{"id":8924,"date":"2023-10-12T15:27:02","date_gmt":"2023-10-12T15:27:02","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/?post_type=chapter&p=8924"},"modified":"2024-10-18T20:58:38","modified_gmt":"2024-10-18T20:58:38","slug":"apportionment-learn-it-5","status":"web-only","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/chapter\/apportionment-learn-it-5\/","title":{"raw":"Apportionment: Learn It 5","rendered":"Apportionment: Learn It 5"},"content":{"raw":"

Huntington-Hill Method<\/h2>\r\n

In 1920, no new apportionment was done, because Congress couldn\u2019t agree on the method to be used. They appointed a committee of mathematicians to investigate, and they recommended the Huntington-Hill Method. They continued to use Webster\u2019s method in 1931, but after a second report recommending Huntington-Hill, it was adopted in 1941 and is the current method of apportionment used in Congress.<\/p>\r\n

The Huntington-Hill Method is similar to Webster\u2019s method, but attempts to minimize the percent differences of how many people each representative will represent.<\/p>\r\n

\r\n
\r\n

Huntington-Hill method<\/h3>\r\n
    \r\n\t
  1. Determine how many people each representative should represent. Do this by dividing the total population of all the states by the total number of representatives. This answer is called the standard divisor<\/strong>.<\/li>\r\n\t
  2. Divide each state\u2019s population by the divisor to determine how many representatives it should have. Record this answer to several decimal places. This answer is called the quota<\/strong>.<\/li>\r\n\t
  3. Cut off the decimal part of the quota to obtain the lower quota, which we\u2019ll call [latex]n[\/latex]. Compute [latex]\\sqrt{n(n+1)}[\/latex], which is the geometric mean<\/strong> of the lower quota and one value higher.<\/li>\r\n\t
  4. If the quota is larger than the geometric mean, round up the quota; if the quota is smaller than the geometric mean, round down the quota. Add up the resulting whole numbers to get the initial allocation<\/strong>.<\/li>\r\n\t
  5. If the total from Step 4 was less than the total number of representatives, reduce the divisor and recalculate the quota and allocation. If the total from step 4 was larger than the total number of representatives, increase the divisor and recalculate the quota and allocation. Continue doing this until the total in Step 4 is equal to the total number of representatives. The divisor we end up using is called the modified divisor<\/strong> or adjusted divisor<\/strong>.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/section>\r\n
    \r\n

    We\u2019ll return to Delaware and apply the Huntington-Hill method. As a reminder, the state of Delaware has three counties: Kent, New Castle, and Sussex. The Delaware state House of Representatives has [latex]41[\/latex] members.<\/p>\r\n

    The populations of the counties are as follows (from the 2010 Census):<\/p>\r\n

    [latex]\\begin{array}{lr} \\text { County } & \\text { Population } \\\\ \\hline \\text { Kent } & 162,310 \\\\ \\text { New Castle } & 538,479 \\\\ \\text { Sussex } & 197,145 \\\\ \\textbf{ Total } & \\bf{ 897,934 }\\end{array}[\/latex]<\/p>\r\n\r\n[reveal-answer q=\"4331\"]Show Solution[\/reveal-answer]
    \r\n[hidden-answer a=\"4331\"]\r\n\r\n

    We begin, as we did with Hamilton\u2019s method, by finding the quotas with the original divisor, [latex]21,900.82927[\/latex].<\/p>\r\n

    [latex]\\begin{array}{lrrccc} \\text { County } & \\text { Population } & \\text{ Quota } & \\text{ Lower Quota } & \\text{ Geom Mean } & \\text{ Initial } \\\\ \\hline \\text { Kent } & 162,310 & 7.4111 & 7 & 7.48 & 7\\\\ \\text { New Castle } & 538,479 & 24.5872 & 24 & 24.49 & 25\\\\ \\text { Sussex } & 197,145 & 9.0017 & 9 & 9.49 & 9\\\\ \\textbf{ Total } & \\bf{ 897,934 } & & & & \\bf{ 41 }\\end{array} [\/latex]<\/p>\r\n

