n   = which push_back?
cap = size of array holding vaector's data
num = number of data items actually in it, num <= cap
wrk = number of data items copied by this push_back
tot = total of all wrks so far

resize vector by adding 1 to cap

n      1   2   3   4   5   6    7    8   9    10
cap 0    1   2   3   4
num 0    1   2   3   4
wrk 0    1   2   3   4
tot 0    1   3   6  10  15   21   28  36   45   55

total effort is proportional to square of amount of data, = N * (N + 1) / 2

adding a number more than 1 does speed things up but it remains quadratic.


resize vector by doubling cap

n      1   2   3   4   5   6    7    8   9    10  ... ... ...  16     17
cap 0    1   2   4   4   8    8    8    8   16                    16     32
num 0    1   2   3   4   5    6    7    8    9
wrk 0    1   2   3   1   5    1    1    1    9   1 1 1 1 1 1 1        17
tot 0    1   3   6   7  12   13   14   15   24                 31     48

total effort is proportional only to amount of data, = 2 * N - 1


These calculations ignore the time taken by new and delete.
If new and delete are taken into account, doubling is even better.

The time an individual use of new or delete takes does not depend
on the size of the array. It is variable but constant on average
if you see what I mean.

with adding 1 a new and delete are required at every step.
adding some larger number such as six would require a new and delete
for every six pushes. So the number of news and deletes is proportional
to the total amount of data whenever growth is by adding.

When doubling the size, news and deletes are only required at
exponentially increasing intervals, after n = 1, 2, 4, 8, 16, 32, 64, etc
so the total number is proportional to the logarithm of n.