"There are 10 kinds of people in the world ...those who understand binary and those who do not"
There are four basic numbering systems that are used in the human, networking, and computer science worlds. Binary, octal, decimal and hexadecimal. Decimal, the numbering system we are most familiar with has a base, or radix, of 10 and to us this is natural. Why 10? Most of us have 10 fingers and that is how it was selected as our numbering system; it's as simple as that. The Yuki people of California on the other hand (pun intended) use an octal numbering system, counting the spaces between their fingers and not the fingers themselves. Pretty cool, huh?
The other systems widely in use in the networking and computer science world are binary, octal and hexadecimal (See table below). In this tutorial we'll concern ourselves only with binary, decimal and hexadecimal, ignoring octal altogether. As a side note, one use of octal numbering is with Linux and UNIX systems to set file permissions using the chmod command.
Yes, we're talking binary here. I start here by asking that you don't think of a binary number as being a just series of 1's and 0's, such as 01000001 (65 decimal), but rather as a valid numbering system. So, why binary and not decimal? Take our previous example of 01000001. It's really a sort of computer Morse code, but instead of dots and dashes, it's the presence or absence of state, on or off; whether it be electrical pulses on an Cat5 cable or the state of transistors within a CPU or memory stick. Individually a bit, that is a 1 or 0, is pretty useless. However, put them together in a byte (8 bits) for example, then you have something. With our previous example, the byte 01000001 represents the character 'A'. With the computer's ability to process and transmit millions of bytes in miliseconds it can give us letters, words, pictures, music, videos and oodles of other things almost instantly. No matter what we type or create via keyboard or mouse, eventually it all gets translated into a numerical series of 1's and 0's to represent our intent. So, you see it IS all just 1's and 0's! Oh, I almost forgot ...a nibble is half a byte, 4 bits.
With the decimal numbering system we have 10 digits to work with, 0 thru 9 and when we count we start at 0, then 1, then 2 and so on until we reach 9. Then what? Well, we've run out of digits. So, we start again at 0, but we place a 1 to the left giving us 10 (ten). And once the 1's and 10's columns reach 9 and 9 (99) they both flip, giving us 100. Same thing goes with binary except we only have 2 digits to work with, 1 and 0.
The principal is the same we start at 0 then 1, then ...well, we've run out of digits haven't we? So, we flip that 1 to 0 and place a 1 to the left giving us 10 (two). Increment by 1 again and we have 11 (three) and once again they both flip and we move into a new column giving us 100 (four). This is probably not so bad for small numbers, but once you get past 8 or so binary digits it becomes unmanageable. I mean what does 10011100 equate to in human terms? Don't even mention that binary numbers are unwieldy; 3,000 is 101110111000 in binary ...geez! So, while it is true that we need to think in binary, we can represent these binary numbers in a more human friendly form. Anyone say decimal? Fortunately we have the ability (and absolute need) to convert between the two.
For this section I'll start by examining and explaining the table below.
First thing to take note of is that there are 8 individual bits giving us a byte. When working with IP addresses, networking folks refer to this as an octet. Obviously, binary numbers can be larger than a simple byte. However, the byte is the most basic representation of data and as such, for the most part, we'll stay at the byte level.
The 'Bit Set' row is for our binary digits, either 0 or 1. Here they are all 1.
The 'Exponent Value' row is how we arrive mathematically at the 'Decimal Value' row. An exponential value such as 23 means 2 x 2 x 2 = 8. By the way, any number with an exponent of 0 ALWAYS equates to 1 (20 = 1 and 4000 also = 1).
The 'Decimal Value' row represents positionally the decimal value of the binary digit in question. It's imperative that you commit this row to memory. Practice by writing it down over and over 'till its part of your DNA :-)
The labels 'MSB' and 'LSB' stand for 'Most Significant Bit' and 'Least Significant Bit' respectively. This lets us know which end of the binary number we are talking about. For example, if I had a binary number of 11010101 and I say 'Starting at bit 1' ...that might be confusing. However, if I say 'Starting with the bit at LSB 1'. Well, you get the picture.
