A scientific approach to XenApp farm sizing

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A scientific approach toXenApp farm sizing

Helge Klein


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Who isHelge Klein?

CTP, MVPAuthor of SetACL and Delprof2Independent consultant and developerArchitect of what later became Citrix Profile Management


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talking about?What is he

Scientifically sound farm sizing methodologyHow to calculate farm capacity


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Methodology

1. Determine capacity of existing farm2. Measure load and identify bottlenecks3. Calculate capacity of new farm


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ExampleDescription Farm capacity

old (total)Farm capacity old (per user)

Factor Farm capacity new (per user)

Farm capacity new (total)

RAM for user sessions [MB] 630,000 315 2.0 630 1,575,000

CPU [normalized cores] 1,250 0.63 0.25 0.16 391

Hard disk [write IOPS] 7,300 3.65 1.5 5.48 13,688

Network [MB/s] 24,000 12.00 0.1 1.20 3,000


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Where to get the

numbers?1. Data collection2. Observation3. Measurements4. Calculation


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Data collection


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for /f %i in (AllFarmServers.txt) do wmic /node:%i cpu get name, maxclockspeed, systemname, description, manufacturer, revision /format:csv >> CPUs.txt

Collect CPU dataCreate AllFarmServers.txt with qfarm

Use resulting list to determine server model


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Srv001,x86 Family 15 Model 4 Stepping 10,GenuineIntel,3400,Intel(R) Xeon(TM) CPU 3.40GHz,1034,Srv001Srv001,x86 Family 15 Model 4 Stepping 10,GenuineIntel,3400,Intel(R) Xeon(TM) CPU 3.40GHz,1034,Srv001Srv001,x86 Family 15 Model 4 Stepping 10,GenuineIntel,3400,Intel(R) Xeon(TM) CPU 3.40GHz,1034,Srv001Srv001,x86 Family 15 Model 4 Stepping 10,GenuineIntel,3400,Intel(R) Xeon(TM) CPU 3.40GHz,1034,Srv001Srv002,x86 Family 6 Model 44 Stepping 2,GenuineIntel,2666,Intel(R) Xeon(R) CPU E5640 @ 2.67GHz,11266,Srv002Srv002,x86 Family 6 Model 44 Stepping 2,GenuineIntel,2666,Intel(R) Xeon(R) CPU E5640 @ 2.67GHz,11266,Srv002Srv002,x86 Family 6 Model 44 Stepping 2,GenuineIntel,2666,Intel(R) Xeon(R) CPU E5640 @ 2.67GHz,11266,Srv002Srv002,x86 Family 6 Model 44 Stepping 2,GenuineIntel,2666,Intel(R) Xeon(R) CPU E5640 @ 2.67GHz,11266,Srv002Srv002,x86 Family 6 Model 44 Stepping 2,GenuineIntel,2666,Intel(R) Xeon(R) CPU E5640 @ 2.67GHz,11266,Srv002Srv002,x86 Family 6 Model 44 Stepping 2,GenuineIntel,2666,Intel(R) Xeon(R) CPU E5640 @ 2.67GHz,11266,Srv002Srv002,x86 Family 6 Model 44 Stepping 2,GenuineIntel,2666,Intel(R) Xeon(R) CPU E5640 @ 2.67GHz,11266,Srv002Srv002,x86 Family 6 Model 44 Stepping 2,GenuineIntel,2666,Intel(R) Xeon(R) CPU E5640 @ 2.67GHz,11266,Srv002Srv003,x86 Family 6 Model 44 Stepping 2,GenuineIntel,2666,Intel(R) Xeon(R) CPU E5640 @ 2.67GHz,11266,Srv003Srv003,x86 Family 6 Model 44 Stepping 2,GenuineIntel,2666,Intel(R) Xeon(R) CPU E5640 @ 2.67GHz,11266,Srv003Srv003,x86 Family 6 Model 44 Stepping 2,GenuineIntel,2666,Intel(R) Xeon(R) CPU E5640 @ 2.67GHz,11266,Srv003Srv003,x86 Family 6 Model 44 Stepping 2,GenuineIntel,2666,Intel(R) Xeon(R) CPU E5640 @ 2.67GHz,11266,Srv003Srv003,x86 Family 6 Model 44 Stepping 2,GenuineIntel,2666,Intel(R) Xeon(R) CPU E5640 @ 2.67GHz,11266,Srv003Srv003,x86 Family 6 Model 44 Stepping 2,GenuineIntel,2666,Intel(R) Xeon(R) CPU E5640 @ 2.67GHz,11266,Srv003Srv003,x86 Family 6 Model 44 Stepping 2,GenuineIntel,2666,Intel(R) Xeon(R) CPU E5640 @ 2.67GHz,11266,Srv003Srv003,x86 Family 6 Model 44 Stepping 2,GenuineIntel,2666,Intel(R) Xeon(R) CPU E5640 @ 2.67GHz,11266,Srv003Srv004,x86 Family 6 Model 26 Stepping 5,GenuineIntel,2666,Intel(R) Pentium(R) III Xeon-Prozessor,6661,Srv004Srv004,x86 Family 6 Model 26 Stepping 5,GenuineIntel,2666,Intel(R) Pentium(R) III Xeon-Prozessor,6661,Srv004Srv004,x86 Family 6 Model 26 Stepping 5,GenuineIntel,2666,Intel(R) Pentium(R) III Xeon-Prozessor,6661,Srv004...

