{"id":8888,"date":"2023-10-11T19:35:18","date_gmt":"2023-10-11T19:35:18","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/?post_type=chapter&#038;p=8888"},"modified":"2024-10-18T20:58:37","modified_gmt":"2024-10-18T20:58:37","slug":"apportionment-learn-it-2","status":"web-only","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/chapter\/apportionment-learn-it-2\/","title":{"raw":"Apportionment: Learn It 2","rendered":"Apportionment: Learn It 2"},"content":{"raw":"<h2>Hamilton\u2019s Method<\/h2>\r\n<p>Alexander Hamilton proposed the method that now bears his name. His method was approved by Congress in 1791, but was vetoed by President Washington. It was later adopted in 1852 and used through 1911.<\/p>\r\n<p>He begins by determining, to several decimal places, how many things each group should get. Since he was interested in the question of Congressional representation, we\u2019ll use the language of states and representatives, so he determines how many representatives each state should get. He follows these steps:<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<div>\r\n<h3>Hamilton\u2019s method<\/h3>\r\n<ol style=\"list-style-type: decimal;\">\r\n\t<li>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 <strong>divisor<\/strong>.<\/li>\r\n\t<li>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 <strong>quota<\/strong>. Since we can only allocate whole representatives, Hamilton resolves the whole number problem, as follows:<\/li>\r\n\t<li>Cut off all the decimal parts of all the quotas (but don\u2019t forget what the decimals were). These are called the <strong>lower quotas<\/strong>. Add up the remaining whole numbers. This answer will always be less than or equal to the total number of representatives (and the \u201cor equal to\u201d part happens only in very specific circumstances that are incredibly unlikely to turn up).<\/li>\r\n\t<li>Assuming that the total from Step 3 was less than the total number of representatives, assign the remaining representatives, one each, to the states whose decimal parts of the quota were largest, until the desired total is reached.<\/li>\r\n<\/ol>\r\n\r\nMake sure that each state ends up with at least one representative!<\/div>\r\n<\/section>\r\n<section class=\"textbox proTip\">\r\n<p>Note on rounding: Today we have technological advantages that Hamilton (and the others) couldn\u2019t even have imagined. Take advantage of them, and keep several decimal places.<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p>The state of Delaware has three counties: Kent, New Castle, and Sussex. The Delaware state House of Representatives has [latex]41[\/latex] members. If Delaware wants to divide this representation along county lines (which is not required, but let\u2019s pretend they do), let\u2019s use Hamilton\u2019s method to apportion them.<\/p>\r\n<p>The populations of the counties are as follows (from the 2010 Census):<\/p>\r\n\r\n[latex]\\begin{array}{lr} \\text { County } &amp; \\text { Population } \\\\ \\hline \\text { Kent } &amp; 162,310 \\\\ \\text { New Castle } &amp; 538,479 \\\\ \\text { Sussex } &amp; 197,145 \\\\ \\textbf{ Total } &amp; \\bf{ 897,934 }\\end{array}[\/latex] [reveal-answer q=\"4331\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"4331\"]\r\n\r\n<ol style=\"list-style-type: decimal;\">\r\n\t<li>\r\n<p>First, we determine the divisor:<\/p>\r\n\r\n[latex]\\frac{897,934}{41} = 21,900.82927[\/latex]<\/li>\r\n\t<li>\r\n<p>Now we determine each county\u2019s quota by dividing the county\u2019s population by the divisor:<\/p>\r\n\r\n[latex]\\begin{array}{lr} \\text { County } &amp; \\text { Population } &amp; \\text{ Quota } \\\\ \\hline \\text { Kent } &amp; 162,310 &amp; 7.4111 \\\\ \\text { New Castle } &amp; 538,479 &amp; 24.5872 \\\\ \\text { Sussex } &amp; 197,145 &amp; 9.0017 \\\\ \\textbf{ Total } &amp; \\bf{ 897,934 } &amp; \\end{array} [\/latex]<\/li>\r\n\t<li>\r\n<p>Removing the decimal parts of the quotas gives:<\/p>\r\n\r\n[latex]\\begin{array}{lrrc} \\text { County } &amp; \\text { Population } &amp; \\text{ Quota } &amp; \\text{ Initial }\\\\ \\hline \\text { Kent } &amp; 162,310 &amp; 7.4111 &amp; 7\\\\ \\text { New Castle } &amp; 538,479 &amp; 24.5872 &amp; 24\\\\ \\text { Sussex } &amp; 197,145 &amp; 9.0017 &amp; 9\\\\ \\textbf{ Total } &amp; \\bf{ 897,934 } &amp; &amp; \\bf{ 40 }\\end{array}[\/latex]<\/li>\r\n\t<li>\r\n<p>We need [latex]41[\/latex] representatives and this only gives [latex]40[\/latex]. The remaining one goes to the county with the largest decimal part, which is New Castle:<\/p>\r\n\r\n[latex]\\begin{array}{lrrcc} \\text { County } &amp; \\text { Population } &amp; \\text{ Quota } &amp; \\text{ Initial } &amp; \\text{ Final } \\\\ \\hline \\text { Kent } &amp; 162,310 &amp; 7.4111 &amp; 7 &amp; 7 \\\\ \\text { New Castle } &amp; 538,479 &amp; 24.5872 &amp; 24 &amp; 25 \\\\ \\text { Sussex } &amp; 197,145 &amp; 9.0017 &amp; 9 &amp; 9 \\\\ \\textbf{ Total } &amp; \\bf{ 897,934 } &amp; &amp; \\bf{ 40 } &amp; \\bf { 41 } \\end{array}[\/latex]<\/li>\r\n<\/ol>\r\n\r\n[\/hidden-answer]<\/section>\r\n<section class=\"textbox tryIt\">\r\n<p>[ohm2_question hide_question_numbers=1]13086[\/ohm2_question]<\/p>\r\n<\/section>\r\n<p>Hamilton\u2019s method obeys something called the <strong>quota rule<\/strong>. The quota rule isn\u2019t a law of any sort, but just an idea that some people, including Hamilton, think is a good one.<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<div>\r\n<h3>quota rule<\/h3>\r\n<p>The <strong>quota rule<\/strong> says that the final number of representatives a state gets should be within one of that state\u2019s quota. Since we\u2019re dealing with whole numbers for our final answers, that means that each state should either go up to the next whole number above its quota, or down to the next whole number below its quota.<\/p>\r\n<\/div>\r\n<\/section>\r\n<h3>Controversy<\/h3>\r\n<p>After seeing Hamilton\u2019s method, many people find that it makes sense, it\u2019s not that difficult to use (or, at least, the difficulty comes from the numbers that are involved and the amount of computation that\u2019s needed, not from the method), and they wonder why anyone would want another method. The problem is that Hamilton\u2019s method is subject to several paradoxes. Three of them happened, on separate occasions, when Hamilton\u2019s method was used to apportion the United States House of Representatives.<\/p>\r\n<ul>\r\n\t<li>\r\n<p>The <strong>Alabama Paradox<\/strong> is named for an incident that happened during the apportionment that took place after the 1880 census. (A similar incident happened ten years earlier involving the state of Rhode Island, but the paradox is named after Alabama.) The post-1880 apportionment had been completed, using Hamilton\u2019s method and the new population numbers from the census. Then it was decided that because of the country\u2019s growing population, the House of Representatives should be made larger. That meant that the apportionment would need to be done again, still using Hamilton\u2019s method and the same 1880 census numbers, but with more representatives. The assumption was that some states would gain another representative and others would stay with the same number they already had (since there weren\u2019t enough new representatives being added to give one more to every state). The paradox is that Alabama ended up losing a representative in the process, even though no populations were changed and the total number of representatives increased.<\/p>\r\n<\/li>\r\n\t<li>\r\n<p>The <strong>New States Paradox<\/strong> happened when Oklahoma became a state in 1907. Oklahoma had enough population to qualify for five representatives in Congress. Those five representatives would need to come from somewhere, though, so five states, presumably, would lose one representative each. That happened, but another thing also happened: Maine gained a representative (from New York).<\/p>\r\n<\/li>\r\n\t<li>\r\n<p>The <strong>Population Paradox<\/strong> happened between the apportionments after the census of 1900 and of 1910. In those ten years, Virginia\u2019s population grew at an average annual rate of [latex]1.07\\%[\/latex], while Maine\u2019s grew at an average annual rate of [latex]0.67\\%[\/latex]. Virginia started with more people, grew at a faster rate, grew by more people, and ended up with more people than Maine. By itself, that doesn\u2019t mean that Virginia should gain representatives or Maine shouldn\u2019t, because there are lots of other states involved. But Virginia ended up losing a representative to Maine.<\/p>\r\n<\/li>\r\n<\/ul>","rendered":"<h2>Hamilton\u2019s Method<\/h2>\n<p>Alexander Hamilton proposed the method that now bears his name. His method was approved by Congress in 1791, but was vetoed by President Washington. It was later adopted in 1852 and used through 1911.