I've been trying for a while to get a lattice-cell-source running with an ordinary cell-source in MCNP. I can get them both working separately but when I try and combine them I get the following fatal error: distribution 1 for cel is the wrong kind
My source code is as follows:
SDEF PAR=SF
CEL=D9 $ Fatal error. distribution 1 for cel is the wrong kind
X D11
Y D12
Z D13
c
DS9 S 4 10
DS11 S 1 14
DS12 S 2 15
DS13 S 3 16
c
c ---- Lattice cell source-----
SI4 L (3<2[-5:5 -5:5 -10:10]<5)
SP4 1 2540r
SI1 -0.12 0.12
SP1 0 1
SI2 -0.12 0.12
SP2 0 1
SI3 -0.12 0.12
SP3 0 1
c
c ---- Separate cell source ---
SI10 L 25
SP10 1
SI14 22.8 27.2
SP14 0 1
SI15 -2.2 2.2
SP15 0 1
SI16 -2.2 2.2
SP16 0 1
Do any of you know how to declare an embedded source along with an ordinary cell source? Image of the lattice and cell source: light blue indicates source material:
Any help much appreciated.
With help from a colleague we have found a solution, included below. It involves including an FCEL dependency on each distribution, as well as calling the cells directly within the CEL distribution without adding another D-card. Then ALL of the cells must be included in the following distributions.
SDEF PAR=SF
CEL=D9
X=FCEL D11
Y=FCEL D12
Z=FCEL D13
c
c
SI9 L (3<2[-5:5 -5:5 -10:10]<5) 25
c Cells 3 and 25 are called (3 is in the lattice notation)
SP9 1 2540r 2079
c assigning importance to each cell: lattice ~55%, PuO ball ~45%
DS11 S 1 2540r 14
c First 2541 cells have D1 attached, the last cell is attached to D14
DS12 S 2 2540r 15
c First 2541 cells have D2 attached, the last cell is attached to D15
DS13 S 3 2540r 16
c First 2541 cells have D3 attached, the last cell is attached to D16
c
c ---- Lattice cell source-----
SI1 -0.12 0.12
SP1 0 1
SI2 -0.12 0.12
SP2 0 1
SI3 -0.12 0.12
SP3 0 1
c
c ---- Separate cell source ---
SI14 22.8 27.2
SP14 0 1
SI15 -2.2 2.2
SP15 0 1
SI16 -2.2 2.2
SP16 0 1
It's always tricky trying to do so without much context.
Related
I am struggling to find the sum of distinct id's value. Example given below.
Week TID Ano Points
1 111 ANo1 1
1 112 ANo1 1
2 221 ANo2 0.25
2 222 ANo2 0.25
2 223 ANo2 0.25
2 331 ANo3 1
2 332 ANo3 1
2 333 ANo3 1
2 999 Ano9 0.25
2 998 Ano9 0.25
3 421 ANo4 0.25
3 422 ANo4 0.25
3 423 ANo4 0.25
3 531 ANo5 0.5
3 532 ANo5 0.5
3 533 ANo5 0.5
From the above data i need to bring the below result. Could anyone help please using some excel formula?
Week Points_Sum
1 1
2 1.50
3 0.75
You say "sum of distinct id's value"? All the IDs are different so I'm assuming you want to sum for each different "Ano" within the week?
=SUM(IF(FREQUENCY(IF(A$2:A$17=F2,MATCH(C$2:C$17,C$2:C$17,0)),ROW(A$2:A$17)-ROW(A$2)+1),D$2:D$17))
confirmed with CTRL+SHIFT+ENTER
where F2 contains a specific week number
Assumes that each "Ano" will always have the same points value
Probably not the most efficient solution... but this array formula works:
= SUMPRODUCT(IF($A$2:$A$15=$F2,$D$2:$D$15),1/MMULT((IF($A$2:$A$15=$F2,$D$2:$D$15)=
TRANSPOSE(IF($A$2:$A$15=$F2,$D$2:$D$15)))+0,(ROW($A$2:$A$15)>0)+0))
Note this is an array formula, so you have to press Ctrl+Shift+Enter after typing this formula instead of just Enter.
