17. EXERCISES

The exercises are numbered according to the numbers of the relevant sections of the manual.

Exercise 2.1 Mean value and variance

In this exercise we use part of the length-frequency data of the coral trout (Plectropomus leopardus) presented in Fig. 3.4.0.2, namely those in the length interval 23-29 cm. These fish are assumed to belong to one cohort. The length-frequencies are presented in Fig. 17.2.1.

Tasks:

Read the frequencies, F(j) from Fig. 17.2.1 and complete the worksheet. Calculate mean, variance and standard deviation.

Worksheet 2.1

j	L(j) - L(j) + dL	F(j)	(j)	F(j) * (j)	(j) -	F(j) * ((j) - )2
1	-				-2.968
2	-				-2.468
3	-				-1.968
4	-				-1.468
5	-				-0.968
6	-				-0.468
7	-				0.032
8	26.5-27.0	6	26.75	160.50	0.532	1.698
9	27.0-27.5	2		54.50	1.032	2.130
10	27.5-28.0	2		55. 50	1.532	4.694
11	28.5-29.0	2		56.50	2.032	8.258
12		1		28.75	2.532	6.411
sums		S F(j) 31
=		s2 =		s =

Fig. 17.2.1 Length-frequency sample

Exercise 2.2 The normal distribution

This exercise consists of fitting a normal distribution to the length-frequency sample of Exercise 2.1, by using the expression:

(Eq. 2.2.1)

for a sufficient number of x-values allowing you to draw the bell-shaped curve.

For your convenience introduce the auxiliary symbols:

so that the formula above can be written

Since A and B do not depend on L and as they are going to be used many times, it is convenient to calculate them separately before-hand.

Tasks:

1) Calculate A and B

B = -1/(2s²) =

2) Calculate Fc(x) for the following values of x:

Worksheet 2.2

x	Fc(x)	x	Fc(x)
22.0		26.0
22.5		26.5
23.0		27.0
23.5		27.5
24.0		28.0
24.5		28.5
25.0		29.0
25.5		29.5

3) Draw the bell-shaped curve on Fig. 17.2.1

Exercise 2.3 Confidence limits

Tasks:

Calculate the 95% confidence interval for the mean value estimated in Exercise 2.1.

Exercise 2.4 Ordinary linear regression analysis

It is often observed that the more boats participate in a fishery the lower the catch per boat will be. This is not surprising when one considers the fish stock as a limited resource which all boats have to share. In Chapter 9 we shall deal with the fisheries theory behind this model.

The data shown below in the worksheet are from the Pakistan shrimp fishery (Van Zalinge and Sparre, 1986).

Tasks:

1) Draw the scatter diagram.
2) Calculate intercept and slope (use the worksheet).
3) Draw the regression line in the scatter diagram.
4) Calculate the 95% confidence limits of a and b.

Worksheet 2.4

		number of boats		catch per boat per year
year	i	x(i)	x(i)2	y(i)	y(i)2	x(i) * y(i)
1971	1	456		43.5		19836.0
1972	2	536		44.6		23905.6
1973	3	554		38.4		21273.6
1974	4	675		23.8		16065.0
1975	5	702		25.2		17690.4
1976	6	730	532900	30.5	930.25
1977	7	750	562500	27.4	750.76
1978	8	918	842724	21.1	445.21
1979	9	928	861184	26.1	681.21
1980	10	897	804609	28.9	835.21
Total		7146		309.5		211099.5
=		=
=
=
					sx =
					sy =
slope:				intercept: =
variance of b:
					sb =
variance of a:
					sa =
Student's distribution: tn-2 = confidence limits of b and a: b - sb * tn-2, b + sb * tn-2 = [________________,________________] a - sa * tn-2, a + sa * tn-2 = [________________,________________]

Exercise 2.5 The correlation coefficient

Refer to Exercise 2.4. Does the correlation coefficient make sense in the example of catch per boat regressed on number of boats? Consider which of the variables is the natural candidate as independent variable. Can we (in principle) decide in advance on the values of one of them?

