In Linux, how does one check high load?

I. Load average
1.1: What is load? What is load average?
Load is the measure of how much the computer is working (Wikipedia: The system load is a measure of the amount of work that a compute system is doing)
Simply put, it is the length of the process queue. Load average is the average load within a period of time (1 minute, 5 minutes, 15 minutes. [Reference: Unix load average Part1: How It Works]

1.2: View command:
W or uptime or procinfo or top

Load average: 0.02, 0.27, 0.17
1 per/minute 5 per/minute 15 per/minute

1.3: how to determine if the system has been over load?
Generally, the system determines the number of CPUs. If the average load is always 1.2, and you have two CPUs. There will be no insufficient CPU. That is, the average load is smaller than the number of CPUs.
1.4: Capacity Planning)
Generally, the average load value of 15 minutes is the first.

1.5: Load misunderstanding:
1: system load is always a problem with performance.
Truth: the high load may be due to CPU-intensive computing.
2: when the system load is high, the CPU capacity is faulty or the number is insufficient.
Truth: high load only indicates that too many queues need to be run. However, tasks in the queue may actually consume CPU or I/0 milk.
3: The system loads high for a long time. First, the CPU is increased.
Truth: load is just a representation, not a substance. When you increase the CPU usage, you will see a temporary decrease in load.

Ii. How to identify system bottlenecks when load average is high.
Is there insufficient CPU or I/O speed or insufficient memory?

2.1: view system load vmstat
Vmstat
Procs ----------- memory ---------- --- swap -- ----- Io ---- System -- ---- CPU ----
R B SWPD free buff cache Si so Bi Bo in CS us Sy ID wa
0 0 100152 2436 97200 289740 0 1 34 45 99 33 0 99 0

Procs
The r column indicates the number of processes that run and wait for the CPU time slice. If the number is greater than 1 for a long time, the CPU is insufficient and the CPU needs to be increased.
Column B indicates the number of processes waiting for resources, such as waiting for I/O or memory switching.
CPU indicates the CPU usage status
The US column shows the percentage of CPU time spent in user mode. When the value of us is high, it indicates that the user process consumes a lot of CPU time. However, if it is longer than 50%, you need to consider optimizing your program.
The Sy column shows the percentage of CPU time consumed by the kernel process. Here, the reference value of US + Sy is 80%. If the value of US + Sy is greater than 80%, there may be insufficient CPU.
The Wa column shows the percentage of CPU time occupied by Io wait. The reference value of WA is 30%. If wa exceeds 30%, the IO wait is serious, which may be caused by a large number of random access to the disk, it may also be caused by the bandwidth bottleneck of the disk or disk access controller (mainly block operations ).
The ID column shows the percentage of time when the CPU is idle.
The system displays the number of interruptions during the collection interval.
The In column indicates the number of device interruptions per second observed at a certain interval.
The CS column indicates the number of context switches generated per second. If CS is much higher than the disk I/O and network information packet rate, further investigation should be conducted.
Memory
The number of memories that SWPD switches to the memory swap area (k indicates ). If the value of SWPD is not 0 or is relatively large, for example, if the value exceeds 100 m, as long as the value of Si and so is 0 for a long time, the system performance is still normal.
Free memory in the current idle page list (k indicates)
The amount of memory that buff uses as the buffer cache. Generally, the buffer is required for reading and writing Block devices.
Cache: The amount of memory that is used as the page cache. Generally, it is used as the cache of the file system. If the cache is large, it indicates that there are many files in the cache. If the Bi in Io is small at this time, it indicates that the file system is more efficient.
Swap
Si refers to the number of switches from memory to memory swap zone.
So the number of memory entries in the memory swap zone.
Io
The total amount of data that bi reads from Block devices (read disk) (KB per second ).
Total data written to Bo Block devices (disk write) (KB per second)
Here we set the reference value of Bi + Bo to 1000. If the value of Bi + Bo is greater than 1000 and the Wa value is large, we should consider Disk Load Balancing. We can use iostat output for analysis.

2.2: View Disk Load iostat
Calculate the disk Io information every 2 seconds until you press Ctrl + C to terminate the program. The-D option indicates the disk information statistics, and the-k indicates that the disk information is displayed in KB per second, -T requires that the time information be printed. 2 indicates that the time is output every 2 seconds. The disk I/O Load Status output for the first time provides statistics about the disk I/O load since the system was started. Each subsequent output is the average Io load between each interval.

# Iostat-x 1 10
Linux 2.6.18-92. el5xen 02/03/2009
AVG-CPU: % USER % nice % System % iowait % steal % idle
1.10 0.00 4.82 39.54 0.07 54.46
Device: rrqm/s wrqm/s r/s w/s rsec/s wsec/s avgrq-SZ avgqu-SZ await svctm % util
SDA 0.00 3.50 0.40 2.50 5.60 48.00 18.48 0.00 0.97 0.97 0.28
SDB 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
SDC 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
SDD 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
SDE 0.00 0.10 0.30 0.20 2.40 2.40 9.60 0.00 1.60 1.60 0.08
SDF 17.40 0.50 102.00 0.20 12095.20 5.60 118.40 0.70 6.81 2.09 21.36
SDG 232.40 1.90 379.70 0.50 76451.20 19.20 201.13 4.94 13.78 2.45 93.16
Rrqm/s: the number of merge read operations per second. That is, Delta (rmerge)/s
Wrqm/S: Number of write operations performed on merge per second. That is, Delta (wmerge)/s
R/S: The number of read I/O devices per second. That is, Delta (Rio)/s
W/s: the number of write I/O devices completed per second. That is, Delta (WIO)/s
Rsec/S: Number of read sectors per second. That is, Delta (rsect)/s
Wsec/S: Number of write sectors per second. That is, Delta (wsect)/s
RKb/s: the number of bytes read per second. It is half of rsect/s because the size of each slice is 512 bytes. (Computing required)
WKB/s: the number of K bytes written per second. Half of wsect/s. (Computing required)
Avgrq-SZ: average data size (slice) of each device I/O operation ). Delta (rsect + wsect)/DELTA (Rio + WIO)
Avgqu-SZ: Average I/O queue length. That is, Delta (aveq)/S/1000 (because aveq is measured in milliseconds ).
Await: average wait time (in milliseconds) for each device I/O operation ). That is, Delta (ruse + wuse)/DELTA (Rio + WIO)
Svctm: Average service time (in milliseconds) for each device I/O operation ). That is, Delta (use)/DELTA (Rio + WIO)
% Util: the percentage of time in one second is used for I/O operations, or the number of I/O queues in one second is not empty. That is, Delta (use)/S/1000 (because the Unit of use is milliseconds)

