Question

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2. Residents in City X tend to read Newspaper A or Newspaper B, or not read any newspaper at all. Statistics shows that, after a year . For people reading newspaper A, 30% of them will switch to reading newspaper B and For people reading newspaper B, 20% of them will switch to reading newspaper A and . For people not reading newspaper, 20% will switch to reading newspaper A and 10% to 10% to not reading newspaper 10% to not reading newspaper. reading newspaper B. The current percentage of residents reading Newspaper A, Newspaper B, and not reading any newspaper are 40%, 50% and 10%, respectively. By hand calculation, what are the percentages of readership after 1 year? Then calculate the expected percentages of readership after 5, 10, 15 and 20 years (you may use any computational software). Do the percentages tend to steady state values?

Answer 1

Let total number of citizens in City X in 0th year is N.

Let number of citizens who read newspaper A in 0th year is A_j.

Let number of citizens who read newspaper B in 0th year is B_j.

Let number of citizens who do not read any newspaper in 0th year is C_j.

j=0,1,2,....

Given:- A₀=0.4N , B₀=0.5N , C₀=0.1N $\Rightarrow$ A₀/N=0.4 , B₀/N=0.5 , C₀/N=0.1

According to the problem :-

A₁ = 0.6A₀ + 0.2B₀ + 0.2C₀

_B1 = 0.3A₀ + 0.7B₀ + 0.1C₀

_C1 = 0.1A₀ + 0.1B₀ + 0.7C₀

$\therefore$  A₁/N = (0.6A₀ + 0.2B₀ + 0.2C₀)/N = 0.6A₀/N + 0.2B₀/N + 0.2C₀/N = 0.6$\times$0.4 + 0.2$\times$0.5 + 0.2$\times$0.1 = 0.36 = 36%

$\therefore$ B₁/N = (0.3A₀ + 0.7B₀ + 0.1C₀)/N = 0.3A₀/N + 0.7B₀/N + 0.1C₀/N = 0.3$\times$0.4 + 0.7$\times$0.5 + 0.1$\times$0.1 = 0.48 = 48%

$\therefore$ C₁/N = (0.1A₀ + 0.1B₀ + 0.7C₀)/N = 0.1A₀/N + 0.1B₀/N + 0.7C₀/N = 0.1$\times$0.4 + 0.1$\times$0.5 + 0.7$\times$0.1 = 0.16 = 16%

The following calculations are done using the following R code :-

For N=100, we see A[6]=33.40160, B[6]=42.76480, C[6]=23.83360

$\therefore$ After 5 years A₅/N=33.40160/100=33.40106% , B₅/N=42.76480/100=42.7948% , C₅/N=23.83360/100=23.8336%

For N=100, we see A[11]=33.33403, B[11]=41.75667, C[11]=24.90930

$\therefore$ After 10 years A₁₀/N=33.33403/100=33.33403% , B₁₀/N=41.75667/100=41.75667% , C₁₀/N=24.90930/100=23.8336%

For N=100, we see A[16]=33.33334, B[16]=41.67371, C[16]=24.99295

$\therefore$ After 15 years A₁₅/N=33.33334/100=33.33334% , B₁₅/N=41.67371/100=41.67371% , C₁₅/N=24.99295/100=24.99295%

For N=100, we see A[21]=33.33333, B[21]=41.66722, C[21]=24.99945

$\therefore$ After 20 years A₂₀/N=33.33333/100=33.33333% , B₂₀/N=41.66722/100=41.66722% , C₂₀/N=24.99945/100=24.99945%