Tasks:

Irrespective of your findings in the first part of the exercise carry out the calculation of the 95% confidence limits of r.

Exercise 2.6 Linear transformations of normal distributions, used as a tool to separate two overlapping normal distributions (the Bhattacharya method)

Fig. 17.2.6A shows a frequency distribution which is the result of two overlapping normal distributions "a" and "b". We assume that the length-frequencies presented in Fig. 17.2.6B are also a combination of two normal distributions. The aim of the exercise is to separate these two components. The total sample size is 398. Assume that each component has 50% of the total or 199. Further assume that the frequencies at the left somewhat below the top are fully representative for component "a", while those at the bottom of the right side are fully representative for component "b".

Fig. 17.2.6A Combined distribution of two overlapping normal distributions

Fig. 17.2.6B Length-frequency sample (assumed to consist of two normal distributions

Tasks:

1) Complete Worksheet 2.6a.

2) Plot D ln F(z) = y' against x + dL/2 = z and decide which points lie on straight lines with negative slopes (see Fig. 2.6.5).

3) On the basis of the plot select the points to be used for the linear regressions. (Avoid the area of overlap and points based on very few observations). Do the two linear regressions and determine a and b.

4) Calculate , s² = -1/b and s = Ö s² for each component.

5) Draw the two plots which represent each distribution in linear form.

6) We now want to convert the straight lines into the corresponding theoretical (calculated) normal distributions. Using Eq. 2.2.1 calculate Fc(x) for both normal distributions for a sufficient number of x-values to allow you to draw the two bell-shaped curves superimposed on Fig. 17.2.6B. Assume n = 199 for both components. (Use the same method as presented in Exercise 2.2). Complete Worksheet 2.6b.

Worksheet 2.6a

interval	x	F(x)	ln F(x)	D ln F(z)	z = x + dL/2
4-5	4.5	2	0.693
				0.916	5
5-6	5.5	5	1.609
				0.875	6
6-7	6.5	12
					7
7-8	7.5	24
8-9	8.5	35
9-10	9.5	42
10-11	10. 5	42
11-12	11.5	46
12-13	12.5	56
13-14	13.5	58
14-15	14.5	45
15-16	15.5	22	3.091
				-1.145	16
16-17	16.5	7	1.946
				-1.253	17
17-18	17.5	2	0.693

Worksheet 2.6b

First component			Second component
B =			B =
x	Fc(x) first	Fc(x) second	x	Fc(x) first	Fc(x) second
1.5			11.5
2.5			12.5
3.5			13.5
4.5			14.5
5.5			15.5
6.5			16.5
7.5			17.5
8.5			18.5
9.5			19.5
10.5			20.5

Exercise 3.1 The von Bertalanffy growth equation

The growth parameters of the Malabar blood snapper (Lutjanus malabaricus) in the Arafura Sea were reported by Edwards (1985) as:

K = 0.168 per year
L_¥ = 70.7 cm (standard length)
t₀ = 0.418 years

Edwards also estimated the standard length/weight relationship for Lutjanus malabaricus:

w = 0.041 * L^2.842 (weight in g and standard length in cm)

as well as the relationship between standard length (S.L.) and total length (T.L.):

T.L. = 0.21 + 1.18 * S.L.

Tasks:

Complete the worksheet and draw the following three curves:

1) Standard length as a function of age
2) Total length as a function of age
3) Weight as a function of age

Worksheet 3.1

age	standard length	total length	body weight	age	standard length	total length	body weight
years	cm	cm	g	years	cm	cm	g
0.5				8
1.0				9
1.5				10
2				12
3				14
4				16
5				(do not use ages above 16 in the graph)
6
7				20
				50

Exercise 3.1.2 The weight-based von Bertalanffy growth equation

Pauly (1980) determined the following parameters for the pony fish or slipmouth (Leiognathus splendens) from western Indonesia:

L_¥ = 14 cm
q = 0.02332
K = 1.0 per year
t₀ = -0.2 year

Tasks:

Complete the worksheet and draw the length and the weight-converted von Bertalanffy growth curves.

Worksheet 3.1.2

age t	length L(t)	weight w(t)	age t	length L(t)	weight w(t)
0			0.9
0.1			1.0
0.2			1.2
0.3			1.4
0.4			1.6
0.5			1.8
0.6			2.0
0.7			2.5
0.8			3.0

Exercise 3.2.1 Data from age readings and length compositions (age/length key)

Consider Table 3.2.1.1 (age/length key) and suppose we caught a total of 2400 fish of the species in question during the cruise from which this age/length key was obtained and that only 439 specimens of Table 3.2.1.1 were aged. The remaining fish were all measured for length. To reduce the computational work of the exercise only a part (386 fish) of this length-frequency sample is used. This part is shown in the worksheet.

Tasks:

Estimate how many of these 386 fish belonged to each of the four cohorts listed in Table 3.2.1.1, by completing the worksheet.

Worksheet 3.2.1

cohort	1982 S	1981 A	1981 S	1980 A		1982 S	1981 A	1981 S	1980 A
length interval	key				number in length sample	numbers per cohort
35-36	0.800	0.200	0	0	53	42.4	10.6	0	0
36-37	0.636	0.273	0.091	0	61	38.8	16.7	5.6	0
37-38					49
38-39					52
39-40					70
40-41					52
41-42	0.222	0.444	0.222	0.111	49	10.9	21.8	10.0	5.4
				total	386	187.2.	133.8

Exercise 3.3.1 The Gulland and Holt plot

Randall (1962) tagged, released and recaptured ocean surgeon fish (Acanthurus bahianus) near the Virgin Islands. Data of 11 of the recaptured fish are shown in the worksheet, in the form of their length at release (column B) and at recapture (column C) and the length of the time between release and recapture (column D).

Tasks:

1) Estimate K and L for the ocean surgeon fish using the Gulland and Holt plot.
2) Calculate the 95% confidence limits of the estimate of K.

Worksheet 3.3.1

A	B	C	D	E	F
fish no.	L(t)	L(t + D t)	D t
	cm	cm	days	cm/year	cm
				(y)	(x)
1	9.7	10.2	53
2	10.5	10.9	33
3	10.9	11.8	108
4	11.1	12.0	102
5	12.4	15.5	272
6	12.8	13.6	48
7	14.0	14.3	53
8	16.1	16.4	73
9	16.3	16.5	63
10	17.0	17.2	106
11	17.7	18.0	111
a (intercept) =				b (slope) =
K =				L¥ =
sb =				tn-2 =
confidence interval of K =

Exercise 3.3.2 The Ford-Walford plot and Chapman's method

Postel (1955) reports the following length/age relationship for Atlantic yellowfin tuna (Thunnus albacares) off Senegal:

age (years)	fork length (cm)
1	35
2	55
3	75
4	90
5	105
6	115

Tasks:

Estimate K and L_¥ using the Ford-Walford plot and Chapman's method.

Worksheet 3.3.2

Plot	FORD-WALFORD		CHAPMAN
t	L(t) (x)	L(t + D t) (y)	L(t) (x)	L(t + D t) - L(t) (y)
1
2
3
4
5
a (intercept)
b (slope)
tn-2
confidence limits of b
K
L¥

Exercise 3.3.3 The von Bertalanffy plot

Cassie (1954) presented the length-frequency sample of 256 seabreams (Chrysophrys auratus) shown in the figure. He resolved this sample into normally distributed components (similar to Fig. 3.2.2.2) using the Cassie method (cf. Section 3.4.3) and found the following mean lengths for four age groups (cf. Fig. 17.3.3.3):

A	B	C	D
age group	mean length (inches)	D L/D t
0	3.22
		2.11	4.28
1	5.33
		2.29	6.48
2	7.62
		2.12	8.68
3	9.74

Note: a Gulland and Holt plot gives (cf. Columns C and D): K = -0.002 and L_¥ = -950 inches, which makes no sense whatsoever.

Tasks:

1) Estimate K from the von Bertalanffy plot.
2) Why does it not make sense to ask you to estimate t₀?

Fig. 17.3.3.3 Length-frequency distribution of 256 sea breams. Arrows indicate mean lengths of age groups as determined by Cassie (1954)

Exercise 3.4.1 Bhattacharya's method

Weber and Jothy (1977) presented the length-frequency sample of 1069 threadfin breams (Nemipterus nematophorus) shown in Fig. 17.3.4.1A. These fish were caught during a survey from 29 March to 1 May 1972, in the South China Sea bordering Sarawak. The lengths measured are total lengths from the snout to the tip of the lower lobe of the caudal fin.

Figs. 17.3.4.1B and 17.3.4.1C show the Bhattacharya plots for the data in Fig. 17.3.4.1A, where B is based on the original data in 5 mm length intervals and C on the same data regrouped in 1 cm intervals. You should proceed with Fig. C for two reasons: 1) because it appears easier to see a structure in Fig. C than in Fig. B and 2) because the number of calculations is much lower.

Tasks:

1) Resolve the length-frequency sample (1 cm groups, Fig. C) into normally distributed components and estimate thereby mean length and standard deviations for each component. Use the four worksheets and plot the regression lines.

2) Estimate L_¥ and K using a Gulland and Holt plot. Draw the plot.

3) Do you think the analysis could have been improved by using Fig. B (5 mm length groups) instead of Fig. C (1 cm groups)?

Fig. 17.3.4.1A Length-frequency sample of threadfin breams. Data source: Weber and Jothy, 1977

Fig. 17.3.4.1B Bhattacharya plot for data in Fig. 17.3.4.1A based on original data, length interval 5 mm

Fig. 17.3.4.1C Bhattacharya plot for data in Fig. 17.3.4.1A based on date regrouped in length intervals of 1 cm (used in the exercise)

Worksheet 3.4.1a

A	B	C	D	E	F	G	H	I
length interval (cm)	N1+	ln N1+	D ln N1+ (y)	L (x)	D ln N1	ln N1	N1	N2+
5.75-6.75	1	0	-	-	-	-	1	0
6.75-7.75	26	3.258	(3.258)	6.75	1.262	-	26	0
7.75-8.75	42#	3.738#	0.480	7.75	0.354	3.738#	42#	0
8.75-9.75	19	2.944	-0.793	8.75	-0.554	3.183	19	0
9.75-10.75	5			9.75
10.75-11.75	15			10.75
11.75-12.75	41			11.75
12.75-13.75	125			12.75
13.75-14.75	135			13.75
14.75-15.75	102			14.75
15.75-16.75	131			15.75
16.75-17.75	106			16.75
17.75-18.75	86			17.75
18.75-19.75	59			18.75
19.75-20.75	43			19.75
20.75-21.75	45			20.75
21.75-22.75	56			21.75
22.75-23.75	20			22.75
23.75-24.75	8			23.75
24.75-25.75	3			24.75
25.75-26.75	1			25.75
Total	1069
a (intercept) =					b (slope) =

Worksheet 3.4.1b

A	B	C	D	E	F	G	H	I
interval	N2+	ln N2+	D ln N2+	L	D ln N2	ln N2	N2	N3+
5.75-6.75
6.75-7.75				6.75
7.75-8.75				7. 75
8.75-9.75				8.75
9.75-10.75				9.75
10.75-11.75				10.75
11.75-12.75				11.75
12.75-13.75				12.75
13.75-14.75				13.75
14.75-15.75				14.75
15.75-16.75				15.75
16.75-17.75				16.75
17.75-18.75				17.75
18.75-19.75				18.75
19.75-20.75				19.75
20.75-21.75				20.75
21.75-22.75				21.75
22.75-23.75				22.75
23.75-24.75				23.75
24.75-25.75				24.75
25.75-26.75				25.75
Total
a (intercept) =				b (slope) =

Worksheet 3.4.1c

A	B	C	D	E	P	G	H	I
interval	N3+	ln N3+	D ln N3+	L	D ln N3	ln N3	N3	N4+
5.75-6.75
6.75-7.75				6.75
7.75-8.75				7.75
8.75-9.75				8.75
9.75-10.75				9.75
10.75-11.75				10.75
11.75-12.75				11.75
12.75-13.75				12.75
13.75-14.75				13.75
14.75-15.75				14.75
15.75-16.75				15.75
16.75-17.75				16.75
17.75-18.75				17.75
18.75-19.75				18.75
19.75-20.75				19.75
20.75-21.75				20.75
21.75-22.75				21.75
22.75-23.75				22.75
23.75-24.75				23.75
24.75-25.75				24.75
25.75-26.75				25.75
Total
a (intercept) =					b (slope) =

Worksheet 3.4.1d

A	B	C	D	E	F	G	H	I
interval	N4+	ln N4+	D ln N4+	L	D ln N4	ln N4	N4	N5+
5.75-6.75				-
6.75-7.75				6.75
7.75-8.75				7.75
8.75-9.75				8.75
9.75-10.75				9.75
10.75-11.75				10.75
11.75-12.75				11.75
12.75-13.75				12.75
13.75-14.75				13.75
14.75-15.75				14.75
15.75-16.75				15.75
16.75-17.75				16.75
17.75-18.75				17.75
18.75-19.75				18.75
19.75-20.75				19.75
20.75-21.75				20.75
21.75-22.75				21.75
22.75-23.75				22.75
23.75-24.75				23.75
24.75-25.75				24.75
25.75-26.75				25.75
Total
a (intercept) =					b (slope) =
					=

Exercise 3.4.2 Modal progression analysis

Fig. 17.3.4.2A shows a time series over twelve months of ponyfish (Leiognathus splendens) from Manila Bay, Philippines, 1957-58. (Data from Tiews and Caces-Borja, 1965; figure redrawn from Ingles and Pauly, 1984). The numbers at the right hand side of the bar diagram indicate the sample sizes, while the height of the bars represents the percentages of the total number per length group.

Fig. 17.3.4.2B shows a time series of six samples of mackerel, (Rastrelliger kanagurta) from Palawan, Philippines, 1965. (Data from Research Division, BFAR, Manila; figure redrawn from Ingles and Pauly, 1984).

Tasks:

1) Fit by eye growth curves to these two time series, trying to follow the modal progression (as was done in Fig. 3.4.2.6). Start by fitting a straight line and then add some curvature to it, but do not be too particular about it. (Actually one should have carried out a Bhattacharya or similar analysis for each sample, but because of the amount of work involved in that approach, we take the easier, but less dependable, eye-fit. This exercise aims at illustrating only the principles of modal progression analysis - not the exact procedure).

2) Read from the eye-fitted growth curves pairs of (t, L) = (time of sampling, length), and use the Gulland and Holt plot to estimate K and L_¥ . Assume that the samples were taken on the first day of the month. Read for Leiognathus splendens only the length for the samples indicated by "*" in Fig. A, as the figure is too small for a precise reading of each month. Use the worksheet.

3) Use the von Bertalanffy plot to estimate t₀.

Worksheet 3.4.2

A. Leiognathus splendens:

		GULLAND AND HOLT PLOT		VON BERTALANFFY PLOT
time of sampling	L(t)	D L/D t		t	- ln (1 - L/L¥ )
1 June
1 Sep.
1 Dec.
1 March
a (intercept)
(slope, -K or K)		L¥ = - a/b =		t0 = - a/b =
L(t) = ___________ [1 - exp (- _______ (t - _________ ))]

Fig. 17.3.4.2A Time series of length-frequencies of ponyfish. Data source: Tiews and Caces-Borja, 1965

B. Rastrelliger kanagurta:

		GULLAND AND HOLT PLOT		VON BERTALANFFY PLOT
time of sampling	L(t)	D L/D t		t	- ln (1 - L/L¥ )
1 Feb
1 March
1 May
1 June
1 July
1 August
a (intercept)
(slope, -K or K)		L¥ = - a/b =		t0 = - a/b =
L(t) = ___________ [1 - exp (- _______ (t - _________ ))]

Fig. 17.3.4.2B Time series of length-frequencies of Indian mackerel. Data source: BFAR, Manila

Exercise 3.5.1 ELEFAN I

This exercise aims at explaining the details of the length-frequency restructuring process. Fig. 17.3.5.1A shows a (hypothetical) length-frequency sample, where the line shows the moving average. The worksheet table shows the calculation procedure and some results. Further explanations are given below for each step of the procedure.

Tasks:

1) Fill in the missing figures in the worksheet table.
2) Draw the bar diagram of the restructured data on the worksheet figure (B).

Worksheet 3.5.1

RESTRUCTURING OF LENGTH FREQUENCY SAMPLE
		STEP 1	STEP 2	STEP 3	STEP 4a	STEP 4b	STEP 5	STEP 6
mid-length L	orig. freq. FRQ (L)	MA (L)	FRQ/MA		zeroes	de-emphasized	points	highest positive points
5	4	4.6 a)	0.870	- 0.197 h)	2	-0.197	-0.109 p)
10	13	4.6			2	0.966 k)		0.966 s)
15	6	4.8 b)	1.250 e)		1	0.123 l)	0.123
20	0	4.0	0	-1.000	1		0
25	1		0.714	-0.341 i)	3	-0.341	-0.188
30	0	0.4	0	-1.000	2
35	0	1.0 c)	0 f)		1	-1.000
40	1		1.000	-0.077	2	-0.077
45	3			1.770 j)	2	1.062 m)	1.062
50	1				1		-0.127 q)
55	0		0	-1.000	1	-1.000	0 r)
60	1	0.4 d)			3	0.523 n)
S =				SP =
(S /12) = M = 1.083 g)				SN =			ASP =
				- SP/SN = R = 0.552 o)

Fig. 17.3.5.1A Hypothetical length-frequency sample. Line indicates moving average over 5 neighbours

Step 1: Calculate the moving average, MA(L) over 5 neighbours.

Examples: (see Fig. 17.3.5.1 A and worksheet table)

MA (5) = (0 + 0 + 4 + 13 + 6)/5 = 4.6 a)

(two zeroes added at start of the sample)

MA (15) = (4 + 13 + 6 + 0 + 1)/5 = 4.8 b)
MA (35) = (1 + 0 + 0 + 1 + 3)/5 = 1.0 c)
MA (60) = (1 + 0 + 1 + 0 + 0)/5 = 0.4 d)

Step 2: Divide the original frequencies, FRQ(L), by the moving average (MA) and calculate their mean value, M:

Examples:

6/4.8 = 1.25 e)
0/1 = 0 f)

(12 = number of length intervals)

Step 3: Divide FRQ/MA by M and subtract 1

Examples:

0.870/1.083 - 1 = -0.197 h)
0.714/1.083 - 1 = -0.341 i)
3.000/1.083 - 1 = 1.770 j)

Step 4a: Count numbers of "zero neighbours" among the four neighbours (two zeroes added to each end of the sample).

Step 4b: De-emphasize positive isolated values: For each "zero-neighbour" the isolated point is reduced by 20%:

and if there are "zero-neighbours" then multiply this value by [1 - 0.2 * (no. of zeroes)]

Examples:

1.610 * (1 - 0.2 * 2) = 0.966 k)
0.154 * (1 - 0.2 * 1) = 0.123 l)
1.770 * (1 - 0.2 * 2) = 1.062 m)
1.308 * (1 - 0.2 * 3) = 0.523 n)

Note: In the most recent version (Gayanilo, Soriano and Pauly, 1988) the de-emphasizing has been made more pronounced by using the factor:

Step 4c: Calculate sum, SP, of positive (restructured) FRQs and calculate sum, SN, of negative (restructured) FRQs and calculate the ratio R = - SP/SN

Example:

SP = 0.966 + 0.123 + 1.062 + 0.523 = 2.674
SN = -0.197 - 1 - 0.340 - 1 - 1 - 0.076 - 0.230 - 1 = -4.845
R = - SP/SN = 2.674/4.845 = 0.552 o)

then multiply this value by R. Values > 0 are not changed.

Examples:

-0.197 * 0.552 = -0.109 p)
-0.231 * 0.552 = -0.123 q)
FRQ (55) = 0 r)
Plot the points in the diagram (Fig. 17.3.5.1B).

Step 6: Calculate ASP (available sum of peaks). Identify the highest point in each sequence of intervals with positive points (a "sequence" may consist of a single interval)

Examples:

0.966 is the highest point in the positive sequence 10-15 cm s)
1.062 is the highest point in the positive sequence 45-45 cm
0.523 is the highest point in the positive sequence 60-60 cm

ASP = 0.966 + 1.062 + 0.523 = 2.551

Fig. 17.3.5.1B Diagram for plotting points obtained after Step 5 (see text)

Exercise 3.5.1a ELEFAN I, continued

This exercise aims at illustrating the importance of the choice of the size of the length interval (cf. Exercise 3.4.1).

Fig. 17.3.5.1C1 shows a length-frequency sample (from Macdonald and Pitcher, 1979) of 523 pike from Heming Lake, Canada, grouped in 2 cm length intervals. There are five cohorts, determined on the basis of age reading of scales with the mean lengths shown in the following table:

age years	mean length cm	standard deviation cm
1	23.3	2.44
2	33.1	3.00
3	41.3	4.27
4	51.2	5.08
5	61.3	7.07

These data put us in a position to test ELEFAN I.

Fig. 17.3.5.1C2 shows the normally distributed components derived from scale readings, and Fig. C3 shows the restructured data.

Except for the largest fish ELEFAN I manages to place the ASPs (indicated by arrows) close to where the "true" mean lengths of the cohorts are, but like all other methods ELEFAN I has difficulties in handling the largest (oldest) fish.

Tasks:

Repeat the restructuring using Worksheet 3.5.1a on the basis of 4 cm intervals (see worksheet figure) instead of 2 cm intervals. Compare the results with those presented in Figs. 17.3.5.1C1 and C2.

Fig. 17.3.5.1C Length-frequency sample of 523 pike (C1), cohorts as derived from age readings (C2) and restructured data of ELEFAN I (C3) for length intervals of 2 cm. Data source: Macdonald and Pitcher, 1979

Fig. 17.3.5.1D Regrouped length-frequency data, 4 cm length intervals (see Fig. 17.3.5.1C)

Worksheet 3.5.1a

RESTRUCTURING OF LENGTH FREQUENCY SAMPLE
		STEP 1	STEP 2	STEP 3	STEP 4a	STEP 4b	STEP 5	STEP 6
mid-length L	orig. freq. FRQ(L)	MA(L)	FRQ/MA		zeroes	de-emphasized	points	highest positive points
20	14
24	32
28	45
32	109
36	115
40	78
44	45
48	29
52	23
56	11
60	12
64	5
68	2
72	1
76	2
S =				SP =
(S /15) = M =				SN =			ASP =
				-SP/SN = R =

Fig. 17.3.5.1E Diagram for plotting points obtained after Step 5 using data from Fig. 17.3.5.1D

Exercise 4.2 The dynamics of a cohort (exponential decay model with variable Z)

Consider a cohort of a demersal fish species recruiting at an age t, which is arbitrarily put to zero. Recruitment is N (0) = 10000.

Tasks:

1) Calculate, using the worksheet, for the first ten half year periods the number of survivors at the beginning of each period and the numbers caught when mortality rates are as shown below:

age group (years)	natural mortality	fishing mortality	Comments
t1 - t2	M	F
0.0-0.5	2.0	0.0	Cohort still on the nursery ground and exposed to heavy predation due to small size
0.5-1.0	1.5	0.0
1.0-1.5	0.5	0.2	Cohort under migration to fishing ground. Some fish escape through meshes
1.5-2.0	0.3	0.4
2.0-2.5	0.3	0.6	Cohort under full exploitation
2.5-3.0	0.3	0.6
3.0-3.5	0.3	0.6
3.5-4.0	0.3	0.6	Predation pressure reduced
4.0-4.5	0.3	0.6
4.5-5.0	0.3	0.6
Recruitment: N (0) = 10000

2) Give a graphical presentation of the results.

Worksheet 4.2

t1 - t2	M	F	Z	e-0.5Z	N(t1)	N(t2)	N(t1) - N(t2)	F/Z	C(t1, t2)
0.0-0.5	2.0	0.0
0.5-1.0	1.5	0.0
1.0-1.5	0.5	0.2
1.5-2.0	0.3	0.4
2.0-2.5	0.3	0.6
2.5-3.0	0.3	0.6
3.0-3.5	0.3	0.6
3.5-4.0	0.3	0.6
4.0-4.5	0.3	0.6
4.5-5.0	0.3	0.6

Exercise 4.2a The dynamics of a cohort (the formula for average number of survivors, Eq. 4.2.9)

Tasks:

Calculate the average number of survivors during the last 3 years for the cohort dealt with in Exercise 4.2 using the exact expression (Eq. 4.2.9) and the approximation demonstrated in Fig. 4.2.3, i.e. calculate N(2.0, 5.0).

Exercise 4.3 Estimation of Z from CPUE data

Assume that in Table 3.2.1.2 the numbers observed are the numbers caught of each cohort per hour trawling on 15 October 1983.

Tasks:

Estimate the total mortality for the stock under the assumption of constant recruitment, using Eq. 4.3.0.3:

Worksheet 4.3

		cohort	1982 A	1982 S	1981 A	1981 S	1980 A 1)
		age t2	1.14	1.64	2.14	2.64	3.14
		CPUE	111	67	40	24	15
cohort	age t1	CPUE
1983 S	0.64	182
1982 A	1.14	111	------
1982 S	1.64	67	------	------
1981 A	2.14	40	------	------	------
1981 S	2.64	24	------	------	------	------
1) A = autumn, S = spring

Exercise 4.4.3 The linearized catch curve based on age composition data

Use the data presented in Table 4.4.3.1 of North Sea whiting (1974-1980).

Tasks:

Estimate Z from the catches of the 1974-cohort after plotting the catch curve. Calculate the confidence limits of the estimate of Z.

Worksheet 4.4.3

age (years) t	year y	C(y, t, t+1)	ln C(y, t, t+1)	remarks
(x)			(y)
0
1
2
3
4
5
6
7	1981	-	-
slope: b =			sb2 = [(sy/sx)2 - b2]/(n-2) =
sb =			sb * tn-2 = ________________ z = _______ ± _______

Exercise 4.4.5 The linearized catch curve based on length composition data

Length-frequency data from Ziegler (1979) for the threadfin bream (Nemipterus japonicus) from Manila Bay are given in the worksheet below, L_¥ = 29.2 cm, K = 0.607 per year.

Tasks:

1) Carry out the length-converted catch curve analysis, using the worksheet.
2) Draw the catch curve.
3) Calculate the confidence limits for each estimate of Z.

Worksheet 4.4.5

L1 - L2	C (L1, L2)	t(L1)	D t			z (slope)	remarks
		a)	b)	c)	(y)
7-8	11						not used, not under full exploitation
8-9	69
9-10	187
10-11	133						?
11-12	114						?
12-13	261						?
13-14	386						?
14-15	445						?
15-16	535						?
16-17	407						?
17-18	428						?
18-19	338						?
19-20	184						?
20-21	73						?
21-22	37						?
22-23	21						?
23-24	19						?
24-25	8						?
25-26	7						too close to L¥
26-27	2