If % util is close to 100%, it indicates that too many I/O requests are generated and the I/O system is fully loaded.
There may be bottlenecks.
When the idle is less than 70% I/O, the load is high. Generally, the read speed is wait.

You can also view the parameters B (number of processes waiting for resources) and Wa in combination with vmstat (percentage of CPU time occupied by I/O wait, higher than 30% when I/O pressure is high)

For more information, see
General:
Svctm <await (because the wait time of the simultaneously waiting request is calculated repeatedly ),
The size of the svctm is generally related to the disk performance: the CPU/memory load will also affect it, and too many requests will indirectly lead to the increase of the svctm.
Await: The size of await generally depends on the service time (svctm), the length of the I/O queue, and the mode in which I/O requests are sent.
If svctm is close to await, I/O has almost no waiting time;
If await is much larger than svctm, it means that the I/O queue is too long and the response time of the application is slow,
If the response time exceeds the allowable range, you can consider replacing a faster disk, adjusting the kernel elevator algorithm, optimizing the application, or upgrading the CPU.
The queue length (avgqu-sz) can also be used as an indicator to measure the system I/O load. However, because avgqu-SZ is based on the average per unit time, therefore, it cannot reflect the instantaneous I/O flood.

A good example for others (I/O system vs. Supermarket Queue)
For example, how can we decide which payment platform to pay when queuing for checkout at the supermarket? First, let's look at the number of people in the queue. is the total number of five people faster than 20? In addition to the number of people, we often look at the number of items purchased by the previous person. If there is a big mom who has purchased food for a week, we can consider changing the team. There is also the cashier's speed. If you have a newbie who doesn't even know about the money, you will have to wait. In addition, the timing is also very important. It may be 5 minutes ago, but the money was crowded, and now people go to the building, but it is refreshing to pay the money. Of course, the premise is that what has been done in the past five minutes is more meaningful than queuing (but I have not found anything boring than queuing ).
I/O systems have many similarities with supermarket Queues:
R/S + w/s is similar to the total number of payers
The average queue length (avgqu-sz) is similar to the number of average queues per unit time.
The average service time (svctm) is similar to the cashier's collection speed
The average wait time (await) is similar to the average wait time of each person
Average I/O data (avgrq-sz) is similar to the average number of items bought by each person
The I/O operation rate (% util) is similar to the time ratio in which someone queues before the cashier.
We can analyze the I/O Request mode and the I/O speed and response time based on the data.
The following is an analysis of the output of this parameter written by someone else.
# Iostat-X 1
AVG-CPU: % USER % nice % sys % idle
16.24 0.00 4.31 79.44
Device: rrqm/s wrqm/s r/s w/s rsec/s wsec/s RKb/s WKB/s avgrq-SZ avgqu-SZ await svctm % util
/Dev/CCISS/c0d0
0.00 44.90 1.02 27.55 8.16 579.59 4.08 289.80 20.57 22.35 78.21 5.00 14.29
/Dev/CCISS/c0d0p1
0.00 44.90 1.02 27.55 8.16 579.59 4.08 289.80 20.57 22.35 78.21 5.00 14.29
/Dev/CCISS/c0d0p2
0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
The iostat output above indicates that 28.57 device I/O operations are performed in seconds: Total IO/S = R/S (read) + w/s (write) = 1.02 + 27.55 = 28.57 (times/second), where the write operation occupies the main body (W: r = 27 ).
On average, only 5 ms is required for each device I/O operation, but 78 MS is required for each I/O request. Why? Because too many I/O requests are sent (about 29 requests per second), assuming these requests are sent at the same time, the average wait time can be calculated as follows:
Average wait time = single I/O service time * (1 + 2 +... + total number of requests-1)/total number of requests
Application to the above example: average wait time = 5 ms * (1 + 2 +... + 28)/29 = 70 ms, which is very close to the average waiting time of 78 MS given by iostat. This in turn indicates that I/O is initiated at the same time.
There are many I/O requests per second (about 29), but the average queue is not long (only about 2). This indicates that the arrival of these 29 requests is uneven, i/O is idle most of the time.
In one second, 14.29% of the time I/O queues have requests. That is to say, the I/O system has nothing to do in 85.71% of the time, all 29 I/O requests are processed within 142 milliseconds.
Delta (ruse + wuse)/DELTA (IO) = await = 78.21 => delta (ruse + wuse)/S = 78.21 * delta (IO) /S = 78.21*28.57 = 2232.8, indicating that I/O requests per second need to wait for a total of 2232.8 Ms.
Therefore, the average queue length should be 2232.8 ms/1000 ms = 2.23, while the average queue length (avgqu-sz) provided by iostat is 22.35. Why ?! Because there is a bug in iostat, The avgqu-SZ value should be 2.23 instead of 22.35.