Hi everyone,

Hi everyone,



Say if I schedule a particular product to run on the machine for 7.67
hours. What is the probability rate that it will fail between 1 to
7.67? I know I need to get the ALPHA and the BETA to complete my
calculations But I don't know how to get it. I know you could use
solver but I never used it before or can I use other formula functions
to get it? Also I cannot post the expected results because I just
don't know what they are. Here is the WEILBULL formula I'm using in
cell E8:

=WEIBULL(7.67,E2,E3,TRUE)-WEIBULL(1,E2,E3,TRUE)

I also posted this question to the link below.

http://www.mrexcel.com/board2/viewto...395&highlight=


Does anyone know how to use solver or other formula functions to get
the ALPHA and the BETA ?

Fin
I'm not sure that you have enough data for a Weibull analysis but here is a
good page by William Dorner on how to use Excel with Weibull analysis.
http://www.qualitydigest.com/jan99/html/weibull.html

In the meantime why not try a simple probability, your data in your post is

Machine Name Run Time Rank
Cast machine-1 1.20 1 Schedule 6
Cast machine-1 1.99 2 P(Failure) 0.55
Cast machine-1 2.12 3 P(Success) 0.45
Cast machine-1 3.01 4
Cast machine-1 4.16 5
Cast machine-1 5.00 6
Cast machine-1 6.01 7
Cast machine-1 7.01 8
Cast machine-1 7.20 9
Cast machine-1 7.20 10
Cast machine-1 7.67 11

Total 52.57
Mean 4.779090909 2.72E-33
SD 2.404006882 1
95%+ 9.587104674
95%- -0.028922856
Skew -0.234037883
Var 5.779249091
Median 5.00

I sorted your data to make it clearer. The standard deviation, did not seem
to give a clear picture. In the end I just used Schedule enter a number,

P(Success) =ROUND(COUNTIF($B$2:$B$12,"<"&$F$2)/COUNT($B$2:$B$12),2)

P(Failure) =ROUND(1-COUNTIF($B$2:$B$12,"<"&$F$2)/COUNT($B$2:$B$12),2)


=WEIBULL(F2,$B$15,$B$16,TRUE) where F2 is the time you want to run the
machine, B15 is the average and B16 is the Standard Deviation. Frankly I
guessed that these are the values you need. it had a propability of 1 for the
machine machine failing after five hours.

I have just remembered you can find the standard deviation of probability.
For a 95% estimate we use 1.96 SDevs. So the Probability of Success would
look like this

Schedule 5
P(Success) 0.55
95% 0.294
P(Suc)hi 0.844
P(Suc)lo 0.256

P(Success) =: =ROUND(1-COUNTIF($B$2:$B$12,"<"&$F$2)/COUNT($B$2:$B$12),2)
95% =: =1.96*SQRT(F3*(1-F3)/COUNT(B2:B12))
P(Success)hi=: =F3+F4
P(Success)lo=: =F3-F4

anyway I guess that you new that 5 hours is all you can hope for

Peter
"Fin Fang Foom" wrote:

Hi everyone,

Hi everyone,



Say if I schedule a particular product to run on the machine for 7.67
hours. What is the probability rate that it will fail between 1 to
7.67? I know I need to get the ALPHA and the BETA to complete my
calculations But I don't know how to get it. I know you could use
solver but I never used it before or can I use other formula functions
to get it? Also I cannot post the expected results because I just
don't know what they are. Here is the WEILBULL formula I'm using in
cell E8:

=WEIBULL(7.67,E2,E3,TRUE)-WEIBULL(1,E2,E3,TRUE)

I also posted this question to the link below.

http://www.mrexcel.com/board2/viewto...395&highlight=


Does anyone know how to use solver or other formula functions to get
the ALPHA and the BETA ?

On Jul 22, 3:32 pm, Billy Liddel
wrote:
Fin
I'm not sure that you have enough data for a Weibull analysis but here is a
good page by William Dorner on how to use Excel with Weibull analysis.http://www.qualitydigest.com/jan99/html/weibull.html

In the meantime why not try a simple probability, your data in your post is

Machine Name Run Time Rank
Cast machine-1 1.20 1 Schedule 6
Cast machine-1 1.99 2 P(Failure) 0.55
Cast machine-1 2.12 3 P(Success) 0.45
Cast machine-1 3.01 4
Cast machine-1 4.16 5
Cast machine-1 5.00 6
Cast machine-1 6.01 7
Cast machine-1 7.01 8
Cast machine-1 7.20 9
Cast machine-1 7.20 10
Cast machine-1 7.67 11

Total 52.57
Mean 4.779090909 2.72E-33
SD 2.404006882 1
95%+ 9.587104674
95%- -0.028922856
Skew -0.234037883
Var 5.779249091
Median 5.00

