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.

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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 ...

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