See working example below. This formula is in cell G2 and dragged down.
For example, if I want to maximize the expectation of returns function
E[r]= w1r1+w2r2 and solve the optimization value for the weight w1 and w2.
The only constraint that you have really given is that w1+w2=1
w1 =.0.25
(,~ -.)w1
0.25 0.75
That takes care of both w1 and w2 given the value of w1.
r1 r2 +/#:* w1 w2 calculates r1w1 + r2w2
r1 =. 5
r2 =.10
(r1,r2) (+/#:* (,-.))w1
8.75
(r1,r2) (+/#:* (,-.))0.9
5.5
(r1,r2) (+/#:* (,-.))0.01
9.95
If you really wanted to maximize you would need to add equations for the value of r1 and r2 and take those into account as well, but perhaps I don't understand your question?
Responding to the comment below: If the constraint of w1+w2=1 still is in play, then the matter just becomes summing the values in r1 and r2, then whichever is bigger should get the w value of 1 and the other will get the w value of 0
r1=.2 4 6 3 2
r2=.2.1 4 6 3 2
r3=.2 4 6 3 2.3
r1 (,-.)#:>/#:(+/#:,.) r2
0 1
r2 (,-.)#:>/#:(+/#:,.) r1
1 0
r3 (,-.)#:>/#:(+/#:,.) r2
1 0
'w1 w2'=.r3 (,-.)#:>/#:(+/#:,.) r2
w1
1
w2
0
'w1 w2'=.r1 (,-.)#:>/#:(+/#:,.) r2
w1
0
w2
1
(r1,.r2) +/#:,#:(+ . *) (0 1) NB. w1=.0 w2=.1
17.1
(r1,.r2) +/#:,#:(+ . *) (1 0) NB. w1=.1 w2=.0
17
(r1,.r2) +/#:,#:(+ . *) (0.5 0.5) NB. w1=.0.5 w2=.0.5
17.05
Based on the follow up comment below I would approach it in one of two ways. I could dig up all my linear programming texts from the 1980's and come up with the definitive mathematical solution (including degenerative cases and local maxima/ minima) or using the same technique as above but for a larger case than n=2. I'm going with the second option.
Let's look first at the r matrix which will be a set of constants. For this example I am taking a random 5 X 10 matrix with values from 1 to 10.
r=. >: ? 5 10 $ 10
r
4 4 8 1 4 3 6 9 6 2
2 6 5 4 4 7 5 10 4 6
2 4 9 10 1 1 9 8 2 7
5 6 5 4 7 9 2 6 10 6
10 3 6 2 10 2 7 10 4 2
Now the trick that I am going to use is that I want to find the column with the highest average to be multiplied by the largest value of w. Easy to do with J using (+/ % #)
(+/ % #) r
4.6 4.6 6.6 4.2 5.2 4.4 5.8 8.6 5.2 4.6
Then find the ranking of the list to be able to reorder the columns of the original r matrix. The leading 7 means that 7 { r is the largest average etc.
\:#:(+/ % #) r
7 2 6 4 8 0 1 9 5 3
I use this to in turn reorder the columns of the matrix r using {"1 since I am working columns. The result is that I have reordered the columns of r so that the column with the largest average is on the left and smallest on the right.
(\:#:(+/ % #) {"1 ]) r
9 8 6 4 6 4 4 2 3 1
10 5 5 4 4 2 6 6 7 4
8 9 9 1 2 2 4 7 1 10
6 5 2 7 10 5 6 6 9 4
10 6 7 10 4 10 3 2 2 2
Once I have that, then the next thing is to develop the w vector. Since I now have all the largest averages on the left I will just maximize the values to the left of w to be as large as possible within the noted constraints.