<\/p>\n<p>He begins by determining, to several decimal places, how many things each group should get. Since he was interested in the question of Congressional representation, we\u2019ll use the language of states and representatives, so he determines how many representatives each state should get. He follows these steps:<\/p>\n<section class=\"textbox keyTakeaway\">\n<div>\n<h3>Hamilton\u2019s method<\/h3>\n<ol style=\"list-style-type: decimal;\">\n<li>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 <strong>divisor<\/strong>.<\/li>\n<li>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 <strong>quota<\/strong>. Since we can only allocate whole representatives, Hamilton resolves the whole number problem, as follows:<\/li>\n<li>Cut off all the decimal parts of all the quotas (but don\u2019t forget what the decimals were). These are called the <strong>lower quotas<\/strong>. Add up the remaining whole numbers. This answer will always be less than or equal to the total number of representatives (and the \u201cor equal to\u201d part happens only in very specific circumstances that are incredibly unlikely to turn up).<\/li>\n<li>Assuming that the total from Step 3 was less than the total number of representatives, assign the remaining representatives, one each, to the states whose decimal parts of the quota were largest, until the desired total is reached.<\/li>\n<\/ol>\n<p>Make sure that each state ends up with at least one representative!<\/p><\/div>\n<\/section>\n<section class=\"textbox proTip\">\n<p>Note on rounding: Today we have technological advantages that Hamilton (and the others) couldn\u2019t even have imagined. Take advantage of them, and keep several decimal places.<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p>The state of Delaware has three counties: Kent, New Castle, and Sussex. The Delaware state House of Representatives has [latex]41[\/latex] members. If Delaware wants to divide this representation along county lines (which is not required, but let\u2019s pretend they do), let\u2019s use Hamilton\u2019s method to apportion them.<\/p>\n<p>The populations of the counties are as follows (from the 2010 Census):<\/p>\n<p>[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<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q4331\">Show Solution<\/button><\/p>\n<div id=\"q4331\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: decimal;\">\n<li>\n<p>First, we determine the divisor:<\/p>\n<p>[latex]\\frac{897,934}{41} = 21,900.82927[\/latex]<\/li>\n<li>\n<p>Now we determine each county\u2019s quota by dividing the county\u2019s population by the divisor:<\/p>\n<p>[latex]\\begin{array}{lr} \\text { County } & \\text { Population } & \\text{ Quota } \\\\ \\hline \\text { Kent } & 162,310 & 7.4111 \\\\ \\text { New Castle } & 538,479 & 24.5872 \\\\ \\text { Sussex } & 197,145 & 9.0017 \\\\ \\textbf{ Total } & \\bf{ 897,934 } & \\end{array}[\/latex]<\/li>\n<li>\n<p>Removing the decimal parts of the quotas gives:<\/p>\n<p>[latex]\\begin{array}{lrrc} \\text { County } & \\text { Population } & \\text{ Quota } & \\text{ Initial }\\\\ \\hline \\text { Kent } & 162,310 & 7.4111 & 7\\\\ \\text { New Castle } & 538,479 & 24.5872 & 24\\\\ \\text { Sussex } & 197,145 & 9.0017 & 9\\\\ \\textbf{ Total } & \\bf{ 897,934 } & & \\bf{ 40 }\\end{array}[\/latex]<\/li>\n<li>\n<p>We need [latex]41[\/latex] representatives and this only gives [latex]40[\/latex]. The remaining one goes to the county with the largest decimal part, which is New Castle:<\/p>\n<p>[latex]\\begin{array}{lrrcc} \\text { County } & \\text { Population } & \\text{ Quota } & \\text{ Initial } & \\text{ Final } \\\\ \\hline \\text { Kent } & 162,310 & 7.4111 & 7 & 7 \\\\ \\text { New Castle } & 538,479 & 24.5872 & 24 & 25 \\\\ \\text { Sussex } & 197,145 & 9.0017 & 9 & 9 \\\\ \\textbf{ Total } & \\bf{ 897,934 } & & \\bf{ 40 } & \\bf { 41 } \\end{array}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm13086\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=13086&theme=lumen&iframe_resize_id=ohm13086&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n<p>Hamilton\u2019s method obeys something called the <strong>quota rule<\/strong>. The quota rule isn\u2019t a law of any sort, but just an idea that some people, including Hamilton, think is a good one.