Question

Two nickel-cadmium batteries provide electrical power to operate a satellite transceiver. If both batteries are operating in parallel, they have an individual failure rate of 0.1 per year. If one fails, the other can operate the transceiver (at a reduced power output). However, the increased electrical demand will triple the failure rate of the remaining battery. Determine the system reliability at 1, 2, 3, 4, and 5 yr. What is the system MTTF?

Answer 1

Answer:

Step-by-step explanation:

1.)

f(t) for each battery = 0.1exp(-0.1t)

F(t) = 1 - exp(-0.1t)

R(t) = P(T > t) = P(Both battery works till at least time t) + 2*P(one battery works till at least time t)P(one battery stops working before time t)

= (exp(-0.1t))2 + 2*(exp(-0.3t) )(1 - exp(-0.1t))

R(t) = exp(-0.2t) + 2exp(-0.3t) - 2exp(-0.4t)

R(1) = 0.95973

R(2) = 0.8693

R(3) = 0.213

R(4) = 0.6479

R(5) = 0.5435

G(t) = 1 -R(t) = 1 - exp(-0.2t) - 2exp(-0.3t) + 2exp(-0.4t)

g(t) = G'(t) = 0.2exp(-0.2t) + 0.6exp(-0.3t) - 0.8exp(-0.4t)

MTTF is the average time to failure. (Also called the mean time to failure, expected time to failure, or average life.)

MTTF =$\int\limits^{infinity}_0 {0.2te^{-0.2t}+0.6te^{-0.3t}-0.8te^{-0.4t}}dt$

= .2 � 25 + 0.6*100/9 - .8 � 6.25

= 6.67