w=. 0.2 0.2 0.2 0.2 0.15 0.01 0.01 0.01 0.01 0.01
#w NB. w1 through w10
10
+/w NB. sum of the values in w
1
>./w NB. largest value in w
0.2
<./w NB. smallest value in w
0.01
Because the r matrix has been reordered using + . * the dot product gives values for w1r1 , w2r2 , w3r3 ... w10r10
(({"1~ \:#: (+/ % #))r) + . * w
1.8 1.6 1.2 0.8 0.9 0.04 0.04 0.02 0.03 0.01
2 1 1 0.8 0.6 0.02 0.06 0.06 0.07 0.04
1.6 1.8 1.8 0.2 0.3 0.02 0.04 0.07 0.01 0.1
1.2 1 0.4 1.4 1.5 0.05 0.06 0.06 0.09 0.04
2 1.2 1.4 2 0.6 0.1 0.03 0.02 0.02 0.02
to actually get the weight of the matrix ravel all the values then sum
+/ , (({"1~ \:#: (+/ % #))r) + . * w
31.22
ID Height Phase Corrected_Height Final
1 0 A 0 0
2 1.2 A 1.2 1.2
3 3.9 A 3.9 3.9
4 5.8 A 5.8 5.8
5 4.6 A NA 7.7
6 7.7 A 7.7 9.3
7 9.3 A 9.3 10.8
..
300 237.5 P 237.5 ..
301 234.7 D 234.7 ..
302 233.3 D 233.3 ..
303 235.1 D NA ..
555 1.0 D 1.0
I have a set of data of similar structure. Calculation of the Phase column was done according to the formula =IF(B2=MAX(B:B);"P";IF(ROW(B2)<MATCH(MAX(B:B);B:B;0);"A";"D"))thanks to #Scott Craner for the solution Naming a behavior in Excel
and to calculate the Corrected_Height column I used =IF(C4="A" & B4>B3;B4; IF(C4="D" & B4<B3;B4;"NA"))). However I did not get the required result. The idea is when in "A" phase, should a lower value arise than the previous one it should change to NA and in "D" phase, should a value be higher than the previous one it should again change to NA. Any suggestion what should I change in the formula? And I also want a final column that gives me the values without NA in it. A,P,and D in phase means Ascent,Peak, and Descent.
The & operator cannot be used to logically and together two conditions in an Excel formula. Instead, use the AND() function:
=IF(AND(C4="A", B4>B3), B4, IF(AND(C4="D", B4<B3), B4, "NA"))
My excel sheet looks like this:
Name C.p Value
a 1 1.75
b 1 2.35
c 1 1.32
d 1 2.45
a 2 2.7
b 2 1.85
c 2 1.9
d 2 2.6
a 3 3.2
b 3 4.5
c 3 9.2
d 3 5.01
Like this 4~5 names 50 ~ 60 check points and values at those check points
I want the excel to look like
C.p a b c d
1 1.75 2.35 1.32 2.45
2 2.7 1.85 1.9 2.6
3 3.2 4.5 9.2 5.01
Here C.p is check point. it is not always 1 2 3 .. it changes values form sheet to sheet
Could Some one help with the code
thank you
If that is the only thing you want to do,You can do it quickly by pivot table in excel itself. You will get some extra columns like Grand Total Which you can remove. As far as effort for removing the unwanted columns to the code it will be quite less.
see the below pic.
I want excel to select the lowest number excluding 0 from the D and G Columns and display it in the H Column. Right now my table looks like this:
A B C D E F G H
1 1 150.00 52 7800.00 0 569.99 0.00 =SMALL((G5,J5),INDEX(FREQUENCY((D1,G1),0),1)+1)
2 0 50.00 52 0.00 0 750.00 0.00 =SMALL((G5,J5),INDEX(FREQUENCY((D2,G2),0),1)+1)
Row 1 works as expected and selects 7800.00 and ignores the 0, but for row 2 selects neither because they're both 0 and displays #NUM!. Any chance that I can fix this?
As Mr SeanC said it was as easy as just using =IFERROR:
=IFERROR(SMALL((G5,J5),INDEX(FREQUENCY((D2,G2),0),1)+1),0)