Home work solutions for Combination_etc and Reliability (MC trip)

[Xiuya Li]

Hi All,

 

Attached are the recent 2 sets of homeworks.

 

It's little tedious and time consuming. Please let me know whether you get the same results for both of these HW.

 

The excel is for MC Trip.

 

Thanks!
Ted Li

[Anand Murugesan]

For MC homework, (f)

I am getting the following answer.  Pl. Comment. 

f. Assuming exponential failures, and using your estimate of λ, and assuming he has already made it all the way through France without a breakdown, what is the conditional probability the motorcycle will have a breakdown in The Netherlands or Belgium?

 

Conditional Probability of failure per KM = λ = 0.00053389.

Remaining distance = 400.

The conditional probability the motorcycle will have a breakdown in The Netherlands or Belgium =  400 * λ = 0.213554

[Xiuya Li]

Hi Anand,

 

I don't think either of us was correct. Here is my modified answer:

 

R(1900) = e^(-λt) = e^(-0.000534*1900) = 0.3625

So: conditional probability = 0.0697/0.3625 = 0.1923 = 19.23%

This new number matches (g) so it should be the right one.

 

Thanks!

Ted Li

 

2011/5/22 Anand Murugesan <notif...@gmail.com>

[Nicholas Woo]

I also got 19.23% for part G.

2011/5/22 Xiuya Li <xiu...@gmail.com>

[Xiuya Li]

Hi All,

 

I mean for both (f) and (g), we shall get 19.23%, they should be the same. My initial answer for (f) was wrong. (f) should NOT be 6.97%.

 

Thanks!

Ted

2011/5/22 Nicholas Woo <nichol...@gmail.com>

[Nicholas Woo]

For F, my solution was:

f(2300) / R(1900) = .0697 / e^-0.00053*1900 = 0.697 / .3653 = .1907 or 19.07%

2011/5/22 Anand Murugesan <notif...@gmail.com>

[Nicholas Woo]

Hi Ted,

I think the solutions to parts (f) and (g) are not the same since in part (g) we are calculating the probability of failure within the first 400KM of the trip (so it doesn't matter where he starts), which I agree is 19.23%. However in part (f), I believe we're supposed to calculate the probability of failure in the last 400KM of the trip given no failures have occurred for the first 1900KM of the trip. I'm not 100% positive, but think the formula is written as f(t)/R(t) such that F(2300) - F(1900) / R(1900) = 19.07%

[Xiuya Li]

Hi Nicholas,

 

I have the exact same formular with yours, I did not list it in my previous email, here is where the 0.0697 come from (from question e):

 

Distance (t1) = 650+350+900 = 1900

Distance (t2) = 2300

 

F(t2) – F(t1) = (1 – e^(-λt2)) – (1 – e^(-λt1)) = e^(-λt1) - e^(-λt2)

=e^(-0.000534*1900) – e^(-0.000534*2300) = 0.0697 = 6.97%

 

R(1900) = e^(-λt) = e^(-0.000534*1900) = 0.3625

So: conditional probability = 0.0697/0.3625 = 0.1923 = 19.23%

 

Our answer have slight difference, I think it's due to calculation rounding error. Could you please double check?

[Nicholas Woo]

Hi Ted,

Thanks, you're correct - the difference was coming from a rounding error. For my R(1900) calculation, I was using lambda = 0.00053 and you were using lambda = 0.000534. Thanks for the catch

[Xiuya Li]

Thank you!

 

Please let me know if you see any other differences in other part of the homeworks.

[Nicholas Woo]

Ted,

What was your rationale behind taking the absolute value of lambda in the first question for part (a)?

The raw regression output is -0.000534 which I interpreted that as the incremental decrease in reliability per 100KM increase. How does that interpretation change if we take the absolute value instead?

2011/5/22 Xiuya Li <xiu...@gmail.com>

[Xiuya Li]

Hi Nicholas,

 

From slides 6 (of 8), we have definition of lambda:

Constant failure proportion of those remaining for each

increment of time (for exponential failures).

This is NOT decrease of reliability, but rather increase (more precisely, positive #) of failure proportion ...

 

Also from the regression formular, we have:

 

ln(# surviving) = ln(total number) - lambda * t

 

When you got - 0.000354 from regression, that "-" belongs to the formular, and you get the net of positive lambda.

