On Mon, Feb 1, 2021 at 6:51 PM David Teigland <teigland(a)redhat.com> wrote:
On Mon, Feb 01, 2021 at 07:18:24PM +0200, Nir Soffer wrote:
> Assuming we could use:
>
> io_timeout = 10
> renewal_retries = 8
>
> The worst case would be:
>
> 00 sanlock renewal succeeds
> 19 storage fails
> 20 sanlock try to renew lease 1/7 (timeout=10)
> 30 sanlock renewal timeout
> 40 sanlock try to renew lease 2/7 (timeout=10)
> 50 sanlock renewal timeout
> 60 sanlock try to renew lease 3/7 (timeout=10)
> 70 sanlock renewal timeout
> 80 sanlock try to renew lease 4/7 (timeout=10)
> 90 sanlock renewal timeout
> 100 sanlock try to renew lease 5/7 (timeout=10)
> 110 sanlock renewal timeout
> 120 sanlock try to renew lease 6/7 (timeout=10)
> 130 sanlock renewal timeout
> 139 storage is back
> 140 sanlock try to renew lease 7/7 (timeout=10)
> 140 sanlock renewal succeeds
>
> David, what do you think?
I wish I could say, it would require some careful study to know how
feasible it is. The timings are intricate and fundamental to correctness
of the algorithm.
Dave
I was taking values also reading this:
https://access.redhat.com/solutions/5152311
Perhaps it needs some review?
Gianluca