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UK Apr2014-Q10-iv

D

deepakraomore

Member
After installing new heating element of excellent condition, how the trnsition matrix is written down?
 
In (iv), all "Failed" heating elements are replaced and the new heating element is in "Excellent" condition (after replacement). So, any transition to "Failed" can be considered as a transition to "Excellent".

For example, using the probabilities in the grid at the start of the question, the first entry in the transition matrix (ie the probability of going from Excellent to Excellent) is 0.5 (Excellent / Excellent) + 0.1 (Excellent / Failed).

The other entries in the matrix can be obtained similarly from the grid of probabilities in the question.
 
Hey Mark,

Thanks for the detailed explanation.

I got caught up in part (vi) of the question. I'm not sure how the coefficients (or weights) of the long-term distribution are arrived at. To be specific, how are the 0.1, 0.2 & 0.5 in the below equation arrived at.

[ 0.1 pi (e) + 0.2 pi (g) + 0.5 pi (p) ].

A stupid question in part (vii) of the same question. Expected cost of failures (i.e. 13500) includes the cost of replacement (50 quid) of the heating element. Why add 1143 separately?

Would appreciate your inputs.

Regards,
Sid
 
0.1 is the probability that an element that is currently excellent fails in the next firing, ie p_EF=0.1. Likewise, 0.2 is p_GF and 0.5 is p_PF.

In (vii), there is an extra cost of replacing parts that are in poor condition. p_EP =0.2 and p_GP=0.3.
 
Julie Lewis said:
0.1 is the probability that an element that is currently excellent fails in the next firing, ie p_EF=0.1. Likewise, 0.2 is p_GF and 0.5 is p_PF.

In (vii), there is an extra cost of replacing parts that are in poor condition. p_EP =0.2 and p_GP=0.3.
Click to expand...
Hi Julie,

Oh crap, I was getting lost for no reason!

Thanks a lot for your assistance.

Regards,
 
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