I want to divide a number (x) into a number (y) of unequal parts ...?

Question

Let's say I want to divide a rope 100 meters long into 25 parts. Each part must have a unique length but the sum of all parts must still be 100 meters.
2 cases are of interest to me:
case 1: i dont care to know or set the size of any parts as long as they sum up to 100 meters and the rate of change should not change from one part to the next (linear?).
case 2: I want to be able to set the size of the biggest part AND the size of the smallest part, both these 2 parts together should not be more than 50 meters or half the total of 100 meters and all the 25 parts together should of course still sum up to 100 meters. In this case the solution should not be linear.
It seems to me reminiscent of some financial formula like npv but I can't figure it out. Help!

Answer 1

I take a_1, a_2, ..., a_25 the 25 parts, a_1 is the smallest, a_25 is the biggest.

If these numbers are, for instance, in a arithmetic progression, we can prove than a_25 is inferior to 8, so, it's a bad method.

I suppose now they are in a geometric progression where the commun ratio is q > 1.

Then we have :

a_25 = a_1 * q^25

and so a_1 + a_25 < 50
a_1 (1 + q^25) < 50
a_1 < 50/(1 + q^25)

the sum is :

a_1 + ... + a_25 = 100

a_1 * (1 - q^26)/(1 - q) = 100

then a_1 = 100(1 - q)/(1 - q^26) < 50/(1 + q^25)

then (1 - q)(1 + q^25)/(1 - q^26) < 1/2

I take f(x) = (1 - x)(1 - x^25)/(1 + x^26)

the graph gives me f(x) < 1/2 for 1/2 < x < 2

then we can take q between 1 and 2

with q = 2 , a_1 is very small and a_25 is very big

with q = 1.1

1 + q^25 = 11.8347
a_1 < 50/11.8347 < 4.224

we can take a_1 = 4
then a_25 = 4 * 1.1^25 = 43.33

the sum is now 4 (1.1^26 - 1)/0.1 > 400 is too much !

After some simulations, It will be difficult to have a_25 not too big and the sum inferior to 100.


Now, if we take a_1 = 0.9, then a_25 = 9.75, the sum will be 98.26, it's better, but a_1 + a_25 is far of 50. Is it good for you ?
We can also modify q (= 1.2 for instance), and find a_1 for a good solution.
Otherwise, another method is perhaps necessary !