AN INTRODUCTION TO ECONOMIC PRINCIPLES OF PRODUCTION 1'
Max R. Langham* 30 or3
1. Profit Maximization with One Input
A farmer making the decision whether he should add a little more fertilizer
to his rice needs two pieces of information. First, he needs to know the
costs associated with adding the extra fertilizer and secondly, he needs to
estimate how much more valuable his rice crop will be as a result of the
added fertilizer. If the added value to the rice crop is larger than the
added costs, a farmer interested in making money will add the fertilizer.
This little example of decision making seems like straightforward common
sense, and it is. It also demonstrates one of the basic principles of
production economics -- the equi-marginal principle. This principle states
that a decision-maker should keep using additional units of a productive
input as long as the use of the added input earns or saves more money than
Let's state the principle a little more precisely with the use of symbols.
We will let X stand for the productive input (fertilizer in the above
example), Y stand for the physical output (rice in the above example), and
*Professor, Department of Food and Resource Economics, University of
Florida, Gainesville, Florida 32611. For seventeen months beginning in
January 1976, the author served as Research Officer for The Agricultural
Dorveolmpont Council, Inc., Singne. rigapor. ng rhi e Lnl. noven jawi hls of 977,
he served as Associate for the Philippines, Tlie Agricultural Development
Council, Inc. and visiting professor in the Department of Agricultural
Economics at the University of the Philippines at Los Banos.
This chapter borrows from sane materials prepared for use in a Farm
Management School held for extension agents at the University of Florida.
the symbol A stand for "the change in something." Putting A and X together
we get AX which is read "the change in X." We will let the letter P stand
for price. ThenPx and Py will denote the prices of X and Y, respectively.
With these symbols we can denote the added cost of the fertilizer in the
above example as PX times AX and the added value of the rice crop as Py
With the use of these symbols the equi-marginal principle says to add the
productive input as long as Py AY > PX AX, where > is read "is greater
than" and the dot between Py and AY means to multiply. This dot is generally
Agricultural production is usually subject to diminishing returns to a variable
input. That is, each added unit of an input, applied in a production situation
where same other inputs are held fixed, yields a smaller increment in output.
As a consequence, if the decision-maker keeps adding input as long as
Py AY > PX AX, the > relationship will continue to weaken and eventually
equality as shown in the following equation will occur:
(1) Py AY = PX AX.
And, if he adds too much Input, Py AY < PXA X, where < is read "is less than."
The principle can be stated in different ways by rearranging terms in equation
(1). An equation is much like a balance scale -- as long as you do the same
thing to both sides you do not affect the balance, i.e., the equality.
Given Py AY = PX AX, divide both sides by AX, to get
Pt, Y PX PX
We are left with,
(2) P Y = PX' an alternative way of stating our principle. The term
5E is called the marginal productivity.
Another alternative is provided by dividing both sides of (2) by A to get
Py = AY
Another alternative is provided by dividing both sides of (2) by Py to get
A fifth way of stating the principle is suggested by the objective
of making money. We want to add the productive input until the
difference between the total amount of money we receive from the
product and the total amount of money we spend in producing it is
a maximum. In symbols this method of stating the principle says
add input until,
(5) Py Y PX X = maximum or as big as possible.
There are other ways we can state the principle, but these five
are sufficient for our purposes. The many alternative ways of
stating the same simple principle emphasizes one of the difficul-
ties in studying economics. The economist uses the statement best
suited for a particular purpose but to the uninitiated stating the
same thing in different ways may seem like unnecessary confusion.
Nevertheless, let us state the last four alternatives in words.
Equation (2). Add productive input until the value of the marginal
product is equal to the price of the input.
P = value of the marginal product
PX = price of input
Equation (3). Add input until marginal revenue is equal to
marginal cost. When prices of inputs and outputs are fixed to
the producer as they normally are in agriculture, marginal revenue
is equal to the price of the product and marginal cost is equal to
price of input divided by the marginal productivity.
P = price of output = marginal revenue, MR, when Py is fixed.
cost per added unit of product = marginal cost, MC.
Equation (4). Add input until the marginal productivity is
equal to the inverse price ratio.
inverse price ratio.
Equation (5). Add input until total returns minus total cost of
input is at a maximum.
