2/8. UNIT COST AND COST EQUATIONS
2.1 Introduction
2.2 Example of Cost Equations
2.3 Applications of Cost Equations
2.2 Example of Cost Equations
2.3 Applications of Cost Equations
2.1 Introduction
The use of breakeven and minimum-cost-point
formulas require the collection of unit costs. Unit costs can be divided into
subunits, each of which measures the cost of a certain part of the total. A
typical unit cost formula might be
X = a + b + c
where X is the cost per unit volume such as
dollars per cubic meter and the subunits a, b, c will deal with distance,
volume, area, or weight. Careful selection of the subunits to express the
factors controlling costs is the key to success in all cost studies.
2.2 Example of Cost Equations
Let us suppose the cost of harvesting from
felling to loading on trucks is being studied. If X is the cost per cubic meter
of wood loaded on the truck, we could represent the total cost per unit as
X = A + B + Q + L
where A would be the cost per unit of felling, B
the cost of bucking, Q the cost of skidding, and L the cost of loading.
To determine the cost per subunit for felling,
bucking, skidding, and loading, the factors which determine production and cost
must be specified. Functional forms for production in road construction and
harvesting are discussed in Sections 4 and 5. Examples for felling and skidding
follow.
For felling, tree diameter may be an important
explanatory variable. For a given felling method, the time required to fell the
tree might be expressed as
T = a + b D2
where T is the time to fell the tree, b is the
felling time required per cm of diameter, D is the tree diameter and
"a" represents the felling time not explained by tree diameter-such
as for walking between trees. The production rate is equal to the tree volume
divided by the time per tree. The unit cost of felling is equal to the cost per
hour of the felling operation divided by the hourly production or
A = C/P = C/(V/T) = C (a
+ B D2)/V
where C is the cost per hour for the felling
method being used, P is the production per hour, V is the volume per tree, and
T is the time per tree. The hourly cost of operation is referred to as the
machine rate and is the combined cost of labor and equipment required for
production. (Machine rates are discussed in Section 3.)
EXAMPLE:
Determine the felling unit cost for a 60 cm tree
if the cost per hour of a man with power saw is $5.00, the tree volume is 3
cubic meters, and the time to fell the tree is 3 minutes plus 0.005 times the
square of the diameter.
T = 3 + .005 (60) (60) = 21 min = .35 hr
P = V/T = 3.0/.35 = 8.57 m3/hr
A = C/P = 5.00/8.57 = $0.58/m3
P = V/T = 3.0/.35 = 8.57 m3/hr
A = C/P = 5.00/8.57 = $0.58/m3
In skidding, for example, if logs were being
skidded directly to a road (Figure 2.1), then the distance skidded is an
important factor and the stump to truck unit cost might be written as
X = A + B + Q + L
X = A + B + F + C(D/2) + L
X = A + B + F + C(D/2) + L
where the skidding subunit Q has been replaced
by symbol F representing fixed costs of skidding such as hooking, unhooking and
decking and C(D/2) represents that part of the skidding cost that varies with
distance. C is the cost of skidding a unit distance such as one meter and D/2
represents the average skidding distance in similar units. It is important to
note that the average skidding cost occurs at the average skidding distance
only when the skidding cost, C does not vary with distance. If C varies with
distance, as for example, with animal skidding where the animal can become
increasingly tired with distance, the average skidding cost does not occur at
the average skidding distance and substantial errors in unit cost calculations
can occur if the average skidding distance is used.
If logs were being skidded to a series of
secondary roads (Figure 2.1) running into a primary road, then the expression
C(D/2) would be replaced by the expression C(S/4) and the cost of truck haul on
the secondary roads would appear as a separate item. In the expression C(S/4),
the symbol S represents the spacing of the secondary roads and the distance S/4
is the average skidding distance if skidding could take place in both
directions. Therefore, the expression C(S/4) would define the variable skidding
cost in terms of spacing of the secondary roads.
Figure 2.1 Nomenclature
for 2-way Skidding to Continuous Landings Among Spur Roads.
A formula for the cost of logs on trucks at the
primary road under these circumstances would be
X = A + B + F + C(S/4) +
L + H(D/2)
where D/2 is the average hauling distance along
the secondary road and H is the variable cost of hauling per unit distance.
The formula can be extended still further to
include the cost of the secondary road system by defining the road construction
cost per meter R, and the volume per square meter, V. Then, the formula becomes
X = A + B + F + C(S/4) +
L + H(D/2) + R/(VS)
2.3 Applications of Cost Equations
In the preceding equation, we have a situation
where as the spacing between skidding roads increases, skidding unit costs
increase, while road unit costs decrease. With the total cost equation, we can
look at the cost tradeoffs between skidding distance and road spacing. Calculus
can be used to derive the formula for road spacing which minimizes costs as
follows:
dX/dS = C/4 - R/(VS2) = 0
or
S = (4R/CV).5
An alternative method is to compare total costs
for various road spacings. The total cost method has become less laborious with
the use of programmable calculators and microcomputers. It provides information
on the sensitivity of total unit cost to road spacing without having to
evaluate the derivative of the cost function.
EXAMPLE:
Given the following table of unit costs, what is
the effect of alternative spur road spacings on the total cost of wood
delivered to the main road if 50 m3 per hectare is
being cut and the average length of the spur road is 2 km. The cost of spur
roads includes landings.
