How to Update value of some matriks element with looping?

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I have condition for my matrix. Matrix x1,x2, and y. When element of y is less than 10, the y element will update will sum of x1 and x2 until the element is 10 or more. This is my code.
clc;
clear;
x1=[1 2 3 4 5 6 7 8 9 10];
x2=[1 2 3 4 5 6 7 8 9 10];
y=[10 9 11 10 11 11 9 10 12 12];
while y<10
y=x1+x2;
end
y
And then, if the problem obove is clear. I want to know how many sum (in this case, how many iteration) to achieve values of 10 or more of each element
  4 Comments
Ameer Hamza
Ameer Hamza on 27 Mar 2020
Edited: Ameer Hamza on 27 Mar 2020
You said it make all elements of y 10 or more but then you said
[10 9 11 10 11 11 9 10 12 12] become [10 12 11 10 11 11 9 10 12 12]
The second vector still has 9.
Eddy Iswardi
Eddy Iswardi on 28 Mar 2020
Yeah. But it's only an example. y(7) will be update to, some ways with y(2). So the vector will get [10 12 11 10 11 11 14 10 12 12]

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Accepted Answer

Ameer Hamza
Ameer Hamza on 28 Mar 2020
Try this:
x1=[1 2 3 4 5 6 7 8 9 10];
x2=[1 2 3 4 5 6 7 8 9 10];
y=[10 9 11 10 11 11 9 10 12 12];
k = 10;
x12 = x1 + x2;
mask = y < k;
y(mask) = ceil(k./x12(mask)).*x12(mask);
Result:
y =
10 12 11 10 11 11 14 10 12 12
  2 Comments
Les Beckham
Les Beckham on 28 Mar 2020
Edited: Les Beckham on 28 Mar 2020
For this part of your question: "I want to know how many sum (in this case, how many iteration) to achieve values of 10 or more of each element", the "iterations" required are given by this part of Ameer's excellent answer:
ceil(k./x12(mask))
If you do this:
iterations = zeros(size(x1));
iterations(mask) = ceil(k./x12(mask))
you will get a vector showing the number of sums required for every element of your original vector with zeros where no iterations (sums) were required.

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