P Y = total returns
P X = total cost of variable input X, also called total
The procedure used in equation (5) is referred to as budgeting by
In the fertilizer example the farmer was making a decision about
fertilizer alone. He was not considering how much to work the
ground, how much seed to use, etc. He was making a decision about
fertilizer with all other inputs held fixed at some level. There-
fore, the difference between total returns and the cost of the
/Budgeting is a convenient method of aiding the decision-
maker when there are only a few production alternatives that need
to be considered in the decision. We will find that linear pro-
gramming is an easier method when many production alternatives
input under consideration is the amount of money left over to pay
for all other inputs that are treated as being fixed. This money
left over to pay fixed inputs is called net returns to fixed inputs
or net returns to fixed factors of production.
Let us look at a problem using the production principle that we
have considered thus far. The data given in Table 1 are based
on a relationship developed at the International Rice Research
Institute.2/ You will note that the blanks for the A's appear
half-way between the interval for which they are computed.
Given some prices we can determine how much nitrogen fertilizer
per hectare the farmer should apply. Let us assume that the
price of rice, Py, is 'l.25 per kilogram and that the price of
nitrogen, PX, is P9.50 per kilogram of active ingredient
(elemental nitrogen). We can now complete Tables 1 and 2.
Exercise 1: Please complete the blanks in Tables 1 and 2. As
you complete the tables look at equations (1) through
(5) to see the relationship between your computations
and the alternative methods. Determine the most
profitable level of fertilization with each method.
Upon the completion of Table 2 we find that each of these methods
leads to the same conclusion. It is most profitable for the farmer
2/ David, C. C., and R. Barker, "The Impact of the New Rice
Technology on Fertilizer Consumption," Resource Paper No. 7, The
International Rice Research Institute, Los Banos, Laguna, Philippines
(December, 1976), p. 6.
Table 1: Relationship of Yield of Modern Varieties of Rice in
Rainfed Fields to Elemental Nitrggen Applications with
Computation of Marginal Producta/
Level Input, Output AX AY AX
Elemental Rough (kgs. (kgs. (kgs.
Nitrogen Rice elemental rough rough
(kg/ha.) (kg/ha.) nitrogen) rice) rice)
1 0 1400.00
5 72.25 14.45
2 5 1472.25
5 66.75 13.35
3 10 1539.00
5 61.25 12.25
4 15 1600.25
5 20 1656.00
6 25 1706.25
7 30 1751.00
8 35 1790.25
9 40 1824.00
10 45 1852.25
11 50 1875.00
12 55 1892.25
13 60 1904.00
14 65 1910.25
15 70 1911.00
16 75 1906.25
a/The data in this table were developed from
matical relationship Y = 1400 + 15X 0.11X2
the following mathe-
which was taken from:
David, C. C., and R. Barker, "The Impact of the New Rice
Technology on Fertilizer Consumption," Resource Paper No. 7,
The International Rice Research Institute, Los Banos, Laguna,
Philippines (December 1976), p. 6.
Table 2: Five Different Methods of Determinating the Most
Profitable Level of Elemental Nitrogen Applications
on Rainfall Rice
lizer Method 1
P AY PX AX
Y0.66 14.45 Y7.60
0.71 13.35 7.60
0.78 12.25 7.60
P yY-P X
to apply 35 kilograms of elemental nitrogen per hectare. The
solution brings out an interesting point -- it does not pay to
fertilize for a maximum output of rice. Indeed it will only
pay to produce at a maximum when fertilizer is free. Method
4 makes it quite clear that the most profitable level of the
input depends on relative prices, i.e. the price ratio between
the price of the input and the price of the output.
Of course, we need to use only one method to arrive at the answer.
Method 5, the budgeting approach based on total returns and total
costs of variable inputs, is the method that is probably most
familiar to us. However, the other methods point to the key con-
sideration. Given the prices of the input and the output, it is
the changes in the input and the output (as we increase the input)
that determine the most profitable level of production.
