Nossi Ch 5 updated

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Nossi Ch 5 updated

  1. 1. A Mathematical View of Our World 1 st ed. Parks, Musser, Trimpe, Maurer, and Maurer
  2. 2. Chapter 5 Apportionment
  3. 3. Section 5.1 Quota Methods <ul><li>Goals </li></ul><ul><ul><li>Study apportionment </li></ul></ul><ul><ul><ul><li>Standard divisor </li></ul></ul></ul><ul><ul><ul><li>Standard quota </li></ul></ul></ul><ul><ul><li>Study apportionment methods </li></ul></ul><ul><ul><ul><li>Hamilton’s method </li></ul></ul></ul>
  4. 4. 5.1 Initial Problem <ul><li>The number of campers in each group at a summer camp is shown below. </li></ul>
  5. 5. 5.1 Initial Problem, cont’d <ul><li>The camp organizers will assign 15 counselors to the groups of campers. </li></ul><ul><li>How many of the 15 counselors should be assigned to each group? </li></ul><ul><ul><li>The solution will be given at the end of the section. </li></ul></ul>
  6. 6. Apportionment <ul><li>The verb apportion means </li></ul><ul><ul><li>“ Assign to as a due portion.” </li></ul></ul><ul><ul><li>“ To divide into shares which may not be equal.” </li></ul></ul><ul><li>Apportionment problems arise when what is being divided cannot be divided into fractional parts. </li></ul><ul><ul><li>An example of apportionment is the process of assigning seats in the House of Representatives to the states. </li></ul></ul>
  7. 7. Apportionment, cont’d <ul><li>The apportionment problem is to determine a method for rounding a collection of numbers so that: </li></ul><ul><ul><li>The numbers are rounded to whole numbers. </li></ul></ul><ul><ul><li>The sum of the numbers is unchanged. </li></ul></ul>
  8. 8. Apportionment, cont’d <ul><li>The Constitution does not specify a method for apportioning seats in the House of Representatives. </li></ul><ul><li>Various methods, named after their authors, have been used: </li></ul><ul><ul><li>Alexander Hamilton </li></ul></ul><ul><ul><li>Thomas Jefferson </li></ul></ul><ul><ul><li>Daniel Webster </li></ul></ul><ul><ul><li>William Lowndes </li></ul></ul>
  9. 9. The Standard Divisor <ul><li>Suppose the total population is P and the number of seats to be apportioned is M . </li></ul><ul><li>The standard divisor is the ratio D = P/M . (handout calls this the ideal ratio .) </li></ul><ul><ul><li>The standard divisor gives the number of people per legislative seat. </li></ul></ul>
  10. 10. Example 1 <ul><li>Suppose a country has 5 states and 200 seats in the legislature. </li></ul><ul><li>The populations of the states are given below. </li></ul><ul><li>Find the standard divisor (ideal ratio). </li></ul>
  11. 11. Example 1, cont’d <ul><li>Solution: The total population is found by adding the 5 state populations. </li></ul><ul><ul><li>P = 1,350,000 + 1,500,000 + 4,950,000 + 1,100,000 + 1,100,000 </li></ul></ul><ul><ul><li>= 10,000,000. </li></ul></ul><ul><ul><li>The number of seats is M = 200 </li></ul></ul>
  12. 12. Example 1, cont’d <ul><li>Solution, cont’d: The standard divisor (ideal ratio) is D = 10,000,000/200 = 50,000. </li></ul><ul><li>Each seat in the legislature represents 50,000 citizens. </li></ul>
  13. 13. The Standard Quota <ul><li>Let D be the standard divisor (ideal ratio). </li></ul><ul><li>If the population of a state is p , then </li></ul><ul><li>Q = p/D is called the standard quota . </li></ul><ul><ul><li>If seats could be divided into fractions, we would give the state exactly Q seats in the legislature. </li></ul></ul>
  14. 14. Example 2 <ul><li>In the previous example, the standard divisor (ideal ratio) was found to be </li></ul><ul><li>D = 50,000. </li></ul><ul><li>Find the standard quotas for each state in the country. </li></ul>
