How can I generate those numbers in Excel.
I have to generate 8 random numbers whose sum is always 320. I need around 100 sets or so.
http://en.wikipedia.org/wiki/User:Skinnerd/Simplex_Point_Picking. Two methods are explained here.
Or any other way so I can do it in Excel.
You could use the RAND() function to generate N numbers (8 in your case) in column A.
Then, in column B you could use the following formula B1=A1/SUM(A:A)*320, B2=A2/SUM(A:A)*320 and so on (where 320 is the sum that you are interested into).
So you can just enter =RAND() in A1, then drag it down to A8. Then enter =A1/SUM(A:A)*320 in B1 and drag it to B8. B1:B8 now contains 8 random numbers that sum up to 320.
Sample output:
I'm a bit late to the game here - but fyi if only integers required then:
=LET(x_,RANDARRAY(8,1,1,1000000,1),y_,ROUND(x_*320/SUM(x_),0),y_)
is somewhat similar to the favourite soln above, albeit parsimonious (formula in single cell required to produce desired array , no helper column). Also addresses insignificant decimal points, albeit you may need to allocate back the deficit / surplus due to the occasional rounding error which may yield a sum total of 321 or 319. Could do this in a random fashion again using something like index(y_,randbetween(1,8))+320-sum(y_) in formula above - or resort to the infamous helper fn..
Someone commented the favourite soln above (and thus mine, since it stems from a similar concept/approach) is not uniform - I'm not sure this was required; a uniform spread would impede the random nature (and is arguably far simpler as you simply divide a sizeable range into distinct octiles, and follow the same approach already laid out here - not sure where/why the notion that a random spread should be arbitrarily/mechanically 'forced' to adopting some type of non-random spread.. anyways... I obviously haven't read the problem properly (ehem).
I'm a bit late to the game here - but fyi if only integers required then:
=LET(x_,RANDARRAY(8,1,1,1000000,1),y_,ROUND(x_*320/SUM(x_),0),y_)
is somewhat similar to the favourite soln above, albeit parsimonious (formula in single cell required to produce desired array , no helper column). Also addresses insignificant decimal points, albeit you may need to allocate back the deficit / surplus due to the occasional rounding error which may yield a sum total of 321 or 319. Could do this in a random fashion again using something like index(y_,randbetween(1,8))+320-sum(y_) in formula above - or resort to the infamous helper fn..
Related
I am calculating the geometric mean of a row in MS Excel by using the GEOMEAN(...) command.
What is the geometric mean: The row could be A1:A10. A geometric mean with
GEOMEAN(A1:A10)
is the product of all 10 cell values (multiplied together) after which the 10th root is taken (mathematically: nth_root(A_1 x A_2 x ... x A_n) ).
The issue: The command GEOMEAN(A1:A10) works fine as long as no cells contain negative values (actually just as long as the product ends up positive). If one cell has a negative value, then taking the root is mathematically an invalid action and Excel gives an error.
The solution: I can work-around this by adding a large enough number such as +1000000 to each value before doing GEOMEAN(A1:A10) and afterwards subtracting -1000000 from the result. This is a mathematical approximation to the pure geometrical mean.
The question: But how do I add +1000000 to each value in Excel? A solution would be to create a whole new extra row where the number is added, and then doing GEOMEAN on this row and subtracting the number from the result. But I would really like to avoid creating a new row, since I have many long data sets to perform this command on.
Is there a way to add the number inside the command itself? To add it onto each value before it is multiplied? Something along the lines of:
GEOMEAN(A1:A10+1000000)-1000000
Solution to avoid the work-around
Based on the answer from and discussion with #ImaginaryHuman072889
It turns out that a working command that avoids any work-around is:
IFERROR(GEOMEAN(A1:A10);-GEOMEAN(ABS(A1:A10)))
If an error are cought by the IFERROR, then we know that a negative result would have appeared, so this is constructed manually in that case.
