Transcript Estate Division (Who`s getting what?)

Apportionment
There are two critical elements in the
dictionary definition of the word apportion
: (1) We are dividing and assigning things,
and (2) we are doing this on a
proportional basis and in a planned,
organized fashion.
 In the next slide we will look at an
example that illustrates the nature of the
problem we are dealing with.

‘Kitchen Capitalism’
Mom has a total of 50 identical pieces of candy, which she is planning
to divide among her five children (this is the division part). She wants
to teach her children about the value of work and about the
relationship between work and reward.
She announces to the kids that the candy is going to be divided at the
end of the week in proportion to the amount of time each of them
spends helping with the weekly kitchen chores–if you worked twice as
long as your brother you get twice as much candy, and so on (this is
the due and proper proportion part).
Unwittingly, mom has turned this division problem into an
apportionment problem.

At the end of the week, the numbers are
in. The Table shows the amount of work
done by each child during the week.

According to the ground rules, Alan, who worked 150 out of a
total of 900 minutes, should get 8 1/3 pieces.

Here comes the problem: Since the pieces of candy are
indivisible, it is impossible for Alan to get his pieces–he can
get 8 pieces (and get shorted) or he can get 9 pieces (and
someone else will get shorted).

A similar problem occurs with each of the other children.
Betty’s exact fair share should be 4 1/3 pieces; Connie’s
should be 9 11/18 pieces; Doug’s, 11 1/3 pieces; and Ellie’s,
16 7/18 pieces.

Because none of these shares can be realized, an absolutely
fair apportionment of the candy is going to be impossible.

What should mom do?
Apportionment- Why and
Definitions!

Apportionment problems can arise in a variety of real-life
applications –
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dividing candy among children,
assigning nurses to shifts,
assigning buses to routes, and so on.
But the gold standard for apportionment applications is
the allocation of seats in a legislature, and thus it is
standard practice to borrow the terminology of
legislative apportionment and apply it to apportionment
problems in general.
Definitions!!!!!!!!!

The basic elements of every apportionment
problem are:
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the “states”,
the “seats”, and
the populations”
The “states”: This is the term we will use to
describe the parties having a stake in the
apportionment. Unless they have specific names
(Azucar, Bahia, etc.), we will use A1, A2,…, AN,
to denote the states.!
More Definitions!!!!!!

The “seats”: This term describes the set of M identical,
indivisible objects that are being divided among the N
states.
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For convenience, we will assume that there are more seats than
there are states, thus ensuring that every state can potentially
get a seat. (This assumption does not imply that every state
must get a seat!)
The “populations”: This is a set of N positive numbers
(for simplicity we will assume that they are whole
numbers) that are used as the basis for the
apportionment of the seats to the states.