 

So from both physical explanation and mathematical interprelation, you should have positive lambda, not negative.

[Ashish Swaroop]

Has anyone done the student OC problem homework Can you share your answer to the last part -
Suppose you are the consumer and your target quality level is 40% with a rejection level of 50%.  Do you like the plan?
and how did you calculate the Pr (0,1 and 2 defective items in sample).

Thanks

[Kartik Ramachandran]

Hi Ted,

I am not sure if the -ve sign goes with the formula. The formula is a+bt...so if b comes out negative, it is because b is negative.

What i think that the Prof has forgot to tell us is that - lambda should be taken as absolute value.

Thanks,

Kartik

2011/5/22 Ashish Swaroop <ashishsw...@gmail.com>

[Xiuya Li]

Hi Kartik,

 

The regression formular is a+bt, but the slides formular is ln(##) = ln(#) - lambda*t, therefore b= - lambda. lambda should be positive by it's definition.

 

Thanks!
Ted

 

 

2011/5/22 Kartik Ramachandran <krama...@gmail.com>

[Xiuya Li]

Hi Ashish,

 

Please see my attachment for details. Let me know your comments.

 

Thanks!

Ted

[Ajay Goyal]

Hi,

For the producer consumer OC Curver problem, how do we interpret the "Target Quality Level" and "Rejection Level"

 

With target quality level 40% and rejection level of 50%

should it be

Pr(m<= 4| p=0.4) for consumer and

Pr(m>=5| p=0.4) for producer

 

thanks

ajay

[Xiuya Li]

NO, Ajay. According to slides we should have the following:

Pr(m<=2 | p=0.5) = (0.179+0.092)/2 = 13.55% (Type II error, consumer's risk).
Pr(m>=3 | p=0.4) = Pr(m>=3 | M=30) = 1 - Pr(m<=2 | M=30) = 1 - 0.3035 = 0.6965 =69.65%  (Type I error, producer's risk)

Since the type I error is greater than type II error, as consumer, I like this plan since I have less risk.

Thanks!
Ted

2011/5/23 Ajay Goyal <ajay...@gmail.com>

[Ashish Swaroop]

Hi Xiuya,

How did you get this calculation -

Pr(m<=2 | p=0.5) = (0.179+0.092)/2

How did you get these numbers 0.179 and 0.092?

regds

2011/5/23 Xiuya Li <xiu...@gmail.com>

[Dinesh Kantaria]

Guys,

I would appreciate some input how to come up with regression result. I am selecting Kms 100-3000 as dependent variable and LN(reliability) for independent variable.

Am I on right track?

Thank you in advance for any input on this.

Dinesh 

2011/5/23 Ashish Swaroop <ashishsw...@gmail.com>

--
Dinesh Kantaria
510-219-9172

[Savitha Gandikota]

Hi Ashish,
I just added up Pr(m=0)+pr(m=1)+Pr(m=2) to reach the same 13.xx%
and with p=0.5.
Similarly 1 -(Prm=0)+pr(m=1)+Pr(m=2) at p=0.4 gave the producers risk 69.xxx%

BTW, during the discussions I noticed people using % for lot of things.
I think one has to be careful in noting when a % was asked vs when
conditional probabilities were asked (which is just a value < 1).

regards,
Savitha

2011/5/23 Ashish Swaroop <ashishsw...@gmail.com>:

>>>>>>>>>>> F(t2) - F(t1) = (1 - e^(-λt2)) - (1 - e^(-λt1)) = e^(-λt1) -
>>>>>>>>>>> e^(-λt2)
>>>>>>>>>>>
>>>>>>>>>>> =e^(-0.000534*1900) - e^(-0.000534*2300) = 0.0697 = 6.97%

[Ajay Goyal]

Hi

Attached is my solution for the OC-Curve homework.

 

thanks

ajay

2011/5/23 Savitha Gandikota <savitha....@gmail.com>

[Xiuya Li]

Hi Ajay,

 

You should use M=37 and M=38 and get the average probability. It will be arround 13.10%. I noticed you can not get result for non-integer population defect, it will use the "floor" of the # you provided. Here we want the # for 37.5 = 0.5* 75, but we have to use average results from 37 and 38.

 

My original post is based on average from defect of 35 and 40, which is little bit far off. 13.10% is more accurate.