Previously we talked about thinking in binary means knowing how to count in binary and that still holds valid, but now, in order to convert between the two, we are going to only concern ourselves with the position of each binary digit as it relates to its decimal equivalent. What?? OK, take for example the binary number 00001010. Referring to the table above just add the positional decimal equivalents where the 1's occur. So, 00001010 means 128 (no), 64 (no), 32 (no), 16 (no), 8 (yes), 4 (no), 2 (yes) 1 (no). Given that, we see that 8 + 2 =10. In other words 00001010 binary is 10 decimal. Here are a few more examples.
11000000 - 128 + 64 = 192
00000001 - = 1
00110011 - 32 + 16 + 2 + 1 = 51
11011011 - 128 + 64 + 16 + 8 + 2 + 1 = 219
00000000 - = 0
11111111 - 128 + 64 + 32 +16 + 8 + 4 + 2 + 1 = 255
If you encounter a binary number less than a byte such as 11001 (25 decimal), just pad the missing binary digits with 0's, making the number 00011001, a complete byte. Also, if you are presented with a binary number larger than a byte then double the decimal number with each subsequent binary position to the left. See table below. Later, when we work with subnetting, we'll use these larger values.
At this point you should be comfortable with converting binary to decimal. Now let's go the other way, decimal to binary. I'll show you two different methods. Use the one that best works for you.
This method is pretty straight forward. The first thing to do is write down the decimal value row previously covered.
Next, somewhere off to the side, write down your decimal number. With a byte, or octet, the smallest number is 0 and the largest 255, so your number must be within this range. Next, follow the procedure below:
Compare the decimal number with the MSB (128). If your number is larger than or equal to 128 then place a 1 under the 128 column, subtract 128 from your number and move to the next position (64). However, if your number is less than 128 then place a 0 under the 128 column and move to the next number (64) without subtracting.
Repeat the above process with 64, 32, 16 etc. until your original number is reduced to zero. If you reach zero before reaching the LSB (1) fill in the remaining columns with 0's. below are a few examples:
This method has you repeatedly dividing a decimal number by 2 and saving the remainder, either 1 or 0, as a binary digit. Keep doing this until, again, your original number has been reduced to 0. It's not as confusing as it sounds. Let's walk through an example:Refer to the example above. Given the decimal number 41 and the divide-and-conquer method, we derive its binary equivalent with the following logic:
41 divided by 2 is 20, remainder 1. Place a 1 in the remainder column and 20 in the division column. This is the LSB and will be the rightmost binary digit.
20 divided by 2 is 10, remainder 0. Place a 0 in the remainder column and 10 in the division column.
10 divided by 2 is 5, remainder 0. Place a 0 in the remainder column and 5 in the division column.
5 divided by 2 is 2, remainder 1. Place a 1 in the remainder column and 2 in the division column.
2 divided by 2 is 1, remainder 0. Place a 0 in the remainder column and 1 in the division column.
1 divided by 2 is 0, remainder 1. Place a 1 in the remainder column and we're done ...
... well almost. Our solution came out to be only 6 binary digits long. Since we are working with bytes, just pad the left with 0's until the binary number is 8 digits in length. One more thing. Your last operation will always be 1/2 = 0, remainder 1. Here a couple more examples:
At times you might have a binary or decimal that is larger than a byte. Don't panic. All you need to do is extend the binary byte row to the left, doubling as you go along.
now becomes the following if we have a 12 bit binary number:
Here is a sample conversion from binary to decimal using process of elimination:
Just when you got your head around binary and decimal, along comes hexadecimal, hex for short. Before, I mentioned that everything in the computing and networking arena is nothing more that 1's and 0's and that axiom remains unchanged. We now know how to convert back and forth between binary (what the computer speaks) and decimal (what we speak), so why do we even need hex? Besides being an efficient numbering system, hexadecimal is widely used to represent to us humans the 1's and 0's that traverse circuits and wires. A few examples where hex numbers are used are colors, memory references, IPv6 addresses and MAC addresses. Let's explore this a little more in detail.