Let us draw a picture of the data in Table 1. A relationship
between an input and an output, such as that shown in Figure 1,
is called a production function. 3The production function tells
3/ The marginal productivity which appears in equations
(2),(3), and (4) can be defined more precisely, with the cal-
culus, as the derivative of Y, The production function, with
respect to X. For example, equation (4) can be rewritten
dx =Py The marginal productivity is therefore simply the
slope of the production function and the point at which the
slope equals the inverse price ratio can be worked out analy-
Figure 1. Relationship of Rough Rice Yields to Nitrogen Fertilizer
. 0 15
0 5 15 2 25 30 5 4 45 5 55 6 65 7 75
0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75
Input of Elemental Nitrogen (kgs./ha.)
us how much output we will get at various levels of input.
We could interpolate between the values given in Table 1 by
connecting the midpoints of the top of the columns with a
smooth curve. Such a curve is called a total product curve.
This curve is simply the graph of the equation given in
footnote a of Table 1.
Figure 2 shows the solutions in Table 2 by methods 1 and 2. These
methods are identical except for a change in scale. The value of
marginal product curve, VMP, is sometimes called the firm's demand
curve for the input because it tells how much of the input the
decision-maker will want for every price of the input. For
example if the price of elemental nitrogen falls to Y6.00 per kg.
as shown by the broken line in Figure 2, the farmer will want to
use a little over 46 kilograms of fertilizer per hectare as shown
by the intersection of the new price line and the VMP curve.
Please draw the solutions shown by methods 3 and
5. Method; 4 will plot similar to methods 1 and
2. In drawing the solutions shown by methods 3
and 5, economists usually put product rather than
input on the horizontal line since these methods
are used when interest is in determining the most
profitable level of output rather than the most
profitable level of input.
II. The Equi-Marginal Principle and Cost Minimization
The problem of keeping costs as low as possible is an important
one for management. For example, let us consider the decision
problem of a farmer who wants to determine his lowest cost ferti-
lizer program for a rice yield of 2000 kg. per hectare. To
simplify the problem we will consider only two fertilizer ingredients,
nitrogen and phosphorus. The data we will use are given in Table
3 and are based in part on experimental data from the International
Figure 2. Optimum Solution to the Farmer's Fertilizer Decision
Problem, Methods 1 and 2.
P AY (Method 1), or value of marginal product
P (Method 2)
P XAX (Method 1),
or P, (Method 2)
5 10 15
20 25 30 35 40 45 50 55 60 65
75 80 85
Input of Nitrogen (kg. of active ingredients/ha.)
Table 3: Combinations of Elemental Phosphate and Nitrogen that
Will Produce 2000 kilograms of Rice per Hectarea/
C Elemental Elemental Nutrient
nation N P205 AX1 AX2 P AX1 PXAX21 Costs
X1 X2 PPXIX+PX2X2
kgs/ha kgs/ha kgs/ha kgs/ha ?
1 65 41
2 70 25
3 75 17
4 80 11
5 85 7
6 90 4
7 95 2
8 100 0
/The values of X2 in this table were obtained by inserting the
specified values of X2 in the following total product function
for rice (where the total product is set at 2000 kilograms) and
solving for the minimum value of X2:
Y = 2000 = 1200 + 14X1 .06X1
+ .03X1X2 .
+ 4X2 .06X2
Rice Research Institute. However, the data are presented here
for pedagogical purposes and should be considered as hypothe-
As the farmer moves from combination one to combination two,
he must add 5 kgs. of nitrogen per hectare. However, he saves
16 kgs. of phosphate per hectare. A manager interested in lower-
ing cost will be willing to switch to combination 2 if the 5 kgs.
of nitrogen are worth less than the 16 kgs. of phosphate, or if
PXI AX < PX2 AX21. The vertical bar on each side of AX2 means
to ignore the minus sign and to consider only the magnitude of
AX2. The symbol, IAX2z is read "the absolute value of AX2".
Let us assume that the price of nitrogen is P9.50 per kg. and
phosphate, P4.40 per kg. Then the 5 kgs. of nitrogen the farmer
gives up by switching from combination one to combination two
costs P47.50. The 16 kgs. of phosphorus he saves is worth P70.40.
Therefore he saves Y22.90 when making the change. If the farmer
is interested in producing the 2000 kgs. of rice for the least
amount of cost he will want to substitute nitrogen for phosphate.