The result


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RAM, NICshard disks

Could be determined via WMI, tooOften knowing the server model is sufficientComponents per model often identical


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Component HP DL360 G4

CPU 2x Xeon dual core 3,0 GHz, no Hyper Threading4 Cores, CPUID: x86 Family 15 Model 4 Stepping 1

RAM 4 GB

NICs 2x Gigabit, 1 connected

RAID Smart Array 6i

Hard disks 2x 70 GB configured as RAID-1, Compaq BF0728A4CB, 3,5“, 15.000 R/Min, Ultra320 SCSI

Spec sheet


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Resulting dataServer model NumberDL360 G7 26DL360 G6 52DL360 G5 70DL360 G4p 60DL360 G4 35


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Observation


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How many users are logged on?Load of CPU, RAM, NICs

Individual processes with a lot of RAM or CPU?


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Physical Disk\% Disk Time

Physical Disk\Avg. Disk Queue Length

„Time“ is equivalent to flickering of hard disk LEDDisk queue length: Number of waiting IOs


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Hypothesis: farm is

memory limited

Limiting factor will differ between farms


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Measurements

Tool: PerfmonNext slides: relevant counters


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Terminal Services\Active SessionsTerminal Services\Inactive Sessions

Terminal Services\Total SessionsSystem\Processes

General system information


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PhysicalDisk(_Total)\% Disk TimePhysicalDisk(_Total)\Avg. Disk Queue Length

PhysicalDisk(_Total)\Disk Reads/secPhysicalDisk(_Total)\Disk Writes/sec

PhysicalDisk(_Total)\Avg. Disk sec/Transfer

Hard disk activityLoad, queue length, operations per second, latency


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Processor(_Total)\% Processor TimeMemory\Available MBytes

Network Interface(*)\Bytes Total/sec

CPU, RAM and networkRAM: total amount must be known!


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logman create counter TSPerf -f csv -cf C:\PerfLogs\Counters.txt -o C:\PerfLogs\Server13.csv -si 60 -rf 24:00:00

AutomationCreate and start data collector set. Format CSV, performance counters are

read from C:\PerfLogs\Counters.txt, output file is C:\PerfLogs\Server013.csv, 60 second sampling interval, duration 24 hours.


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logman create counter TSPerf -f csv -cf C:\PerfLogs\Counters.txt –o C:\PerfLogs\Server13.csv -si 60 -rf 24:00:00-s Server13

Execution on remote computer Server13


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for /f %i in (Servers.txt) do logman create counter TSPerf -f csv -cf C:\PerfLogs\Counters.txt -o C:\PerfLogs\%i.csv -si 60 -rf 24:00:00 -s %i

Many serversOne computer name per line in Servers.txt


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Analyzingthe measured data


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CPU and sessions

Server 37

Moderate load during logon phase,

afterwards even less

A lot of overcapacity


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CPU and sessions

Server 89


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Overlaying the CPU load of

many servers

Easily verify the analyis


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HDD and sessions

Server 37

Moderate load, peaks during logon phase

Full load at aprox. 200


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HDD and sessions

Server 89


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Overlaying the HDD load of

many servers



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RAM and sessions

Server 37

Continually increasing load, maximum in the

afternoon

Available RAM must not go near zero

(because of disk cache)

