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The Huntington-Hill Method – Version 2Lecture 24Section 4.5
Robb T. Koether
Hampden-Sydney College
Wed, Oct 18, 2017
Robb T. Koether (Hampden-Sydney College) The Huntington-Hill Method – Version 2 Wed, Oct 18, 2017 1 / 21
1 Verson 2
2 Assignment
Robb T. Koether (Hampden-Sydney College) The Huntington-Hill Method – Version 2 Wed, Oct 18, 2017 2 / 21
Outline
1 Verson 2
2 Assignment
Robb T. Koether (Hampden-Sydney College) The Huntington-Hill Method – Version 2 Wed, Oct 18, 2017 3 / 21
The Huntington-Hill Method
Initially, every state gets a quota q = 1 (as required by theConstitution).Then divide each state’s population p by D =
√q(q + 1), where q
is that state’s current quota (initially D =√
2).The state with the largest such quotient gets one more seat, soadd 1 to its quota q.Repeat the previous 2 steps until all the seats have beenapportioned.
Robb T. Koether (Hampden-Sydney College) The Huntington-Hill Method – Version 2 Wed, Oct 18, 2017 4 / 21
Example
Example (Example – Version 2)The populations of three states are 3,7 and 10 million people,respectively.The total number of seats apportioned to those states is 7.Use Version 2 to determine how many seats each state shouldget.
Robb T. Koether (Hampden-Sydney College) The Huntington-Hill Method – Version 2 Wed, Oct 18, 2017 5 / 21
Example
Example (Example – Version 2)
State Population (p) Seats (q) D =√
q(q + 1) p/DA 3 1
√1 · 2 = 1.41 3√
2= 2.12
B 7 1√
1 · 2 = 1.41 7√2= 4.94
C 10 1√
1 · 2 = 1.41 10√2= 7.07
Robb T. Koether (Hampden-Sydney College) The Huntington-Hill Method – Version 2 Wed, Oct 18, 2017 6 / 21
Example
Example (Example – Version 2)
State Population (p) Seats (q) D =√
q(q + 1) p/DA 3 1
√1 · 2 = 1.41 3√
2= 2.12
B 7 1√
1 · 2 = 1.41 7√2= 4.94
C 10 1√
1 · 2 = 1.41 10√2= 7.07
Robb T. Koether (Hampden-Sydney College) The Huntington-Hill Method – Version 2 Wed, Oct 18, 2017 7 / 21
Example
Example (Example – Version 2)
State Population (p) Seats (q) D =√
q(q + 1) p/DA 3 1
√1 · 2 = 1.41 3√
2= 2.12
B 7 1√
1 · 2 = 1.41 7√2= 4.94
C 10 2√
1 · 2 = 1.41 10√2= 7.07
Robb T. Koether (Hampden-Sydney College) The Huntington-Hill Method – Version 2 Wed, Oct 18, 2017 8 / 21
Example
Example (Example – Version 2)
State Population (p) Seats (q) D =√
q(q + 1) p/DA 3 1
√1 · 2 = 1.41 3√
2= 2.12
B 7 1√
1 · 2 = 1.41 7√2= 4.94
C 10 2√
2 · 3 = 2.44 10√6= 4.08
Robb T. Koether (Hampden-Sydney College) The Huntington-Hill Method – Version 2 Wed, Oct 18, 2017 9 / 21
Example
Example (Example – Version 2)
State Population (p) Seats (q) D =√
q(q + 1) p/DA 3 1
√1 · 2 = 1.41 3√
2= 2.12
B 7 1√
1 · 2 = 1.41 7√2= 4.94
C 10 2√
2 · 3 = 2.44 10√6= 4.08
Robb T. Koether (Hampden-Sydney College) The Huntington-Hill Method – Version 2 Wed, Oct 18, 2017 10 / 21
Example
Example (Example – Version 2)
State Population (p) Seats (q) D =√
q(q + 1) p/DA 3 1
√1 · 2 = 1.41 3√
2= 2.12
B 7 2√
1 · 2 = 1.41 7√2= 4.94
C 10 2√
2 · 3 = 2.44 10√6= 4.08
Robb T. Koether (Hampden-Sydney College) The Huntington-Hill Method – Version 2 Wed, Oct 18, 2017 11 / 21
Example
Example (Example – Version 2)
State Population (p) Seats (q) D =√
q(q + 1) p/DA 3 1
√1 · 2 = 1.41 3√
2= 2.12
B 7 2√
2 · 3 = 2.44 7√6= 2.85
C 10 2√
2 · 3 = 2.44 10√6= 4.08
