2 Throughput-Delay Plots Need for Speed Benchmarking Paradox Paradox Resolved 3 Storage Performance Throughput Latency Concurrency 4 Conclusion c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 2 / 34
An immutable law of performance 1 1 If your data don’t fit LL, change your data! c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 3 / 34
An immutable law of performance 1 2 Why is it important? 1 If your data don’t fit LL, change your data! c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 3 / 34
An immutable law of performance 1 2 Why is it important? L = λW proven 1961 1 If your data don’t fit LL, change your data! c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 3 / 34
An immutable law of performance 1 2 Why is it important? L = λW proven 1961 Algebraic simplification 1 If your data don’t fit LL, change your data! c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 3 / 34
An immutable law of performance 1 2 Why is it important? L = λW proven 1961 Algebraic simplification Cross-checking J.D.C. Little’s lore (in his own words): perfdynamics.blogspot.com/2011/07/ 1 If your data don’t fit LL, change your data! c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 3 / 34
queueing theory LL relates inventory and manufacturing cycle time John Little (now 84) is not a computer performance analyst Prof. Little did not invent his own law LL was known to A. K. Erlang more than 100 years ago There are actually two versions of Little’s law A Paradox 1 LL expresses the fact that R decreases with increasing X 2 Benchmarks show R increases with increasing throughput X c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 4 / 34
2 Resolve the XR paradox by introducing 3D version of LL 3 Apply LL to understand IOPS bottleneck c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 5 / 34
In steady state, the mean rate of arrival (λ) of customers into a system is equal to the mean output rate or throughput (X) of customers departing the system. λ = X (1) The total number of customers, requests, processes, threads (N) in the system is given by: N = λR = XR (2) where R is the mean total time spent in the system. Classic Little’s law N is the mean number of customers/requests in residence. c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 6 / 34
If the system is like a grocery store, the device level is like a checkout lane. Please link back to the page you downloaded this from, or just link to parkablogs.blogspot.com Please link back to the page you downloaded this from, or just link to parkablogs.blogspot.com Please link back to the page you downloaded this from, or just link to parkablogs.blogspot.com Please link back to the page you downloaded this from, or just link to parkablogs.blogspot.com Please link back to the page you downloaded this from, or just link to parkablogs.blogspot.com At any device (labelled k = 1, 2, . . .), equation (2) yields the local number of customers/requests (Qk ) enqueued: Qk = λRk (3) where Rk is the time in residence at the device. Rk is defined as the sum of the service time (Sk ) at the cashier and the time (Wk ) spent waiting to get serviced by the cashier: Rk = Wk + Sk (4) The total number, N, in the global system (2) is the sum of all the customers/requests enqueued at each device: N = Q1 + Q2 + · · · + Qk (5) c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 7 / 34
utilization of the device comes from (3) by ignoring the waiting time contribution. Logically, this is equivalent to letting W → 0: Qk = λRk = λ(Wk + Sk ) → λSk (6) We changed the right side of (6), so the left side must also be changed. But to what? It has to be number (like N) and Qk can be unbounded: Qk < ∞ (but not infinite). Call the “new” number ρk (to agree with queueing literature) so that (6) becomes: ρk = λSk (7) Since the cashier cannot service more than one customer at a time: ρk < 1 (8) or ρk < 100%, on average. Little’s utilization law The utilization ρk is the mean number of customers/requests in service at device k. c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 8 / 34
Law 2 Throughput-Delay Plots Need for Speed Benchmarking Paradox Paradox Resolved 3 Storage Performance Throughput Latency Concurrency 4 Conclusion c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 9 / 34
Driving on the freeway at 60 mph. At that speed, you travel a mile a minute. How far will you travel in 15 minutes? c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 10 / 34
Driving on the freeway at 60 mph. At that speed, you travel a mile a minute. How far will you travel in 15 minutes? Answer In a quarter of an hour you will travel one quarter the distance you would have covered in an hour. Therefore, in 15 minutes you will travel 15 miles. Congratulations! You just used LL without realizing it. Let X be the speed, R the elapsed time and N the miles covered: N = X R 15 miles = 60 mph × 15 60 hours c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 10 / 34
Related Example Now, suppose it’s an emergency and you need to cover the same distance in 10 minutes. How fast do you need to go? c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 11 / 34