  15. 15. Example 2, cont’d <ul><li>Solution: Divide the population of each state by the standard divisor. </li></ul><ul><ul><li>State A: Q = 1,350,000/50,000 = 27 </li></ul></ul><ul><ul><li>State B: Q = 1,500,000/50,000 = 30 </li></ul></ul><ul><ul><li>State C: Q = 4,950,000/50,000 = 99 </li></ul></ul><ul><ul><li>State D: Q = 1,100,000/50,000 = 22 </li></ul></ul><ul><ul><li>State E: Q = 1,100,000/50,000 = 22 </li></ul></ul><ul><ul><li>Wow! They all came out whole numbers! </li></ul></ul>
  16. 16. Example 2, cont’d <ul><li>This solution, with all standard quotas being whole numbers, is not typical. </li></ul><ul><li>Note that the sum of the quotas is 27 + 30 + 99 +22 + 22 = 200, the total number of seats. </li></ul><ul><li>The standard quotas indicate how many seats each state should be assigned. </li></ul>
  17. 17. Apportionment, cont’d <ul><li>Typically, the standard quotas will not all be whole numbers and will have to be rounded. </li></ul><ul><li>The various apportionment methods provide procedures for determining how the rounding should be done. </li></ul>
  18. 18. Hamilton’s Method <ul><li>Find the standard divisor (ideal ratio). </li></ul><ul><li>Determine each state’s standard quota. </li></ul><ul><ul><li>Round each quota down to a whole number. </li></ul></ul><ul><ul><ul><li>Each state gets that number of seats, with a minimum of 1 seat. </li></ul></ul></ul><ul><li>Leftover seats are assigned one at a time to states according to the size of the fractional parts of the standard quotas. </li></ul><ul><ul><li>Begin with the state with the largest fractional part. </li></ul></ul>
  19. 19. Example 3 <ul><li>A country has 5 states and 200 seats in the legislature. </li></ul><ul><li>Apportion the seats according to Hamilton’s method. </li></ul>
  20. 20. Example 3, cont’d <ul><li>Solution: the standard divisor (ideal ratio) is found: </li></ul><ul><li>Then the standard quota for each state is found. </li></ul><ul><ul><li>For example: </li></ul></ul>
  21. 21. Example 3, cont’d <ul><li>Solution, cont’d: All of the standard quotas are shown below. </li></ul>
  22. 22. Example 3, cont’d <ul><li>Solution, cont’d: The integer parts of the standard quotas add up to 26 + 30 + 98 + 22 + 22 = 198. </li></ul><ul><ul><li>A total of 198 seats have been apportioned at this point. </li></ul></ul><ul><ul><li>There are 2 seats left to assign according to the fractional parts of the standard quotas. </li></ul></ul>
  23. 23. Example 3, cont’d <ul><li>Solution, cont’d: Consider the size of the fractional parts of the standard quotas . </li></ul><ul><ul><li>State C has the largest fractional part, of 0.7 </li></ul></ul><ul><ul><li>State A has the second largest fractional part, of 0.4. </li></ul></ul><ul><ul><li>The 2 leftover seats are apportioned to states C and A. </li></ul></ul>
  24. 24. Example 3, cont’d <ul><li>Solution, cont’d: The final apportionment is shown below. </li></ul>
  25. 25. Quota Rule <ul><li>Any apportionment method which always assigns the whole number just above or just below the standard quota is said to satisfy the quota rule . </li></ul><ul><li>Any apportionment method that obeys the quota rule is called a quota method . </li></ul><ul><ul><li>Hamilton’s method is a quota method. </li></ul></ul>
  26. 26. 5.1 Initial Problem Solution <ul><li>A camp needs to assign 15 counselors among 3 groups of campers. </li></ul><ul><li>The groups are shown in the table below. </li></ul>
  27. 27. Initial Problem Solution, cont’d <ul><li>First the counselors will be apportioned using Hamilton’s method. </li></ul><ul><li>The standard divisor (ideal ratio) is: </li></ul><ul><ul><li>This indicates that 1 counselor should be assigned to approximately every 12.67 campers. </li></ul></ul>
  28. 28. Initial Problem Solution, cont’d <ul><li>Next, calculate the standard quota for each group of campers. </li></ul>
