how many number of values when added becomes equal to B(Known value)

I have to find the X value for which \sum_(i=1)^X (i * a_i ) >= B
or ( 1*a_1 + 2*a_2 + --- + X*a_x >= B ) where a_i & B are known.
For finding the X value; is there any explicit formula? If anyone knows, please answer.
Thank you very much, kalpana

 Respuesta aceptada

Assume 'a' is a row vector.
x = 1:size(a,2);
X = x(cumsum(x.*a)>=B);

5 comentarios

Thank you for your answer. But, i am looking for any explicit formula for finding X. If you know, please answer.
The values in that final X are the ones for which your inequality holds. It is as "explicit" as it is possible to be, unless perhaps you are looking for, say, the first X for which that is true in which case you would choose the first element in X.
In equation ( 1*a_1 + 2*a_2 + --- + X*a_x >= B ) , i have to find X ( of course, the first X value that satisfies the equation ). Is there any formula for finding X ? ( why i want this is, i am trying to reduce the number of flops for finding X. So, if there is expression for finding X, then i could reduce no. of flops ).
I've already told you. The first X that satisfies your inequality is the first element in my X. It is of course X(1).
Thank you for your answer sir.

Iniciar sesión para comentar.

Más respuestas (1)

ES
ES el 28 de Jul. de 2014
Immediate Response would be, you can run a loop and sum it up and check if the sum is greater than or equal to B at every iteration and break out of loop when it is satisfied.
A more measured approach will be, to use summation by parts. For example if length of a_i is 100, you could first check if summation(a_i*X) for X=1:50. If B is smaller, then you can run the summation for X=1:75. If B is greater, then you can run for X=1:25 and proceed so on.. But for this to work, a_i should be monotonically increasing.
Analysis of a_i (if it is a series or so) could also be useful.

1 comentario

Thank you for answering. Your suggestion is useful.
Can we have some explicit formula for this if a_i's random numbers.--kalpana

Iniciar sesión para comentar.

Categorías

Más información sobre Mathematics en Centro de ayuda y File Exchange.

Etiquetas

Preguntada:

el 28 de Jul. de 2014

Comentada:

el 30 de Jul. de 2014

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by