One would follow the same line of reasoning to determine if
combination three were cheaper than combination two. And, if
PX AX1 < P X2 AX2 between combinations two and three it would
pay him to change to combination three. In fact, he would continue
the substitution process until,
(6) PX1 AX1 = Px2 AX2
It would be foolish to substitute beyond this point because the
amount he would have to pay would be greater than the amount he
Exercise 3: Please complete the blanks in Table 3 and determine
the least cost combination of inputs by the relation-
ship expressed in equation (6) and by the budgeting
procedure given in the last column.
Again the budgeting procedure is quite straightforward and quite
familiar to most of us. One simply adds together the cost of the
two fertilizer ingredients for the various combinations of inputs
and then chooses that one which is smallest. This procedure
offers an easy method of solving small problems. For larger prob-
lems linear programming provides an easier, more accurate method
of solving for a least cost combination of inputs.
If we divide both sides of equation (6) first by AX1 and secondly
by PX2, we get
PX 1 X2JAX21 AX2I PX
Px2 = ^ Axl or ~x~ = x
PX 41 X I X1 X2
In economic literature the absolute value signs are generally
omitted. If one does this and writes the relationship as it is
in equation (7), he must remember to ignore the minus signs. If
one does not ignore the negativity of AX2, he must add a minus sign
to the inverse price ratio on the right side to make the two sides
of (7) equal to each other.
(7) AX2 PX
Equations (6) and (7) look very similar to equations (1) and (4),
respectively. Interestingly enough the profit maximization problem
is conceptually equivalent to the cost minimization problem. In a
later lesson we will look briefly at the idea of linear programming.
Linear programming is an easy way to solve minimization and maxi-
mization problems -- especially when the problems are large. And
just as we see here, the linear programming procedure for solving
a profit maximization problem is similar to that for solving a cost
The AX2 is called the marginal rate of substitution of X1 for X2. If
we plot the values of X1 and X2 given in Table 3 on a graph with
X2 on the vertical line and XI on the horizontal line and connect
them with a smooth line,we get a curved line shown in Figure 3.
This line is called an isoproduct line. Iso- means equal so the
line gets its name from the fact.that it represents an equal
amount of product -- in this case 2000 kgs. of rice.
The marginal rate of substitution of Xi for X2, X2 is actually
the slope of the isoproduct line. The isoproduct curve slopes
downward to the right so it has a negative slope.
Figure 3. Graphic Solution
Given in Table 3
to the Cost Minimization Problem
.2000 kg. isoproduct
curve for rice,
slope = AX2
isocost line \
(1), slope =
X1 (kgs. of elemental
20 30 40 50 60 70 80 90 100
A note on the idea of slope -- The average slope between two points
is defined as the vertical change in distance divided by the
horizontal change in distance. We are all quite familiar with
this definition. We say that a hill is steep if we go up a lot
in comparison with the horizontal distance we travel. If we
are going uphill (or if a line curves.up viewing from left to
right) the slope is positive; if we go downhill, the slope is
negative. A horizontal line has zero slope.
The slope of the isoproduct curve gets less steep as one views it
from left to right. Therefore we say that there is a diminishing
marginal rate of substitution along the curve.
All inputs do not combine at a diminishing rate of substitution.
Certain inputs may combine at a constant rate of substitution.
If they do, the isoproduct curves are straight lines and we say
that the inputs are perfect substitutes. Certain feed grains
are perfect or nearly perfect substitutes for each other. Inputs
may also combine in rigid proportions. In this case, only one
combination of the inputs will work. An example of this situation
is provided by a farmer and a carabao. It takes one man for each
caraboa working. Adding more carabao hours without adding more
men does nothing to production. Adding more men to work carabao
without adding more carabao also contributes nothing to production.
We see then that the left hand side of equation (7) is the average
slope of the isoproduct curve between the points for which AX2 and
AX1 are computed. Similarly, the right hand side of equation (7)
is the slope of a line. The inverse price ratio (with a minus
sign attached) is the slope of the isocost curve. The isocost
curve is a line which gives the various combinations of the two
fertilizer ingredients that can be purchased for a given amount
of money or cost. Let us look at one of these isocost lines.