Robb T. Koether (Hampden-Sydney College) The Huntington-Hill Method – Version 2 Wed, Oct 18, 2017 12 / 21
Example
Example (Example – Version 2)
State Population (p) Seats (q) D =√
q(q + 1) p/DA 3 1
√1 · 2 = 1.41 3√
2= 2.12
B 7 2√
2 · 3 = 2.44 7√6= 2.85
C 10 2√
2 · 3 = 2.44 10√6= 4.08
Robb T. Koether (Hampden-Sydney College) The Huntington-Hill Method – Version 2 Wed, Oct 18, 2017 13 / 21
Example
Example (Example – Version 2)
State Population (p) Seats (q) D =√
q(q + 1) p/DA 3 1
√1 · 2 = 1.41 3√
2= 2.12
B 7 2√
2 · 3 = 2.44 7√6= 2.85
C 10 3√
2 · 3 = 2.44 10√6= 4.08
Robb T. Koether (Hampden-Sydney College) The Huntington-Hill Method – Version 2 Wed, Oct 18, 2017 14 / 21
Example
Example (Example – Version 2)
State Population (p) Seats (q) D =√
q(q + 1) p/DA 3 1
√1 · 2 = 1.41 3√
2= 2.12
B 7 2√
2 · 3 = 2.44 7√6= 2.85
C 10 3√
3 · 4 = 3.46 10√12
= 2.88
Robb T. Koether (Hampden-Sydney College) The Huntington-Hill Method – Version 2 Wed, Oct 18, 2017 15 / 21
Example
Example (Example – Version 2)
State Population (p) Seats (q) D =√
q(q + 1) p/DA 3 1
√1 · 2 = 1.41 3√
2= 2.12
B 7 2√
2 · 3 = 2.44 7√6= 2.85
C 10 3√
3 · 4 = 3.46 10√12
= 2.88
Robb T. Koether (Hampden-Sydney College) The Huntington-Hill Method – Version 2 Wed, Oct 18, 2017 16 / 21
Example
Example (Example – Version 2)
State Population (p) Seats (q) D =√
q(q + 1) p/DA 3 1
√1 · 2 = 1.41 3√
2= 2.12
B 7 2√
2 · 3 = 2.44 7√6= 2.85
C 10 4√
3 · 4 = 3.46 10√12
= 2.88
Robb T. Koether (Hampden-Sydney College) The Huntington-Hill Method – Version 2 Wed, Oct 18, 2017 17 / 21
Which Version to Use?
Suppose we had 3 states, with populations 2, 5, and 8 million, and100 seats to apportion.Which method would be faster?
Why?
Robb T. Koether (Hampden-Sydney College) The Huntington-Hill Method – Version 2 Wed, Oct 18, 2017 18 / 21
Which Version to Use?
Suppose we had 3 states, with populations 2, 5, and 8 million, and100 seats to apportion.Which method would be faster?Why?
Robb T. Koether (Hampden-Sydney College) The Huntington-Hill Method – Version 2 Wed, Oct 18, 2017 18 / 21
Which Version to Use?
Suppose we had 8 states, with populations 1, 2, 4, 5, 8, 10, 13,and 14 million, and 9 seats to apportion.Which method would be faster?
Why?Work this example with M = 12.
Robb T. Koether (Hampden-Sydney College) The Huntington-Hill Method – Version 2 Wed, Oct 18, 2017 19 / 21
Which Version to Use?
Suppose we had 8 states, with populations 1, 2, 4, 5, 8, 10, 13,and 14 million, and 9 seats to apportion.Which method would be faster?Why?
Work this example with M = 12.
Robb T. Koether (Hampden-Sydney College) The Huntington-Hill Method – Version 2 Wed, Oct 18, 2017 19 / 21
Which Version to Use?
Suppose we had 8 states, with populations 1, 2, 4, 5, 8, 10, 13,and 14 million, and 9 seats to apportion.Which method would be faster?Why?Work this example with M = 12.
Robb T. Koether (Hampden-Sydney College) The Huntington-Hill Method – Version 2 Wed, Oct 18, 2017 19 / 21
Outline
1 Verson 2
2 Assignment
Robb T. Koether (Hampden-Sydney College) The Huntington-Hill Method – Version 2 Wed, Oct 18, 2017 20 / 21
Assignment
AssignmentChapter 4 Exercises 49, 50. Use Version 2 with M = 10.
Robb T. Koether (Hampden-Sydney College) The Huntington-Hill Method – Version 2 Wed, Oct 18, 2017 21 / 21