BUT: This does not take into account the case mentioned by #ImaginaryHuman072889, though, because Excel seems to forbid any negative numbers involved and not just if the inner product is negative. For example, both GEOMEAN(-2,-2) as well as GEOMEAN(-2,-2,-2) give errors in Excel, even though they both should be mathematically valid, giving the results 2 and -2, respectively. To overcome this Excel-issue, we can simply write out the exact same command line manually:
IFERROR(PRODUCT(A1:A10)^(1/COUNTA(A1:A10));-(PRODUCT(ABS(A1:A10))^(1/COUNTA(A1:A10)))))
I add this solution to aid any by-comers who have the same issue. This mathematically works, but the fact that -2 and -2 have the geometrical mean 2 does seem a bit odd and not at all like any useful value of a "mean". It is still mathematically legal as far as I can find (WolframAlpha has no issue with it and the Wikipedia article never mentions a sign).
Your "workaround" of doing this:
GEOMEAN(A1:A10+1000000)-1000000
Is completely wrong. This is absolutely not equal to GEOMEAN(A1:A10).
Simple counter-example:
GEOMEAN({2,8}) returns the value of 4, which is the geometric mean of 2 and 8.
GEOMEAN({2,8}+1)-1 is equal to GEOMEAN({3,9})-1 which is approximately 4.196.
What is a valid workaround is if you multiply each value inside GEOMEAN by a certain value, then divide the result by that value.
Simple example:
GEOMEAN({2,8}*3)/3 is equal to GEOMEAN({6,24})/3 which is 4.
However, this method of multiplying by a constant does not help your situation, since this won't get rid of negative values.
Mathematically speaking, the geometric mean of a positive number and a negative number is an imaginary number, which is presumably why Excel cannot handle it.
Example:
2*-8 = -16
sqrt(-16) = 4i
Therefore, 4i is the geometric mean of 2 and -8. Notice how it has the same magnitude as GEOMEAN({2,8}), just that it is an imaginary number.
All that said... here is what I recommend you doing:
I suggest you return two results, one result is the magnitude of the geometric mean and the other is the phase of the geometric mean.
Formula for magnitude:
= GEOMEAN(ABS(A1:A10))
(Note, this is an array formula, so you'd have to press Ctrl+Shift+Enter instead of just Enter after typing this formula.) The use of ABS converts all negative numbers to positive before the GEOMEAN calculation, guaranteeing a positive geometric mean.
Formula for phase, I would just do something like this:
= IF(PRODUCT(A1:A10)>=0,"Real","Imaginary")
Which obviously returns Real if the geometric mean is a real number and returns Imaginary if the geometric mean is an imaginary number.
EDIT
Technically speaking, some of what I said wasn't completely precise, although the magnitude formula above still stands.
Some things I want to clarify:
If PRODUCT(data) is positive (or zero), then the geometric mean of data is positive (or zero).
If PRODUCT(data) is negative and if the number of entries in data is odd, then the geometric mean of data is negative (but still real).
If PRODUCT(data) is negative and if the number of entries in data is even, then the geometric mean of data is imaginary.
That said... if you want these formulas to be a bit more technically accurate, I would modify to this:
Adjusted formula for magnitude:
= GEOMEAN(ABS(A1:A10))*IF(AND(PRODUCT(A1:A10)<0,MOD(COUNT(A1:A10),2)=1),-1,1)
Adjusted formula for phase:
= IF(AND(PRODUCT(A1:A10)<0,MOD(COUNT(A1:A10),2)=0),"Imaginary","Real")
If the geometric mean is real, it returns the precise geometric mean (whether it is positive or negative), and if the geometric mean is imaginary, it returns a positive real value with the correct magnitude.
So, I just found the answer - although I have no idea why this works.
Doing GEOMEAN(A1:A10+1000000)-1000000 is actually possible. But by pressing enter and error #VALUE is displayed. You must click control+shift+enter to have the actual result displayed.
According to this: https://www.mrexcel.com/forum/excel-questions/264366-calculating-geometric-mean-some-negative-values.html
If anyone has an explanation for this, I am very interested.
