Question

假设邮政管理处正在使用64台借机服务器,在t4(32轨)和t8(64轨)一栏之间业绩差异如何? 该手册指出,t4比t8更有效率,但如果基础服务器为64比值,是否存在着实际业绩差异(从(1)小盘、(2)记忆和(3)储存而言)?

Answer 1

in terms of (1) cpu, (2) memory and (3) storage

::

64个比值是32个比值的两倍。
64个比值是32个比值的两倍。
64个比值是32个比值的两倍。

我回顾一下,有几条基准的堆积器read。编制一个表格,填补100万分。然后发现、增加、群体、加入等。我不记得这些具体内容,但使用无动于衷8比无动于4。

test=# create table int4_test (id int primary key);
CREATE TABLE
test=# create table int8_test (id bigint primary key);
CREATE TABLE
test=# insert into int4_test select i from generate_series(1,1000000) i;
INSERT 0 1000000
test=# insert into int8_test select i from generate_series(1,1000000) i;
INSERT 0 1000000
test=# vacuum analyze;
VACUUM
test=# 	iming on
Timing is on.
test=# select sum(i.id) from int4_test i natural join int4_test j where i.id % 19 = 0;
     sum     
-------------
 26315710524
(1 row)

Time: 1364.925 ms
test=# select sum(i.id) from int4_test i natural join int4_test j where i.id % 19 = 0;
     sum     
-------------
 26315710524
(1 row)

Time: 1286.810 ms
test=# select sum(i.id) from int8_test i natural join int8_test j where i.id % 19 = 0;
     sum     
-------------
 26315710524
(1 row)

Time: 1610.638 ms
test=# select sum(i.id) from int8_test i natural join int8_test j where i.id % 19 = 0;
     sum     
-------------
 26315710524
(1 row)

Time: 1554.066 ms

test=# select count(*) from int4_test i natural join int4_test j where i.id % 19 = 0;
 count 
-------
 52631
(1 row)

Time: 1244.654 ms
test=# select count(*) from int4_test i natural join int4_test j where i.id % 19 = 0;
 count 
-------
 52631
(1 row)

Time: 1247.114 ms
test=# select count(*) from int8_test i natural join int8_test j where i.id % 19 = 0;
 count 
-------
 52631
(1 row)

Time: 1541.751 ms
test=# select count(*) from int8_test i natural join int8_test j where i.id % 19 = 0;
 count 
-------
 52631
(1 row)

Time: 1519.986 ms

Answer 2

在储存和记忆方面,答案是显而易见的: INT8与INT4一样大两倍,因此它两次使用储存和两次记忆。

在计算性能方面,我怀疑计算结果在64台轨道机器上根本不会改变,在某些情况下,INT4在32台轨道机器上可能更有效率。尽管除非你在这些国际会计科目上做复杂的数学(而不仅仅是在序列上使用),但计算差异可能为零或几乎为零。

当你开始与你的信使做复杂的事时,这实际上不是数据库的绩效问题。

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