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<blockquote data-quote="mm1352000" data-source="post: 925459" data-attributes="member: 82144"><p>I *think* setting min and max to the same value gives you the best possible buffer size in both cases.</p><p> </p><p>Consider the following situation.</p><p>You have a 4GB RAM drive and 2 clients.</p><p>File size is set to 240 MB.</p><p>Assume the ideal where partitoned disk size = physical disk size.</p><p>One client starts timeshifting and timeshifts for long enough to fill up their back buffer then pauses.</p><p>At some later point, the other client starts timeshifting.</p><p> </p><p>What happens?</p><p> </p><p>If you set min = max = 8:</p><p>When client B starts timeshifting half the disk space will still be available regardless of how long client A has been paused. Client A will overwrite their own back buffer when they pause, up until the point when when they finish overwriting the 7th buffer file at which point data will start to be lost. Client A and B share the available disk space as equals, and they have the longest possible back and pause buffers within their allocated space.</p><p> </p><p>If you set min = 8 and max = 16:</p><p>When client B starts timeshifting they may not have much of the disk available, depending on how long client A has been paused. Client A can use the whole disk eventually. I'm not certain whether client B can reuse client A's back buffer files, but somehow I doubt it. In other words, client A pausing for long enough *may* prevent client B from starting to timeshift.</p><p> </p><p>If you set min = 4 and max = 8:</p><p>When client B starts timeshifting half the disk space will still be available regardless of how long client A has been paused. When paused, client A will create up to 4 new buffer files before overwriting their back buffer.</p><p> </p><p>***Disclaimer: I have not tested the above scenarios. I'm going on code alone, and I could easily be wrong.***</p><p> </p><p>Personally with the given assumptions (in particular, perfect sharing of the disk space) I'm going to go for min = max every time.</p><p>Min = 8 + max = 16 seems non-ideal if the clients really are "symetrical" in terms of use. Maybe you could bump the max a little bit (min = 8 + max = 12), but I think every step you bump it up gives more potential for one client to impact on another.</p><p>Min = 4 + max = 8 doesn't make much sense to me as it limits the size of the back buffer for no apparent gain.</p><p> </p><p></p><p>Yes, you're right - I think I saw that at the same time as you. <img src="data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7" class="smilie smilie--sprite smilie--sprite1" alt=":)" title="Smile :)" loading="lazy" data-shortname=":)" /></p></blockquote><p></p>
[QUOTE="mm1352000, post: 925459, member: 82144"] I *think* setting min and max to the same value gives you the best possible buffer size in both cases. Consider the following situation. You have a 4GB RAM drive and 2 clients. File size is set to 240 MB. Assume the ideal where partitoned disk size = physical disk size. One client starts timeshifting and timeshifts for long enough to fill up their back buffer then pauses. At some later point, the other client starts timeshifting. What happens? If you set min = max = 8: When client B starts timeshifting half the disk space will still be available regardless of how long client A has been paused. Client A will overwrite their own back buffer when they pause, up until the point when when they finish overwriting the 7th buffer file at which point data will start to be lost. Client A and B share the available disk space as equals, and they have the longest possible back and pause buffers within their allocated space. If you set min = 8 and max = 16: When client B starts timeshifting they may not have much of the disk available, depending on how long client A has been paused. Client A can use the whole disk eventually. I'm not certain whether client B can reuse client A's back buffer files, but somehow I doubt it. In other words, client A pausing for long enough *may* prevent client B from starting to timeshift. If you set min = 4 and max = 8: When client B starts timeshifting half the disk space will still be available regardless of how long client A has been paused. When paused, client A will create up to 4 new buffer files before overwriting their back buffer. ***Disclaimer: I have not tested the above scenarios. I'm going on code alone, and I could easily be wrong.*** Personally with the given assumptions (in particular, perfect sharing of the disk space) I'm going to go for min = max every time. Min = 8 + max = 16 seems non-ideal if the clients really are "symetrical" in terms of use. Maybe you could bump the max a little bit (min = 8 + max = 12), but I think every step you bump it up gives more potential for one client to impact on another. Min = 4 + max = 8 doesn't make much sense to me as it limits the size of the back buffer for no apparent gain. Yes, you're right - I think I saw that at the same time as you. :) [/QUOTE]
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