I sorted your data to make it clearer. The standard deviation, did not seem
to give a clear picture. In the end I just used Schedule enter a number,

P(Success) =ROUND(COUNTIF($B$2:$B$12,"<"&$F$2)/COUNT($B$2:$B$12),2)

P(Failure) =ROUND(1-COUNTIF($B$2:$B$12,"<"&$F$2)/COUNT($B$2:$B$12),2)

=WEIBULL(F2,$B$15,$B$16,TRUE) where F2 is the time you want to run the
machine, B15 is the average and B16 is the Standard Deviation. Frankly I
guessed that these are the values you need. it had a propability of 1 for the
machine machine failing after five hours.

I have just remembered you can find the standard deviation of probability.
For a 95% estimate we use 1.96 SDevs. So the Probability of Success would
look like this

Schedule 5
P(Success) 0.55
95% 0.294
P(Suc)hi 0.844
P(Suc)lo 0.256

P(Success) =: =ROUND(1-COUNTIF($B$2:$B$12,"<"&$F$2)/COUNT($B$2:$B$12),2)
95% =: =1.96*SQRT(F3*(1-F3)/COUNT(B2:B12))
P(Success)hi=: =F3+F4
P(Success)lo=: =F3-F4

anyway I guess that you new that 5 hours is all you can hope for

Peter

"Fin Fang Foom" wrote:
Hi everyone,

Hi everyone,

Say if I schedule a particular product to run on the machine for 7.67
hours. What is the probability rate that it will fail between 1 to
7.67? I know I need to get the ALPHA and the BETA to complete my
calculations But I don't know how to get it. I know you could use
solver but I never used it before or can I use other formula functions
to get it? Also I cannot post the expected results because I just
don't know what they are. Here is the WEILBULL formula I'm using in
cell E8:

=WEIBULL(7.67,E2,E3,TRUE)-WEIBULL(1,E2,E3,TRUE)

I also posted this question to the link below.

http://www.mrexcel.com/board2/viewto...395&highlight=

Does anyone know how to use solver or other formula functions to get
the ALPHA and the BETA ?


Thank You so much for the help Billy Liddel!


I tried formulas with my data set and its not coming out as i
expecting.

Here my data set with your data calculations.

History of the Hours
Machine Ran
Machine 1 7.01
Machine 1 4.16
Machine 1 7.67
Machine 1 5
Machine 1 2.12
Machine 1 6.01
Machine 1 7.2

( Resutls )
Schedule 7.67
P(Success) 0.14
95% 25.71%
P(Suc)hi 39.71%
P(Suc)lo -14.00%


( Formulas )
Schedule 7.67
P(Success) =ROUND(1-COUNTIF($B$3:$B$9,"<"&$B$12)/COUNT($B$3:$B$9),
2)
95% =1.96*SQRT(B13*(1-B13)/COUNT(B3:B9))
P(Suc)hi =B13+B14
P(Suc)lo =B14-B15

That you can see its not coming out right.

Then I used these formulas and it looks like its coming out right but
I'm not sure.

mean 5.595714286
sigma 1.980949748
alpha 2.615140091
beta 18.89546699
P(Suc)hi 90.97%


( Formulas )
mean AVERAGE(B3:B9)
sigma STDEV(B3:B9)
alpha D2*SQRT(EXP(GAMMALN(1+2/D3))-EXP(GAMMALN(1+1/D3))^2)
beta D2*EXP(GAMMALN(2+3/D3))

P(Suc)hi =1-WEIBULL(7.67,D4,D5,1)


Can verify this?

Fin
I did not agree with the percentages you got for the new data but at a quick
glance the weibull looks ok. However, I think that we have both got our
terminology wrong.
Weibull is the probability of failure. so the 91% weibull would mean that
out of 20 days we could expect 2 full days production. 20-.91*20.

This is similar to the figures to the siple probabilty I arrived at. (it's
always a good idea to have a simple canculation to go on)
My figures a
P(success) E 20 runs
P 11.11% 2
SD95% 14.52% 3
P(hi) 25.63% 5
P(lo) 0.00% 0
-3.41%
I used zero for the lo P as you can't have minus machine working.
Combining tthe two sets of figures gives:

P(success) E 20 runs
P1 14.29% 3
SD95% 25.92%
Prob hi 40.21% 8
Prob(lo) 0.00% 0
-11.64%

This is slightly higher than the weibull figures but within a decent range.

When I get time to analyse the weibull figures I'll post back

Regards
Peter
"Fin Fang Foom" wrote:

Thank You so much for the help Billy Liddel!


I tried formulas with my data set and its not coming out as i
expecting.

Here my data set with your data calculations.

