20%
01.06.2024
in the /usr/lib/systemd/system/sysstat-collect.timer file. You can set a scan every five minutes by entering *:00/05 instead of the default setting *:00/10:
[Unit]
Description=Run system activity accounting
20%
05.09.2011
-user@nebula-cloud-server:~$ onevm create kvmVM.template
02 nebula-user@nebula-cloud-server:~$ onevm list
03 ID NAME STAT CPU MEM HOSTNAME TIME
04 0 kvmVM pend 0 0 00 00:00
20%
16.05.2013
Recently, a customer asked me what was going on with his system. All of a sudden, he no longer had an eth0; instead, he was seeing strange names like em1 or p3p1 at the console. He wanted to know ... Ethernet devices in Linux have always been called eth0 and nothing else. All of a sudden, this universal truth has lost its validity, and Linux administrators need to understand why and how.
20%
14.03.2013
: 0
16 initial apicid : 0
17 fdiv_bug : no
18 hlt_bug : no
19 f00f_bug : no
20 coma_bug : no
21 fpu : yes
22 fpu_exception : yes
23 cpuid level : 10
24 wp
20%
16.08.2018
: The top command on a Linux system.
15:28:23 up 1 day, 20:10, 3 users, load average: 0.10, 0.14, 0.13
The second line of the display lists the aggregate state of the system's processes – 205 in all
20%
11.04.2016
-enterprise-3.8.1-ubuntu-14.04-amd64/packages/ubuntu-14.04-amd64
sudo dpkg -i pe-cloud-provisioner-libs_0.3.2-1puppet1_amd64.deb
sudo dpkg -i pe-cloud-provisioner_1.2.0-1puppet1_all.deb
vSphere Setup
After
20%
07.06.2019
(dayOfYear):as.factor(wday)Monday 16.64 18 8.382 < 2e-16 ***
s(dayOfYear):as.factor(wday)Saturday 11.29 18 3.307 3.00e-09 ***
s(dayOfYear):as.factor(wday)Sunday 12.92 18 4.843 1.02e-13 ***
---
Signif. codes: 0
20%
17.06.2011
,200, comprising 55 different commands, were issued. The system, a server with 768MB RAM and a Pentium 3 CPU, took a total of 22 seconds to answer them, the longest response took 32 milliseconds, the shortest
20%
11.06.2014
Version
CentOS
6.5
Ganglia
3.6.0
Ganglia web
3.5.12
Confuse
2.7
RRDtool
1.3.8
Before installing any binaries, I try
20%
17.06.2017
-1) = 0.25 * (a(1:n-2,2:n) + a(3:n,2:n) + a(2:n,1:n-2) + a(2:n,3:n))
Using forall, the same can be written as:
forall (i=2:n-1, j=2:n-1) a(i,j) = 0.25*(a(i-1,j) + a(i+1,j) + a(i,j-1) + a(i,j+1