    This gives the required total, so we\u2019re done.<\/p>\r\n\r\n[\/hidden-answer]<\/section>\r\n

    In this case, the apportionment produced by the Huntington-Hill method was the same as those from Webster\u2019s method.<\/p>\r\n

    \r\n

    [ohm2_question hide_question_numbers=1]13233[\/ohm2_question]<\/p>\r\n<\/section>\r\n

    Let's look at an example where the Huntington-Hill method produces a different result than the Hamilton method.<\/p>\r\n

    \r\n

    Consider a small country with [latex]5[\/latex] states, two of which are much larger than the others. We need to apportion [latex]70[\/latex] representatives. We will apportion using both Webster\u2019s method and the Huntington-Hill method.<\/p>\r\n

    [latex]\\begin{array}{lr} \\text { State } & \\text { Population } \\\\ \\hline \\mathrm{A} & 300,500 \\\\ \\mathrm{B} & 200,000 \\\\ \\mathrm{C} & 50,000 \\\\ \\mathrm{D} & 38,000 \\\\ \\mathrm{E} & 21,500 \\end{array} [\/latex]<\/p>\r\n\r\n[reveal-answer q=\"4332\"]Show Solution[\/reveal-answer]
    \r\n[hidden-answer a=\"4332\"]\r\n\r\n

    Let's being by finding the original divisor and quotas.<\/p>\r\n

    The total population is [latex]610,000[\/latex]. Dividing this by the [latex]70[\/latex] representatives gives the divisor: [latex]8714.286[\/latex].<\/p>\r\n

    Dividing each state\u2019s population by the divisor gives the quotas.<\/p>\r\n

    [latex]\\begin{array}{lrr} \\text { State } & \\text { Population } & \\text { Quota } \\\\ \\hline \\text { A } & 300,500 & 34.48361 \\\\ \\text { B } & 200,000 & 22.95082 \\\\ \\text { C } & 50,000 & 5.737705 \\\\ \\text { D } & 38,000 & 4.360656 \\\\ \\text { E } & 21,500 & 2.467213 \\end{array} [\/latex]<\/p>\r\n

    Webster\u2019s Method<\/h4>\r\n

    Using Webster\u2019s method, we round each quota to the nearest whole number.<\/p>\r\n

    [latex]\\begin{array}{lrrr} \\text { State } & \\text { Population } & \\text { Quota } & \\text { Initial } \\\\ \\hline \\mathrm{A} & 300,500 & 34.48361 & 34 \\\\ \\mathrm{B} & 200,000 & 22.95082 & 23 \\\\ \\mathrm{C} & 50,000 & 5.737705 & 6 \\\\ \\mathrm{D} & 38,000 & 4.360656 & 4 \\\\ \\mathrm{E} & 21,500 & 2.467213 & 2 \\end{array} [\/latex]<\/p>\r\n

    Adding these up, they only total [latex]69[\/latex] representatives, so we adjust the divisor down. Adjusting the divisor down to [latex]8700[\/latex] gives an updated allocation totaling [latex]70[\/latex] representatives.<\/p>\r\n

    [latex]\\begin{array}{lrrr} \\text { State } & \\text { Population } & \\text { Quota } & \\text { Initial } \\\\ \\hline \\mathrm{A} & 300,500 & 34.54023 & 35 \\\\ \\mathrm{B} & 200,000 & 22.98851 & 23 \\\\ \\mathrm{C} & 50,000 & 5.747126 & 6 \\\\ \\mathrm{D} & 38,000 & 4.367816 & 4 \\\\ \\mathrm{E} & 21,500 & 2.471264 & 2 \\end{array}[\/latex]<\/p>\r\n

