Assumptions:
At the start, the number of juveniles and adults are not equal
Each adult has 1 calf per year ???? (your problem is not clear on that)
Major flaw: When do adults no longer reproduce or die?
current juvenile females --- x
current adult females ----- y
after 1 year:
juves = y , half of surviving became adults + each of the adults had a calf (juvenile)
= y
adults = y + x/2 <----- current adults + half of juves
= (x + 2y)/2
ratio is y : (x + 2y)/2
= 2y : x+2y
after year 2:
juves = (x + 2y)/2 <--- this will always be the previous adults
adults = (x + 2y)/2 + y/2
= (x + 3y)/2
ratio = x+2y : x + 3y
after year 3:
juves = (x+3y)/2
adults = (x+3y)/2 + (x+2y)/4
= (3x + 8y)/4
continue this to see if you can see a pattern
This would be nice to see develop in some kind of computer simulation
I might work on it a bit
Develop an age-structured population model for elk, where you divide the population to two cohorts: juvenile females and adult females. Juveniles do not reproduce; they reach reproductive maturity in 1 year. The probability of juveniles to reach adulthood is 50%. Adult females have on average 1 calf. a: Compute a long term proportion between (juveniles/adults) in this population. b: Will the population survive in the long term? c: How many calves would a female elk have to have every year for population to survive?
4 answers
Made up this rinky-dink program in an ancient computer language, (which I still like for stuff like this)
10 input j,a
20 for n = 1 to 10
30 newj = a : newa = a+j/2
40 j = newj : a = newa
60 print n-1,j,a,j/a
70 next n
here are the results for 5 different inputs of j (juveniles) and a (adults)
? 16,32
0 32 40 0.8
1 40 56 0.714286
2 56 76 0.736842
3 76 104 0.730769
4 104 142 0.732394
5 142 194 0.731959
6 194 265 0.732075
7 265 362 0.732044
8 362 494.5 0.732053
9 494.5 675.5 0.73205
? 100,20
0 20 70 0.285714
1 70 80 0.875
2 80 115 0.695652
3 115 155 0.741935
4 155 212.5 0.729412
5 212.5 290 0.732759
6 290 396.25 0.731861
7 396.25 541.25 0.732102
8 541.25 739.375 0.732037
9 739.375 1010 0.732054
? 20,100
0 100 110 0.909091
1 110 160 0.6875
2 160 215 0.744186
3 215 295 0.728814
4 295 402.5 0.732919
5 402.5 550 0.731818
6 550 751.25 0.732113
7 751.25 1026.25 0.732034
8 1026.25 1401.875 0.732055
9 1401.875 1915 0.73205
the ratio seems to approach 1:0.73205
no matter how many juveniles and adults we start with
10 input j,a
20 for n = 1 to 10
30 newj = a : newa = a+j/2
40 j = newj : a = newa
60 print n-1,j,a,j/a
70 next n
here are the results for 5 different inputs of j (juveniles) and a (adults)
? 16,32
0 32 40 0.8
1 40 56 0.714286
2 56 76 0.736842
3 76 104 0.730769
4 104 142 0.732394
5 142 194 0.731959
6 194 265 0.732075
7 265 362 0.732044
8 362 494.5 0.732053
9 494.5 675.5 0.73205
? 100,20
0 20 70 0.285714
1 70 80 0.875
2 80 115 0.695652
3 115 155 0.741935
4 155 212.5 0.729412
5 212.5 290 0.732759
6 290 396.25 0.731861
7 396.25 541.25 0.732102
8 541.25 739.375 0.732037
9 739.375 1010 0.732054
? 20,100
0 100 110 0.909091
1 110 160 0.6875
2 160 215 0.744186
3 215 295 0.728814
4 295 402.5 0.732919
5 402.5 550 0.731818
6 550 751.25 0.732113
7 751.25 1026.25 0.732034
8 1026.25 1401.875 0.732055
9 1401.875 1915 0.73205
the ratio seems to approach 1:0.73205
no matter how many juveniles and adults we start with
Spent a lot of time answering this question.
No response form the poster, not even sure if he/she even looked at it.
Frustrating, will remember that name for future.
No response form the poster, not even sure if he/she even looked at it.
Frustrating, will remember that name for future.
Hey i'm sorry for response from my side. just now i have seen. i'm very happy with your answer and thanks for spending lot of time to solve my question. tq so much
1 answer