Calculate with exponents. A lot of back-of-the-envelope calculations are done with just coefficients and exponents, e.g.
Latency Comparison Numbers
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Source: https://gist.github.com/BlackHC/2d0a3a21542b524a7cf2f8eac977481e
Benchmarks for read: https://ssd.userbenchmark.com/, https://hdd.userbenchmark.com/
L1 cache reference 0.5 ns
Branch mispredict 5 ns
L2 cache reference 7 ns 14x L1 cache
Mutex lock/unlock 25 ns
Main memory reference 100 ns 20x L2 cache, 200x L1 cache
Compress 1K bytes with Snappy 3,000 ns 3 µs
Read 1 MB sequentially from memory 20,000 ns 20 us .02 ms ~50GB/s DDR5
Read 1 MB sequentially from NVMe 100,000 ns 100 us .1 ms ~10GB/sec NVMe, 5x memory
Read 1 MB sequentially from SSD 300,000 ns 300 µs .3 ms ~3GB/sec SSD, 15x memory, 3x NVMe
Round trip within same datacenter 500,000 ns 500 us .5 ms
Read 1 MB sequentially from HDD 6,000,000 ns 6,000 µs 6 ms ~150MB/sec, 300x memory, 60x NVMe, 20x SSD
Send 1 MB over 1 Gbps network 10,000,000 ns 10,000 us 10 ms
Disk seek 10,000,000 ns 10,000 µs 10 ms 20x datacenter roundtrip
Send packet CA->Netherlands->CA 150,000,000 ns 150,000 us 150 ms
Notes
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1 ns = 10^-9 seconds
1 us = 10^-6 seconds = 1,000 ns
1 ms = 10^-3 seconds = 1,000 us = 1,000,000 ns
Cost Numbers
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Approximate numbers that should be consistent between Cloud providers.
What Amount $/Month
CPU 1 $10
Memory 1 GB $1
SSD 1 GB $0.1
HDD 1 GB $0.01
S3, GCS 1 GB $0.01
Network 1 GB $0.01
- 1 request per second = 100k requests / day (exact 1 req/s = 86.4k req/day)
- 1 request per second = 2.5M requests / month
- 10 requests per second = 1M requests per day (exact 11.6 req/s = 1M req/day)
- 40 requests per second = 100 million requests per month
- 400 requests per second = 1 billion requests per month
- 6-7 world-wide round trips per second
- 2000 round trips per second within a data center
- 100k commands per second in an in-memory single-threaded data store
- It’s typically the case that we can ignore any memory latency as soon as I/O is involved in a 1Gbps network,
in cloud datacenters bandwidth is capped depending on the instance type, from the
Google Cloud
docs there are different
limits for ingress and egress, for simplicity let’s assume 10Gbps for both.
- C4 and C4A lowest egress is 10Gbps
- C4 and C4A highest egress is 100Gbps
- Writes are 40 times more expensive than reads, therefore architect for scaling writes!
Exercises
We get better at using this table by practicing, https://sirupsen.com/napkin/ has lots of exercises with different difficulty levels. The following exercises are a warmup to the ones in other places.
Let’s assume a data store with the following types:
- In-memory data store: state stored in RAM memory (volatile).
- Persistent data store: state stored in disk (non volatile).
The data store can be located:
- in-process: in the same computer.
- out-of-process: in a different computer (so there’s the need of packet transimission over the network).
Read 1MB from an out-of-process data store, consider both in-memory and persistent caches (SSD), assume a 1Gbps and a 10Gbps network.
- 1Gbps
- (in memory)
0.02 ms/MB (read from memory) + 10^1 ms (transmission) = 10.02 ms
- (persistent)
0.3 ms/MB (read from SSD) + 10^1 ms (transmission) = 10.3 ms
- (in memory)
- 10Gbps
- (in memory)
0.02 ms/MB (read from memory) + 10^0 ms (transmission) = 1.02 ms
- (persistent)
0.3 ms/MB (read from SSD) + 10^0 ms (transmission) = 1.3 ms
- (in memory)
Read 5GB from HDD, SSD and RAM then write 5GB to the same medium. Assume no network IO needed
Read 5GB:
- (memory)
5*10^3 MB * 0.02 ms/MB (memory read) = 100ms = 0.1s
- (SSD)
5*10^3 MB * 0.3 ms/MB (SSD read) = 1500 ms = 1.5s
- (HDD)
5*10^3 MB * 6 ms/MB (HDD read) = 30000 ms = 30s
Write 5GB, let’s assume that a write is 40x slower than a read:
- (memory)
40 (write penalty) * 0.1s (read) = 4s
- (SSD)
40 (write penalty) * 1.5s = 60s
- (HDD)
40 (write penalty) * 30s = 1200s
Store information about 2B users including basic info and a profile picture
- Basic info: name (20 chars), dob (10 chars), email (20 chars) = 50 bytes,
- Profile picture: 100 KB,
Your SSD-backed database has a usage-pattern that rewards you with a 80% page-cache hit-rate (i.e. 80% of disk reads are served directly out of memory instead of going to the SSD). The median number of pages (e.g. InnoDB pages in MySQL) read to serve a query is 50 . What is the expected average query time from your database?
The default size of a page in InnoDB is 16KB , for each query we read 50 pages, 50 * 0.8 = 40 are read from memory and 10 from SSD
- 40 pages read from memory:
40 * 16KB * 0.02 ms/MB = 640KB * 10^-3 MB/KB * 0.02 ms/MB = 0.0128 ms
- 10 pages read from SSD:
10 * 16KB * 0.3 ms/MB = 160KB * 10^-3 MB/KB * 0.3 ms/MB = 0.048ms
In real life we just round the numbers, 1ms tops for the sum. It’s typically the case that we can ignore any memory latency as soon as I/O is involved for low Gbps (1GB).
- 1Gbps
- 50 pages (50 * 16KB = 800KB) transmitted in about 10ms,
1ms (read pages) + 10ms (transmission) = 11ms
- 50 pages (50 * 16KB = 800KB) transmitted in about 10ms,
- 10Gbps
1ms (read pages) + 1ms (transmission) = 2ms
How many commands-per-second can a simple, in-memory, single-threaded data store do? Assume that the commands don’t do any server side processing. e.g. Reading data is just reading data from the memory/disk and isn’t applying any algorithms on it.
I/O controls the number of ops/s, assuming that we transmit 1KB
Amount of computing power to process 1PB everyday, assume that the time required for the computation of 1MB is 0.1s
10^9 MB * 10^01 s/MB = 10^8 MB
- The above has to be computed everyday or in
10^5 s
10^8 s * 10^-5 day/s = 10^3 days
We would need
We have 3 storage devices, a 128GB DRAM as a 1st level cache, a 600GB flash memory as a 2nd level cache and a rotational disk for storage. With a random read workload, the rotational disk delivers 2000 reads/s with an 8 KB I/O size. How much time would it take to warm both caches in the ideal scenario?
- Throughput in terms of data transmitted over time:
2000 reads/s * 8 KB = 16 Mb/s
. - 1st level cache:
- Time to fill out the cache:
128 GB / 16 MB/s = 8000 s = ~2.3h
- Time to fill out the cache:
- 2nd level cache:
- Time to fill out the cache:
600 GB / 16 MB/s = 38400 s = ~10.67h
- Time to fill out the cache: