Question

(1)The field test data in respect of 172 components is as given below. In the life-testing of 100 specimens of a particular device, the number of failures during each time interval of twenty hours is shown in Table below. Estimate and Plot: the hazard function, failure density and reliability function. Time/Hours Failure 0-1000 59 1000-2000 24 2000-3000 3000-4000 4000-5000 5000-6000 29 30 17 13

Answer 1

(1).

Answer:

Given that,

$\rightarrow$ The field test data in respect of 172 components is as given below.

In the life-testing of 100 specimens of a particular device, the number of failures during each time interval of twenty hours is shown in Table below.

Estimate and Plot: the hazard function, failure density and reliability function:

Given data,

Size of the sample (N_0)=172

Time/Hours	0-1000	1000-2000	2000-3000	3000-4000	5000-6000
Failure	59	24	29	30	17	13

N_f(t) $\rightarrow$ The number of components that have failed by time t.

N_f(0)=0

N_f(1000)=0+59

N_f(2000)=59+24=83

N_f(3000)=83+29=112

Similarly remaining cumulative frequencies.

N_s(t) $\rightarrow$ The number of components that are performing their intended function at time t.

N_s(0)=172

N_s(1000)=172-59=113

N_s(2000)=113-24=89

N_s(3000)=89-29=60

Similary remaining entries.

R(t) $\rightarrow$ Reliability at time t.

R(t)=N_s(t)/N_0

R(0)=N_s(0)/N_0=172/172=1

R(1000)=N_s(1000)/N_0=113/172=0.656

R(2000)=89/172=0.517

R(3000)=60/172=0.348

Similarly remaining entries,

f(t) $\rightarrow$ the value of failure density at time t.

$f(t)=\frac{N_f(t+\Delta t)-N_f(t)}{N_0\Delta t}$

It = 1000

t=0, 1000, 2000,.................

N_0=172

$f(0)=\frac{N_f(0+1000)-N_f(0)}{172\times 1000}=\frac{N_f(1000)}{172\times 1000}=\frac{59}{172\times 1000}=0.00034$

f(1000) = N (2000) - N/(1000) 172 x 1000 83 - 59 172 x 1000 0.00013

$f(2000)=\frac{N_f(3000)-N_f(2000)}{172\times 1000}=\frac{112-83}{172\times 1000}=0.00016$

$f(3000)=\frac{N_f(4000)-N_f(3000)}{172\times 1000}=\frac{142-112}{172\times 1000}=0.00017$

Similarly remaining entries,

$\lambda$(t) $\rightarrow$ the value of Hazard rate at time, t

$\lambda (0)=\frac{N_f(0+1000)-N_f(0)}{1000\times N_s(0)}=\frac{59-0}{1000\times 172}=0.00034$

$\lambda (1000)=\frac{N_f(2000)-N_f(1000)}{1000\times N_s(1000)}=\frac{83-59}{1000\times 113}=0.00021$

$\lambda (2000)=\frac{N_f(3000)-N_f(2000)}{1000\times N_s(2000)}=\frac{112-83}{1000\times 89}=0.00032$

$\lambda (3000)=\frac{N_f(4000)-N_f(3000)}{1000\times N_s(3000)}=\frac{142-112}{1000\times 60}=0.00050$

Similarly remaining entries,

Estimation of R(t), f(t) and $\lambda (t)$

Time (hr)	Number of failures (f)	Cumulative failures N_f(t)	Number of survivors N_s(t)	Relaibility R(t)	Failure density f(t)	Hazard Rate $\lambda (t)$
0	0	0	172	1	0.00034	0.00034
1000	59	59	113	0.656	0.00013	0.00021
	24	83	89	0.517	0.00016	0.00032
	29	112	60	0.348	0.00017	0.00050
	30	142	30	0.174	0.00009	0.00056
	17	159	13	0.075	0.00007	0.001
	13	172	0	0	-	-

1000 2000 300 4000 5000 6000 ARCH) +Reliability 0.8 0.6 204 toga IDO CO MONO KO 50 cm t time Reliability ^fct) dood 2008 Paco Tacool • Time 0 1000 2000 3000 4000 5000 GOCO (1) Failue density a act) 0.00 Hazard rate 0.0005 0.0006 0.0004 Octo! time Ct) → Hazard Rate

Note:

As per Chegg Answering Policy, 1st question has been answered.