Question

take yz =82

Failure data for a cutting tool insert (shown above) is as follows: hours of machining until insert failure: (50+YZ/10), 68, 72, 79, 84, 84, 90, 92, 93, (100+YZ/10), (110 +YZ/10) Assuming that the cutting tool insert is at the constant failure rate phase: I Calculate the mean time to failure (i.e. average of times), and calculate the failure rate. (4)
PROBLEM 2 (continued) III For these tool inserts, what is the 65 hour reliability? Compare this to the 90 hour reliability, assuming constant failure rate.(4) IV Using the 65 hour reliability, and given that the inserts are sampled and tested in batches of eight (8) inserts: What is the likelihood of 2 or more of the eight inserts NOT meeting the 65 hour mission time? (4) V Using the 90 hour reliability, and given that the inserts are sampled and tested in batches of eight inserts: What is the likelihood of 2 or more of the eight inserts NOT meeting the 90 hour mission time? (3)

Answer 1

Mean time to failure = Total hours of operation ÷ Total assets in use

count	hours before insert failure	substituting yz value
1	50+yz/10	58.2
2	68	68
3	72	72
4	79	79
5	84	84
6	90	90
7	92	92
8	93	93
9	100+ yz/10	108.2
10	110+yz/10	118.2
	Sum	862.6
n=	10
MTTF	862.6/10	86.26	hours

MTTF = 862.6/10 = 86.26 hours

Failure rate = 1/ MTTF = 1/86.26 = 0.116

Reliability at 65 hours = total working hours / no of insert failure under 65 hours

R 60= 862.6 / 1 ( there is only one instance below 65 hours ) = 862.6 or we can say 100%

Reliability at 90 hours = total working hours / no of insert failure under 90 hours

R 60= 862.6 / 6 ( there is 6 instance below 90 hours ) = 143.766 hours or we can say 16.66%

Part 4

We have Reliability at 65 hours = 100 % or 1

Given n = 8

k = 2

We know that reliability R(t) = e-^t  

= 0.116

t= 65 hours

= 34.11% probability

Now for 90 hours

Given n = 8

k = 2

We know that reliability R(t) = e-^t  

= 0.166

t= 65 hours

= 52.655 %

.