A binary byte is exactly eight digits long, e.g. 00101110 and a hex byte number is exactly two digits in length, e.g. 6A. A decimal byte on the other hand is anywhere between one and three digits long, e.g. 4, 64 or 233. Yes, you could pad these with 0's making 004, 064 and 233, but this is still not as efficient or as practical as hex. (as we'll see when we start converting hex and binary numbers). Download myextended ASCII table; it will help with the following illustration and serve as a useful refernece for your studies:
This screenshot is a memory dump of my computer. Imagine if those numbers were represented in binary!
Let's get on with it.
Below is a hex table with decimal equivalents:
With hex, the first 10 digits, 0 thru 9, are the same as decimal. However, 10 thru 15 decimal are represented as a single hex digit A thru F. The number 5 in decimal is 5 in hex and the number 13 in decimal is D in hex. What happens when we reach F (15 decimal) and need to increment by 1 is that, just like with all other numbering systems, we flip that to 0 and place a one to the left. 0xF + 0x1 = 0x10. What is this 0x business? Placing 0x in front of a hex number is the typical way to indicate it's really hex. It makes it clear that 11 (decimal) and 0x11 (hex) are two different numbers entirely. So, from here on out I'll use the 0x to indicate hexadecimal numbers. Sometimes you'll see the hex numbers 0xA thru 0xF written in lower case (0xa thru 0xf). There's nothing wrong with that and it does not change the numerical value of the number in question. I use caps in this how-to.
To convert a binary byte to hex byte, first split the binary number into two nibbles, treating them as separate numbers, and then compute the hex equivalent for each half. Finally, concatenate the two hex numbers into a single solution. This works perfectly because a nibble has a value from 0 thru 15 or 0x0 thru 0xF.
Here's some examples:
The conversion process from hex to binary is not much different, we just reverse the process. A picture is worth a thousand words. Here's 4,000 of them.
Once in a while you may find yourself presented with a binary or hex number larger than a byte. The methods are the same, just expanded. If you have a binary number to convert then you may have to pad the MSB with 0's in order to have a binary number that falls on nibble boundaries, multiples of 4. For example, the binary number 1001100101 won't convert correctly, because it's ten digits in length. Add two 0's making the number 001001100101. A few illustrations:
Up to now we've been able to accomplish our conversions armed with only paper and pencil. It's time to break out your calculator for converting decimal to hex and vice versa. Take Look at these decimal / hex equivalents:
16 = 0x10 | 34 = 0x22 | 214 = 0xD6 | 175 = 0xAF
Here's how we get from the decimal value on the left of the = sign to its hexadecimal equivalent on the right. Remember the divide and conquer method we used for converting decimal to binary? We are going to use the same method except that we are going to be dividing by 16 rather than 2. The upside is that the process is a lot shorter.
225 divided by 16 is 14, remainder 0x1. Place a 0x1 in the remainder column and 14 in the division column. This is our rightmost hex digit.
14 divided by 16 is 0, remainder 14. Since we are converting to hex we use 0xE. Place 0xE in the remainder column and we're done.
If you have a small decimal number such as 12, you'll have your answer on the first divide. 12 / 16 = 0, remainder 12 or 0xC. Place a leading 0 in from of the C to keep things uniform. So, the answer would be 0x0C
How easy is that? Here are a few more:
It's no different with numbers larger than a byte:
Whew! Nearly done. This is our last number conversion segment, hexadecimal to decimal. With a hex number each position represents a power of 16 and what we do is calculate each position's decimal equivalent and then add them all up. We are going to use a large hex number to illustrate this concept. Given the hex number 0xA59C we find its decimal equivalent with the following logic.
