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ElaKiri Talk!
Sri Lanka's Fertility Rate Falls Below Replacement Level
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<blockquote data-quote="Hankook" data-source="post: 30015083" data-attributes="member: 580311"><p>If we assume zero mortality rates, no migration, and only fertility rates influence population growth, we can estimate when Sri Lankan Moors might surpass the Sinhalese in population percentage.</p><p></p><p>Here’s a simplified approach to the calculation:</p><p></p><p>Initial Populations and Fertility Rates:</p><p></p><p>Sinhalese: 75%</p><p>Moors: 9%</p><p>Sinhalese fertility rate: 2.3</p><p>Moor fertility rate: 3.3</p><p>Population Growth Model:</p><p></p><p>The population of a group grows according to the formula P(t) = P_0 \times (1 + r)^tP(t)=P </p><p>0</p><p> ×(1+r) </p><p>t</p><p> , where P(t)P(t) is the population at time tt, P_0P </p><p>0</p><p> is the initial population, and rr is the growth rate (which in this simplified model is \text{fertility rate} - 1fertility rate−1).</p><p>Growth Rates:</p><p></p><p>Sinhalese growth rate: 2.3 - 1 = 1.32.3−1=1.3</p><p>Moors growth rate: 3.3 - 1 = 2.33.3−1=2.3</p><p>Projection:</p><p>Let P_{\text{Sinhalese}}(t)P </p><p>Sinhalese</p><p> (t) and P_{\text{Moors}}(t)P </p><p>Moors</p><p> (t) represent the population proportions of Sinhalese and Moors at time tt, respectively. We have:</p><p></p><p>P_{\text{Sinhalese}}(t) = 0.75 \times (1 + 1.3)^t</p><p>P </p><p>Sinhalese</p><p> (t)=0.75×(1+1.3) </p><p>t</p><p> </p><p></p><p>P_{\text{Moors}}(t) = 0.09 \times (1 + 2.3)^t</p><p>P </p><p>Moors</p><p> (t)=0.09×(1+2.3) </p><p>t</p><p> </p><p>We want to find tt when P_{\text{Moors}}(t)P </p><p>Moors</p><p> (t) exceeds P_{\text{Sinhalese}}(t)P </p><p>Sinhalese</p><p> (t).</p><p></p><p>Solving:</p><p>Set up the inequality:</p><p></p><p>0.09 \times (1 + 2.3)^t > 0.75 \times (1 + 1.3)^t</p><p>0.09×(1+2.3) </p><p>t</p><p> >0.75×(1+1.3) </p><p>t</p><p> </p><p></p><p>0.09 \times 3.3^t > 0.75 \times 2.3^t</p><p>0.09×3.3 </p><p>t</p><p> >0.75×2.3 </p><p>t</p><p> </p><p></p><p>\frac{0.09}{0.75} > \left(\frac{2.3}{3.3}\right)^t</p><p>0.75</p><p>0.09</p><p> >( </p><p>3.3</p><p>2.3</p><p> ) </p><p>t</p><p> </p><p></p><p>0.12 > \left(\frac{2.3}{3.3}\right)^t</p><p>0.12>( </p><p>3.3</p><p>2.3</p><p> ) </p><p>t</p><p> </p><p></p><p>\left(\frac{2.3}{3.3}\right)^t < 0.12</p><p>( </p><p>3.3</p><p>2.3</p><p> ) </p><p>t</p><p> <0.12</p><p>Taking natural logs on both sides to solve for tt:</p><p></p><p>\log\left(\left(\frac{2.3}{3.3}\right)^t\right) < \log(0.12)</p><p>log(( </p><p>3.3</p><p>2.3</p><p> ) </p><p>t</p><p> )<log(0.12)</p><p></p><p>t \cdot \log\left(\frac{2.3}{3.3}\right) < \log(0.12)</p><p>t⋅log( </p><p>3.3</p><p>2.3</p><p> )<log(0.12)</p><p></p><p>t > \frac{\log(0.12)}{\log\left(\frac{2.3}{3.3}\right)}</p><p>t> </p><p>log( </p><p>3.3</p><p>2.3</p><p> )</p><p>log(0.12)</p><p> </p><p>Plugging in the values:</p><p></p><p>\log(0.12) \approx -0.921</p><p>log(0.12)≈−0.921</p><p></p><p>\log\left(\frac{2.3}{3.3}\right) \approx -0.274</p><p>log( </p><p>3.3</p><p>2.3</p><p> )≈−0.274</p><p></p><p>t > \frac{-0.921}{-0.274} \approx 3.36</p><p>t> </p><p>−0.274</p><p>−0.921</p><p> ≈3.36</p><p>So, in this simplified model, the Moors would surpass the Sinhalese in about 3 to 4 periods of growth, or roughly 3 to 4 cycles of population doubling.