- For \(N=28\), the Hamilton method results in A: 5, B: 11, C: 2, D: 10.
- Comparing to \(N=27\) (A: 4, B: 11, C: 2, D: 10), no campus lost projectors, so the Alabama paradox did not occur.
- For \(N=29\), the Hamilton method results in A: 5, B: 12, C: 2, D: 10.
- Comparing to \(N=28\) (A: 5, B: 11, C: 2, D: 10), no campus lost projectors, so the Alabama paradox did not occur.
- Final answers: \((a) \text{ A: 5, B: 11, C: 2, D: 10, Paradox: No}; (b) \text{ Paradox: No}\).
Explanation
- Apportionment for 28 Projectors
To apportion the projectors using the Hamilton method, we first identify the populations of the four campuses: \(P_A = 824\), \(P_B = 1926\), \(P_C = 316\), and \(P_D = 1728\). The total population is \(P_{total} = 824 + 1926 + 316 + 1728 = 4794\). The Hamilton method involves calculating the standard quota \(q_i = \frac{P_i}{P_{total}} \times N\) for each campus, assigning the integer part \(\lfloor q_i \rfloor\), and distributing any remaining projectors to campuses with the largest fractional parts \(f_i = q_i - \lfloor q_i \rfloor\).
For \(N = 28\), the standard quotas are:
- Campus A: \(q_A = \frac{824}{4794} \times 28 \approx 4.813\)
- Campus B: \(q_B = \frac{1926}{4794} \times 28 \approx 11.249\)
- Campus C: \(q_C = \frac{316}{4794} \times 28 \approx 1.846\)
- Campus D: \(q_D = \frac{1728}{4794} \times 28 \approx 10.093\)
The initial integer parts are \(4, 11, 1, 10\), which sum to \(26\). We need to distribute \(28 - 26 = 2\) additional projectors based on the largest fractional parts: \(0.813\) (A), \(0.249\) (B), \(0.846\) (C), and \(0.093\) (D). The two largest fractions belong to Campus C (\(0.846\)) and Campus A (\(0.813\)). Adding these, the apportionment for \(N = 28\) is: Campus A: \(5\), Campus B: \(11\), Campus C: \(2\), and Campus D: \(10\). Comparing this to the \(N=27\) apportionment (A: 4, B: 11, C: 2, D: 10), no campus received fewer projectors, so the Alabama paradox did not occur.
- Apportionment for 29 Projectors
For \(N = 29\), the standard quotas are:
- Campus A: \(q_A = \frac{824}{4794} \times 29 \approx 4.985\)
- Campus B: \(q_B = \frac{1926}{4794} \times 29 \approx 11.651\)
- Campus C: \(q_C = \frac{316}{4794} \times 29 \approx 1.912\)
- Campus D: \(q_D = \frac{1728}{4794} \times 29 \approx 10.453\)
The initial integer parts are \(4, 11, 1, 10\), which sum to \(26\). We need to distribute \(29 - 26 = 3\) additional projectors based on the largest fractional parts: \(0.985\) (A), \(0.651\) (B), \(0.912\) (C), and \(0.453\) (D). The three largest fractions are from Campus A (\(0.985\)), Campus C (\(0.912\)), and Campus B (\(0.651\)). Thus, the apportionment for \(N=29\) is: Campus A: \(5\), Campus B: \(12\), Campus C: \(2\), and Campus D: \(10\). Comparing this to the \(N=28\) apportionment (A: 5, B: 11, C: 2, D: 10), no campus lost any projectors. Therefore, the Alabama paradox did not occur.
Examples
The Hamilton method and the study of paradoxes like the Alabama paradox are crucial in political science and social choice theory for ensuring fair representation. For example, in legislative bodies, it is vital to understand how seat distributions change as the total size of the assembly grows. By using these mathematical methods, governments can ensure that no state or region is unfairly penalized by a mere increase in the number of representatives, promoting stability and equitable governance in real-world democratic processes.