Assume that we have Y500 and we want to determine the different
combinations of X1 and X2 that we can buy with it. The equation
of the line we want is:
9.5X1 + 4.4X2 = P500
If we subtract 9.5X1 from both sides of the equation, we have
9.5X1 + 4.4X2 = 500
-9.5X1 = -9.5XI
4.4X2 = 500 -9.5X1
Now, if we divide both sides by 4.4, we have
4.4X2 500 9.5X1
4.4 4.4 4.4
X2 = 113.64 2.16X1
If we put this line of Figure 3 we get isocost line (1). When
XI = 0, we see that X2 = 113.64, the beginning of the line. If
Xl = 1, then X2 = 113.64 2.16 111.48. Each time we make X1
larger by 1 we make X2 smaller by 2.16. Therefore the slope of
the isocost line is equal to X or in this case -. = -2.16.
This isocost line doesn't tell us much new, it just gives us a
picture of what is happening. We all know that if elemental
nitrogen costs P9.50 per kg. and we have Y500 we can buy 52.63
kgs. if we spend it all on nitrogen. If we spend it all on
phosphate we can buy 500 + 4.4 or 113.64 kgs.
The P500 isocost line does not touch the isoproduct line. There-
fore, Y500 is not enough money to buy the amount of fertilizer
needed for 2000 kgs. of rice. We need to determine how much
more money is needed to obtain the least cost combination of
fertilizer ingredients that will permit us to obtain a 2000 kg.
yield. The least amount of money that will permit the farmer
to get a 2000 kg. yield will be that amount represented by an
isocost line just barely touching the isoproduct curve. The
P600 isocost line (isocost line 2) determined in the same way
we determined the Y500 curve, is X2 = 136.36 2.16X1.
The amount of money represented by this line is still not sufficient
to get a 2000 kg.yield. We note however that isocost lines (1)
and (2) are parallel, that is their slopes are equal. Indeed, for
a given set of prices all isocost lines will be parallel. There-
fore, the isocost line we want is that one which is parallel to
(1) and (2) and which just touches the isoproduct curve. It is
the isocost line for y = 770. The equation for this line is,
X2 = 175 2.16XI
When two lines touch but do not intersect at a point their slopes
are equal at that point. At point A the slope of the isoproduct
curve is equal to the slope of the isocost line or as stated in
AX2 P X1
The value of X1 and X2 at point A are 69 and 26, respectively.-
One would not want to move to a higher isocost line since the J
farmer would be spending more money for fertilizer than necessary
to get 2000 kgs. of rice.
In summary we can state that the least cost combination of inputs
for a given amount of product occurs at that combination where the
marginal rate of substitution is equal to the inverse price ratio
(minus signs ignored). Graphically the solution occurs where the
isocost line just touches (is tangent to) the isoproduct line.
Although you could not draw a picture of it, this conclusion is
true for any number of inputs. When dealing with many inputs it
is easiest to state the least cost principle using the form of
equation (6). For example, the least cost combination of four
inputs would be, ignoring signs, that combination where,
PX AX = PX2AX2 = Px3AX3 = P xAX4.
4/ These values of X1 and X2 were obtained from usinq the
equation in footnote a to '['able 3 and the prices of elenonLal
N and P20O; and were rounded to the nearest kilogram. The mini-
mum cost of Y770 was obtained by rounding the cost of 69 kilo-
grams of Xi and 26 kilograms of X2 to the nearest peso.
III. Profit Maximization With Two or More Products
Enterprise combination is one of the most important problems a
farm manager faces. He owns or controls a certain amount of
productive inputs which can generally be used on a number of dif-
ferent crops. A farmer interested in making profits will want
to put his resources in their most profitable use.
For example, assume that a farmer has 80 kilograms of elemental
nitrogen and that he wants to know how much of it he should use
on his hectare of rainfed palay and how much he should set aside
for the corn which he plans to plant as a second crop. The
farmer's production functions are given in Table 4. Here the symbol
X1 is used to represent the nitrogen. The production functions
given in Table 4 can be used to determine the combinations of Y1 and
Y2 that can be used with the elemental nitrogen available. This
is left as an exercise to be completed and entered in Table 5. The
combination entered in Table 5 represent the alternative courses
of action available to the farmer under the given set of circumstances.
Again the most profitable way to use the nitrogen resource can be
determined in several different ways. The budgeting approach indi-
cated in the last column of Table 5 is the easiest and most common'
sense approach. Since the same amount of nitrogen (80 kgs.) and
therefore the same cost are used for all possible combinations of
palay and corn all we need to do is maximize equation (8).
