Source

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Available Bandwidth isnt the full story

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0.01 0.1 1 10 100

CDF

BTC / Pathload rate

BTC vs Pathload: 15 RON hosts -- one test per path

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Conclusion

Theres a lot you can learn with simple probes

packet pair packet trains regular interval, constant packet size

Just by looking at packet delay variations you can determine

Path Capacity

Common Packet Sizes

Available Bandwidth

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