TABLE 2.1 Table of costs
by activity for the road spacing example.
Activity
|
Unit
|
Cost
|
Fell
|
$/m3
|
0.50
|
Buck
|
$/m3
|
0.20
|
Skid
|
$/m3
|
2.00
(fixed cost)
|
Skid
|
$/m3-km
|
2.50 (variable cost)
|
Load
|
$/m3
|
0.80
|
Transport
|
$/m3-km
|
0.15
|
Roads
|
$/km
|
2000
|
Since only the skidding costs and spur road
costs are affected by the road space
g, the total unit cost can be expressed as
X = A + B + F + C(S/4) +
L + H(D/2) + R/(VS)
X + 0.50 + 0.20 + 2.00 + C(S/4) + 0.80 + .15 (1) + R/(VS)
X = 3.65 + C(S/4) + R/(VS)
X + 0.50 + 0.20 + 2.00 + C(S/4) + 0.80 + .15 (1) + R/(VS)
X = 3.65 + C(S/4) + R/(VS)
To evaluate different road spacings, we vary the
spur road spacing S and calculate the total unit costs (Table 2.2). It is
important to use dimensionally consistent units. That is, if the left side of
the equation is in $/m3, the right side of the equation must be in $/m3. This is most easily
done if all volumes, costs and distances are expressed in meters; such as
volume cut per m2, skidding cost per m3 per meter, and
road cost per meter. For example, the total cost for a spur road spacing of 200
meters is 3.65 + (2.5/1000) (200/4) + (2000/1000)/[(50/10000) (200)] or $5.78
per m3.
TABLE 2.2 Total unit
cost as a function of road spacing.
Spur Road Spacing, m
|
Total Unit Cost, $/m3
|
200
|
5.78
|
400
|
4.90
|
600
|
4.69
|
800
|
4.65
|
1000
|
4.68
|
1200
|
4.73
|
1400
|
4.81
|
1600
|
4.90
|
1800
|
5.00
|
2000
|
5.10
|
The road spacing which minimized total cost
could be interpolated from the table or calculated from the formula
S = (4R/CV) .5
S = 800 m.
When costs have been collected in a form which
permits unit costs to be developed from them, not only is it possible to
predict costs, it is also possible to adjust conditions so that minimum cost
can be achieved. Too often, recorded costs are only "experience
figures". They are usually made available in a form which can be used to
predict costs only under conditions that closely conform to those existing
where and when the recorded costs were collected. This is not true of unit
costs, which can be fitted into the framework of many different harvesting
situations and can be made to tell the story of the future as well as that of
the past.
A wide range of cost control formulas can be
derived. Typical problems include:
1. The economic location of roads and landings.
- The calculation of the optimal spacing between spur roads and landings
subject to one-way skidding, two-way skidding, skidding on slopes, linear and
nonlinear skidding cost functions.
2. The economic service standard for roads. -
The comparison of the benefits of lower haul costs and road maintenance costs
as a function of increased initial investment. The calculation of the optimal
length of swing roads as a function of the tributary volume.
3. The economic selection of equipment for road
systems fixed by topography or other factors. - The identification of the
breakeven points between alternative skidding methods which have different
fixed and variable operating costs.
4. The economic spacing of roads which will be
served by two types of skidding machines. - For example, machines used to skid
sawtimber and to relog for fuelwood.
5. The economic spacing of roads which will be
reused in future time periods.
Another important application of unit costs is
in choosing between alternative harvesting systems.
EXAMPLE:
A forest manager is developing an area and is
trying to decide between harvesting methods. He has two choices of skidding
systems (small or large), two choices of road standards (high or low), and two
choices of trucks (small or large). If larger skidding equipment is selected to
bring the logs to the landing, he can still choose to buck them into smaller
logs on the landing. We assume that bucking on the landing will not affect log
quality or yield.
The managers staff has developed the relevant
unit costs, which are summarized in Table 2.3 and Table 2.4. What should he do?
TABLE 2.3 Unit costs for
options of using small equipment and large equipment.
Small Equipment
$/m3 |
Large Equipment
$/m3 |
|
Fall,
buck
|
0.70
|
0.50
|
Skid
|
1.70
|
2.55
|
Load
|
1.00
|
0.80
|
Transport
|
1/
|
1/
|
Unload
|
0.40
|
0.30
|
Process
|
-
|
0.05 2/
|
1/See Table 2.4 for
transport costs as a function of road standard. Wood for large system could be
bucked on landing for $0.15/m3 and loaded on
small trucks.
2/Large logs must be bucked at mill.
TABLE 2.4 Unit costs for
road and transport options using small and large equipment.
Small Equipment
$/m3 |
Large Equipment
$/m3 |
||
Road
|
|||
High Standard
|
1.30
|
1.30
|
|
Low
Standard
|
1.00
|
1.00
|
|
Transport
|
|||
High
Standard
|
3.50
|
3.00
|
|
Low Standard
|
4.00
|
3.40
|
These choices can be viewed as a network (Figure
2.2). You can verify that the least cost path is obtained by using the larger
skidding equipment and trucks and constructing the higher standard road. The
total unit cost will be $8.50 per m3. A key point is the
ease at which these problems can be analyzed, once the unit costs have been
derived. In turn, the derivation of the unit costs is facilitated by having
machine rates available (Section 3).
Figure 2.2 Network
Diagram for Equipment Choice
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