On Mon, Feb 1, 2021 at 6:51 PM David Teigland <teigl...@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
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