<\/p>\n<section class=\"textbox keyTakeaway\">\n<div>\n<h3>quota rule<\/h3>\n<p>The <strong>quota rule<\/strong> says that the final number of representatives a state gets should be within one of that state\u2019s quota. Since we\u2019re dealing with whole numbers for our final answers, that means that each state should either go up to the next whole number above its quota, or down to the next whole number below its quota.<\/p>\n<\/div>\n<\/section>\n<h3>Controversy<\/h3>\n<p>After seeing Hamilton\u2019s method, many people find that it makes sense, it\u2019s not that difficult to use (or, at least, the difficulty comes from the numbers that are involved and the amount of computation that\u2019s needed, not from the method), and they wonder why anyone would want another method. The problem is that Hamilton\u2019s method is subject to several paradoxes. Three of them happened, on separate occasions, when Hamilton\u2019s method was used to apportion the United States House of Representatives.<\/p>\n<ul>\n<li>\n<p>The <strong>Alabama Paradox<\/strong> is named for an incident that happened during the apportionment that took place after the 1880 census. (A similar incident happened ten years earlier involving the state of Rhode Island, but the paradox is named after Alabama.) The post-1880 apportionment had been completed, using Hamilton\u2019s method and the new population numbers from the census. Then it was decided that because of the country\u2019s growing population, the House of Representatives should be made larger. That meant that the apportionment would need to be done again, still using Hamilton\u2019s method and the same 1880 census numbers, but with more representatives. The assumption was that some states would gain another representative and others would stay with the same number they already had (since there weren\u2019t enough new representatives being added to give one more to every state). The paradox is that Alabama ended up losing a representative in the process, even though no populations were changed and the total number of representatives increased.<\/p>\n<\/li>\n<li>\n<p>The <strong>New States Paradox<\/strong> happened when Oklahoma became a state in 1907. Oklahoma had enough population to qualify for five representatives in Congress. Those five representatives would need to come from somewhere, though, so five states, presumably, would lose one representative each. That happened, but another thing also happened: Maine gained a representative (from New York).<\/p>\n<\/li>\n<li>\n<p>The <strong>Population Paradox<\/strong> happened between the apportionments after the census of 1900 and of 1910. In those ten years, Virginia\u2019s population grew at an average annual rate of [latex]1.07\\%[\/latex], while Maine\u2019s grew at an average annual rate of [latex]0.67\\%[\/latex]. Virginia started with more people, grew at a faster rate, grew by more people, and ended up with more people than Maine. By itself, that doesn\u2019t mean that Virginia should gain representatives or Maine shouldn\u2019t, because there are lots of other states involved. But Virginia ended up losing a representative to Maine.<\/p>\n<\/li>\n<\/ul>\n","protected":false},"author":15,"menu_order":14,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Math in Society (Lippman)\",\"author\":\"Mike Kenyon & David Lippman\",\"organization\":\"LibreTexts Mathematics\",\"url\":\"https:\/\/math.libretexts.org\/Bookshelves\/Applied_Mathematics\/Math_in_Society_(Lippman)\/04%3A_Apportionment\/4.02%3A_Hamiltons_Method\",\"project\":\"4.2: Hamilton\\'s Method\",\"license\":\"cc-by-sa\",\"license_terms\":\"Access for free at https:\/\/math.libretexts.org\/Bookshelves\/Applied_Mathematics\/Math_in_Society_(Lippman)\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":90,"module-header":"learn_it","content_attributions":null,"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/8888"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":8,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/8888\/revisions"}],"predecessor-version":[{"id":12746,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/8888\/revisions\/12746"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/parts\/90"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/8888\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/media?parent=8888"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapter-type?post=8888"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/contributor?post=8888"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/license?post=8888"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}