We will use p1, p2,…, pN, to denote the state’s respective
populations and P to denote the total population P = p1 + p2
+…+ pN.
Apportionment
Apportionment – the distribution of an item fairly
among a group.
Ideal ratio (Standard Divisor)Total population
Number of seats
Quota - individual population
ideal ratio
Apportionment
A school has the following student
populations: senior – 435, juniors 525,
sophomores – 581, and freshmen – 653.
There are 40 seats in the student council.
a.) Find the ideal ratio (round to the
nearest hundredth)
b.) Find the quota for each class
c.) Find the apportioned number of seats
for each of the four methods (Hamilton,
Jefferson, Webster, and Hill)
Methods of Apportionment
Hamilton Method – award the remaining seat(s)
to the choice with the highest decimal in the
quota. If there are more than one seat to
award, then highest decimal first and work your
way in descending order until all seats are
awarded.
Hamilton’s Method
Class (#)
Sr- 435
Jr- 525
So- 581
Fr- 653
Total 2194
Quota
Lower Quota
Initial
Apportionment
Residue
FINAL
Apportionment
Ideal Ratio =
Hamilton uses the highest decimal from the quota to decide who
receives the next seat.
Jefferson Method – Find the adjusted ratio for
each choice and the adjusted ratio closest to
the ideal ratio receives the seat. If more
than one seat is being awarded, then start
with choice whose adjusted ratio is closest to
ideal ratio and work your way out until all
available seats have been awarded.
adjusted ratio = Individual population
truncated quota + 1
Jefferson’s Method
Class (#)
Sr- 435
Jr- 525
So- 581
Fr- 653
Total 2194
Quota
Lower Quota
Initial
Apportionment
Adjusted Ratio
FINAL
Apportionment
Ideal Ratio =
Adj. Ratio - Divide the class population by the Lower Quota+ 1
Webster Method 1.) Find the quota for each state.
2.) Round each quota.
3.) Find the Adjusted Ratio… (next slide)
4.) Decide whether the house is okay,
overfilled, or underfilled.
Find Adjusted Ratio:
adjusted ratio = Individual population
arithmetic mean
(In this case, the arithmetic mean will be the
same as Lower Quota + .5)
Overfilled – The state with the smallest adjusted
ratio will lose a seat.
Underfilled – The state with the largest adjusted
ratio will gain a seat.
Webster’s Method
Class (#)
Sr- 435
Jr- 525
So- 581
Fr- 653
Quota
ROUNDED
Quota
Initial
Apportionment
Adjusted Ratio
FINAL
Apportionment
Ideal Ratio =
Divide the class population by the truncated value + . 5
Total 2194
Hill Method –
1.) Find the quota for each state.
2.) Round each quota, using Geometric
mean.
3.) Decide whether the house is okay,
overfilled, or underfilled.
adjusted ratio = Individual population
geometric mean
______________________
The geometric mean is: √ (Lower Quota) X (Lower Quota + 1)
Overfilled – The state with the smallest adjusted
ratio will lose a seat.
Underfilled – The state with the largest adjusted
ratio will gain a seat.
Hill’s Method
Class (#)
Sr- 435
Jr- 525
So- 581
Fr- 653
Total 2194
Quota
Geometric
Mean
ROUNDED
Quota
Initial
Apportionment
Adjusted Ratio
FINAL
Apportionment
Ideal Ratio =
Divide the class population by the Square root of LQ value times the
LQ value plus one
COMPARING ALL 4 METHODS:
Ideal Ratio =
Class (#)
Quota
Lower Quota
Hamilton
Jefferson
Webster
Hill
Sr- 435
Jr- 525
So- 581
Fr- 653
Total 2194
Apportionment
The new island state of Bennettland has 30 seats to
award the 6 counties that make up the state’s House
of Representatives. Here are the populations of the 6
counties:
A – 1000, B – 255, C – 2339,
D – 311, E – 634, F – 531.
a.) Find the ideal ratio
b.) Find the quota for each county
c.) Find the number of apportioned seats in all
four methods.
Induction
Steps for solution:
1.) Plug in a value for the variable and check to
see if the equation works for a value.
2.) Write the original equation
3.) Rewrite equation and change the variable to
the variable k.
4.) Write the equation in terms of k+1
5.) Substitute the right side of the equation in
terms of k into the beginning of the left side of
the equation in terms of k+1. Solve (Make both
sides look the same, show your work)
Solve by Mathematical Induction
1.) 1 + 2 + 3 + … + n = n (n+1)
2
Step 1
1 + 2 + 3 + … + n = n ( n+1)
2
Plug in any value for n to see if the equation works for a value.
Set n = 4
So,
1+2+3+4=
4 (4 + 1)
2
10
=
20
2
10
=
10
√ works
On the left side the value for n represent the number of terms you
are going to use to find the value, I chose n=4, so I used the
first four terms. The last term on the left side tells you how the
pattern is formed.
Steps 2 and 3
1 + 2 + 3 + … + n = n ( n+1)
2
1 + 2 + 3 + … + n = n ( n+1)
2
1 + 2 + 3 + … + k = k ( k+1)
2
Steps 4
1 + 2 + 3 + … + n = n ( n+1)
2
1 + 2 + 3 + … + k = k ( k+1)
2
1 + 2 + 3 + … + k + (k+1) = (k+1) ( (k+1) +1)
2
Notice the left side the k+1 is placed at the end
and on the right side the variables k are
changed to k+1
Step 5
1 + 2 + 3 + … + n = n ( n+1)
2
1 + 2 + 3 + … + k = k ( k+1)
2
1 + 2 + 3 + … + k + (k+1) = (k+1) ( (k+1) +1)
2
k ( k+1) + (k+1) = (k+1) ( (k+2)
2
2
Solve by using Mathematical Induction.
1 + 4 + 9 + … + n2 = n(n + 1)(2n + 1)
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