1 The Apportionment Problem October 9, 2016 The Apportionment Problem An Apportionment Method M is a process (a function) that inputs a census (a list of n state populations (p1 , p2 . . . . , pn ) and a number h (the size of the house of representatives) and outputs a list of apportionments (A1 , . . . , An ) with A1 + + An = h, where Ak is the number of seats assigned to state k. M (p1 , , pn ; h) = (A1 , , An ) Basic Terms Input data h = # of seats in the house pk = population of state k p = total population = p1 + + pn s = standard divisor = qk = k th standard quota = lk = k th lower quota = qk rounded down = qk = qk rounded up = qk Calculations uk = k th upper quota p h pk s (ave district size) = ( ppk )h Apportionment Ak = # seats to state k It is rather easy to see that (except in extremely rare cases in which all qk s are integers) l 1 + + l n < q 1 + + q n = h < u 1 + + un so that assigning every state its lower quota gives too few seats and assigning every state its upper quota gives too many. A quota method M is a method with Ak = lk or uk for every k. Each state gets its lower quota, then some method decides which states get one more. The book discusses eight apportionment methods of which we will consider four in detail. The Hamilton method , the most natural, is a quota method that was the first method considered. (The Louwdes method is a variant of the Hamilton method in which the fractional k part qk lk is replaced by the proportional fractional part qkql .) k The Jefferson, Adams and Webster methods are divisor methods using other rounding functions1 . (The Hill and Dean methods are the other two standard divisor methods using different rounding functions.) Finally, the Balinski-Young method is a non-divisor variation of the Jefferson method that has certain theoretical advantages. The Hill method has been used for most of the last century in the US but almost always gives the same values as the Webster method. 1 See the file on Divisor methods 2 The Apportionment Problem Hamiton Method Definition: The Hamilton method H is a quota method that first assigns each state its lower quota lk : the remaining h(l1 + +lk ) seats are then given, one each, to the states with the largest values of qk lk ; i.e., the largest fractional parts of qk . Denote by hk the Hamilton apportionment for state k. Example: Suppose that a country has three states with populations 264, 361 and 375 and a house of size 10: i.e., n = 3 and (p1 , p2 , p3 ; h) = (264, 361, 375; 10). k pk q k l k uk q k l k h k 1 264 2.64 2 3 .64 3 2 361 3.61 3 4 .61 3 3 375 3.75 3 4 .75 4 p=1000 s=100 s= 1000 10 = 100 q1 = q2 = q3 = 8 264 100 361 100 375 100 h=10 = 2.64, l1 = 2.64 = 2, q1 l1 = .64 = 3.61, l2 = 3.61 = 3, q2 l2 = .61 = 3.75, l3 = 3.75 = 3, q3 l3 = .75 After assigning each state its lower quota, there remain two seats to distributethey go to States 1 and 3 as indicated. Thus, H(264, 361, 378; 10) = (3, 3, 4). The Apportionment Problem 3 The Hamilton Paradoxes An unexpected and unwelcome feature of the Hamilton method is the manner in which apportionments change when the input data is changed. The following examples show an apportionment with data that is then changed in three ways (changes indicated in boldface), leading to what are known as the paradoxes. In all three cases, State 1 unreasonably loses a seat. The New Seat Paradox or Alabama Paradox : a seat is added to the house; The Population Paradox : there is a population shift: State 1 grows and State 2 shrinks; The New State Paradox or Oklahoma Paradox : a new state joins the union. Basic H(1,450,000, 3,400,000, 5,150,000; 10) = (2, 3, 5). k pk q k l k uk q k l k h k 1 1, 450, 000 1.45 1 2 .45 2 2 3, 400, 000 3.40 3 4 .40 3 3 5, 150, 000 5.15 5 6 .15 5 p=10,000,000 9 h=10 s=1,000,000 Add a seat H(1,450,000, 3,400,000, 5,150,000; 11) = (1, 4, 6). k pk qk l k uk q k l k h k 1 1, 450, 000 1.595 1 2 .595 1 2 3, 400, 000 3.740 3 4 .740 4 3 5, 150, 000 5.665 5 6 .665 6 p=10,000,000 9 h=11 s=909,090 Population shift H(1,470,000, 3,380,000, 5,150,000; 10) = (1, 4, 5). k pk q k l k uk q k l k h k 1 1, 470, 000 1.55 1 2 .55 1 2 3, 380, 000 3.56 3 4 .56 4 3 4, 650, 000 4.89 4 5 .89 5 p=9,500,000 8 h=10 s=950,000 Add a state H(1,450,000, 3,400,000, 5,150,000, 2,600,000; 13) = (1, 4, 5, 3). k pk q k l k uk q k l k h k 1 1, 450, 000 1.50 1 2 .50 1 2 3, 400, 000 3.51 3 4 .51 4 3 5, 150, 000 5.31 5 6 .31 5 4 2, 600, 000 2.68 2 3 .68 3 p=12,600,000 11 h=13 s=969,231 For these reasons, most experts reject the Hamilton method for apportioning the U. S. House of Representatives. In fact, the first ever presidential veto vetoed the use of Hamilton's method. The bill was returned to Congress and the new bill, sent to and signed by Washington, set the use of Jefferson's method. The arguments were not mathematical but political. 