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Conference rusure::math

Title:	Mathematics at DEC

Moderator:	RUSURE::EDP

Created:	Mon Feb 03 1986
Last Modified:	Fri Jun 06 1997
Last Successful Update:	Fri Jun 06 1997
Number of topics:	2083
Total number of notes:	14613

653.0. "SA482 word problem." by CHOVAX::YOUNG (Back from the Shadows Again,) Tue Jan 20 1987 13:19

    Here's an interesting and yet practical word problem:
    
    A customer buys 2 SA482's (SA482 = 4 x RA82).  He uses volume shadowing
    to shadow each unit of the first SA482 with each correponding unit
    od the second.  He then mounts all 4 shadow volumes and binds them
    together into one large logical volume.
    
    Q:  What is the Mean Time to Service Failure (MTSF)?
    
    Assume:
    
    	-- That Mean Time to Unit Failure (MTUF) = MTBF of an RA82
    		= 24 months = 720 days.
    
    	-- That if a unit fails that it can be serviced without affecting
    		any other units.
    
    	-- That if a unit fails that the mean time to complete restoration
    		of the unit is 2 days.
    
    
    --  Barry

T.R Title User Personal
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653.1 SSDEVO::LARY Wed Jan 21 1987 02:43 16

T.R	Title	User	Personal Name	Date	Lines
653.1		SSDEVO::LARY		`Wed Jan 21 1987 02:43`	16
	I believe an exact solution to this problem requires using Markov Chains, which I am thoroughly incompetent in, so here's an inexact solution; it will be interesting to see how close the exact solution is. With 8 RA81's you will get a failure every 90 days on the average. Only the failure of both members of a shadow pair produces a failure, and the chance of that specific drive failing during the 2 day repair period is 1/360, so you will get a Service Failure once in 360*90 days or around once every 89 years. If you keep a spare drive in your back pocket and wheel it into place when the failure is detected, reducing the down time (but not the repair time) to 3 hours, the answer goes to 1420 years and becomes even more inexact; the Markov Chain problem also becomes more complicated. Anyone have a solution to that one, too?

I believe an exact solution to this problem requires using Markov Chains,
which I am thoroughly incompetent in, so here's an inexact solution; it
will be interesting to see how close the exact solution is.

With 8 RA81's you will get a failure every 90 days on the average. Only the
failure of both members of a shadow pair produces a failure, and the chance
of that specific drive failing during the 2 day repair period is 1/360, so
you will get a Service Failure once in 360*90 days or around once every 89
years.

If you keep a spare drive in your back pocket and wheel it into place when
the failure is detected, reducing the down time (but not the repair time)
to 3 hours, the answer goes to 1420 years and becomes even more inexact;
the Markov Chain problem also becomes more complicated. Anyone have a
solution to that one, too?