EXAMPLE 20 Rounding Summands

Let’s use the Hill-Huntington method to round each summand in

Each summand is the quota, for the standard divisor is 1. The rounding points are as follows:

Each summand is greater than its corresponding rounding point, so all are rounded up, resulting in a too large sum: . We’ll need to choose a divisor larger than 1; let’s try 2. The apportionment quotients are

The first two apportionment quotients are between 0 and 1, and so they are rounded to 1; the third is below the rounding point between 1 and 2, so it too rounds to 1. The sum, 3, is too small.

Let’s try again with a smaller divisor, . The new apportionment quotients are

The first two apportionment quotients are still between 0 and 1, so they round to 1. The third is greater than the rounding point between 1 and 2, so it rounds to 2. Therefore the Hill-Huntington rounding of this sum is