Related Example Now, suppose it’s an emergency and you need to cover the same distance in 10 minutes. How fast do you need to go? The answer may not be so obvious, but not to worry. We can still use LL. Answer N = X R 15 miles = X × 10 60 hours Solving for X: X = 15 × 6 = 90 mph c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 11 / 34
Related Example Now, suppose it’s an emergency and you need to cover the same distance in 10 minutes. How fast do you need to go? The answer may not be so obvious, but not to worry. We can still use LL. Answer N = X R 15 miles = X × 10 60 hours Solving for X: X = 15 × 6 = 90 mph Theorem (Inverse Proportion of LL) To reduce the delay R (elapsed time), the speed X must be increased. c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 11 / 34
1 N 15 N 50 0 5 10 15 R 0 5 10 15 X N 1 N 15 N 50 0 5 10 15 X 0 5 10 15 R Example was for the N = 15 miles curve Time for N = 15 miles is reduced by going from green to red dot Different distance means a different curve Curves are symmetric about the diagonal Can flip X and R axes w/o changing the curves Independent variable goes on x-axis c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 12 / 34
X QPS 0.0 0.5 1.0 1.5 2.0 R s 0 50 100 N Three variables (like PVT in chemistry) 3D surface Like a cone but not rotationally symmetric about apex Square edges cause hyperbolic contours c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 14 / 34
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ctual data (with and without FIO) 5 10 15 20 25 30 X 0.0 0.5 1.0 1.5 R Extracted data SQL Server RDBMS: Measure X in QPS and R in s at each load (N) Two curves: before (red) and after (blue) application of FIO device Manually extracted pertinent data points c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 15 / 34
Paradox 1 LL says R decreases with increasing X (3D contour lines) 2 Benchmarks show R increases with increasing throughput X 5 10 15 20 25 30 X 0.0 0.5 1.0 1.5 R Extracted data 0 10 20 30 40 50 X 0.0 0.5 1.0 1.5 2.0 R Data moves on LL contours The Resolution Superimpose LL 3D contours onto 2D benchmark data. c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 16 / 34
24 26 28 30 X QPS 0.6 0.8 1.0 1.2 1.4 R s Theorem (Gunther 2012) All benchmark data “moves” along LL contours. c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 17 / 34
Law 2 Throughput-Delay Plots Need for Speed Benchmarking Paradox Paradox Resolved 3 Storage Performance Throughput Latency Concurrency 4 Conclusion c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 18 / 34
a query requires the execution of 100 K instructions on the CPU. The CPU can execute 10 GIPS. 1 IPQ: 100 K = 100 × 103 instruction per application query 2 IPS: 10 GIPS = 10 × 109 cpu instructions per second The throughput (or request rate) for queries is: λQPS = IPS IPQ = 10 × 109 100 × 103 = 1010 105 = 100, 000 The steady state assumption (1) tells us: λQPS = 100 KQPS = XQPS (9) A maximum of 100 KQPS can be processed c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 19 / 34
further that within the query instructions a single IO is issued. The CPU thread must wait before the rest of the query instructions can be completed. This creates a nice convenience since λIOPS ≡ λQPS . λIOPS = QPS IOPQ = 105 1 = 100, 000 λIOPS = 100 KIOPS = XIOPS (10) Device IOPS But this is aggregate IOPS. How many IOPS can a single disk do? c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 20 / 34
Seagate Barracuda 7200 RPM disk is capable of about 100 IOPS. Follows from combined seek time and RPS time being on the order of 10 ms. Hence: IOPS = 1 0.010 = 100 (11) Simple arithmetic suggests that 1000 Seagate Barracudas would needed to accommodate the 100 KIOPS aggregate throughput being considered here. Caveat emptor Note that (11) is a rearrangement of the LL utilization law (7): λIOPS = ρ Sdisk (12) with ρ = 1. Hence, it is the theoretical maximum possible IOPS that this disk can support. In practice, the sustainable IOPS rate will be considerably lower. c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 21 / 34
device is capable of responding to an IO request in 1 ms (10x Seagate Barracuda), the processor needs to issue 100 concurrent IO requests to the storage system so that it can complete 100 KQPS. If the storage device were 10 times faster (e.g., SSD), then the processor would only need to be handing a 10th as many IO requests, or just 10 concurrent requests. Sdisk = 10−3 s Sssd = 10−4 s Applying the LL utilization law (7): ρdisk = λIOPS Sdisk = 105 × 10−3 = 100 (13) Suggests we need more than 100 spindles. Similary, for faster SSD devices: ρssd = λIOPS Sssd = 105 × 10−4 = 10 (14) Latency Latency is an ill-defined word that means different things to different technical people. Need the more exacting language of queueing theory to see where different latencies arise. c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 22 / 34
not deterministic, we represent CPU and storage as a queueing network with two stages: Scpu Sdev Snk ! Src ! Queries are sourced by the application at an aggregate request rate of λ = 100 KQPS and the CPU issues IO requests at the rate of 100 KIOPS. However, from (13) we know ρdisk = 100 or 10,000% !! Trouble This violates the utilization bound ρdisk < 1 given by (8). We already suspected we would need at least 100 spindles from (13). But how should the disks be arranged to give the correct latencies? c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 23 / 34