  29. 29. Initial Problem Solution, cont’d <ul><li>A total of 3 + 5 + 6 = 14 counselors have been assigned so far. </li></ul><ul><li>The 1 leftover counselor is assigned to the 6 th grade group. </li></ul>
  30. 30. Initial Problem Solution, cont’d <ul><li>The final apportionment according to Hamilton’s method is: </li></ul><ul><ul><li>3 counselors for 4 th grade </li></ul></ul><ul><ul><li>5 counselors for 5 th grade </li></ul></ul><ul><ul><li>7 counselors for 6 th grade </li></ul></ul>
  31. 31. Section 5.2 Divisor Methods <ul><li>Goals </li></ul><ul><ul><li>Study apportionment methods </li></ul></ul><ul><ul><ul><li>Jefferson’s method </li></ul></ul></ul><ul><ul><ul><li>Webster’s method </li></ul></ul></ul>
  32. 32. 5.2 Initial Problem <ul><li>Suppose you, your sister, and your brother have inherited 85 gold coins. </li></ul><ul><li>The coins will be divided based on the number of hours each of you have volunteered at the local soup kitchen. </li></ul><ul><li>How should the coins be apportioned? </li></ul><ul><ul><li>The solution will be given at the end of the section. </li></ul></ul>
  33. 33. Jefferson’s Method <ul><li>Suppose M seats will be apportioned. </li></ul><ul><ul><li>Choose a number, d , called the modified divisor . </li></ul></ul><ul><ul><li>For each state, compute the modified quota , which is the ratio of the state’s population to the modified divisor: </li></ul></ul>
  34. 34. Jefferson’s Method, cont’d <ul><li>Cont’d: </li></ul><ul><ul><li>If the integer parts of the modified quotas for all the states add to M , then go on to Step 2. Otherwise go back to Step 1, part (a) and choose a different value for d . </li></ul></ul><ul><li>Assign to each state the integer part of its modified quota. </li></ul>
  35. 35. Example 1 <ul><li>Use Jefferson’s method to apportion 200 seats to the 5 states in the example from Section 5.1. </li></ul>
  36. 36. Example 1, cont’d <ul><li>Solution: Recall the standard divisors and the apportionment found using Hamilton’s method. Note: total delegates = 200 </li></ul>
  37. 37. Example 1, cont’d <ul><li>Solution, cont’d: In Jefferson’s method all the (modified) quotas will be rounded down. </li></ul><ul><ul><li>Note that if all the standard quotas were rounded down, the total would be only 198 seats. </li></ul></ul><ul><ul><li>The modified quotas need to be slightly larger than the standard ones. </li></ul></ul>
  38. 38. Example 1, cont’d <ul><li>Solution: Recall the standard divisors and the apportionment found using the rounded down method. Note: total delegates = 198 </li></ul>22 22 98 30 26 Rounded-down method
  39. 39. Example 1, cont’d <ul><li>Solution, cont’d: For the modified quotas to be larger, the modified divisor needs to be smaller than the standard divisor of 50,000. </li></ul><ul><ul><li>A good guess for a modified divisor can be found by dividing the largest state’s population by 1more, 2 more, 3 more,…, than the integer part of its standard quota. </li></ul></ul><ul><ul><ul><li>Start with 1 more, and keep going until you find one that works. </li></ul></ul></ul>
  40. 40. Example 1, cont’d <ul><li>Solution, cont’d: The largest state has a population of 4,935,000 and its standard quota has an integer part of 98. </li></ul><ul><ul><li>A possible modified divisor is: </li></ul></ul>
  41. 41. Example 1, cont’d <ul><li>Solution, cont’d: We will try a modified divisor of d = 49,848. </li></ul><ul><li>The modified quota for each state is calculated. </li></ul><ul><ul><li>For example, the modified quota of state A is: </li></ul></ul>
  42. 42. Example 1, cont’d <ul><li>Solution, cont’d: When the modified quotas are rounded down they add to 199. </li></ul><ul><ul><li>The modified divisor needs to be even smaller. </li></ul></ul>
  43. 43. Example 1, cont’d <ul><li>Solution, cont’d: The largest state has a population of 4,935,000 and its standard quota has an integer part of 98. </li></ul><ul><ul><li>A second possible modified divisor is: </li></ul></ul>
  44. 44. Example 1, cont’d <ul><li>Solution, cont’d: New modified quotas are calculated. Sum of delegate = 200 </li></ul>
  45. 45. Example 1, cont’d <ul><li>Solution, cont’d: Now when the modified quotas are rounded down they add to 26 + 30 + 100 + 22 + 22 = 200. </li></ul><ul><li>Since the sum of the rounded modified quotas equals the number of seats, the apportionment is complete. </li></ul>
  46. 46. Divisor Methods <ul><li>Any apportionment method that uses a divisor other than the standard divisor is called a divisor method . </li></ul><ul><ul><li>Jefferson’s method is a divisor method. </li></ul></ul><ul><ul><li>Webster’s method, which will be studied next, is also a divisor method. </li></ul></ul>
  47. 47. Webster’s Method <ul><li>Suppose M seats are to be apportioned. </li></ul><ul><ul><li>Choose a number, d , called the modified divisor. </li></ul></ul><ul><ul><li>For each state, calculate the modified quota, mQ . </li></ul></ul>
  48. 48. Webster’s Method, cont’d <ul><ul><li>If when the modified quotas are rounded normally, their sum is M , then go on to Step 2. Otherwise go back to Step 1, part (a) and choose a different value for d . </li></ul></ul><ul><li>Assign to each state the integer nearest its modified quota. </li></ul>
  49. 49. Example 3 <ul><li>Suppose a retail store needs to apportion 54 sales associates to three stores. </li></ul>
  50. 50. Example 3 <ul><li>Solution: First determine the standard divisor by dividing the total customer base by the number of sales associates. </li></ul><ul><li>(27,500+40,000+65,000)/54 = 2454 </li></ul><ul><li>There needs to be approximately 1 sales associate for every 2454 customers. </li></ul>
  51. 51. Example 3 Jefferson’s Method <ul><li>Solution, cont’d: The standard quotas are calculated, as shown in the table below. </li></ul><ul><ul><li>Note that the rounded quotas add to 53. </li></ul></ul>
  52. 52. Example 3 Jefferson’s method <ul><li>Solution, cont’d: The sum of the rounded standard quotas was too small, so the modified divisor needs to be a little larger than the standard divisor. </li></ul><ul><ul><li>Find a new modified divisor using the guess-and-check method. </li></ul></ul><ul><ul><li>First, try </li></ul></ul>
  53. 53. Example 3 Jefferson’s method <ul><li>Solution, cont’d: The modified divisor is appropriate because the rounded modified quotas add to 54. </li></ul>
  54. 54. 5.2 Initial Problem Solution <ul><li>You, your sister, and your brother will divide 85 gold coins based on the number of hours you have each volunteered at the soup kitchen. </li></ul>
  55. 55. Initial Problem Solution, cont’d <ul><li>The standard divisor is found by dividing the total number of hours worked by the number of gold coins. </li></ul><ul><li>(72+43.5+34.5)/85 coins = 1.76 </li></ul><ul><li>You should each get about 1.76 coins for every hour you have worked. </li></ul>
  56. 56. Initial Problem Solution, cont’d <ul><li>The standard quotas for each person are shown in the table. </li></ul>
  57. 57. Initial Problem Solution, cont’d <ul><li>Consider the apportionment for Jefferson’s method: </li></ul><ul><ul><li>Rounding down the standard quotas yields a sum of 40 + 24 + 19 = 83, which is too small. </li></ul></ul><ul><ul><li>The modified divisor must be smaller. </li></ul></ul>
  58. 58. Initial Problem Solution, cont’d <ul><li>Suppose we try a modified divisor of d = 1.74. (This is a guess, but a method to find a guess is largest pop./(quota +1)) </li></ul>
  59. 59. Initial Problem Solution, cont’d <ul><li>The rounded-down modified quotas add to 85. </li></ul><ul><li>The final apportionment, using Jefferson’s method, is: </li></ul><ul><ul><li>You will receive 41 coins. </li></ul></ul><ul><ul><li>Your sister will receive 25 coins. </li></ul></ul><ul><ul><li>Your brother will receive 19 coins. </li></ul></ul>