    Huntington-Hill Method<\/h4>\r\n

    Using the Huntington-Hill method, we round down to find the lower quota, then calculate the geometric mean based on each lower quota. If the quota is less than the geometric mean, we round down; if the quota is more than the geometric mean, we round up.<\/p>\r\n

    [latex]\\begin{array}{lrrrrr} \\text { State } & \\text { Population } & \\text { Quota } & \\begin{array}{r} \\text { Lower } \\\\ \\text { Quota } \\end{array} & \\begin{array}{r} \\text { Geometric } \\\\ \\text { Mean } \\end{array} & \\text { Initial } \\\\ \\hline \\mathrm{A} & 300,500 & 34.48361 & 34 & 34.49638 & 34 \\\\ \\mathrm{B} & 200,000 & 22.95082 & 22 & 22.49444 & 23 \\\\ \\mathrm{C} & 50,000 & 5.737705 & 5 & 5.477226 & 6 \\\\ \\mathrm{D} & 38,000 & 4.360656 & 4 & 4.472136 & 4 \\\\ \\mathrm{E} & 21,500 & 2.467213 & 2 & 2.44949 & 3 \\end{array} [\/latex]<\/p>\r\n

    These allocations add up to [latex]70[\/latex], so we\u2019re done.<\/p>\r\n

    Notice that this allocation is different than that produced by Webster\u2019s method. In this case, state E got the extra seat instead of state A.<\/p>\r\n\r\n[\/hidden-answer]<\/section>","rendered":"

    Huntington-Hill Method<\/h2>\n

    In 1920, no new apportionment was done, because Congress couldn\u2019t agree on the method to be used. They appointed a committee of mathematicians to investigate, and they recommended the Huntington-Hill Method. They continued to use Webster\u2019s method in 1931, but after a second report recommending Huntington-Hill, it was adopted in 1941 and is the current method of apportionment used in Congress.<\/p>\n

    The Huntington-Hill Method is similar to Webster\u2019s method, but attempts to minimize the percent differences of how many people each representative will represent.<\/p>\n

    \n
    \n

    Huntington-Hill method<\/h3>\n
      \n
    1. Determine how many people each representative should represent. Do this by dividing the total population of all the states by the total number of representatives. This answer is called the standard divisor<\/strong>.<\/li>\n
    2. Divide each state\u2019s population by the divisor to determine how many representatives it should have. Record this answer to several decimal places. This answer is called the quota<\/strong>.<\/li>\n
    3. Cut off the decimal part of the quota to obtain the lower quota, which we\u2019ll call [latex]n[\/latex]. Compute [latex]\\sqrt{n(n+1)}[\/latex], which is the geometric mean<\/strong> of the lower quota and one value higher.<\/li>\n
    4. If the quota is larger than the geometric mean, round up the quota; if the quota is smaller than the geometric mean, round down the quota. Add up the resulting whole numbers to get the initial allocation<\/strong>.<\/li>\n
    5. If the total from Step 4 was less than the total number of representatives, reduce the divisor and recalculate the quota and allocation. If the total from step 4 was larger than the total number of representatives, increase the divisor and recalculate the quota and allocation. Continue doing this until the total in Step 4 is equal to the total number of representatives. The divisor we end up using is called the modified divisor<\/strong> or adjusted divisor<\/strong>.<\/li>\n<\/ol>\n<\/div>\n<\/section>\n
      \n

      We\u2019ll return to Delaware and apply the Huntington-Hill method. As a reminder, the state of Delaware has three counties: Kent, New Castle, and Sussex. The Delaware state House of Representatives has [latex]41[\/latex] members.<\/p>\n

      The populations of the counties are as follows (from the 2010 Census):<\/p>\n

      [latex]\\begin{array}{lr} \\text { County } & \\text { Population } \\\\ \\hline \\text { Kent } & 162,310 \\\\ \\text { New Castle } & 538,479 \\\\ \\text { Sussex } & 197,145 \\\\ \\textbf{ Total } & \\bf{ 897,934 }\\end{array}[\/latex]<\/p>\n