(10 x 163) + (5 x 162) + (9 x 161) + (12 x 160) = 43,296. Here's how it breaks down:
...And a few more:
I encourage you to practice, practice, practice. Download this worksheet to work thru various conversion exercises.
Parting shot - If you are an aspiring network or systems admin then make sure you understand without question binary and decimal conversion through and through.
Coming soon - Part II Understanding IP addresses, subnet masks and subnetting.
./Robert
浠ュ墠涓鐩翠笉澶細鐢ㄨ繖涓弬鏁般傜幇鍦ㄨ鐪熺爺絀朵簡涓涓媔ostat錛屽洜涓哄垰濂芥湁鍙伴噸瑕佺殑鏈嶅姟鍣ㄥ帇鍔涢珮,鎵浠ユ斁涓婃潵鍒嗘瀽涓涓?涓嬮潰榪欏彴灝辨槸IO鏈夊帇鍔涜繃澶х殑鏈嶅姟鍣?/p>
$iostat -x 1 Linux 2.6.33-fukai (fukai-laptop) _i686_ (2 CPU) avg-cpu: %user % nice %system %iowait %steal %idle 5.47 0.50 8.96 48.26 0.00 36.82 Device: rrqm /s wrqm /s r /s w /s rsec /s wsec /s avgrq-sz avgqu-sz await svctm %util sda 6.00 273.00 99.00 7.00 2240.00 2240.00 42.26 1.12 10.57 7.96 84.40 sdb 0.00 4.00 0.00 350.00 0.00 2068.00 5.91 0.55 1.58 0.54 18.80 |
rrqm/s: 姣忕榪涜 merge 鐨勮鎿嶄綔鏁扮洰錛堝綋緋葷粺璋冪敤闇瑕佽鍙栨暟鎹殑鏃跺欙紝VFS灝嗚姹傚彂鍒板悇涓狥S錛屽鏋淔S鍙戠幇涓嶅悓鐨勮鍙栬姹傝鍙栫殑鏄浉鍚孊lock鐨勬暟鎹紝FS浼氬皢榪欎釜璇鋒眰鍚堝茍Merge錛夈傚嵆 delta(rmerge)/s
wrqm/s: 姣忕榪涜 merge 鐨勫啓鎿嶄綔鏁扮洰銆傚嵆 delta(wmerge)/s
r/s: 姣忕瀹屾垚鐨勮 I/O 璁懼嬈℃暟銆傚嵆 delta(rio)/s
w/s: 姣忕瀹屾垚鐨勫啓 I/O 璁懼嬈℃暟銆傚嵆 delta(wio)/s
rsec/s: 姣忕璇繪墖鍖烘暟銆傚嵆 delta(rsect)/s
wsec/s: 姣忕鍐欐墖鍖烘暟銆傚嵆 delta(wsect)/s
rkB/s: 姣忕璇籏瀛楄妭鏁般傛槸 rsect/s 鐨勪竴鍗婏紝鍥犱負姣忔墖鍖哄ぇ灝忎負512瀛楄妭銆?闇瑕佽綆?
wkB/s: 姣忕鍐橩瀛楄妭鏁般傛槸 wsect/s 鐨勪竴鍗娿?闇瑕佽綆?
