RELIABILITY CALCULATION - EXPONENTIAL FAILURE MODEL

The best way to see how an application might work is to consider
an example. It all depends upon requirements, so what if an MTBF 
of 10 years for a TV set is required to give 80% chance of home
viewing in 10 years. The 20% risk of not seeing a TV program is
fullfilled by number 5: 2 set are always on and a third is off
and standing by for operation in the event of TV set failure.
Also, how about a satellite communication channel amplifier.
The same would apply, except an amplifier MTBF would be quite large.

 1  y =  Function of exp(-x/10)          2 required of 3, 3on 
 2  y =  Function of exp(-x/10)          1 required of 1, 1on
 3  y =  Function of exp(-x/10)          1 required of 2, 2on
 4  y =  Function of exp(-x/10)          1 required of 3, 3on 
 5  y =  Function of exp(-x/10)          1 required of 2, 2on, 1sb 

Year   2of3  ::  1of1  ::  1of2  ::  1of3  ::1:2on1off
  1  0.974556::0.904837::0.990944::0.999138::0.999698
  2  0.913337::0.818731::0.967141::0.994044::0.997812
  3  0.833296::0.740818::0.932825::0.982589::0.993303
  4  0.745598::0.670320::0.891311::0.964167::0.985585
  5  0.657378::0.606531::0.845182::0.939084::0.974410
  6  0.572985::0.548812::0.796429::0.908151::0.959768
  7  0.494878::0.496585::0.746574::0.872421::0.941816
  8  0.424254::0.449329::0.696761::0.833015::0.920823
  9  0.361486::0.406570::0.647840::0.791018::0.897124
 10  0.306432::0.367879::0.600424::0.747420::0.871094
 11  0.258643::0.332871::0.554939::0.703087::0.843120
 12  0.217506::0.301194::0.511670::0.658753::0.813584
 13  0.182337::0.272532::0.470790::0.615017::0.782856
 14  0.152439::0.246597::0.432384::0.572356::0.751282
 15  0.127143::0.223130::0.396473::0.531138::0.719178
 16  0.105827::0.201897::0.363031::0.491633::0.686831
 17  0.087926::0.182684::0.331994::0.454028::0.654497
 18  0.072938::0.165299::0.303274::0.418442::0.622400
 19  0.060420::0.149569::0.276766::0.384940::0.590732
This page hosted by Get your own Free Home Page
1