Exam 156-315.80 topic 1 question 32 discussion

should be fw ctl affinity -l -a -v -r.

Had this question today in exam.

Cmd usage: ＃fw ctl affinity List affinity (specific target or all): fw ctl affinity -l [ <proc_selection> | [-r] [-a] ] [-v] -a - All, list also targets without specific affinity. -r - Reverse, list targets ordered by cpu. -v - Verbose, print more info. -q - Quiet, suppress errors. proc_selection: { -p <pid> | -n <name> | -k <instance_id> | -i <interface> | -d} (in case of virtual device, no identifier is needed, current context will be used) -p - Process ID (will not survive reboot). -n - CP daemon name (e.g. fwd). -k - User space instance number. -i - Interface name. Sample output of the cmd: ＃fw ctl affinity -l -a -v -r CPU 0: eth0 (irq 19) eth1 (irq 16) eth2 (irq 17) eth3 (irq 18) eth4 (irq 19) eth5 (irq 16) fw_1 CPU 1: fw_0 All: mpdaemon fwd rtmd pdpd in.asessiond lpd pepd rad vpnd cprid cpd

""l ""a ""r ""v ....... This is a bit of a copy-paste from another source guys (e.g. speaking with moderators... bit more professionalism pls...10x

good answer is B - fw ctl affinity -l -a -v -r