4.6: Lowndes’ Method
- Page ID
- 34195
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)William Lowndes (1782-1822) was a Congressman from South Carolina (a small state) who proposed a method of apportionment that was more favorable to smaller states. Unlike the methods of Hamilton, Jefferson, and Webster, Lowndes’s method has never been used to apportion Congress.
Lowndes believed that an additional representative was much more valuable to a small state than to a large one. If a state already has 20 or 30 representatives, getting one more doesn’t matter very much. But if it only has 2 or 3, one more is a big deal, and he felt that the additional representatives should go where they could make the most difference.
Like Hamilton’s method, Lowndes’s method follows the quota rule. In fact, it arrives at the same quotas as Hamilton and the rest, and like Hamilton and Jefferson, it drops the decimal parts. But in deciding where the remaining representatives should go, we divide the decimal part of each state’s quota by the whole number part (so that the same decimal part with a smaller whole number is worth more, because it matters more to that state).
- 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 divisor.
- Divide each state’s 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.
- Cut off all the decimal parts of all the quotas (but don’t forget what the decimals were). Add up the remaining whole numbers.
- Assuming that the total from Step 3 was less than the total number of representatives, divide the decimal part of each state’s quota by the whole number part. Assign the remaining representatives, one each, to the states whose ratio of decimal part to whole part were largest, until the desired total is reached.
We’ll do Delaware again. We begin in the same way as with Hamilton’s method:
\(\begin{array}{lrrc}
\text { County } & \text { Population } & \text{ Quota } & \text{ Initial }\\
\hline
\text { Kent } & 162,310 & 7.4111 & 7\\
\text { New Castle } & 538,479 & 24.5872 & 24\\
\text { Sussex } & 197,145 & 9.0017 & 9\\
\textbf{ Total } & \bf{ 897,934 } & & \bf{ 40 }\end{array}\)
Solution
We need one more representative. To find out which county should get it, Lowndes says to divide each county’s decimal part by its whole number part, with the largest result getting the extra representative:
\(\begin{array}{lr} \text {Kent: } & 0.4111/7 \approx 0.0587 \\ \text{New Castle: } & 0.5872/24 \approx 0.0245 \\ \text{ Sussex: } & 0.0017/9 \approx 0.0002 \\ \end{array}\)
The largest of these is Kent’s, so Kent gets the \(41^{\text{th}}\) representative:
\(\begin{array}{lrrcc}
\text { County } & \text { Population } & \text{ Quota } & \text{ Initial } & \text{ Ratio } & \text{ Final } \\
\hline
\text { Kent } & 162,310 & 7.4111 & 7 & 0.0587 & 8 \\
\text { New Castle } & 538,479 & 24.5872 & 24 & 0.0245 & 24 \\
\text { Sussex } & 197,145 & 9.0017 & 9 & 0.0002 & 9 \\
\textbf{ Total } & \bf{ 897,934 } & & \bf{ 40 } & & \bf{ 41 }\end{array}\)
Rhode Island, again beginning in the same way as Hamilton:
\(\begin{array}{lrrc}
\text { County } & \text { Population } & \text{ Quota } & \text{ Initial }\\
\hline \text { Bristol } & 49,875 & 3.5538 & 3 \\
\text { Kent } & 166,158 & 11.8395 & 11 \\
\text { Newport } & 82,888 & 5.9061 & 5 \\
\text { Providence } & 626,667 & 44.6528 & 44\\
\text { Washington } & 126,979 & 9.0478 & 9\\
\textbf{ Total } & \bf{ 1,052,567 } & & \bf{ 72 }\end{array}\)
Solution
We divide each county’s quota’s decimal part by its whole number part to determine which three should get the remaining representatives:
\(\begin{array}{lr} {\text {Bristol: }} & {0.5538/3 \approx 0.1846} \\ {\text{Kent: }} & {0.8395/11 \approx 0.0763} \\ {\text{Newport: }} & {0.9061/5 \approx 0.1812} \\ {\text{Providence: }} & {0.6528/44 \approx 0.0148} \\ {\text{Washington: }} & {0.0478/9 \approx 0.0053} \\ \end{array}\)
The three largest of these are Bristol, Newport, and Kent, so they get the remaining three representatives:
\(\begin{array}{lrrcc}
\text { County } & \text { Population } & \text{ Quota } & \text{ Initial } & \text{ Ratio } & \text{ Final } \\
\hline \text { Bristol } & 49,875 & 3.5538 & 3 & 0.1846 & 4 \\
\text { Kent } & 166,158 & 11.8395 & 11 & 0.0763 & 12 \\
\text { Newport } & 82,888 & 5.9061 & 5 & 0.1812 & 6 \\
\text { Providence } & 626,667 & 44.6528 & 44 & 0.0148 & 44 \\
\text { Washington } & 126,979 & 9.0478 & 9 & 0.0053 & 9 \\
\textbf{ Total } & \bf{ 1,052,567 } & & \bf{ 72 } & & \bf{ 75 }\end{array}\)
As you can see, there is no “right answer” when it comes to choosing a method for apportionment. Each method has its virtues, and favors different sized states.