1. 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. 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. In the next slide we will look at an example that illustrates the nature of the problem we are dealing with.

2. ‘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.

3. At the end of the week, the numbers are in. The Table shows the amount of work done by each child during the week. At the end of the week, the numbers are in. The Table shows the amount of work done by each child during the week.

4. According to the ground rules, Alan, who worked 150 out of a total of 900 minutes, should get 8 1/3 pieces. 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). 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. 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. Because none of these shares can be realized, an absolutely fair apportionment of the candy is going to be impossible. What should mom do? What should mom do?

5. Apportionment- Why and Definitions! Apportionment problems can arise in a variety of real-life applications – Apportionment problems can arise in a variety of real-life applications – dividing candy among children, dividing candy among children, assigning nurses to shifts, assigning nurses to shifts, assigning buses to routes, and so on. 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. 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.

6. Definitions!!!!!!!!! The basic elements of every apportionment problem are: The basic elements of every apportionment problem are: the “states”, the “states”, the “seats”, and the “seats”, and the populations” 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.! 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.!

7. More Definitions!!!!!! The “seats”: This term describes the set of M identical, indivisible objects that are being divided among the N states. The “seats”: This term describes the set of M identical, indivisible objects that are being divided among the N states. 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!) 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. 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. We will use p1, p2,…, pN, to denote the state’s respective populations and P to denote the total population P = p1 + p2 +…+ pN.

8. Apportionment Apportionment – the distribution of an item fairly among a group. Ideal ratio (Standard Divisor)- Total population Total population Number of seats Number of seats Quota - individual population ideal ratio ideal ratio

9. 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)

10. 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.

11. Hamilton’s Method Ideal Ratio = Hamilton uses the highest decimal from the quota to decide who receives the next seat. Class (#)Sr- 435Jr- 525 So- 581Fr- 653Total - 2194 Quota Lower Quota Initial Apportionment Residue FINAL Apportionment

12. 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 truncated quota + 1

13. Jefferson’s Method Ideal Ratio = Adj. Ratio - Divide the class population by the Lower Quota+ 1 Class (#)Sr- 435Jr- 525 So- 581Fr- 653Total - 2194 Quota Lower Quota Initial Apportionment Adjusted Ratio FINAL Apportionment

14. 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.

15. Find Adjusted Ratio: adjusted ratio = Individual population arithmetic mean 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.

16. Webster’s Method Ideal Ratio = Divide the class population by the truncated value +. 5 Class (#)Sr- 435Jr- 525 So- 581Fr- 653Total - 2194 Quota ROUNDED Quota Initial Apportionment Adjusted Ratio FINAL Apportionment

17. 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.

18. 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.

19. Hill’s Method Ideal Ratio = Divide the class population by the Square root of LQ value times the LQ value plus one Class (#)Sr- 435Jr- 525 So- 581Fr- 653Total - 2194 Quota Geometric Mean ROUNDED Quota Initial Apportionment Adjusted Ratio FINAL Apportionment

20. COMPARING ALL 4 METHODS: Ideal Ratio = Class (#)Sr- 435Jr- 525 So- 581Fr- 653Total - 2194 Quota Lower Quota Hamilton Jefferson Webster Hill

21. 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.

23. 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)

24. Solve by Mathematical Induction 1.) 1 + 2 + 3 + … + n = n (n+1) 2

25. 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.

26. 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

27. 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

28. 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 1 + 2 + 3 + … + k + (k+1) = (k+1) ( (k+1) +1) 2 k ( k+1) + (k+1) = (k+1) ( (k+2) 2 2 k ( k+1) + (k+1) = (k+1) ( (k+2) 2 2Solve…

29. Solve by using Mathematical Induction. 1 + 4 + 9 + … + n 2 = n(n + 1)(2n + 1) 6