Presentation is loading. Please wait.

Presentation is loading. Please wait.

Quiz.

Similar presentations


Presentation on theme: "Quiz."β€” Presentation transcript:

1 Quiz

2 Huntington-Hill Method
The Huntington-Hill method of apportionment is the currently used method of apportionment in the United States Congress. It involves what is called the geometric mean between two numbers. Geometric Mean of two numbers 𝑛 π‘Žπ‘›π‘‘ π‘š is π‘›βˆ™π‘š .

3 Huntington-Hill Method
Find the standard divisor. Find the standard quotas. For each standard quota, find the geometric mean between its upper and lower quotas. If the standard quota is greater than the geometric mean round up. If the standard quota is less than the geometric mean round down. Find a modified divisor that perfectly apportions the seats.

4 Example #1 The United States of Canines has a senate with 100 seats to give out to four states. Find the standard divisor and standard quotas. Find the geometric means for the upper and lower quotas and determine an initial apportionment. Find a modified divisor that gives a perfect apportionment.

5 Example #1 Solution Standard Divisor: 100,000 100 =1,000 𝑝𝑒𝑝𝑠/π‘ π‘’π‘Žπ‘‘
Standard Quotas: A: =86.9 B: C: D:3.36 Geometric Means: 𝐴: 86βˆ™87 = B: 4βˆ™5 =4.472 C: 5βˆ™6 = D: 3βˆ™4 =3.464 Initial Apportionment: A: 87 B: 4 C: 5 D: 3 Total: 99

6 Example #1 Solution We are one seat short, so we reduce our divisor.
Modified Divisor: 990 Quotas: A: B: C: D: 3.394 Notice we have to recalculate our geometric mean for A as it went up a number. The rest remain the same.

7 Example #1 Solution Quotas: A: 87.793 B: 4.369 C: 5.455 D: 3.394
Geometric Means: A: 87βˆ™88 = B: 4.472 C: D: 3.464 Apportionment: A: 88 B: 4 C: 6 D: 3 Total: 100. WOOOHOOO!

8 Example #2 Five classes being taught and apportioned based on how many students sign up. Algebra: 36 Geometry: 61 Calculus: 3

9 Example #2 Solution Standard Divisor: =20 𝑠𝑑𝑒𝑑𝑒𝑛𝑑𝑠/π‘ π‘’π‘Žπ‘‘ Standard Quotas: Algebra: 1.8 Geometry: 3.05 Calc: .15 Geometric Means: Algebra: 1βˆ™2 =1.414… Geometry: 3βˆ™4 =3.464… Calc: 0βˆ™1 =0.000

10 Example #2 Solution If the standard quota is higher than the geometric mean, we round up! Standard Quotas: Algebra: 1.8 Geometry: Calc: .15 Geometric Means: Algebra: Geometry: Calc: 0 Round Up Round Down Round Up Initial Apportionment: Algebra: 2 Geometry: 3 Calc: 1

11 Example #2 Solution Our apportionment is over five classes so we must make our standard divisor bigger. Modified Divisor: 21 Algebra: 1.7 Geometry:2.9 Calc: 0.14 Geometric Means: A: G: C: 0 Apportionment: A: 2 G: 3 C: 1 Notice our geometric mean for Geometry changed.

12 Example #2 Solution Modified Divisor: 25 Algebra: 1.44 G: 2.44 C: .12
Geometric Means: A: G: 2.45 C: 0 Round Up Down Up Apportionment: A: 2 G: 2 C: 1


Download ppt "Quiz."

Similar presentations


Ads by Google