avgrq-sz: 騫沖潎姣忔璁懼I/O鎿嶄綔鐨勬暟鎹ぇ灝?(鎵囧尯)銆俤elta(rsect+wsect)/delta(rio+wio)
avgqu-sz: 騫沖潎I/O闃熷垪闀垮害銆傚嵆 delta(aveq)/s/1000 (鍥犱負aveq鐨勫崟浣嶄負姣)銆?br />await: 騫沖潎姣忔璁懼I/O鎿嶄綔鐨勭瓑寰呮椂闂?(姣)銆傚嵆 delta(ruse+wuse)/delta(rio+wio)
svctm: 騫沖潎姣忔璁懼I/O鎿嶄綔鐨勬湇鍔℃椂闂?(姣)銆傚嵆 delta(use)/delta(rio+wio)
%util: 涓縐掍腑鏈夌櫨鍒嗕箣澶氬皯鐨勬椂闂寸敤浜?I/O 鎿嶄綔錛屾垨鑰呰涓縐掍腑鏈夊灝戞椂闂?I/O 闃熷垪鏄潪絀虹殑銆傚嵆 delta(use)/s/1000 (鍥犱負use鐨勫崟浣嶄負姣)
濡傛灉 %util 鎺ヨ繎 100%錛岃鏄庝駭鐢熺殑I/O璇鋒眰澶錛孖/O緋葷粺宸茬粡婊¤礋鑽鳳紝璇ョ鐩?br />鍙兘瀛樺湪鐡墮銆?br />idle灝忎簬70% IO鍘嬪姏灝辮緝澶т簡,涓鑸鍙栭熷害鏈夎緝澶氱殑wait.
鍚屾椂鍙互緇撳悎vmstat 鏌ョ湅鏌ョ湅b鍙傛暟(絳夊緟璧勬簮鐨勮繘紼嬫暟)鍜寃a鍙傛暟(IO絳夊緟鎵鍗犵敤鐨凜PU鏃墮棿鐨勭櫨鍒嗘瘮,楂樿繃30%鏃禝O鍘嬪姏楂?/font>)
鍙﹀ await 鐨勫弬鏁頒篃瑕佸鍜?svctm 鏉ュ弬鑰冦傚樊鐨勮繃楂樺氨涓瀹氭湁 IO 鐨勯棶棰樸?br />avgqu-sz 涔熸槸涓仛 IO 璋冧紭鏃墮渶瑕佹敞鎰忕殑鍦版柟錛岃繖涓氨鏄洿鎺ユ瘡嬈℃搷浣滅殑鏁版嵁鐨勫ぇ灝忥紝濡傛灉嬈℃暟澶氾紝浣嗘暟鎹嬁鐨勫皬鐨勮瘽錛屽叾瀹?IO 涔熶細寰堝皬.濡傛灉鏁版嵁鎷跨殑澶э紝鎵岻O 鐨勬暟鎹細楂樸備篃鍙互閫氳繃 avgqu-sz × ( r/s or w/s ) = rsec/s or wsec/s.涔熷氨鏄錛岃瀹氶熷害鏄繖涓潵鍐沖畾鐨勩?/strong>
鍙﹀榪樺彲浠ュ弬鑰?/strong> 涓句竴涓緥瀛愶紝鎴戜滑鍦ㄨ秴甯傛帓闃?checkout 鏃訛紝鎬庝箞鍐沖畾璇ュ幓鍝釜浜ゆ鍙板憿? 棣栧綋鏄湅鎺掔殑闃熶漢鏁幫紝5涓漢鎬繪瘮20浜鴻蹇惂? 闄や簡鏁頒漢澶達紝鎴戜滑涔熷父甯哥湅鐪嬪墠闈漢璐拱鐨勪笢瑗垮灝戯紝濡傛灉鍓嶉潰鏈変釜閲囪喘浜嗕竴鏄熸湡椋熷搧鐨勫ぇ濡堬紝閭d箞鍙互鑰冭檻鎹釜闃熸帓浜嗐傝繕鏈夊氨鏄敹閾跺憳鐨勯熷害浜嗭紝濡傛灉紕頒笂浜嗚繛 閽遍兘鐐逛笉娓呮鐨勬柊鎵嬶紝閭e氨鏈夌殑絳変簡銆傚彟澶栵紝鏃舵満涔熷緢閲嶈錛屽彲鑳?5 鍒嗛挓鍓嶈繕浜烘弧涓烘?zhèn)g殑鏀秼茟謴板Q岀幇鍦ㄥ凡鏄漢鍘繪ゼ絀猴紝榪欐椂鍊欎氦嬈懼彲鏄緢鐖藉晩錛屽綋鐒訛紝鍓嶆彁鏄偅榪囧幓鐨?5 鍒嗛挓閲屾墍鍋氱殑浜嬫儏姣旀帓闃熻鏈夋剰涔?(涓嶈繃鎴戣繕娌″彂鐜頒粈涔堜簨鎯呮瘮鎺掗槦榪樻棤鑱婄殑)銆?/p>
