Transcript Lecture 30 Fish Mgmt II.ppt

Constant Quota fishing at levels approaching the MSY shortens the biomass
range the population will recover, and the likelihood of entering the danger zone
increases. Once the danger zone is entered fishing must stop or be severely
curtailed
dB
dt
Danger zone
Stable biomass range
r0K
4
Catch rate
K/2
r0K
4
K
Maximum Sustainable Yield (MSY)
B
What would happen to a population at K/2 subjected to a harvest rate of
dB
dt
r0K
4
C
K/2
r0K
4
K
B
is called the Maximum Sustainable Yield (MSY)
r0K
4
The logistic model
dB
 B
 r 0 B 1  
dt
 K
dB
dt
r0K
4
*
C
K/2
K
B
An operational example
1.2
1
A lake that hasn’t been fished
for 20 yr is opened for fishing
and annual creel surveys
show that CPUE is declining
rapidly
0.8
CPUE
Kg/hr
0.6
0.4
0.2
0
0
5
10
15
20
25
3000
2500
•After a 9-yr moratorium the
population has gradually
bounced back, and the fishery
is opened again with more
restrictive limits
2000
Catch
Kg/yr
•The estimated total catch
varies from year to year but
the trend rapidly becomes
apparent and the lake is
closed
1500
1000
500
0
0
5
10
15
Year
20
25
•After signs of tapering
off limits are tightened
even further.
1.2
1
How can such a population
be managed with the logistic
model?
We don’t know B, K or r0
0.8
CPUE
Kg/hr
0.6
0.4
0.2
0
0
5
10
15
20
25
3000
We can also assume that the
population is at K to start with
2500
2000
Catch
Kg/yr
We can assume that CPUE
is a linear function of
Biomass (eg B=q*CPUE
Where q is called the
“catchability” factor
1500
1000
500
0
0
5
10
15
Year
20
25
•The catch (C) in the first year was 1020 kg, at the rate of 1.0 kg/fisherman hr.
•The following year the CPUE went down to 0.9 kg/fisherman hr.
•Assuming that the population was at its K, and that the CPUE reflects the
biomass linearly, B=q *CPUE, then dB/dt at K will be 0, and the entire catch will
be subtracted from the biomass present.
•That is we assume that there will be no significant recruitment or growth
response to compensate for thinning within the first season
•The response to the first year’s thinning will be reflected in the next year’s
catch data
•assume the change in CPUE reflects the change in B,
(D CPUE)/CPUE = (D B)/B, 0.1/1=1000/K. Therefore K =10,000 kg
•And CPUE =q B, that is 1.0 kg/hr =q * 10,000 kg,
• so q =1kg/hr/10000 kg = 0.0001kg/hr/kg
•The next year the catch rises to 1050 kg, and the CPUE falls to 0.83 kg/hr.
This catch appears to be unsustainable at this B level, but how much would
have been sustainable?
•The next year the catch rises to 1050 kg, and the CPUE falls to 0.83 kg/hr.
•This catch appears to be unsustainable at this B level, but how much would have
been sustainable?
•Using the catchability estimate q we can translate the drop in CPUE and estimate
that the biomass dropped from 9000 to 8300 kg (a drop of 700), and reason that if a
catch of 1050 caused a drop of 700, then a catch of 350 would have been
sustainable at that level of biomass.
•We can then repeat this for every year, including the years where the fishery is
closed
•CPUE can still be estimated from catch and release fishing if the fishery is closed.
•In this way each year’s catch combined with the change that takes place in CPUE
the following year can allow you to estimate the sustainable catch for that year
3000
Actual catches in kg/yr
Estimates of
sustainable catch
2500
By comparing the catch
for each year to the
change in biomass
(estimated from change in
CPUE) we can estimate
the catch that would have
been sustainable each
year
2000
Catch
Kg/yr 1500
1000
500
0
0
5
10
15
20
25
For the first 10 yr catches
were unsustainably high
and the population
crashed,
1.2
1
0.8
CPUE
0.6
Kg/hr
After the new limits were
put in they tended to
oscillate around the
estimated catch.
0.4
0.2
0
0
5
10
15
20
25
Fit to the logistic curve from the lake time series
dB/dt
1200
MSY
1000
(r0K)/4=990
800
600
400
200
K/2 = 5000
0
-200
0
2000
4000
6000
Biomass
If (r0K)/4 = 990 kg/yr and K=10000 kg,
then r0=0.39 kg*kg-1yr-1
8000
10000
12000
Suppose we have a lake with a population of 10,000 kg of pike where fishing
has not been allowed for at least 20 years.
Fishing is then opened up for several years and the population is knocked back
to 5,000 kg. Assume that growth follows the logistic curve.
(a) What would the biomass be after 10 yr if r0 were 0.2 kg/kg/yr
(b) If after closure it recovers to 9,800 kg in 10 years. Find r0?
K
1  e  r 0 ( t i )
since the population was not pushed back
past the inflection point ( K/ 2), we don' t
Bt 
need to worry about i and let it  0.
10000

 8,807
 0.2 (10)
1 e
K
1  e  r 0 ( t i )
Bt
1

x
 r 0 ( t i )
K 1 e
1  x 
 ln 
x 

r0 
t
1  0.98 
 ln 
0.98 

r0 
 0.4kg / kg / yr
10
Bt 
(c) What should be the maximum sustainable yield of this fishery?
(d) If the population were reduced to 2,000 kg, would a catch of 500 kg/yr be
sustainable?, 1000kg?
r0K
MSY 
4
0.4kg  kg 1 yr 1  10,000kg

4
 1,000kg / yr
dB
 B
 r 0 B 1  
dt
 K
 2,000kg 
 0.4kg  kg 1 yr 1  2,000kg 1 

10
,
000
kg


 800  0.8
 640kg / yr
Constant Quota fishing at levels approaching the MSY shortens the biomass
range the population will recover, and the likelihood of entering the danger zone
increases. Once the danger zone is entered fishing must stop or be severely
curtailed
dB
dt
Danger zone
Stable biomass range
r0K
4
Catch rate
K/2
r0K
4
K
Maximum Sustainable Yield (MSY)
B
The red zone is much bigger if the curve is strongly skewed right
Solutions that are being considered by fish managers
What if we consider a fishery based on constant effort rather than constant quota
The catch isn’t as great but the likelihood of entering the red zone is lower
dB
dt
Catch = qB
Slope=q
r0K
4
Stable range
K/2
C
K
B