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05-03-2017 11:56 PM

Hi,

I have a question regarding the capabilites of proc iml. Suppose I have a matrix of probabilites, then want to multiply each cell directly by a scalar matrix, so that only some of the probabilites go up while some remain the same. For example:

Now, I'd like to multiply the other values in each row by a scalar in order for each row to still sum to one. But the scalar also needs to be computed for each row.

This can be accomplished with excel's solver, as shown:

Would this be possible to accomplish in IML, or would I need to look to a different procedure, or is there a better way to go about this?

Attached is code to set up the beginning matrices in SAS:

```
proc iml;
A = { 0.20 0.15 0.30 0.35,
0.15 0.25 0.35 0.25,
0.10 0.20 0.50 0.20,
0.25 0.25 0.30 0.20};
B = A[ ,+];
C = { 1.0 1.2 1.0 1.0,
1 1 1.2 1.0,
1 1 1 1.2,
1.2 1 1 1};
D = A#C;
E = D[ ,+];
```

Thank you,

Justin

Accepted Solutions

Solution

05-05-2017
11:17 AM

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Posted in reply to jl1005

05-05-2017 02:34 AM

The problem is that (H ^= 0) first evaluates to a matrix of zeros and ones, so when this is applied to D with elementwise multiplication it either preserves the existing values in D, or sets them to zero.

It is possible to get the Inverse Scalar Matrix using a small trick. First subtract one from G, next calculate H as you have above, finally add back the one. The overall effect is to place a one, rather than a zero, at every location in H corresponding to one of the modified values.

```
F = (D#(C^=1))[,+];
G = (1-F)/(E-F);
H = (G-1)#(C=1) + 1;
J = D # H;
print H [l='Inverse Scalar Matrix'], J [l='Finished Matrix'];
```

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Posted in reply to jl1005

05-04-2017 03:33 AM

I think you are over complicating things by suggesting you need to use the Excel solver. Once you know the sum of values in each row that have been increased (F), then it is a simple calculation to find the row scalars (G).

```
F = (D#(C^=1))[,+];
G = (1-F)/(E-F);
print G;
```

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Posted in reply to IanWakeling

05-04-2017 02:33 PM - edited 05-04-2017 02:33 PM

That works great; thank you.

Now to apply that to the new matrix, so that the values of G multiply all the other cells in D, I tried:

```
H = G#(C=1); /*To create similar scalar matrix */
J = D#(H ^= 0);
```

But J isn't showing the values I'd like.

Solution

05-05-2017
11:17 AM

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Posted in reply to jl1005

05-05-2017 02:34 AM

The problem is that (H ^= 0) first evaluates to a matrix of zeros and ones, so when this is applied to D with elementwise multiplication it either preserves the existing values in D, or sets them to zero.

It is possible to get the Inverse Scalar Matrix using a small trick. First subtract one from G, next calculate H as you have above, finally add back the one. The overall effect is to place a one, rather than a zero, at every location in H corresponding to one of the modified values.

```
F = (D#(C^=1))[,+];
G = (1-F)/(E-F);
H = (G-1)#(C=1) + 1;
J = D # H;
print H [l='Inverse Scalar Matrix'], J [l='Finished Matrix'];
```

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Posted in reply to jl1005

05-04-2017 09:08 AM

It looks like you want SOLVE() function.

```
proc iml;
A = { 0.20 0.15 0.30 0.35,
0.15 0.25 0.35 0.25,
0.10 0.20 0.50 0.20,
0.25 0.25 0.30 0.20};
C = { 1.0 1.2 1.0 1.0,
1 1 1.2 1.0,
1 1 1 1.2,
1.2 1 1 1};
D=A#C;
Y=j(nrow(A),1,1);
X=solve(D,Y);
want=D#X`;
print want (want[,+]);
quit;
```

WANT

0.2467064 0.0667512 0.3381478 0.3483947 1

0.1850298 0.0927101 0.4734069 0.2488533 1

0.1233532 0.074168 0.5635796 0.2388992 1

0.3700595 0.0927101 0.3381478 0.1990827 1