I/O 緋葷粺涔熷拰瓚呭競鎺掗槦鏈夊緢澶氱被浼間箣澶? r/s+w/s 綾諱技浜庝氦嬈句漢鐨勬繪暟 鎴戜滑鍙互鏍規(guī)嵁榪欎簺鏁版嵁鍒嗘瀽鍑?I/O 璇鋒眰鐨勬ā寮忥紝浠ュ強 I/O 鐨勯熷害鍜屽搷搴旀椂闂淬?/p>
涓嬮潰鏄埆浜哄啓鐨勮繖涓弬鏁拌緭鍑虹殑鍒嗘瀽 涓婇潰鐨?iostat 杈撳嚭琛ㄦ槑縐掓湁 28.57 嬈¤澶?I/O 鎿嶄綔: 鎬籌O(io)/s = r/s(璇? +w/s(鍐? = 1.02+27.55 = 28.57 (嬈?縐? 鍏朵腑鍐欐搷浣滃崰浜嗕富浣?(w:r = 27:1)銆?/p>
騫沖潎姣忔璁懼 I/O 鎿嶄綔鍙渶瑕?5ms 灝卞彲浠ュ畬鎴愶紝浣嗘瘡涓?I/O 璇鋒眰鍗撮渶瑕佺瓑涓?78ms錛屼負浠涔? 鍥犱負鍙戝嚭鐨?I/O 璇鋒眰澶 (姣忕閽熺害 29 涓?錛屽亣璁捐繖浜涜姹傛槸鍚屾椂鍙戝嚭鐨勶紝閭d箞騫沖潎絳夊緟鏃墮棿鍙互榪欐牱璁$畻: 騫沖潎絳夊緟鏃墮棿 = 鍗曚釜 I/O 鏈嶅姟鏃墮棿 * ( 1 + 2 + … + 璇鋒眰鎬繪暟-1) / 璇鋒眰鎬繪暟 搴旂敤鍒頒笂闈㈢殑渚嬪瓙: 騫沖潎絳夊緟鏃墮棿 = 5ms * (1+2+…+28)/29 = 70ms錛屽拰 iostat 緇欏嚭鐨?8ms 鐨勫鉤鍧囩瓑寰呮椂闂村緢鎺ヨ繎銆傝繖鍙嶈繃鏉ヨ〃鏄?I/O 鏄悓鏃跺彂璧風殑銆?/p>
姣忕鍙戝嚭鐨?I/O 璇鋒眰寰堝 (綰?29 涓?錛屽鉤鍧囬槦鍒楀嵈涓嶉暱 (鍙湁 2 涓?宸﹀彸)錛岃繖琛ㄦ槑榪?29 涓姹傜殑鍒版潵騫朵笉鍧囧寑錛屽ぇ閮ㄥ垎鏃墮棿 I/O 鏄┖闂茬殑銆?/p>
涓縐掍腑鏈?14.29% 鐨勬椂闂?I/O 闃熷垪涓槸鏈夎姹傜殑錛屼篃灝辨槸璇達紝85.71% 鐨勬椂闂撮噷 I/O 緋葷粺鏃犱簨鍙仛錛屾墍鏈?29 涓?I/O 璇鋒眰閮藉湪142姣涔嬪唴澶勭悊鎺変簡銆?/p>
delta(ruse+wuse)/delta(io) = await = 78.21 => delta(ruse+wuse)/s =78.21 * delta(io)/s = 78.21*28.57 = 2232.8錛岃〃鏄庢瘡縐掑唴鐨処/O璇鋒眰鎬誨叡闇瑕佺瓑寰?232.8ms銆傛墍浠ュ鉤鍧囬槦鍒楅暱搴﹀簲涓?2232.8ms/1000ms = 2.23錛岃?iostat 緇欏嚭鐨勫鉤鍧囬槦鍒楅暱搴?(avgqu-sz) 鍗翠負 22.35錛屼負浠涔?! 鍥犱負 iostat 涓湁 bug錛宎vgqu-sz 鍊煎簲涓?2.23錛岃屼笉鏄?22.35銆?/p>
