Population dynamics with Matrices

• A is the population projection matrix

• Leslie 1945 summarized the existing theory at the time for populations with a certain age structure. Each age was one unit of time apart

• F is the stage specific Fecundity.• G is the survival from stage i to stage i+1

• Lefkovitch (1965) proposed that the population stages need not have the same duration and that some in a given stage will survive and stay in the same stage after one year (or time interval).

• Lefkovitch (1965) proposed that the population stages need not have the same duration and that some in a given stage will survive and stay in the same stage after one year (or time interval).

• In the above P1, P2, P3, P4 is the probability that females in stages 1-4 will remain in the same stage the following year.

Northern Spotted Owl

Northern Spotted Owl

• http://www.fs.fed.us/psw/rsl/projects/wild/lamberson1.PDF • ROLAND H. LAMBERSON, ROBERT McKELVEY, BARRY R. NOON,

CURTIS VOSS, 1992. A Dynamic Analysis of Northern Spotted Owl• Viability in a Fragmented Forest Landscape*. Conservation Biology• Volume 6, No. 4, December 1992• Or http://www.fs.fed.us/psw/publications/documents/gtr-133/chap8.pdf

• For the questions to follow we will assume a Lefkovitch population projection matrix structure as shown above

4 years of population data for the spotted owl is shown below.

• Using the 1991 to 1992 data what is the fecundity F of the pairs? (F2=0)

• Assume that P1=P2=0 i.e. Owls in stage 1 or 2 automatically advance to the next stage and that P3=0.94 i.e. 94% survival rate of mating pairs.

4 years of population data for the spotted owl is shown below.

• Using the 1991 to 1992 data what is the fecundity F of the pairs? (F2=0)• F=F3=33/88=0.38• Assume that P1=P2=0 i.e. Owls in stage 1 or 2 automatically advance to the next stage and that P3=0.94 i.e. 94% survival rate of mating pairs.

4 years of population data for the spotted owl is shown below.

• Using the 1991 to 1992 data what is the value of G1? G1 is the fraction of stage 1 individuals advancing to stage 2.

• Assume that P1=P2=0 i.e. Owls in stage 1 or 2 automatically advance to the next stage and that P3=0.94 i.e. 94% survival rate of mating pairs.

4 years of population data for the spotted owl is shown below.

• Using the 1991 to 1992 data what is the value of G1? G1 is the fraction of stage 1 individuals advancing to stage 2. • G1=7/36=0.19• Assume that P1=P2=0 i.e. Owls in stage 1 or 2 automatically advance to the next stage and that P3=0.94 i.e. 94% survival rate of mating pairs.

4 years of population data for the spotted owl is shown below.

• Using the 1991 to 1992 data what is the value of G2? G2 is the fraction of stage 2 individuals advancing to stage 3.

• Assume that P1=P2=0 i.e. Owls in stage 1 or 2 automatically advance to the next stage and that P3=0.94 i.e. 94% survival rate of mating pairs.

4 years of population data for the spotted owl is shown below.

• Using the 1991 to 1992 data what is the value of G2? G2 is the fraction of stage 2 individuals advancing to stage 3. • G2=(87-88*.94)/9=0.48• Assume that P1=P2=0 i.e. Owls in stage 1 or 2 automatically advance to the next stage and that P3=0.94 i.e. 94% survival rate of mating pairs.

• Four points are worth noting here about the eigenvalues, r for population projection matrices

Nt+1=ANt:

• When r=1.0 the exponential term is a constant term, • when r less than 1.0 the exponential term eventually goes

to zero• if r is greater than 1.0 will be exponential growth.• If r is a complex number this corresponds to oscillations

Question

Using a difference equation

Nt+1=Ant

The dominant eigenvalue is =1.04.

What is the implied population rate of increase?

Will this population grow or get smaller?

Question

Using a difference equation

Nt+1=Ant

The dominant eigenvalue is =1.04.

What is the implied population rate of increase?

4% increase each year

Question

Using a flow equation

The dominant eigenvalue is r=.02. What is the implied population rate of increase?

ANdt

dN

Four points are worth noting here about the eigenvalues, r , for transport matricesIn flow equations like above :

When r=0 the exponential term is a constant term, when r is negative the exponential term eventually goes to zero if r is positive there will be exponential growth.If r is a complex number this corresponds to oscillations

Question

Using a flow equation

The dominant eigenvalue is r=.02. What is the implied population rate of increase?

2% increase each year

ANdt

dN

What is the transpose of the matrix below?

153

726

241

A

What is the transpose of the matrix below?

172

524

361TA

The population projection matrix and initial population are shown below. What is the population after 1 year?

____

____

____

0

0

10

N

110

115.

241

1

0

N

A

The population projection matrix and initial population are shown below. What is the population after 1 year?

Assume N1=AN0

_0

5

10

0

0

10

N

110

115.

241

1

0

N

A

The last four years of a long population model simulation are shown below.

• What is the dominant eigenvalue for this population? And what is the percent growth rate?

The last for years of a long population model simulation are shown below.

• What is the dominant eigenvalue for this population? 1.11

• And what is the percent growth rate? 11 %

• Deborah T.Crouse, L.B. Crowder, and H. Caswell. 1987. A stage-based population Model for Loggerhead Sea Turtles and implications for conservation. Ecology, 68 (5), 1412 1423.