</p><p></p><p>Assuming a 10-year cycle for simplicity, it would take approximately 30 to 40 years for the Moors to surpass the Sinhalese in population percentagewise.</p></blockquote><p></p>
[QUOTE="Hankook, post: 30015083, member: 580311"] If we assume zero mortality rates, no migration, and only fertility rates influence population growth, we can estimate when Sri Lankan Moors might surpass the Sinhalese in population percentage. Here’s a simplified approach to the calculation: Initial Populations and Fertility Rates: Sinhalese: 75% Moors: 9% Sinhalese fertility rate: 2.3 Moor fertility rate: 3.3 Population Growth Model: The population of a group grows according to the formula P(t) = P_0 \times (1 + r)^tP(t)=P 0 ×(1+r) t , where P(t)P(t) is the population at time tt, P_0P 0 is the initial population, and rr is the growth rate (which in this simplified model is \text{fertility rate} - 1fertility rate−1). Growth Rates: Sinhalese growth rate: 2.3 - 1 = 1.32.3−1=1.3 Moors growth rate: 3.3 - 1 = 2.33.3−1=2.3 Projection: Let P_{\text{Sinhalese}}(t)P Sinhalese (t) and P_{\text{Moors}}(t)P Moors (t) represent the population proportions of Sinhalese and Moors at time tt, respectively. We have: P_{\text{Sinhalese}}(t) = 0.75 \times (1 + 1.3)^t P Sinhalese (t)=0.75×(1+1.3) t P_{\text{Moors}}(t) = 0.09 \times (1 + 2.3)^t P Moors (t)=0.09×(1+2.3) t We want to find tt when P_{\text{Moors}}(t)P Moors (t) exceeds P_{\text{Sinhalese}}(t)P Sinhalese (t). Solving: Set up the inequality: 0.09 \times (1 + 2.3)^t > 0.75 \times (1 + 1.3)^t 0.09×(1+2.3) t >0.75×(1+1.3) t 0.09 \times 3.3^t > 0.75 \times 2.3^t 0.09×3.3 t >0.75×2.3 t \frac{0.09}{0.75} > \left(\frac{2.3}{3.3}\right)^t 0.75 0.09 >( 3.3 2.3 ) t 0.12 > \left(\frac{2.3}{3.3}\right)^t 0.12>( 3.3 2.3 ) t \left(\frac{2.3}{3.3}\right)^t < 0.12 ( 3.3 2.3 ) t <0.12 Taking natural logs on both sides to solve for tt: \log\left(\left(\frac{2.3}{3.3}\right)^t\right) < \log(0.12) log(( 3.3 2.3 ) t )<log(0.12) t \cdot \log\left(\frac{2.3}{3.3}\right) < \log(0.12) t⋅log( 3.3 2.3 )<log(0.12) t > \frac{\log(0.12)}{\log\left(\frac{2.3}{3.3}\right)} t> log( 3.3 2.3 ) log(0.12) Plugging in the values: \log(0.12) \approx -0.921 log(0.12)≈−0.921 \log\left(\frac{2.3}{3.3}\right) \approx -0.274 log( 3.3 2.3 )≈−0.274 t > \frac{-0.921}{-0.274} \approx 3.36 t> −0.274 −0.921 ≈3.36 So, in this simplified model, the Moors would surpass the Sinhalese in about 3 to 4 periods of growth, or roughly 3 to 4 cycles of population doubling. Assuming a 10-year cycle for simplicity, it would take approximately 30 to 40 years for the Moors to surpass the Sinhalese in population percentagewise. [/QUOTE]
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