svctm 涓鑸灝忎簬 await (鍥犱負鍚屾椂絳夊緟鐨勮姹傜殑絳夊緟鏃墮棿琚噸澶嶈綆椾簡)錛宻vctm 鐨勫ぇ灝忎竴鑸拰紓佺洏鎬ц兘鏈夊叧錛孋PU/鍐呭瓨鐨勮礋鑽蜂篃浼氬鍏舵湁褰卞搷錛岃姹傝繃澶氫篃浼氶棿鎺ュ鑷?svctm 鐨勫鍔犮俛wait 鐨勫ぇ灝忎竴鑸彇鍐充簬鏈嶅姟鏃墮棿(svctm) 浠ュ強 I/O 闃熷垪鐨勯暱搴﹀拰 I/O 璇鋒眰鐨勫彂鍑烘ā寮忋傚鏋?svctm 姣旇緝鎺ヨ繎 await錛岃鏄?I/O 鍑犱箮娌℃湁絳夊緟鏃墮棿錛涘鏋?await 榪滃ぇ浜?svctm錛岃鏄?I/O 闃熷垪澶暱錛屽簲鐢ㄥ緱鍒扮殑鍝嶅簲鏃墮棿鍙樻參錛屽鏋滃搷搴旀椂闂磋秴榪囦簡鐢ㄦ埛鍙互瀹硅鐨勮寖鍥達紝榪欐椂鍙互鑰冭檻鏇存崲鏇村揩鐨勭鐩橈紝璋冩暣鍐呮牳 elevator 綆楁硶錛屼紭鍖栧簲鐢紝鎴栬呭崌綰?CPU銆?br />闃熷垪闀垮害(avgqu-sz)涔熷彲浣滀負琛¢噺緋葷粺 I/O 璐熻嵎鐨勬寚鏍囷紝浣嗙敱浜?avgqu-sz 鏄寜鐓у崟浣嶆椂闂寸殑騫沖潎鍊鹼紝鎵浠ヤ笉鑳藉弽鏄犵灛闂寸殑 I/O 媧按銆?/p>
鍒漢涓涓笉閿欑殑渚嬪瓙.(I/O 緋葷粺 vs. 瓚呭競鎺掗槦)
騫沖潎闃熷垪闀垮害(avgqu-sz)綾諱技浜庡崟浣嶆椂闂撮噷騫沖潎鎺掗槦浜虹殑涓暟
騫沖潎鏈嶅姟鏃墮棿(svctm)綾諱技浜庢敹閾跺憳鐨勬敹嬈鵑熷害
騫沖潎絳夊緟鏃墮棿(await)綾諱技浜庡鉤鍧囨瘡浜虹殑絳夊緟鏃墮棿
騫沖潎I/O鏁版嵁(avgrq-sz)綾諱技浜庡鉤鍧囨瘡浜烘墍涔扮殑涓滆タ澶氬皯
I/O 鎿嶄綔鐜?(%util)綾諱技浜庢敹嬈懼彴鍓嶆湁浜烘帓闃熺殑鏃墮棿姣斾緥銆?/p>
# iostat -x 1
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
select * from user_tab_subpartitions t where t.table_name = upper('tablename');
select * from user_tab_partitions t where t.table_name = upper('tablename');