4 The Apportionment Problem Jefferson Method The Jefferson method of apportionment J replaces the standard divisor s with an appropriately chosen modified divisor dJ (interpreted as district size), replaces the standard quota qk with the modified quota mk and then rounds all mk 's down to get the apportionments. Precisely, Definition: Suppose (p1 , , pn ; h) is a census and a house size. For any divisor d, called the modified divisor, compute the quotients sum J(d) = jp k 1 d + pk , d called the modified quotas, and compute the jp k 2 d + + jp k n d . If jdJ k is such that J(dJ ) = h, then the Jefferson apportionment is (j1 , j2 , . . . , jn ) where pk jk = dJ . This value if dJ is the Jefferson modified divisor . Denote by jk the Jefferson apportionment for state k. Expressed this way, the process is one of guessing, testing and revising. It's not hard to show that such a dJ always exists except for the unlikely event that two populations are exactly the same. Furthermore, it is clear that since J(s) = l1 + + ln < h the modified divisor dJ must be less than s to increase the sum. The technique of critical divisors, described later, shows how to determine the Jefferson apportionment directly. Finding a Jefferson Modified Divisor: In this example, the initial divisor of 1,000,000,(the standard divisor) leads to an apportionment of 9 seats; reducing the divisor to 800,000 leads to 11 seats and finally d = 850, 000 gives the desired apportionment of 10. Thus dJ = 850,000 is a valid Jefferson modified divisor. (In fact, any d in the range 800,000 < d 883,333 will work). divisor 1,000,000 800,000 850,000 k pk qk lk mk mk mk mk = jk 1 1, 500, 000 1.50 1 1.87 1 1.75 1 2 3, 200, 000 3.20 3 4.00 4 3.26 3 3 5, 300, 000 5.30 5 6.62 6 6.24 6 seats 9 11 10 Thus J (1,500,000, 3,200,000, 5,300,000; 10) = (1, 3, 6). 5 The Apportionment Problem Adams Method The Adams method of apportionment A replaces the standard divisor s with an appropriately chosen modified divisor dA (interpreted as district size), replaces the standard quota qk with the modified quota mk and then rounds all mk 's up to get the apportionments. Precisely, Definition: Suppose (p1 , , pn ; h) is a census and a house size. For any divisor d, called the modified divisor, compute the quotients pdk , called the modified quotas, and compute the sum lp m lp m lp m 2 n 1 + + + . A(d) = d d d \u0006 pk \u0007If dA is such that A(dA ) = h, then the Adams apportionment is (a1 , a2 , . . . , an ) where ak = . This value if dA is the Adams modified divisor . Denote by ak the Adams apportionment d for state k. Expressed this way, the process is one of guessing, testing and revising. It's not hard to show that such a dA always exists except for the unlikely event that two populations are exactly the same. Furthermore, it is clear that since A(s) = u1 + + un > h the modified divisor dA must be greater than s to decrease the sum. The technique of critical divisors, described later, shows how to determine the Adams apportionment directly. Finding an Adams Modified Divisor: In this example, the initial divisor of s = 1,000,000,(the standard divisor) leads to an apportionment of 12 seats; increasing the divisor to 1,400,000 leads to 9 seats and finally d = 1,100,000 gives the desired apportionment of 10. Thus dA = 1,100,000 is a valid value for aA . (In fact, any d in the range 1,066,667 < d < 1,325,000 will work). divisor 1,000,000 1,400,000 1,100,000 k pk qk uk mk mk mk mk = ak 1 1, 500, 000 1.50 2 1.07 2 1.36 2 2 3, 200, 000 3.20 4 2.28 3 2.90 3 3 5, 300, 000 5.30 6 3.79 4 4.82 5 seats 12 9 10 Thus, A(1,500,000, 3,200,000, 5,300,000; 10) = (2, 3, 5). It's worth noting that in this example,the Jefferson method comparatively favors the largest state and the Adams method favors the smallest state. This is a general feature of the methods: Jefferson favors large states and Adams favors small states. Not coincidentally, Thomas Jefferson was from the (then) large state of Virginia and John Adams from the small state of Massachusetts. An alternative method of determining the Adams apportionment is to first give each state one seat and then apportion the remaining h n seats by the Jefferson method. This works because a Jefferson modified divisor for h n seats works as an Adams modified divisor for h seats. 6 The Apportionment Problem Webster Method The Webster method of apportionment W replaces the standard divisor s with an appropriately chosen modified divisor dW (interpreted as district size), replaces the standard quota qk with the modified quota mk and then rounds all mk 's from the average to get the apportionments. \u0006 \u0007 Definition: The value of x rounded from the average is x = x 12 . In other words, if the fractional part of x is .5 or larger, then x = x; otherwise, x = x Precisely, Definition: Suppose (p1 , , pn ; h) is a census and a house size. For any divisor d, called the modified divisor, compute the quotients sum W (d) = pk , d called the modified quotas, and compute the p2 pn p1 + + + . d d d If dW is such that A(dW ) = h, then the Webster apportionment is (a1 , a2 , . . . , an ) where ak = dpWk . This value if dW is the Webster modified divisor . Denote by wk the Webster apportionment foe state k. Expressed this way, the process is one of guessing, testing and revising. It's not hard to show that such a dW always exists except for the unlikely event that two populations are exactly the same. The technique of critical divisors, described later, shows how to determine the Webster apportionment directly. Finding a Webster Modified Divisor: In this example, the initial divisor of 1,000,000,(the standard divisor) actually works as a Webster modified divisor. So we let dW = 1,000,000. k 1 2 3 divisor pk 1, 500, 000 3, 200, 000 5, 300, 000 seats 1,000,000 qk qk = wk 1.50 2 3.20 3 5.30 5 10 Thus W(1,500,000, 3,200,000, 5,300,000; 10) = (2, 3, 5). It is often the case that dW = s, but it could be that either s < dW or dW < s. The first occurs if W (s) < h and the second if dw > s