History of the Hours
Machine Ran
Machine 1 7.01
Machine 1 4.16
Machine 1 7.67
Machine 1 5
Machine 1 2.12
Machine 1 6.01
Machine 1 7.2

( Resutls )
Schedule 7.67
P(Success) 0.14
95% 25.71%
P(Suc)hi 39.71%
P(Suc)lo -14.00%


( Formulas )
Schedule 7.67
P(Success) =ROUND(1-COUNTIF($B$3:$B$9,"<"&$B$12)/COUNT($B$3:$B$9),
2)
95% =1.96*SQRT(B13*(1-B13)/COUNT(B3:B9))
P(Suc)hi =B13+B14
P(Suc)lo =B14-B15

That you can see its not coming out right.

Then I used these formulas and it looks like its coming out right but
I'm not sure.

mean 5.595714286
sigma 1.980949748
alpha 2.615140091
beta 18.89546699
P(Suc)hi 90.97%


( Formulas )
mean AVERAGE(B3:B9)
sigma STDEV(B3:B9)
alpha D2*SQRT(EXP(GAMMALN(1+2/D3))-EXP(GAMMALN(1+1/D3))^2)
beta D2*EXP(GAMMALN(2+3/D3))

P(Suc)hi =1-WEIBULL(7.67,D4,D5,1)


Can verify this?

On Jul 23, 4:50 am, Billy Liddel
wrote:
Fin
I did not agree with the percentages you got for the new data but at a quick
glance the weibull looks ok. However, I think that we have both got our
terminology wrong.
Weibull is the probability of failure. so the 91% weibull would mean that
out of 20 days we could expect 2 full days production. 20-.91*20.

This is similar to the figures to the siple probabilty I arrived at. (it's
always a good idea to have a simple canculation to go on)
My figures a
P(success) E 20 runs
P 11.11% 2
SD95% 14.52% 3
P(hi) 25.63% 5
P(lo) 0.00% 0
-3.41%
I used zero for the lo P as you can't have minus machine working.
Combining tthe two sets of figures gives:

P(success) E 20 runs
P1 14.29% 3
SD95% 25.92%
Prob hi 40.21% 8
Prob(lo) 0.00% 0
-11.64%

This is slightly higher than the weibull figures but within a decent range.

When I get time to analyse the weibull figures I'll post back

Regards
Peter



"Fin Fang Foom" wrote:
Thank You so much for the help Billy Liddel!

I tried formulas with my data set and its not coming out as i
expecting.

Here my data set with your data calculations.

History of the Hours
Machine Ran
Machine 1 7.01
Machine 1 4.16
Machine 1 7.67
Machine 1 5
Machine 1 2.12
Machine 1 6.01
Machine 1 7.2

( Resutls )
Schedule 7.67
P(Success) 0.14
95% 25.71%
P(Suc)hi 39.71%
P(Suc)lo -14.00%

( Formulas )
Schedule 7.67
P(Success) =ROUND(1-COUNTIF($B$3:$B$9,"<"&$B$12)/COUNT($B$3:$B$9),
2)
95% =1.96*SQRT(B13*(1-B13)/COUNT(B3:B9))
P(Suc)hi =B13+B14
P(Suc)lo =B14-B15

That you can see its not coming out right.

Then I used these formulas and it looks like its coming out right but
I'm not sure.

mean 5.595714286
sigma 1.980949748
alpha 2.615140091
beta 18.89546699
P(Suc)hi 90.97%

( Formulas )
mean AVERAGE(B3:B9)
sigma STDEV(B3:B9)
alpha D2*SQRT(EXP(GAMMALN(1+2/D3))-EXP(GAMMALN(1+1/D3))^2)
beta D2*EXP(GAMMALN(2+3/D3))

P(Suc)hi =1-WEIBULL(7.67,D4,D5,1)

Can verify this?- Hide quoted text -

- Show quoted text -

Thank You Billy Liddel!

Please let me know if the WEIBULL Function is the right scenario I
should be using to get the correct probability rate failure of the
machine.

Fin

I have slavishly copied the method by Dave ? using your figures. I came up
with Beta 2.296404 and alpha 6.485752 giving a weibull probability of failure
for 6.67 hours of 77%.

If you like, I'll send you my workings. peter_athertonAThotmail.com. do the
obvious with the AT.

Is it the best way? I don't know, what do you intend doing with the result.
For instance, how many widgits are produced during a run is it always the
same proportion of units produced by the hours or when there is a quick
failure has a lower number than expected been produced?

Ma


Can verify this?- Hide quoted text -

- Show quoted text -

Thank You Billy Liddel!

Please let me know if the WEIBULL Function is the right scenario I
should be using to get the correct probability rate failure of the
machine.