random number generation for known sum

i what do generate some 15 random numbers that sum for each separate value of say, 60 , 70, 40, 65. How would i go abouts this?
sum = [60 70 40 65];
n=15;
b = [ ];
for i = 1:1:n,
a = randi([1,??]);
b = [b a];
end

 Akzeptierte Antwort

Azzi Abdelmalek
Azzi Abdelmalek am 27 Aug. 2013

1 Stimme

n=60;
m=1:n;
a=m(sort(randperm(60,15)));
b=diff(a);
b(end+1)=60-sum(b)
sum(b)

13 Kommentare

thanks Azzi, i get an error
Error in ==> Untitled at 3
a=m(sort(randperm(60,15)));
You must be using an older MATLAB that does not have that enhancement to randperm(). Try
t = randperm(n);
a=m(sort(t(1:15));
harley
harley am 27 Aug. 2013
thanks
Arnab Pal
Arnab Pal am 4 Nov. 2018
I want to generate the same, but I need Non-integer numbers. Is it possible?
Bruno Luong
Bruno Luong am 4 Nov. 2018
Bearbeitet: Bruno Luong am 4 Nov. 2018
@Arnab
sumtarget = 60;
n = 5;
x = diff([0,sort(randperm(sumtarget+n-1,n-1)),sumtarget+n])-1
Arnab Pal
Arnab Pal am 4 Nov. 2018
Bearbeitet: Arnab Pal am 4 Nov. 2018
Sir, It is generating the integer numbers only.
Bruno Luong
Bruno Luong am 4 Nov. 2018
Ah sorry some how I read non-negative.
For random non-negative floating numbers, you need using Roger Stafford FEX
Tejas
Tejas am 13 Jul. 2020
Is there an easy way to control the maximum value each number in 'b' can have? Say, each value in 'b' must be less than or equal to 7, and still the 15 numbers in 'b' should add up to 60.
Walter Roberson
Walter Roberson am 13 Jul. 2020
Roger's FEX contribution, the a and b parameters are lower bound and upper bound. You can use zeros for the lower bound if that is appropriate for your situation.
Tejas
Tejas am 13 Jul. 2020
Bearbeitet: Tejas am 13 Jul. 2020
I want the numbers to be integers, as in the original question. Roger's FEX contribution seems to work with real numbers.
Bruno Luong
Bruno Luong am 13 Jul. 2020
Bearbeitet: Bruno Luong am 13 Jul. 2020
The easiest way is perhaps using Roger FEX function, then do some sort of "integering" the float solution
sumatarget = 60
n = 15;
ub = 7;
x = floor([0; cumsum(randfixedsum(n,1,sumatarget,0,ub))]);
x(end) = sumatarget; % prevent floating point error
r = diff(x)
The distribution might be not perfectly uniform but possibly close enough and suitable for what ever you want to do with it.
Tejas
Tejas am 14 Jul. 2020
Integering the solution from Roger's function works very well for me. I do not require the numbers to be perfectly uniform. Thanks!

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