  60. 60. Initial Problem Solution, cont’d <ul><li>Now apportion the coins using Webster’s method: </li></ul><ul><ul><li>When the standard quotas are rounded normally, they add to 86. </li></ul></ul><ul><ul><li>The modified divisor needs to be slightly larger than the standard divisor so that the sum of the rounded quotas will be smaller. </li></ul></ul>
  61. 61. Initial Problem Solution, cont’d <ul><li>Try using a modified divisor of d = 1.77. </li></ul>
  62. 62. Initial Problem Solution, cont’d <ul><li>The normally-rounded modified quotas add to 85. </li></ul><ul><li>The final apportionment, using Webster’s method, is: </li></ul><ul><ul><li>You will receive 41 coins. </li></ul></ul><ul><ul><li>Your sister will receive 25 coins. </li></ul></ul><ul><ul><li>Your brother will receive 19 coins. </li></ul></ul><ul><li>In this case, both apportionments are the same. </li></ul>
  63. 63. Section 5.3 Flaws of the Apportionment Methods <ul><li>Goals </li></ul><ul><ul><li>Study the Alabama paradox </li></ul></ul><ul><ul><li>Study the population paradox </li></ul></ul>
  64. 64. Apportionment Problems <ul><li>No apportionment method is free of flaws. </li></ul><ul><li>Circumstances that can cause apportionment problems include: </li></ul><ul><ul><li>A reapportionment based on population changes. </li></ul></ul><ul><ul><li>A change in the total number of seats. </li></ul></ul><ul><ul><li>The addition of one or more new states. </li></ul></ul>
  65. 65. The Quota Rule <ul><li>Recall that the quota rule says that each state’s apportionment should be equal to the whole number just below or just above the state’s standard quota. </li></ul><ul><ul><li>Every quota method satisfies the quota rule. </li></ul></ul><ul><ul><li>No divisor method can always satisfy the quota rule. </li></ul></ul><ul><li>Both quota and divisor methods may have flaws. </li></ul>
  66. 66. The Alabama Paradox <ul><li>In 1880 it was discovered that if the number of seats in the House of Representatives was increased from 299 to 300 then Alabama would be apportioned one fewer seat than before, using Hamilton’s method. </li></ul><ul><li>The possibility that the addition of one legislative seat will cause a state to lose a seat is called the Alabama paradox . </li></ul>
  67. 67. Population Paradox <ul><li>The population paradox can occur when the population in two states increases. </li></ul><ul><ul><li>The legislature is reapportioned based on a new census and a seat is switched from one state to the other. </li></ul></ul><ul><ul><li>The paradox occurs when the faster-growing state is the one that loses the seat. </li></ul></ul>
  68. 68. Paradox Summary <ul><li>The three types of paradoxes studied here are summarized below. </li></ul>
  69. 69. Paradox Summary, cont’d <ul><li>The different types of problems that can occur with the various apportionment methods are summarized below. </li></ul>
  70. 70. Paradox Summary, cont’d <ul><li>As seen in the table, all 4 apportionment methods either violate the quota rule or allow paradoxes to occur. </li></ul><ul><li>Mathematicians Michel L. Balinski and H. Peyton Young proved there is no apportionment method that obeys the quota rule and always avoids the three paradoxes. </li></ul>
  71. 71. Ch 5 Assignment <ul><li>The Chapter 5 assignment is on the 2 handouts given in class. Please pay attention to the handwritten notes about questions that only need to be read and ones that may be omitted. </li></ul><ul><li>Reminder: for the Midterm you may use quizzes, but not cell phones or computers. You will want to bring a calculator. </li></ul>

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