Suppose that a population of 525,477 solid-state devices is burned in at elevated temperature, and the number of failed devices are weeded out as a function of time is recorded in the following table. Determine the failure rates, λ(t), at the indicated times.

Question

Accepted Answer

Results of the calculations are tabulated.

Time (h)	Number failed, f(t)	Cumulative failures ,F(t)	[latex]\lambda(t)(h^{-1})[/latex]	Number of device hours	FITs
1	6253	6253	0.0119	[latex]525,477 imes 1[/latex]	[latex]1.19 imes 10^{7}[/latex]
2	1034	7287	0.00199	[latex]519,224 imes 2[/latex]	[latex]9.96 imes 10^{5}[/latex]
3	617	7904	0.00119	[latex]518,190 imes 3[/latex]	[latex]3.97 imes 10^{5}[/latex]
4	419	8323	0.000810	[latex]517,573 imes 4[/latex]	[latex]2.02 imes 10^{5}[/latex]
5	502	8825	0.000971	[latex]517,154 imes 5[/latex]	[latex]1.94 imes 10^{5}[/latex]
6	401	9226	0.000777	[latex]516,652 imes 6[/latex]	[latex]1.29 imes 10^{5}[/latex]
7	297	9523	0.000577	[latex]516,252 imes 7[/latex]	[latex]8.22 imes 10^{4}[/latex]
8	214	9737	0.000415	[latex]515,954 imes 8[/latex]	[latex]5.18 imes 10^{4}[/latex]
9	206	9943	0.000400	[latex]515,740 imes 9[/latex]	[latex]4.44 imes 10^{4}[/latex]
10	193	10,136	0.000375	[latex]515,534 imes 10[/latex]	[latex]3.74 imes 10^{4}[/latex]

Sample calculation at 5 h using Eqn(4.5) :

[latex]\lambda (t)= \frac{F(t + \Delta t)-F(t)}{\Delta t (1-F(t))}=\frac{f(t)}{1-F(t)}=\frac{f(t)}{R(t)}[/latex] (4.5)

[latex]\lambda (t)=502/525,477 imes (1-8323/525,477)^{-1} =0.0009710/h[/latex]

Because the failure rates are very low, a new numerically larger unit, the FIT, has been defined and is widely used in the microelectronics industry. The FIT is not an acronym, but a contraction of “failure unit”; it equals the number of failures in [latex]10^{9}[/latex] device-hours. For example, 1 FIT is one failure in [latex]10^{9}[/latex] device-hours, one failure in [latex]10^{7}[/latex] devices after 100 h of operation, or one failure in [latex]10^{6}[/latex] devices after 1000 h, etc.
Employing this definition, failure rates in terms of FITs were generated, and the values are displayed in the last column of the above table. After 5 h, for example, [latex]\lambda (t)=(502/517,154 imes 5) imes 10^{9}=1.94 imes 10^{5}[/latex] FITs.

Question 4.1: Suppose that a population of 525,477 solid-state devices is ...

Related Answered Questions

Verified Answer:

Consider a 50-pin module such that the probability of failure upon inserting it into a mating connector is 1.8 × 10^-3 % per pin. What is the probability of failure of the module? ...

Verified Answer:

Verified Answer:

Suppose the population of devices is described by the life distribution F(t) = 1 – exp(-0.0001t). 1. What is the probability that a new device will fail after 1000 h? After 5000 h? Between these times? 2. What is the failure rate at 1000 h? After 5000 h? ...

Verified Answer:

Tantalum capacitors were tested at elevated temperature, and the failure times were Weibull distributed with scale parameter α = 2.5 × 10^5 h and a shape parameter β = 0.5. If the AF is estimated to be 1 × 10^4 , what is the probability that the capacitors will last 1 year? ...

Verified Answer: