FORESTRY
DEPARTMENTAL SERIES NO. 5
SEPTEMBER 1972
BASIC
StIRUC12,
FOR TIIE
ADVANCE \IENT CI
AOFDSCIENCE
+1V AND A R T S
3-P
Sampling
Agricultural Experiment Station
AUBURN UNIVERSITY
R. Dennis Rouse, Director Auburn, Alabama

Basic 3-P Sampling
EVERT W. JOHNSON*
INTRODUCTION
SAMPLING with probability proportional to prediction,
or 3-P Sampling, often is viewed as an esoteric procedure
that has no application to the real world of forest inven-
tory. This is unfortunate because the concept is far less
difficult to master than is imagined and, more important,
it can lead to very large reductions in inventory costs,
particularly when used in conjunction with point-sampling
and existing computer programs (4,5,7,11). This publica-
tion was written in the hope of breaking down the barrier
of misunderstanding surrounding the procedure and, per-
haps, encouraging its general adoption by persons engaged
in forest inventory operations.
In practice, 3-P sampling has been almost always as-
sociated with the use of dendrometers for upper stem
measurements and with computer programs designed to
process the large volumes of data that are generated by
the procedure. The existing literature almost invariably
incorporates these items into the discussion of 8-P sam-
pling, perhaps tending to obscure the sampling procedure
itself. In actuality, the sampling procedure is independent
of both the dendrometer and computer and even has po-
tential uses in many non-timber inventory situations. Be-
cause of this, dendrometry and computer programs will
not be mentioned in this discussion except in the conclud-
ing section. The discussion will be as informal as feasible
and will be carried out within the context of forest inven-
tory. The procedure that forms the core of the discussion
is a primitive form of 8-P sampling, which itself probably
would have little general application. However, if a per-
son understands this simple procedure he should have lit-
tle difficulty understanding the more complex adaptations
that are coming into use.
FRAME OF REFERENCE
Assume a partial cut is to be carried out and an estimate
must be made of the volume or value of timber which has
been marked for sale. Before the cut, each tree in the
sale area must be visited, examined, judged, and then
marked or left alone. This marking procedure yields no
information on the volume marked or left. Some addi-
tional procedure is needed.
It would be possible to keep a "cut and leave" tally
where a record is made of the number of trees in each
* Professor, Department of Forestry.
category, usually by d.b.h. class. 
If the trees are large and
valuable and the stand density is 
appropriately low, this
procedure might be justified. However, 
in stands of small
trees or trees of relatively low value, particularly 
where
the stands are dense, the cost of even such a simple op-
eration might be excessive.
Whenever the cost of a complete enumeration 
becomes
excessive, the tendency is to look toward sampling for the
needed information. One solution might be to divorce
the data gathering part of the problem from the marking
and to carry out an inventory using either fixed-radius plots
or point-sampling after the marking is completed. 
Under
appropriate statistical control, such an inventory 
probably
would yield usable results and still be relatively inexpen-
sive.
Another approach to the problem might be to sample
the trees as they are marked. In this procedure, the ulti-
mate sampling unit is the tree, not a piece of land surface
which may or may not bear trees. Let us say that a cruise
intensity of 10 per cent has been deemed appropriate 
in
this case. This means that one of every 10 trees would be
selected for sampling. The sampling could be systematic,
in which every tenth tree would be measured, regardless
of size or value. This approach, of course, would yield a
valid estimate of average tree size, or value, provided
bias were not present.
1 
Bias could easily be introduced
by the marker consciously or subconsciously choosing bet-
ter-than-average (or worse-than-average) trees as the sam-
ple trees. This could have a serious effect on the accuracy
of the inventory. A further problem with systematic sam-
pling is that with it valid variability statistics cannot be
obtained and such statistics are essential for valid interval
estimates.
Use of random sampling would eliminate much of the
problem of bias. Insofar as this inventory problem is con-
cerned, the mechanics of random sampling would be sim-
ple and straightforward. The marker-cruiser could carry
1 Bias occurs when the mean of all possible estimates of a
given population's parameters (mean, total, variance, etc.),
based on samples of a given size, does not equal the param-
eter. Where the population is not known in sufficient detail so
that its component units can be listed so as to construct a
frame, and if the sampling design is such that the sample 
size
is indeterminate prior to the sampling, bias is apt to be pres-
ent. These conditions exist in the problem used as the basis
for this discussion, consequently, some bias may be present 
in
all cases (8,9). However, this bias will be ignored in the dis-
cussion since it probably would be small and, under the 
con-
ditions of the problem, unavoidable.
a pouch on his belt in which there were 10 marbles, 9
white and 1 black. After each tree was marked, the
cruiser would stir up the marbles in the pouch and make
a random draw. If the marble were white the tree would
not be measured and the man would move on. If the
marble were black, however, the tree would be measured.
In either case the marble would be returned to the pouch,
making it ready for the next draw. Such a procedure
would result in approximately 10 per cent of the marked
trees being measured. The statistics of the inventory
would be computed conventionally:
2
n
1 Xi
X = i=1
n
where: i = mean tree volume or value
Xi = volume or value of the ith tree, and
n = number of trees measured;
n
s
2 
= i=1 (Xi - )2
n-l
where: s
2 
= variance of sample;
s = 
+  
/S
2
where:s = standard deviation of sample;
C.V. = s/x
where: C.V. = coefficient of variation;
s2 n
S2X - (1- )
n N
where: s27 = variance of sample mean, and
N = total number of trees in population;
s = + ,/S2X
where: s = standard error of sample mean;
SE= !_ts
where: SE
t
= sampling error for mean, and
= Student's t value at the desired prob-
ability level and with n- 1 degrees of
freedom;
SET = N(SE)
where: SET - sampling error for total;
T = N
where: T = total volume or value of timber marked
for cutting.
same data, is shown in Table 2. As can be 
seen, this is a
more cumbersome and less efficient procedure than the
one conventionally used and would rarely, if ever, 
be used
in a real-life situation. However, the basic 
idea behind it
is fundamental to the discussion of 3-P sampling 
and,
consequently, must be understood.
TABLE 1. DATA AND COMPUTATIONS FROM A 
SIMPLE RANDOM
SAMPLE USING CONVENTIONAL PROCEDURES
Sample tree no. Measured 
tree value
Dol.
11.50
4.50
2.00
7.50
3.00
12.50
2.50
3.00
4.00
Total 50.50
x = 50.50/9 - $5.611 mean value per tree
T = 85 ($5.611)- $476.93 estimated total value of marked
trees
s2 =15.7361
s = ? $3.97 per tree
C.V. = 0.707 or 70.7%
s., ? $1.25 per tree
SE, for trees, at the 95% probability level = ? $2.88 per
tree
SE, for total value, at the 95% probability level - 85(2.88)
- $245.07
TABLE 2. DATA AND COMPUTATIONS FROM A SIMPLE RANDOM
SAMPLING USING THE MODIFIED COMPUTATION PROCEDURE
Sample 
Weight
tree no.
Measured
tree
value
Dol.
Blow-up
factor
Estimate o
total valu
Dol.
1-....-- 1 11.50 85/1=85 11.50(85)
2--...-- 1 4.50 85/1= 85 4.50 (85) =
3-...-.. 1 2.00 85/1=85 2.00 (85) =
4 -....- 1 7.50 85/1= 85 7.50 (85) =
5 -..... 1 3.00 85/1- 85 3.00(85) = -
6 1 12.50 85/185 12.50 (85) =1,(
7------...--..- 1 2.50 85/1-85 2.50(85)= 1
8 ...... 1 3.00 85/1= 85 3.00 (85) =
9 ...... 1 4.00 85/1= 85 4.00 (85) =
Total.... . 4,
T - 4292.50/9= $476.94 total value of marked trees
s 
= 
113,693.4028
s = ? $337.18 per estimate of total
C.V. = 337.18/476.94 = 0.707 or 70.7%
sx ?+ $106.27 per estimate of total
SE, for total, at 95% level of probability = ? $245.07
f
Dol.
977.50
382.50
170.00
637.00
255.00
062.50
212.50
255.00
340.00
292.50
To illustrate this, assume that on a certain tract 85 trees
have been marked for cutting. Assume further that 9 of
these trees were selected for measurement using the ran-
dom sampling procedure described above. The resulting
data are shown in Table 1 along with the computation of
the estimate of the total value of the marked timber
($476.93 ?+ $245.07, at the 95% level of probability).
It is possible to obtain these same results using a some-
what different computation procedure. This, 
using the
2 This is merely a brief review of elementary sampling theory.
Most foresters learned this theory when they were students,
but a few were not taught application of the theory.
[41
Note the order of events in the two procedures. In the
first, or conventional, method the mean value of the
sample trees was computed first and then this was "blown-
up" to an estimate of the total value by multiplying the
mean value by the total count of trees in the population.
In this case the blowing-up process came after the com-
putation of the mean. In the second method, the value of
each sample tree was first blown-up to an estimate of the
total value and then these estimates of the total were
averaged. The sequences of averaging and blowing-up
are reversed. The results, however, are identical, within
rounding.
Fundamental to the whole process, regardless of the
procedure used, is the blow-up factor. It is the reciprocal
of the probability of a given tree being chosen for meas-
urement in a single random draw from the population.
3
In this case, the tree has 1 chance in 85 of being chosen
if only one random draw is made, so the reciprocal of this
probability, 85, is the blow-up factor. Thus, in the con-
ventional procedure, the mean tree value is blown-up by
85 to arrive at the estimate of the total. In the second
procedure, each sample tree value is independently
blown-up by 85 to an estimate of the total, then these
estimates are averaged. In the conventional procedure,
the value of each sample tree is used as an estimate of the
mean value. Each tree may have a value that is greater
or smaller than the mean, but these estimates are exactly
correct on the average across all trees in the population.
Likewise, in the second procedure the estimates of the
total value, obtained from the sample trees, may be indi-
vidually too large or too small, but they are exactly cor-
rect on the average across all members of the population.
It should be noted that in the calculation of the mean
in both procedures, each observation or data item has a
weight of one. This fact of equal weights for the indi-
vidual observations is characteristic of simple random sam-
pling with equal probability.
This random sampling procedure is sound, unbiased,
and practical, but it is inefficient. Each tree, regardless
of size or value, has the same probability of being chosen.
Consequently, trees of low value (small or defective trees,
both with low volumes) are likely to be oversampled while
highly valued trees are undersampled.
To overcome this problem, some method must be de-
vised which would sample high-value trees more heavily
than low-value trees. This, in short, would involve variable
probability. Sampling with probability proportional to size
(Bitterlich or point-sampling) would not be applicable
since such an inventory could not be carried out until after
the marking was completed. A further consideration is
that value may or may not be closely correlated to size.
For example, a 12-inch black walnut or black cherry
would be worth a great deal more than a 12-inch post oak.
Size also does not take into consideration defect and its
relation to merchantability.
List sampling would not be appropriate since it could
not be carried out until the marking was completed and
a tentative value assigned to each tree. The trees would
have to be marked with identifying numbers so that,
if chosen for measurement from the list, they could be
recovered. Finally, the field work of locating the chosen
trees would be extremely laborious.
3-P SAMPLING
The idea of a tentative value placed on each tree can
be used as the basis of a stratification scheme that can
solve this inventory problem. A set of tree value classes
* A further example of this principle might be a situation in
which a 1,000-acre tract has been divided into 5,000 square,
1/5-acre, sample plots. Assume that one plot will be drawn
at random from the 5,000 and the volume on it determined.
Its probability of selection is 1/5,000. If one multiplies the
volume on this plot by 5,000 (which is the reciprocal of
1/5,000) the product will be an estimate of the total volume.
This estimate may be too high or too low but the average of
all possible such estimates (5,000 of these) would be correct.
or strata which are arbitrary but still representative of the
tree population involved can be developed. To each of
these classes a probability of being chosen is assigned.
4
These probabilities would be proportional to the class
values so that high-value classes would have a relatively
high probability and low-value classes would have a rela-
tively low probability of being chosen. Each tree, follow-
ing marking, would be assigned to one of these classes.
Then, right on the spot, a random draw using the proba-
bility appropriate to that class would be made-to see if
the tree was to be measured.
Such a classification must be quick and easy to carry
out and it must not require measurements, because much
of the idea is to minimize measurement operations. The
only logical procedure is to use ocular estimations of the
value. Each tree would be assigned to a class using the
marker's subjective judgment on size, quality, probable
utility, local markets, and so forth. It is recognized that
ocular estimating is subject to considerable random and
systematic error when used for volume estimation. How-
ever, when it is used to assign a tree to a value class the
only thing affected by such errors is the probability of the
tree being chosen. If a low-value tree is erroneously as-
signed to a high-value class it merely means that particular
tree has a greater chance of being sampled than it should.
This would have little or no effect on the total volume
estimates, but it would have an impact on the sampling
error. This is explained later.
The procedure described above is 3-P sampling. Let
us examine the idea within the context of an example.
Assume that a series of four tree value classes has been
established: (A) $10 per tree; (B) $8 per tree; (C) $6
per tree; and (D) $3 per tree. Assume further that the
following probabilities of being chosen have been assigned
to the classes: (A) 4 in 40; (B) 3 in 40; (C) 2 in 40;
and (D) 1 in 40. A pouch containing 40 marbles is made
up for each of the four classes. In the A class pouch, 4
marbles are black and 36 white. In the B class pouch, 3
are black and 37 white. In the C class pouch, 2 are black
and 88 white, and in the D class pouch only 1 of the 40
marbles is black. The marker-cruiser carries all four of
these pouches on his belt.
When the marker selects a tree, he classifies it ocularly
by value class. He then stirs the marbles in the appropriate
pouch and makes a random draw of one marble. If it is
white, he records that a tree has been marked and de-
notes the class to which it is assigned, then he moves to
the next tree. If the marble is black, the tree is measured
and its size or value is recorded. The class to which it was
assigned is also recorded. In either case, the marble is
returned to the proper pouch so that all is ready for the
next drawing. Notice that in this procedure the probabil-
ity of a given tree being selected for measurement changes
from class to class and a high-value tree has a greater
chance of being measured than does a low-value tree.
Thus, the important problem of preferential sampling has
been solved.
COMPUTATIONS
Because of the varying probabilities used in this ap-
proach, the conventional computational procedures shown
^ The probabilities assigned are discussed at length in a fol-
lowing section of this publication.
[5]
earlier cannot be used. If 3-P data were used in 
such a
computational procedure, the total volume 
or value would
be strongly biased upward because 
proportionally more
high-value than low-value trees 
would be sampled. How-
ever, the second procedure described earlier 
for simple
random sampling can be used provided 
one change is
made. For example, assume that the same 85 trees 
cruised
by simple random sampling procedure 
were sampled ac-
cording to the four classes described above. 
Further as-
sume that the resulting data were as 
shown in Table 3.
The computations are shown in Table 4. 
In Table 4 the
blow-up factor is not constant across all trees. 
This is a
consequence of the fact that the probabilities 
of the trees
being chosen are not constant. A tree 
in class A has a
probability of being chosen that is 4 
times as great as
does a tree in class D. In other words, 
the probabilities
are weighted and the trees in class A 
have a weight of 4.
The trees in class B have a weight 
of 3, those in class C
have a weight of 2, while those in class 
D have a weight
of 1.
TABLE 3. DATA FROM A 3-P SAMPLE
Class 
Proba- 
Wt.
Class bility units/
ity tree
No.
A ... 4:40 4
B _.. 3:40 3
C -.. 2:40 2
D ... 1:40 1
Total
Trees Trees 
Tree values
as- meas- b
signed 
ured 
obtained
No. No. Dol.
15 2 10.50, 11.50
20 2 6.00, 8.00
20 0
30 1 2.00
85 5 ......
Total
weight
4(15) 60
3(20)=60
2(20)=40
1(30)=30
190
TABLE 4. COMPUTATION OF 3-P SAMPLE
Sam-
ple
tree
no.
Meas-
Wt. ured
ltree
value
Blow-up
factor
Estimate of total value
Dol. Dol. Dol.
1__ .... 4 10.50 190/4= 47.500 10.50( 
47.500) = 498.75
2 --- 4 11.50 190/4= 47.500 11.50( 47.500) 
= 546.25
3 3 6.00 190/3 63.333 6.00( 
63.333) = 380.00
4 --------. 3 8.00 190/3- 63.333 8.00( 
63.333)-=506.67
5 . 1 2.00 190/1190.000 2.00(190.000) 
- 80.00
Total 
2,311.67
T 2311.67/5 $462.33 total value of marked 
trees
s2 = 5975.1343
s ?+ $77.30 per estimate of total
C.V. - 16.7%
s2- 1124.7593
sX ___ 4 $33.54 per estimate of total
SE, for total, at 95% level of probability =- 
$93.10
To compute probabilities in a case such 
as this, it is
necessary to sum the weights which were 
assigned and
use this sum as the denominator in the probability 
fraction.
In the case of Table 2, where the probabilities 
are constant
across all the trees, each tree has 
a weight of 1 and the
sum of the weights is 85. As a result, 
the probability of
any individual tree being chosen in 
a single random draw
is 1/85 and the blow-up factor is equal 
to 85. In the case
of the 3-P sample shown in Table 3, 15 
trees are assigned
to class A. Each of these trees has 
a weight of 4 so the
sum of weights for class A is 60. 
Correspondingly, the sum
for class B is 60; for class C, 40; and 
for class D, 80. The
grand total of these weights is 190, which 
is used to cal-
culate the probabilities. Consequently 
the probability of
[6]
a class A tree being chosen is 4/190, the probability 
of
a class B tree is 3/190, and so forth. The 
reciprocals of
these probabilities are the blow-up factors. 
When one
considers the calculation procedures 
in Tables 2 and 4 in
light of these ideas, it becomes obvious 
that, fundament-
ally, they are exactly alike. One simply 
has to use the
proper probabilities, whether they are 
constant or variable.
All the terms used until the next section 
had been com-
mon before 3-P sampling was developed. No specialized
3-P sampling terms have yet appeared. 
For ready refer-
ence, terms are defined in the Appendix.
GETTING AWAY FROM THE POUCHES
In most cases, pouches of marbles would 
not be practi-
cal since many classes would be needed. 
The only truly
practical solution is to use a set of random 
numbers. How-
ever, to set the stage for a description 
of how those ran-
dom numbers should be set up and used, 
it appears
desirable to describe a technique based 
on the use of a
deck of cards. This also will provide a vehicle 
for intro-
ducing some of the special 
terminology and symbology
associated with 3-P 
sampling.
The deck would be made up of cards bearing two 
kinds
of labeling. First, there would be 
a card representing each
of the classes which are to be recognized. 
Each of these
cards would bear an integer. The magnitudes 
of these
integers should be correlated with the volumes or values
thought to be associated with the classes. 
For example,
a deck that would yield results analagous to those 
obtained
using the four pouches would have a 
card with a 4 (for
class A), a 3 (for class B), a 
2 (for class C), and a 1 (for
class D). Note that these values 
correspond to the weights
previously described. In 3-P terminology, there would be
K cards bearing integers. In this case, K would be equal
to 4.
All the remaining cards would be alike. They 
represent
rejection of the tree and are analagous to the white mar-
bles in the pouches. Therefore, they are called "rejection
cards." These cards may be blank or they may bear
symbols such as asterisks, black balls, zeroes, 
groups of
X's, or whatever is desired. These rejection cards serve
to control the probabilities of trees being chosen. There
are Z such cards. Regardless of weight, 
the greater the
value of Z the smaller is the probability 
of any tree being
chosen and vice versa. Z, which can be any 
desired whole
number, must equal at least 1 and probably 
should be
greater than 1. More will be said about the magnitude 
of
Z later in the discussion. In the case of the deck 
for the
four pouch problem, Z would be equal to 36. 
This would
bring the deck size up to 40, the same magnitude as the
number of marbles in each pouch. The reasoning behind
this will be made clear after the use of the 
deck has been
explained.
The deck is used as follows. As before, each 
tree is
visited, examined, judged, and either 
marked or not
marked. If marked, it is then ocularly assigned 
to a class,
and the class is recorded. Remember that 
now the class
identification is in terms of the numbers mentioned 
in the
paragraph about the K cards. In standard 
3-P sampling
terminology this class label is called a 
KPI number. The
deck is now thoroughly shuffled and a 
card is drawn at
random from the deck. If the card bears 
a rejection sym-
bol, the tree is rejected and the cruiser moves 
on. If the
card bears a number (called a KI value) that is larger
than the class label (KPI value), the tree also is rejected.
If, however, the card bears a number (KI value) that is
equal to or smaller than the class label (KPI value), the
tree is measured and evaluated. The resulting volume or
value magnitude is called a YI value. Every tree in the
forest has a YI value that is unknown until the tree is
measured and evaluated.
How does probability enter into this procedure? To
demonstrate this, let us compare the 40-card deck with
the four pouches of marbles. Table 5 shows how the
probabilities are determined by the deck. Note that these
probabilities are identical with those associated with the
pouches of marbles. An examination of this table also
should reveal why the deck contains 36 rejection cards.
TABLE 5. How THE PROBABILITY OF A TREE BEING SAMPLED
VARIES ACCORDING TO CLASS (KPI VALUE)
USING THE 40 CARD DECK
KPI
m
I KPI
m
, KPI
KPI
= class label, for class A
for class B
for class C
for class D
= 4
= 3
S2
S 1
= sum of weights =
blow-up factor, for class A = 190/4
for class B = 190/3 -
for class C = 190/2 =
for class D = 190/1=
190
47.500
63.333
95.000
190.000
= number of sample trees
YI = value of the Ith tree
m
TI = KPI (YI)
KPI
estimate of the total volume or
value on tract, based on the Ith
tree above
Action taken Probability of
Measurement 
Rejection
When a tree is assigned to class (KPI)1:
SNo measurement
4 > KPI No measurement
3 > KPI No measurement
2 > KPI No measurement
1 KPI Measure 1/40
1/40
A tree in class (KPI)1 has 1
being measured.
When a tree is assigned to class (KPI)2:
No measurement
4 > KPI 
No measurement
3 > KPI No measurement
2- KPI Measure 1/40
1 < KPI Measure 1/40
2/40
36/40
1/40
1/40
1/40
39/40
chance in 40 of
36/40
1/40
1/40
38/40
A tree in class (KPI)2 has 2 chances in 40 of
being measured.
When a tree is assigned to class (KPI)3:
* No measurement
4 > KPI No measurement
When a tree is a,
4 KPI
3 < KPI
2 < KPI
1 <KPI
36/40
1/40
Measure 1/40
Measure 1/40
Measure 1/40
3/40 37/40
A tree in class (KPI)3 has 3 chances in 40 of
being measured.
ssigned to class (KPI)4:
No measurement 36/40
Measure 1/40
Measure 1/40
Measure 1/40
Measure 1/40
4/40 36/40
A tree in class (KPI)4 has 4 chances in 40 of
being measured.
From this point on the computations would be exactly
the same as those shown in Table 4. In terms of the sym-
bols used by Grosenbaugh (1,2,7), the computations
would take the following form:
K = number of classes and class cards = 4
Z = number of rejection cards = 36
m = number of trees in population = 85
n
T TI_= mean estimate of the total volume or
n value on the tract = $462.33
PRECISION AND BIAS OF 3-P SAMPLES
In actual practice it nearly always would be desirable
to use more than four classes. The actual number depends
on several factors. To illustrate these factors, let us con-
sider the 85-tree population of the four-class problem.
Let us assume that the true values of the 15 marked trees
in Class A average $12.00 per tree, the 20 trees in Class
B average $8.00 per tree, the 20 trees in Class C average
$5.00 per tree, and the 30 trees in Class D average $3.00
per tree. Under these circumstances the actual total value
of the marked trees would be $530.00. Now, instead of
four classes, let us establish enough classes (KPI val-
ues) so that there would be a class (KPI value) for
every possible tree value when those values are expressed
in cents. This would require somewhat more than 1,500
classes (KPI values). If all 85 trees were correctly as-
signed to these classes, the sum of the weights would be:
m
I KPI - 15(1,200) + 20(800) + 20 (500) + 80 (800)
= 53,000
The 15 in the first numerical term of the equation repre-
sents the 15 trees in old Class A and the 1,200 represents
the average value in cents. The remaining terms are equiv-
alent. The computations for the 5 tree sample used be-
fore would be as shown in Table 6.
Notice that in Table 6 each estimate of the total value
is exactly correct. There is no variability. The variance
and coefficient of variation are both equal to zero. Why
did this occur? First, the classes were numerous enough
for a class or KPI value to be assigned expressing a value
in terms of cents, the smallest value unit. Second, these
class (KPI) values were exactly the same as the actual
tree (YI) values, therefore were perfectly correlated with
them.
Such perfect correlation rarely exists. A more realistic
situation is to assume that the class (KPI) values were
in terms of dollars and that each tree had been correctly
classified. Now there would be approximately 15 classes.
The sum of the weights would be:
[ 7]
Card
3 KPI
2 < KPI
1 <KPI
m
Y KPI = 15(12) + 20(8) + 20(5) + 30(3)
530
The computations are shown in Table 7. Notice that when
the correlation between class (KPI) value and actual (YI)
value is not perfect, as in trees 1 and 2, the individual
estimates of the total value vary in magnitude, causing the
variance and C.V. to take on values other than zero. Fur-
thermore, as can be seen if one compares the C.V. values
in Tables 4 and 7, the closer the KPI and YI values come
to agreement the smaller the C.V. and, consequently, the
greater the precision of the cruise. As a matter of fact, as
can be seen in Table 8, the C.V. of the (YI/KPI) ratios
is exactly equal to the C.V. of the estimates of the total.
This all indicates that a designer of a 3-P sampling scheme
should make every effort to keep the (YI/KPI) ratio as
nearly constant as possible. The greater his success in
accomplishing this the better will be the precision of the
cruise.
TABLE 6. COMPUTATION OF TOTAL VALUE USING KPI
VALUES EQUIVALENT TO TREE VALUES IN CENTS
Sample 
Weight (or 
Measured
tree no. KPI value) 
)tree value
Dol.
1,050
1,150
-. 600
800
200
10.50
11.50
6.00
8.00
2.00
Blow-ups Estimates
2
factor v of totalue
Dol.
50.476 530.00
46.087 530.00
88.833 530.00
66.250 530.00
265.000 530.00
2,650.00
T = 2,650.00/5 = $530.00 total value of marked trees
s=0
C.V. = 0
st = 0
s = 0
SE = 0
1e.g., 53,000/1,050 
= 50.476, etc.
2 e.g., $10.50 (50.476) = $530.00, etc.
TABLE 7. COMPUTATION OF TOTAL VALUE USING KPI
VALUES EQUILAVENT TO TREE VALUES IN DOLLARS
Sample Weight 
(or teeaue 
Blow-up
1  
Es imats
tree no. KPI value) tree value factor 
v of totale( YI ) ctr value
11
11
6
8
2
Dol.
10.50
11.50
6.00
8.00
2.00
48.182
48.182
88.333
66.250
265.000
Dol.
505.91
554.09
530.00
530.00
530.00
2,650.00
T = 2,650.00/5-= $530.00 total volume of marked trees
s" = 290.1641
s= ? $17.03 per estimate of total
C.V. = 0.032 or 3.2%
s = 54.6205
sx = ? $7.39 per estimate of total
SE, for total, at 95 % level of probability= -?- $20.52
1 e.g., 530/11= 48.182, etc.
2 e.g., $10.50 
(48.182) = $505.91, 
etc.
"Unbiased" means that on the average, over all possible
attempts, the cruiser's mean assignment (KPI) value is equal
to the true mean KPI value.
[8]
TABLE 8. COMPUTATION OF THE COEFFICIENT OF VARIATION
OF THE (YI/KPI) RATIOS WHEN THE KPI VALUES ARE
EQUIVALENT TO TREE VALUES IN DOLLARS
Ratio
10.50/11 - 0.9545 
X = 1.0000
11.50/11 - 1.0455 s
2
= 0.001033
6.00/6 = 1.0000 
s =? 0.0321
8.00/8 -= 1.0000 
CV = 0.0321 or 3.2%
2.00/2 = 1.0000
A cruiser who is unbiased, but erratic in assigning
trees to classes (KPI values) will make an unbiased esti-
mate of the true total volume or value of the marked
trees but his estimate will be low in precision because the
(YI/KPI) ratios vary excessively.
When the cruiser is biased and consistently places trees
in classes (KPI values) that are too low or too high, the
estimate of the total may or may not be biased depending
on the nature of the bias. If the cruiser's bias can be
described as a ratio, either high or low, the resulting 3-P
estimate of the total will be unbiased. For example, in the
case of an unbiased cruiser an individual blow-up factor
would be:
m
SKPI
KPI
In the case of a cruiser biased so as to consistently class
trees a certain percentage too high or too low, an indi-
vidual blow-up factor would be:
m
I (KPI)* (Bias%)
(KPI)* (Bias%)
which may be transformed to:
m
(Bias%) (I KPI)
(Bias%) ( KPI)
in which case the bias terms cancel out and the usual
blow-up factor would be left unchanged.
If the cruiser's bias is a constant number of classes
(KPI values) too low or too high
6
, the resulting estimate
of the total will be biased. In such a case an individual
blow-up factor would be:
m
S(KPI) + (Bias)
(KPI) + (Bias)
There is no way by which the bias could be cancelled out
or otherwise removed in this situation.
The effect of those two types of bias are demonstrated
in Tables 9 and 10. The basic data in both cases are from
the 85-tree population used previously. The entire popu-
lation has been included in the calculations because only
in this way can the bias be determined. The basic class
(KPI) system used in these examples is that associated
with dollar values. In Table 9 it is assumed that the
cruiser consistently assigned class (KPI) values that were
10 per cent too high, while in Table 10 the assumption is
6 This is also true if the bias is otherwise additive instead of
multiplicative. This would occur whenever the discrepancies
between YI and KPI values do not cancel out.
1
2 ..
3
4
5
Total_
2
3
4 -..
5
Total
TABLE 9. COMPUTATION OF TOTAL VALUE WHEN CRUISER
WAS CONSISTENTLY BIASED 10% HIGH ON
CLASS ASSIGNMENTS
True Blow-Estimate
Old weight Biased' Trees 
Measured 2 f- Esta
class (or KPI weight Trees 
value fauptor vttal
No. Dol. Dol.
12
8
5
3
13.2 15
8.8 20
5.5 20
3.3 30
85
45050.09/85
12.00
8.00
5.00
3.00
44.167
66.250
106.000
176.667
7,950.06
10,600.00
10,600.00
15,900.03
45,050.09
$530.00 estimated total value of
marked trees
530.00 true total value
$ 0.00 Bias in value
s
2  
G5
s0
C.V. = 0.0
Se.g., 12 + 12(0.1) 13.2, etc.
' 1 KPI = 15(13.2) + 20(8.8) + 20(5.5) + 30(3.3) _
583
e.g., 583/13.2 = 44.167, etc.
4
e.g., 15 trees ($12.00) (44.167) =7950.06, etc.
5
Note that: $12.00( 44.167) - $530.00
$ 8.00( 66.250) - $530.00
$ 5.00(106.000) = $530.00
$ 3.00(176.667) = $530.00, there are no dif-
ferences.
TABLE 10. COMPUTATION OF TOTAL VALUE WHEN CRUISER
WAS CONSISTENTLY ONE CLASS HIGH ON CLASS ASSIGNMENTS
True
Old weight Biased'
class (or KPI weight
value)
12
8
5
3
13
9
6
4
43536.22/85 =
Trees 
Measured
value
No. Dol.
15 12.00
20 8.00
20 5.00
30 3.00
85 ..
Blow- Estimate
up
2 3  
of total
factor value
47.308
68.333
102.500
153.750
Dol.
8,515.44
10,933.28
10,250.00
13,837.50
43,536.22
$512.19 estimated total value of
marked trees
530.00 true total value
-$ 17.81 Bias in value
s' = 1759.66625
s ? $41.95
C.V. = 0.0819 or 8.19%
e.g., 12 + 1 =13, etc.
m
_ KPI= 15(13) + 20(9) + 20(6) + 30(4)
e.g., 615/13= 47.308, etc.
e.g., 15 trees ($12.00) (47.308) - 8515.44, etc.
Note that: $12.00( 47.308) - $567.70
$ 8.00( 68.333) = $546.66
$ 5.00(102.500) = $512.50
$ 3.00(153.750) = $461.25
615
that the cruiser consistently assigned class (KPI) values
that were one class too high. To save space all 85 trees
are not listed. Instead the counts and mean values as-
sociated with the four original classes are used. This
makes it necessary to use non-integer KPI values in Table
9, which should be acceptable since they are, in essence,
means.
As can be seen in Table 9, a percentage bias has no
effect on either the estimated total or the C.V. In spite
C91:
A ...
B ---
C .
D
Total
T
of the fact that the class (KPI) values are not the same
as the actual (YI) values, the (YI/KPI) ratio remains
constant since the KPI values are expanded by a constant
multiplier.
An additive bias, however, as shown in Table 10,
changes both the estimated total and the C.V. In this
case the (YI/KPI) ratio is changed because of the bias,
resulting in considerable alterations in estimates of the
total. It should be noted that the sign of the error in the
estimated total is the opposite of the sign of the original
cruiser's bias.
The evidence of Tables 9 and 10 indicates that persons
charged with assigning class (KPI) values in the course
of a 3-P inventory should be given training prior to doing
the work and that every effort should be made during that
training to break any tendency on the part of the cruiser
to consistently assign classes a set number of units too
high or too low.
Bias may occur in a 3-P sampling scheme from faulty
design as well as from cruiser error. For example, assume
that in the case of the situation shown in Table 7 the
largest class (KPI value) was set at 12. This class would
then be open-ended and any tree too large or too valu-
able to fall in class (KPI value) 11 would be assigned to
class (KPI value) 12, even if it should have been assigned
to a class with a larger KPI value. In this case the
(YI/KPI) ratio would not be reasonably constant within
class (KPI value) 12 and it would usually be higher
(thus biased) than the ratios associated with the other
classes. The effect can be seen in Table 11. It is as-
sumed that two more of the 15 trees in original class A
are included in the sample and their true (YI) values
were $20 and $25, respectively. This is not reasonable
under the original assumptions made about the population
of trees since these trees deviate so strongly from the class
mean of $12. However, this only exaggerates the bias
and does not invalidate the idea being illustrated.
The upward bias under these circumstances is clearly
evident. The effect of the open-ended class on the C.V.
TABLE 11. COMPUTATION OF TOTAL VALUE USING KPI
VALUES EQUIVALENT TO TREE VALUES IN DOLLARS
WITH AN OPEN-ENDED CLASS
Sample Weight Measured Blow-up Estimate
tree no. (or KPI tree value Blfactor total value of
value) ( YI) factor total value
Dol. Dol.
1-....... 11 10.50 48.182 505.91
2 -....... 11 11.50 48.182 554.09
3........ 6 6.00 88.333 530.00
4-....... 8 8.00 66.250 530.00
5-....... 2 2.00 265.000 530.00
6 12 20.00 44.167 883.33
7 12 25.00 44.167 1,104.17
Total 4,637.50
T- 4637.50/7 - $662.50 estimated total value of marked
trees
530.00 true total value
+$132.50 Bias
s
= 
55463.3640
C.V. 0.3555 or 35.55%
m
' KPI = 530.
2 e.g., 530/11 = 48.182, etc.
* e.g., $10.50(48.182) = $505.91, etc.
A
B ---
C -- -
D
Total
T =
is also evident. These results emphasize that 
it would be
better to have too many potential classes (KPI values),
some of which would not be used, than it would be to
have too few so that open-endedness 
occurs.
It would be possible to define a maximum class or KPI
value and state prior to sampling that all the trees falling
into this class would be measured (i.e., the sampling 
in
this class would be 100 per cent). The trees included 
in
this class would be completely excluded 
from the 3-P
sample and would be considered as a separate 
part of the
inventory problem. This procedure would 
eliminate the
problem of an open-ended class.
CONTROL OF THE PROBABILITY
OF SELECTION
In 3-P sampling the sampling intensity itself is subject
to probability since no one knows prior to the cruise how
many trees will be selected. If the same trees were
cruised several times, the number of trees selected for
measurement would vary.
Basically, the rate of sampling is dependent on two
quantities: K, the number of classes (KPI values); and
Z, the number of rejection cards in the deck. If K is held
constant and Z is increased the likelihood of drawing a
rejection card will increase. However, the magnitude of
Z does not influence the relative probability of choosing
a tree of a given class compared to the other classes 
(i.e.,
the relative weights of the classes remain constant). In
terms of the original four classes, regardless of how large
Z is, the trees in class A would be four times as likely to
be selected for measurement as the trees in class D. In
essence, Z acts as a dilutant.
The expected number of trees to be sampled (ESN)
can be computed as follows:'
m
I KPI
ESN = K
K+Z
m
In any given situation the sum of weights (I KPI)
would not be known prior to the marking. Consequently,
an estimate of the sum would have to be obtained from
a light pre-sample. For this purpose a set of fixed-radius
plots (or prism points) would be distributed over the
area and each tree in each plot that is likely to be chosen
for cutting at the time of marking would be assigned to
a class (KPI value). There would be a sum of such KPI
values from each plot. These sums would be averaged
and put on a per acre basis, as is done with tree volumes.
This per acre value would be multiplied by the number of
m
acres to arrive at the estimate of I KPI needed in the cal-
culations.
K would be set by the inventory designer so that the
7
ESN - KPI _ KPI + KP2 +...+ KPM
K+Z K+Z K+Z K+Z
KPI is a probability (e.g., when KPI - 4,
Each K+Z 
K - 4, andZ = 
36,
KPI 4 = 0.1).
K+Z 40
In the case of the 85-tree population there would be 85 such
ratios.
desired range in classes (KPI values) would be 
available,
preferably without any open-ended classes.
Z would then control ESN and determine the propor-
tion of the population that would be considered for sam-
pling. For example, using the original 4-class problem:
Z = 36
m
2 KPI = 190
ESN = 190/(4+36) = 4.75 or 5 trees
Had Z equalled 96:
ESN = 190/(4+96) -= 1.90 or 2 trees
Had Z equalled 6:
ESN - 190/(4+6) 19.0 or 19 trees.
To determine the magnitude of Z, it would be neces-
sary to estimate the number of sample trees needed to
achieve a specified level of precision at a specified proba-
bility level. This could be done by using standard form-
ulae:
8
t
2 
S
2
n =
AE
2
where: n = sample size
t = Student's t at the specified probability
level and for the appropriate degrees of
freedom
s
2 
= variance of T values
or
t
2 
C.V.2n -
AE
2
where: n = as above
t = as above
C.V. = coefficient of variation of T values
AE = allowable sampling error as a proportion
of T.
As an example, assume that an inventory is being de-
signed with 12 classes (KPI values) so that K equals 12.
Also assume that a pre-sample has been carried out and
the total number of trees is estimated to be 85 and the
m
sum of the weights (I KPI) is estimated to be 580. The
C.V. is found to be 0.08214. The allowable error is ? 2
per cent at the 95 per cent level of probability. Under
these circumstances, n is found as follows, using iteration
to reconcile n and the appropriate value of t:
First sample, t for 00 d.f. = 1.960
t2CV2 - (1.960)2 (0.08214)2 -
9.
n - - -- 9.92or"10trees
AE
2  
(0.02)2
Second estimate, t for 10-1 d.f. = 2.262
(2.262)
2 
(0.03214)2
n = = 13.2 or 14 trees
(0.02)2
8 These formulae are used 
since the populations usually 
are
so large that corrections for finite populations would not be
needed.
[ 10 ]
Third estimate, t for 14-1 d.f. = 2.160
(2.160)2 (0.03214)2
n = = 12.05 or 
13 trees
(0.02)2
Fourth estimate, t for 13-1 d.f. = 2.179
(2.179)2 (0.03214)2
n = - 12.26 or 13 trees
(0.02)2
In this case, n stabilizes at 13 trees. To 
obtain Z, n is
substituted for ESN in the ESN formulae:
m
n KPI
n
K+Z
and the equation is solved 
for Z
m
Z 
KPI 
-
nK
Zn
Substituting the data from 
above:
530 =13(12) 
28.77 or 29
Z 
28.77 or 29
13
To meet the inventory specifications, 
the deck of cards
would contain 12 number 
cards and 29 rejection cards.
When designing a 3-P inventory 
there are two rules
which must be observed concerning 
the relative magnitude
of ESN (the expected size 
of sample). These are:
1) The product of ESN and the largest class 
(KPI)
m
value should be less than the sum of weights 
(I KPI).
Applying this to the above example:
m
(ESN) (largest 
KPI) = 13(12) 
= 156; 1 KPI 
= 530.
Hence, this requirement has 
been met.
2) The square of the ratio of 
Z to K should be greater
than (4/ESN) - (4/m). e.g.,
(Z/K)2 = (29/12)2 
= 
5.84
(4/ESN) - (4/m) (4/13) 
- (4/85) = 0.2606
5.84>0.2606, so this requirement 
also has been met.
WHAT HAPPENS WHEN NO TREES
ARE SELECTED
It is possible in a cruise of this type 
that no trees at all
would happen to be selected 
for measurement. This
could lead to the erroneous conclusion 
that the volume
or value was zero. Should this occur, 
the cruise must be
repeated. If the provision is made in the inventory 
design
that a repeat cruise would be made in the 
case of no tree's
being selected, the probabilities of selection are changed.
The corrections necessary to maintain exact probabilities
are beyond the scope of this discussion. Reference is made
to Grosenbaugh (6) and Space (10).
If one wishes, the same tract could be 
cruised several
times simultaneously by using interpenetrating 
samples.
In such a case, each tree is assigned to a class (KPI
value) in the normal manner, but instead of comparison
with a single randomly drawn (KI) value this KPI value
is then compared with several randomly drawn (KI)
values. Each of these represents a separate cruise. The
data of the several cruises must be kept separate and
analyzed separately. This approach materially reduces
the probability of ending up with the patently false esti-
mate of zero volume or value. Again, the step-wise pro-
cedures are beyond the scope of this publication.
USING RANDOM NUMBERS RATHER
THAN A DECK
A deck of cards such as that described earlier is not a
practical tool in the woods. A practical substitute for the
deck is a set of random numbers which vary in magni-
tude in the same manner as the numbered cards in the
deck. Diluting these random numbers are rejection sym-
bols which correspond to the rejection cards. These ran-
dom numbers and the rejection symbols are generated by
a computer using Grosenbaugh's RN3P program. The
computer print-out from this program can be cut into
strips, which are joined end-to-end and rolled on spools
for use in a special random number dispenser that has
been adequately described by Grosenbaugh (2) and Me-
savage (7).
WHERE DOES THE DENDROMETER FIT IN?
The 3-P sampling procedure is designed to select the
trees that are to be measured or evaluated so that an
efficient estimate of the total volume or value of the
whole population can be made. Nowhere in the 3-P pro-
cedure are any requirements stated as to how the trees
that are selected must be measured or evaluated. This
evaluation procedure could consist of an ocular estimate;
it could be based on a local, standard, or form-class vol-
ume table; or it could consist of a series of measurements
made on the upper stem using the dendrometer. Though
the dendrometer approach is preferred, it is not the only
one that could be used.
The reason the dendrometer approach is preferred is
that the evaluation of each tree is based on data from that
tree alone. There is no recourse to volume tables, which
had to be constructed from data obtained from other trees
that might or might not have been similar to the trees
being sampled. With the dendrometer procedure, meas-
urement and estimation errors are minimized and consid-
erably more accurate results can be obtained.
IN SUMMARY
The pure 3-P sampling design 
used as the foundation
for this discussion is the simplest and probably the least
useful that might be employed. A far more useful design
involves the combination of 3-P sampling with point sam-
pling which can be used for several types of inventories,
including C.F.I. Grosenbaugh and his colleagues have
developed a number of different designs. Grosenbaugh's
STX computer program is sufficiently flexible to handle
all of them. The combining of 3-P sampling with den-
drometry and computer programming has made available
to the forestry profession a powerful and versatile tool
for inventory operations of all kinds. Foresters should
take advantage of this tool whenever its use seems ad-
vantageous.
[11]
LITERATURE CITED
(1) GROSENBAUGH, L. R. 1963. Some Suggestions 
for
Better Sample Tree Measurement. Proc. Soc. Amer.
For., pp. 36-42.
(2) ------------------------------------. 1965. Three-pee Sam pling
Theory and Program 'THRP' for Computer Genera-
tion of Selection Criteria. USDA, Forest Service Re-
search Paper PSW-21. 53 p.
(3) ---------. 1967. The Gains From Sam-
ple-Tree Selection with Unequal Probabilities. For-
estry 65:203-206.
(4) -- ---------------------------------. 1967. STX -FO RTRA N -4
Program for Estimates of Tree Populations from 3P
Sample Tree Measurements. USDA, Forest Serv-
ice Research Paper P5W-13 (Revised). 76 p.
(5) ------------------------------------ . 1968. Sample Tree 
Meas-
urement: A New Science. Forest Farmer, Dec.,
pp. 10-11.
(6) -----------------------------------. 1971. STX 1-11-71 for Den-
drometry of Multistage 3P Samples. USDA, Forest
Service. 63 p.
(7) MESAVAGE, C. 1971. STX Timber Estimating With
3-P Sampling and Dendrometry. USDA, Forest
Service, Agr. Handbook No. 415.
(8) SCHREUDER, H. T., J. SEDRANSK, AND K. D. WARE.
1968. 3-P Sampling and Some Alternatives, I. For-
est Science 14:429-454.
(9 ) -------------------------------- ---------A N D
D. A. HAMILTON. 1971. 3-P Sampling and Some
Alternatives, II. Forest Science 17:103-118.
(10) SPACE, J. C.. 1971. Field Instructions, Three-P For-
est Inventory. USDA, Forcst Service, State & Private
Forestry -SE Area. 18 p.
(11) STEBER,. G. D. AND J. C. SPACE. 1972. New Inven-
tory System Sweeping the South. J. Forestry 70:
76-79.
APPENDIX
Meanings of symbols conventionally included in the
3-P sampling literature.
K The number of classes (or KPI values) to which
trees can be assigned.
Z = The number of rejection cards in a 3-P sampling
deck.
KPI = Class label. It is a whole number used-to iden-
tify a class. This number is equal in magnitude
to the weight assigned to the class.
KI A value obtained from a random draw from the
3-P sampling deck provided an integer card is
drawn.
YI - The actual value or volume of the tree in ques-
tion. Every tree has a YI value but it is. un-
known until after the tree has been measured
and evaluated.
m= The number of trees in the population being
sampled.
n= The number of trees sampled.
m
I KPI= The sum of the weights of all the trees in the
population being sampled.
m
r KPI =
KPI =
The blow-up factor for the Ith tree.
TI = The estimate of the total value or volume in the
population based on the Ith tree alone.
= The mean estimate of the total value or volume
based on all the trees in the sample.
ESN The expected number of sample trees.