I want to calculate the response time across multiple subjects with one restriction: only response time from the correct trials should be included in the average. The structure of my data looks like below (for simplicity, I show only 3 subjects and 10 trials, in reality, I have many more)
I would like to get average of RT across subj1, subj2, and subj2 for each of the trials. Only correct trials are included in the average. 0 and 1 are used to denote incorrect and correct trials, respectively. For instance, for cell G2, I would only include B2 and D2 in the average, F2 is left out since the ACC for that trial from that subject is 0. I imagined using If AND function to include the appropriate RT but with many subjects, this becomes very clumsy. Does anyone have a clever solution to this?
Since 0 * anything = 0, G2 = SUM(A2*B2,C2*D2,E2*F2)/SUM(A2,C2,E2)
You can do this with AVERAGE and an array formula which can be easily extended to larger ranges, i.e.
=AVERAGE(IF((RIGHT(B$1:F$1,2)="RT")*(A2:E2=1),B2:F2)
confirm with CTRL+SHIFT+ENTER
...or even simpler with AVERAGEIFS like this
=AVERAGEIFS(B2:F2,A2:E2,1,B$1:F$1,"*RT")
note the “offset” ranges
I have a excel sheet with a few formulas like this:
A1,A2,A3= 0.13,1.25,2.21
A4: =(A1*A2) =0.16 ( 2 decimal points)
A5: =(A2*A3) =2.76 ( 2 decimal points)
A6: =SUM(A4;A5) =2.93 ( 2 decimal points )
And i want to show 0.16+2.76=2.92
well, there's my problem in bold. i want to add the values from the cells, not the formuls result. How can i do that ? Thank you
Presumably you're working with money which is why you need this.
One way to resolve this is to use =ROUND(A1*A2, 2) etc. and base your subsequent calculations from that.
Do be aware though that you will still occasionally get spurious results due to Excel using a 64 bit IEEE754 floating point double to represent numbers. (Although it does have some extremely clever circumvention techniques - see how it evaluates 1/3 + 1/3 + 1/3 - it will not resolve every possible oddity). If you're building an accounting-style sheet you are best off working in pence, and dividing the final result.
Round the values before you sum, ie:
=ROUND(A1*A2,2)
=ROUND(A2*A3,2)
You could wrap your formulas with the ROUND function:
=ROUND(A1*A2,2)
This will give you 0.16 as opposed to 0.163. Do this for each of your calculations and you'll only be calculating everything to two decimal places. Although I'm not sure why you'd want to do that.
So I'm creating a spreadsheet that determines the cost of materials and the number of each material needed in order to complete a desire project using input from myself. Right now the desired project is a wall that is 250x9 that requires replace all the 4x8 sheets of wood with OSB and install Vinyl Siding. The issue I'm running into is I cannot get it to always round up. By that I mean even if the value is 1.1 it should round up. In this specific case I am buying nails for my nail gun in a box of 2,000 and each sheet of OSB will have 32 nails in it. If 250x9 area requires 70.3125 sheets of OSB it means I still have to buy 71 sheets of OSB. If that OSB is 71 sheets then it require that I have 2272 Nails then the result is I need 1.125 Boxes of nails. However I can't seem to get it to show this as 2 boxes because again I still need to purchase more than one box to complete the project. So with that being said if I take the number of OSB needed 70.3125 and I place it in a formula with a roundup function it still rounds down (gives me a headache that there is a roundup and a rounddown function and it will still round down on me. Perhaps it is the way I am using it in the formula that is incorrect, I'm not sure. So let me translate the formula's used and you can let me know if I'm doing something wrong or if there is a function or set of functions that I can use to solve this issue.