High load


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RAM and sessions

Server 89


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Overlaying the memory load

of many servers



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Network and sessions

Server 37

200 = 2 MB/s

Average rate < 200 KB/s

Very low load, a lot of overcapacity


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Network and sessions

Server 89


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Overlaying the network load

of many servers



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Hypothesis confirmed: farm is limited by

available memoryCPU load: low, network: negligiblehard disk: moderate


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Calculatingfarm capacity

Description Farm capacity old (total)

Farm capacity old (per user)






Network [MB/s] 24,000 12.00 0.1 1.20 3,000


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How to compare performance of different CPUs?Benchmarking is difficultBetter: Moore’s law (doubling of performance every 18-24 months)Surprisingly accurate (amongst other things because it is a self-fulfilling prophecy)

NormalizingCPU performance


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Performance after time (in months)Assumed performance doubling every 21 monthsOldest CPU in farm = 1,0

𝐿(𝑡 )=𝑒0,033𝑡


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CPU Cores First sold Performance index Performance index per core

Xeon 7100 „Tulsa“ 3,0 GHz

2 08/2006 1,0 1,0

Xeon 7100 „Tulsa“ 3,4 / 3,6 GHz

2 03/2007 1,26 1,26

Xeon E5430 / E5440

4 11/2007 1,64 0,82

Xeon X5550 4 03/2009 2,78 1,39

Xeon E5640 4 03/2010 4,13 2,06


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CPU Number of CPUs

Number of cores

Number of normalized CPUs

Number of normalized cores

Xeon 3 GHz 70 140 70 140

Xeon 3,4/3,6 GHz

119 238 150 300

Xeon E5430/E5440

70 280 115 230

Xeon X5550 67 268 186 370

Xeon E5640 26 104 107 210

Total 352 1030 628 1250


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Farm capacity:

1250 normalized CPUs

0.63 CPUs / user


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Hard disk performance =

IOPSWith many concurrent accesses transfer rate is mostly irrelevantMore important: IOPS (operations per second)Exact number depends on measurement method do not believe vendors


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PhysicalDisk(_Total)\Disk Reads/secPhysicalDisk(_Total)\Disk Writes/sec

IOPS measurement with PerfmonRead and write IOPS may be very different


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Read and write IOPS

Average: ~15

Including spikes: 30


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Farm capacity:

7300 IOPS

3,7 IOPS / user


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RAM – how muchdo we have?

We need total RAM that is available for user sessions


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Server model RAM / server [GB] #Servers RAM / server total [GB]

DL360 G4 3,5 35 122,5

DL360 G4p 3,5 60 210

DL360 G5 3,25 70 227,5

DL360 G6 3,49 52 181,48

DL360 G7 3,49 26 90,74

Total 243 832


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Description Memory [MB]Session 0 (OS + Presentation Server + antivirus) 325

Kernel (pooled + non-pooled) 175File system cache 400Total 900

Calculate RAM used by the OS (Server 2003)


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Farm capacity:

630 GB

315 MB / user

25% Overhead because 900 MB used by OS


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Network – too much for HDX?

1 Gigabit/second per server~ 100 MB/s


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Farm capacity:

24 GB/s

12 MB/s / user

Extreme overcapacity


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Futurecapacity

Description Farm capacity old (total)

Farm capacity old (per user)






Network [MB/s] 24,000 12.00 0.1 1.20 3,000


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Parameters. Calculation includes:Changed user count, different memory requirements of new OS,server replacement, capacity increase for disaster recovery

Parameter ValueFarm users old (CCU) 2,000

Farm users new (CCU) 2,500

RAM for OS new [MB] 1,300

CPU performance new [normalized cores]

1.9

IOPS per disk new 300

Reserve for disaster recovery 1.5


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Servers are memory limited after setting the amount of RAM,the other parameters can be calculated

Virtual server

RAM [MB] 32.768

User sessions 49.9

CPU cores (physical) 4.1

IOPS 273

NICs 0.6


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After setting the number of VMs per host the other parameterscan be calculated.Excel sheet available upon request.

Physical serversVMs per host 3 #Virtual

servers50

RAM [MB] 101,376 #Hosts 25

CPU cores 12 Users / host

150

Hard disks 3

NICs 2


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Thank you!Helge Klein

Documents

A scientific approach to XenApp farm sizing