## 195 Reputation

16 years, 36 days

## Thank you! That did the trick!...

Thank you! That did the trick!

## Very interesting...

First off, thanks for your work and response. Much appreciated. I didn't consider the adjacency matrix approach. In this case, I have 14 nodes / vertices and 16 arcs / edges; the circuit commences from node 1.

I think that, between your solution and that of mmcdara, I'll have a better insight into the treatment of such problems ..... all good fun!

Thanks again ...

Thanks for this - much appreciated.

I am considering case 2 so I think this should do the trick!

## Interesting!...

Thanks for the suggestion - most interesting.

Actually, I initially required a surface approximation; the interpolation is interesting ... I never considered that option but will certainly look into this option.

Thanks again ...

## Thank you!...

This is appreciated .... many thanks, Carl.

## @tomleslie  Thanks for this ....&n...

Thanks for this ....  this input and guidance is much appreciated.

When I get this resolved, I'll post my results.

Thanks for this.

## Resolved...

@Rouben Rostamian

It appears that importing the source data from Excel requires a conversion to vector.

Using the Assistant to import the source file, I used the routine ..

> convert (Import("c:\\ ....etc), Vector)

and

> convert (Import("c:\\ ...etc), Matrix)

and that did the job!

Thanks again.

## Thanks...

Yes ... this should output a finite sum of terms. In this case, the sum is within square brackets and so, is a list ... therein lies the problem.

Now, I have imported the source data tables from Excel, so I'm wondering if the problem lies there. The matrices appear to be ok ...

I'll battle on!

## This is sorted...

To be clear ...

vv originally provided the solution .... and it works.

Since then, I have tested it in several applications to my great satisfaction.

## Table attached...

Markiyan,

Here's the table in question.

Thanks for pursuing this!

Z-Chart_&_Loss_Function.pdf

## L(z)...

Hi Markiyan.

I like that ...!

The function in question, L(z), is the standard loss function and comprises both the density and distribution functions of the standard normal distribution.There are tables of L(z) values readily available given (for instance) z in the range [-3,3].

Given a known L value, I was seeking to extract the associated z-value and our friend has successfully supplied a routine to do just that.

I hope that clarifies matters.

## That's it!...

@vv

Perfect .... this addresses the problem. Nice routine.

The numerical results align to the tabulated values.

Thanks so much.

## Slick!...

Now, that's efficiency!

I'm obliged .... thanks.

## Of course!...

I'm very grateful .... thank you!

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