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ayman
Level II

Redundancy Analysis

Hello, 

 

I am trying to perform a redundancy analysis. Is JMP able to do it? 

 

Thanks 

3 ACCEPTED SOLUTIONS

Accepted Solutions
mzwald
Staff

Re: Redundancy Analysis

Short answer is yes, because it's math and JMP can do math.

Longer answer is JMP doesn't have a dedicated redundancy analysis platform, but there are other platforms which can facilitate it such as the Reliability Block Diagrams which can model K out of M failure modes in systems. 

 

Below I show how you can use the JMP scripting language to do any type of custom analysis that JMP doesn't do out of the box.

Mark

View solution in original post

Re: Redundancy Analysis

Also, while I am not familiar with this method, my brief research indicates that it might be a multivariate method that extends multiple regression and principle components analysis to both multiple predictors and multiple responses. It might be similar to partial least squares (PLS), which is available in JMP. See the JMP documentation about this platform.

 

There are myriad multivariate methods that have been independent developed in myriad disciplines.

View solution in original post

mzwald
Staff

Re: Redundancy Analysis

To show the flexibility of the JMP scripting language, I created a simple script which will compute the yield of a system which consists of M blocks where K out of M need to be functional for the system to be considered functional. 

Names Default to Here(1);

K = 8;  // At least K out of M blocks need to be functional
M = 10; // Total number of blocks in system 

Y.block = 0.90;  // Probability of one block being functional

BinomialCoeff = Function( {n, k}, // calc coefficients of binomial expansion
    Return( Factorial(n)/Factorial(k)/Factorial(n-k) ); 
); 

Y.system = 0;  
for (i = 0, i <= (M-K), i++,
	Y.system = Y.system + BinomialCoeff(M,i)*Y.block^(M-i)*(1-Y.block)^i;
);

write("System yield of ",K," out of ",M," blocks is ",Y.system);

So in this example, if each block has 90% yield and you have 10 blocks where 8 need to be functional for the system to work, the system yield will be 93% (or 74% if 9 out of 10, or 35% if 10 out of 10).

Mark

View solution in original post

4 REPLIES 4
mzwald
Staff

Re: Redundancy Analysis

Short answer is yes, because it's math and JMP can do math.

Longer answer is JMP doesn't have a dedicated redundancy analysis platform, but there are other platforms which can facilitate it such as the Reliability Block Diagrams which can model K out of M failure modes in systems. 

 

Below I show how you can use the JMP scripting language to do any type of custom analysis that JMP doesn't do out of the box.

Mark

Re: Redundancy Analysis

Also, while I am not familiar with this method, my brief research indicates that it might be a multivariate method that extends multiple regression and principle components analysis to both multiple predictors and multiple responses. It might be similar to partial least squares (PLS), which is available in JMP. See the JMP documentation about this platform.

 

There are myriad multivariate methods that have been independent developed in myriad disciplines.

mzwald
Staff

Re: Redundancy Analysis

To show the flexibility of the JMP scripting language, I created a simple script which will compute the yield of a system which consists of M blocks where K out of M need to be functional for the system to be considered functional. 

Names Default to Here(1);

K = 8;  // At least K out of M blocks need to be functional
M = 10; // Total number of blocks in system 

Y.block = 0.90;  // Probability of one block being functional

BinomialCoeff = Function( {n, k}, // calc coefficients of binomial expansion
    Return( Factorial(n)/Factorial(k)/Factorial(n-k) ); 
); 

Y.system = 0;  
for (i = 0, i <= (M-K), i++,
	Y.system = Y.system + BinomialCoeff(M,i)*Y.block^(M-i)*(1-Y.block)^i;
);

write("System yield of ",K," out of ",M," blocks is ",Y.system);

So in this example, if each block has 90% yield and you have 10 blocks where 8 need to be functional for the system to work, the system yield will be 93% (or 74% if 9 out of 10, or 35% if 10 out of 10).

Mark

ayman
Level II

Re: Redundancy Analysis

Thank you so much for the constructive responses.