Hello, I\"m stuck on some math homework for a mathematical modelingclass. The qu

Question

Hello, I"m stuck on some math homework for a mathematical modelingclass. The question is as follows:

A guinea pig population has a two-year life cycle. At the end ofthe first year 70% of the females give birth to a liter. At the endof their second year of life 40% of females produce a liter. Theaverage liter size is two animals and male and females are born inroughly equal portion. Find a model for the population. Approximatethe growth rate of the population and predict what will happen.Given: Population last year was 120, and the previous year 105.

My current model is currently
p(n)=.7p(n-1)*2+.4p(n-2)*2

With .7 and .4 representing the growth rates and the 2 representingthe number of offspring per liter.
Is my model accurate?
Thank you a lot for any help!

Explanation / Answer

Let p[n] stand for the number of pigs born at the endof year n. We'll assume that the pigs live exactly 2 yearsand that all the two year olds die immediately after the two yearold females give birth. We'll also assume that the population iscounted at the end of each year immediately after the new pigsare born and two year olds have died. This means that the number ofpigs in the population at the end of year n willbe p[n]+p[n-1]. Our formula for the recursion is then:    p[n] = 2(7/10)(1/2)p[n-1] +2(4/10)(1/2)p[n-2] Note the 1/2 because the females are one-half the popultionand only the females give birth to two new pigs each. To start things off new need to know p[0] and p[-1]. One wayto find those values is to solve the following for an integersolution:    120 =p[0]+p[-1] AND105 = p[-1]+p[-2] AND p[0] = 2 (7/10)(1/2)p[-1]+2 (4/10)(1/2) p[-2] I won't get into how to solve the linear diophantine equation,but the only solution it has in integers is p[0]=60, p[-1]=60 andp[-2]=45. This means that the the starting populations are 120 and105 in the two prior years as the problem requires. With those starting values for p[0] and p[-1] you can solvethe recursion to obtain the general solution:    p[n] =(30/209)((209-15209)((1/20)(7-209))n +((1/20)(7+209))n(209+15209)) I realize that you may not know how to solve the recursion toobtain that solution. But getting into to how to do that is way toomuch to write here. I guess I'll just assume that you know how todo this or that your professor doesn't require you to actuallysolve the recursion. One way or the other you can now apply thesolution, or use the original formula, to generate a table ofvalues of p[n] for n=1,2,3,.... The first few entries in that tableare:    60.,66.,70.2,75.54,80.958,86.8866,93.2038,99.9973,107.28,115.095,123.478,132.473,142.122 Given those values the population at the end of years1,2,3,... is the sum of these numbers in pairs or:    126.,136.2,145.74,156.498,167.845,180.09,193.201,207.277,222.374,238.573,255.951,274.595 You can find the growth rate by fitting a model of the formb(1+r)n. Do that, or just divided consecutive populationtotals, and you get r = 7.285% annually. I should point out that regardless of the values we select forp[-1] and p[0] to get the process started, the ultimate growth rateafter a sufficiently long time is still r = 7.285% annually. Infact, the exact value of r is:    r = (-13+209)/20 I won't get into how to solve the linear diophantine equation,but the only solution it has in integers is p[0]=60, p[-1]=60 andp[-2]=45. This means that the the starting populations are 120 and105 in the two prior years as the problem requires. With those starting values for p[0] and p[-1] you can solvethe recursion to obtain the general solution:    p[n] =(30/209)((209-15209)((1/20)(7-209))n +((1/20)(7+209))n(209+15209)) I realize that you may not know how to solve the recursion toobtain that solution. But getting into to how to do that is way toomuch to write here. I guess I'll just assume that you know how todo this or that your professor doesn't require you to actuallysolve the recursion. One way or the other you can now apply thesolution, or use the original formula, to generate a table ofvalues of p[n] for n=1,2,3,.... The first few entries in that tableare:    60.,66.,70.2,75.54,80.958,86.8866,93.2038,99.9973,107.28,115.095,123.478,132.473,142.122 Given those values the population at the end of years1,2,3,... is the sum of these numbers in pairs or:    126.,136.2,145.74,156.498,167.845,180.09,193.201,207.277,222.374,238.573,255.951,274.595 You can find the growth rate by fitting a model of the formb(1+r)n. Do that, or just divided consecutive populationtotals, and you get r = 7.285% annually. I should point out that regardless of the values we select forp[-1] and p[0] to get the process started, the ultimate growth rateafter a sufficiently long time is still r = 7.285% annually. Infact, the exact value of r is:    r = (-13+209)/20

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