Circling Back To Littles Law Now that we have tools to gather information.

Circling Back To Little’s Law

Now that we have tools to gather information

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Latency –A measure of time delay experienced in a system,the precise definition of which depends on the system and the time being measured. In storage, latency is generally referred to as response time, in ms.

Throughput – The amount of material or items passing through a system or process. In storage, IO/s in units of 4k

Latency & Throughput

Random SQL SERVER example: http://www.sql-server-performance.com/2003/2000io-config-sannas/

Latency starts to spike as near saturation

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Latency & Throughput

Latency starts to spike as near saturation

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Disk IOPS versus Latency

throughput

A

LATENCY

LATENCY

throughputB

True in Real Life Too

Little’s Law Review & Example

• Little’s Law:• Restated: N = L * W

N = # Cars in JamT = Lanes (Throughput)Wait = time from A->B

• Assume 4 cars arrive every second (lanes)• A->B is 30 seconds • N = 4*30 = 120

€

L = λ ×W

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Little’s Law - ReviewWe can use this with Latency & Throughput on a Netapp system too.

Standard version:

Re-written for Netapp:

Translating into IO terms:

N = # of outstanding IOsT = Throughput of IOsR = Response time of each IO

€

L = λ ×W

€

N = T × R

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Little’s Law - An Example

• Typical situation:– An user complains of poor performance:

My dd/cp/tar/Oracle query (for example: full table scan) etc. process isn’t fast enough

– A casual look at sysstat shows the filer is not very busy

– NetApp Service returns with a statement of “thread-limited”

• What does this mean?

Little’s Law - An ExampleCompute

Wait for Storage

Read Request

Read Request

Read Request

Read RequestDat

a Re

turn

Dat

a Re

turn

Dat

a Re

turn

TimeIn this example, the process is either computing or reading. It is always busy. But the CPU and the storage are not, on average, fully used.

Client side tools would be needed to determine this: debugger, strace, dtrace, etc.

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Little’s Law - An Example

Using stats show volume:

volume:dwhprod1:san_read_data:28828868b/s volume:dwhprod1:san_read_latency:4.23ms volume:dwhprod1:san_read_ops:653/s

How many threads (on average) are running here?

From Little’s Law:

(N threads) / (service time per op) = throughput

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How many threads (on average) are running here? (N threads) / (service time per op) = throughput N threads = throughput × (service time)

Service Time: volume:dwhprod1:san_read_latency:4.23ms

Throughput: volume:dwhprod1:san_read_ops:653/s

Little’s Law - An Example

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Little’s Law - An Example

How many threads (on average) are running here? throughput × (service time) N threads 653 × .00423 2.8

What are the performance implications of having only 2.8 concurrent requests (on average)?

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Little’s Law - An Example

This example is a concurrency-limited workload– Each thread is always busy– Not enough threads to keep the system busy

Implications:– Storage system not fully utilized– High I/O wait times at the server

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Little’s Law - An ExampleSolution:• Add more threads

– Sometimes you cannot, for example if there is a mapping of 1 thread to each application user, you cannot increase the user population

– Fix Client Inefficiencies• FCP/iSCSI - Increase queue depth• NFS - Poor IO concurrency due to inefficient NFS client design, use an

updated NFS client or 3rd party product (ex. Oracle DirectNFS)and/or

• Make the IO subsystem/disks faster– Including fixing client filesystem caching– PAM/Hybrid Aggregates