(8) is that we need many (q) disks operating in parallel. Scpu ! !/q !/q !/q !/q Sink ! Source ! Parallel disks divide the total throughput (λ) into q substreams, each load-balanced with equal rate λ/q. Moreover, considering (13), we can write: ρdisk = 100 q < 1 (15) LL tells us we actually need more than q = 100 disks to satisfy the utilization bound. Disk Arrays This is why typical storage subsystems are configured as arrays. c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 24 / 34
time) for a query: SCPU = IPQ IPS = 10−5 seconds (16) i.e., 10 µs per query . The mean CPU utilization is: ρCPU = λQPS SCPU = 105 × 10−5 = 1 which is right on the edge of the utilization bound. Scpu Scpu Scpu Snk Src So, we need more than one core or execution unit. Duo-core LL tells us we need a duo-core, at least, to meet the utilization bound. c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 25 / 34
the storage system is capable of responding to an IO request in 1 ms · · · · · · If the storage were 10 times faster in responding with I/O requests... These numbers become Sdev in the following diagram. Sdev Sdev Sdev Scpu Scpu ! Src ! Snk !/q !/q !/q !/q ! We use this queueing model to examine both latency and concurrency effects. “Infinite IOPS” is represented by 1000 parallel storage devices. c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 26 / 34
storage system is capable of responding to I/O requests in 1/1,000th of a second, then the CPU will need to issue N = 100 concurrent requests · · · · · · If the storage were 10 times faster then the processor would only need to be handing 1/10th as many concurrent requests, or just N = 10 concurrent requests. Latency Concurrency Device (#) Service Residence Qk N CPU (2) 0.00001 0.0000133333 1.33333 1.33333 Disk (1000) 0.001000 0.001111 0.1111 111.1 SSD (1000) 0.000100 0.0001010 0.01010 10.10 FIOa (1000) 1.000 × 10−6 1.000 × 10−6 0.0001 0.1000 FIOb (1) 1.000 × 10−6 1.111 × 10−6 0.1111 0.1111 The overall time in the system, per LL in eqn. (2), is the sum of the CPU residence time (1st row, 3rd column) and the residence time of an IO at the respective storage device. With 1000 disks, N = 111.1 concurrent IOs. With 1000 SSDs, N = 10.1 concurrent IOs. With 1000 FIOs, N = 0.1 concurrent IOs. But wait! It gets even better... c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 27 / 34
2 Throughput-Delay Plots Need for Speed Benchmarking Paradox Paradox Resolved 3 Storage Performance Throughput Latency Concurrency 4 Conclusion c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 28 / 34
33% bigger than query execution time, SCPU . In general, this time can be reduced further with more cores. Disk All 1000 disks have S = 1 ms service time. Residence time is twice the service time. Concurrent IO threads Nio = 111. These threads also have to be managed by the OS (not shown). Threads management also uses up CPU cycles (not shown). Response time = 0.000013 + 0.001111 is dominated by disk latency. SSD Faster “SSD” (10x) with nominal S = 0.1 ms service time. Residence time is now close to service time. Concurrency is also reduced by 10x to N = 10 threads. Response time = 0.000013 + 0.0001010 still dominated by storage latency. FIOa Fusion flash service time S = 1 microsecond. Residence time is equal to the device service time. Concurrent IO threads N = 0.1 are negligible. Response time = 0.000013 + 0.000001 is now CPU-bound. FIOb Bigger message: Don’t need 1000 Fusion flash devices. Small NFIOa = 0.1 means a single FIO device has same IO concurrency. A single Fusion card can replace 1000 standard devices! SAN in your hand c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 29 / 34
R). LL has 2 versions: N = XR (with waiting) and ρ = XS (no waiting). Assume no bandwidth limit and choose throughput target (here, 100 KQPS). With current tech, LL tells us we need parallel devices (disk array, multicore). Storage “latency” (service times) orders of magnitude longer than CPU execution times. The number of outstanding IOs determines the the total (response) time in the system to complete each application query: R = W + S. Rstor Rcpu so, storage latency dominates system response time. If can make Rstor Rcpu, then outstanding IOs become negligible. Application query times determined soley by the CPU execution time. A CPU-bound application is always the optimal goal. Fusion-io also eliminates IO controller latency: all data gets closer to CPU. c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 31 / 34
Relative Technology Persistent Controller Device Cost Disk Yes High High Low SSD Yes High Low High Fusion-IO Yes Low Low High RAM No Low Low Highest c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 32 / 34
Valley, California www.perfdynamics.com perfdynamics.blogspot.com twitter.com/DrQz facebook.com/Performance-Dynamics -Company [email protected] +1-510-537-5758 c 2012 Performance Dynamics Little’s Law in 3D and Storage Performance August 7, 2012 34 / 34
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