=SUM(((B30*C30)+(B35*C35)+(E30*F30)+(H30*I30))/(E9*G9))
This says that if I added Wall1 L*W with Wall2 L*W with Wall3 L*W with Wall4 L*W and divide it by OSB H*W I get the number of sheets needed. Which in this case is 2250/32 basically. But its programmed in a manner that I can input the information for individual walls to different area's and get it to spit out the total SqFt for each wall and give an individual breakdown per wall of material needed with cost associated per sq ft of material bleh bleh bleh. The point is I take the result that is the 70.3125 and I move it to a different workbook and I say "Sheets OSB Needed" and in that box I have
=ROUNDUP(Sheet1!A9,1)
Whereas I'm asking it to roundup A9 which is the result of the above formula by intervals of 1. But the output is still 70 instead of 71. and much the same case with the nails needed. Which can be calculated in a few different manners but regardless the amount of nails needed divided by 2000 would output the decimal answer which yields a value of less than 1.5 and it too provides me with a value of 1 instead of 2 with much the same formula. I could achieve my desired result I suppose with Trunc and Mod functions collaborating using multiple cells to output the different portions of the data. But is there a way to do this that doesn't involve so many cells being used up?
C7
=Trunc(A9)
Removes Decimal from 70.3125
C8
=MOD(A9)
Outputs decimals from 70.3125
C9
=IF(C8<1,"1",C8)
If Decimals are < a whole number make it a whole number
C3
=SUM(C7+C9)
Add the whole number to the Trunc Number to get value desired.
Which I'm already seeing an issue with this if there is no decimals in the sheets needed then wouldn't it always add one because the decimal place would be 0? How can I handle this issue? Isn't there an easier way to do this or a way to code it so that its all nested into one calculation or at least mostly all into one calculation without making a circular reference of some sort?
You need to change the second parameter to a 0 ROUNDUP(70.3125, 1) is 70.3 the 3 must be getting dropped elsewhere or lost in formatting.
ROUNDUP(70.3125, 0) will give 71.
The second parameter of round up is the decimal place. So to round to integers it should be 0 not 1
I have a requirement in Excel to spread small; i.e. pennies, monetry rounding errors fairly across the members of my club.
The error arises when I deduct money from members; e.g. £30 divided between 21 members is £1.428571... requiring £1.43 to be deducted from each member, totalling £30.03, in order to hit the £30 target.
The approach that I want to take, continuing the above example, is to deduct £1.42 from each member, totalling £29.82, and then deduct the remaining £0.18 using an error spreading technique to randomly take an extra penny from 18 of the 21 members.
This immediately made me think of Reservoir Sampling, and I used the information here: Random selection,
to construct the test Excel spreadsheet here: https://www.dropbox.com/s/snbkldt6e8qkcco/ErrorSpreading.xls, on Dropbox, for you guys to play with...
The problem I have is that each row of this spreadsheet calculates the error distribution indepentently of every other row, and this causes some members to contribute more than their fair share of extra pennies.
What I am looking for is a modification to the Resevoir Sampling technique, or another balanced / 2 dimensional error spreading methodology that I'm not aware of, that will minimise the overall error between members across many 'error spreading' rows.
I think this is one of those challenging problems that has a huge number of other uses, so I'm hoping you geniuses have some good ideas!
Thanks for any insight you can share :)
Will
I found a solution. Not very elegant, through.
You have to use two matrix. In the first you get completely random number, chosen with =RANDOM() and in the second you choose the n greater value
Say that in F30 you have the first
=RANDOM()
cell.
(I have experimented with your sheet.)
Just copy a column of n (in your sheet 8) in column A)
In cell F52 you put:
=IF(RANK(F30,$F30:$Z30)<=$A52, 1, 0)
Until now, if you drag left and down the formulas, you have the same situation that is in your sheet (only less elegant und efficient).
But starting from the second row of random number you could compensate for the penny esbursed.
In cell F31 you put:
=RANDOM()-SUM(F$52:F52)*0.5
(pay attention to the $, each random number should have a correction basated on penny already spent.)
If the $ are ok you should be OK dragging formulas left and down. You could also parametrize the 0.5 and experiment with other values. With 0,5 I have a error factor (the equivalent of your cell AB24) between 1 and 2