Total returns = PY YI + PyY2.
Table 4: Hypothetical Production Functions for the Production
of Palay and Corn Double Cropped on Rainfed Fields
XI Y1 Y2
(kgs. of elemental (kgs. of palay/ha. (kgs. of corn/ha, for
nitrogen/ha.) for given level of N) given level of N)
0 1400 700
10 1540 1000
20 1655 1300
30 1750 1550
40 1825 1750
50 1875 1900
60 1905 1975
70 1910 2000
80 1905 2010
Table 5: Combinations of Palay and Corn that can be Produced with
80 kgs. of Elemental Nitrogen on a Hectare of Rainfed
S Y2 pAY1 P AY2 P Y1+P Y2
AY1 AY22 Y Y2 Y1 22
Combi-(kgs. of (kgs.
nation Palay) of corn) (kgs.) (kgs.) (kgs.)
1 1905 700
2 1910 1000
3 1905 1300
8 _____ _______
The opportunity cost principle can also be used to obtain the
solution. This principle is another form of an equi-marginal
principle. The opportunity cost principle states that for profit
maximization the marginal value product of a resource used on one
activity must be equal to the marginal value product of the resource
used in all other alternatives. Again this is just common sense.
It states that a manager should use each additional unit of a pro-
ductive input where it will return the most. In symbols the
principle states to produce that combination of products where,
(9) AY AY2
AX1 2 AX1
We can multiply both sides of equation (9) by AX1 and get another
form of the principle,
(10) Py AY1 = P2 AY2
In equation (10) we have eliminated AX1. This procedure assumes
that AX1 is the same size on both sides of equation (9). When Y1
and Y2 compete for a fixed amount of a resource, the only way we
can change Y1 is to change Y2 in the opposite direction -- so again
we must ignore the signs of the change. If we divide both sides by
P and by AY1 we can get another familiar form of the principle
(again ignoring minus signs),
(11) AY1 PY2
Again we see that equations (10) and (11) are identical in form
to equations (6) and (7), respectively. Equations (10) and (11)
are also identical in form to (1) and (4), respectively.
Exercise 4: Assume that the farm price of palay is P1.20 per
kg. and the price of corn 41.70 per kg. Please
complete Table 5. Circle the most profitable
combination of palay and corn to produce.
As before, we can draw a picture of this solution. Such a picture
has been started in Figure 4. The left hand side of equation (11)
is called the marginal rate of substitution of Y1 for Y2. It is
the slope of a line called the production possibilities curve. It
gets its name from the fact that this curve shows all possible
combinations of the two products, in this case palay and corn,
that can be produced from a fixed amount of elemental nitrogen.
The right hand side of (11) with a minus sign attached is the
slope of a line called the isorevenue line. The isorevenue line
gets its name from the fact the line represents all combinations
of the two products that will yield a given amount of revenue.
The point fartherest from the origin where the production possibili-
ties curve just touches the isorevenue line represents that combina-
tion of products which can be produced and which yields the greatest
amount of return.
Exercise 5: Please complete the production possibilities curve in.
Figure 4 and draw the isorevenues curve that represents
the maximum profit solution determined in Table 5.
Figure 4. Graphic Solution to the Profit Maximization Problem given
in Table 5
1400 1500 1600 1700 1800 1900
Y1 (kgs. of palay per ha.)
Note that the production possibilities curve starts decreasing
at first, i.e. the products are competitive. Where the produc-
tion possibilities curve is curved upward to the right (both
products are increasing together)'the products are said to be
complementary. When one product increases as the other de-
creases, the products are said to be competitive. When one
product remains constant as the other increases, the products
are said to be supplementary.
> 0, or positive, the products are complementary
= 0, the products are supplementary
--- < 0, or negative, the products are competitive.
The profit maximizing solution always occurs in the competitive
area. Because, if you can get more of both products or more of
one and the same amount of the other without additional cost, it
will always pay you to do so. The example in Figure 4 shows that
the farmer will get more palay if he uses some nitrogen on corn.
The recommendation of growing crops in a rotation derives in part
from benefits due to complementarities and supplementarities.
IV. Vertical Combination of Enterprises
The principle of cost minimization and profit maximization can
be put together in certain kinds of decision problems to give
added insight into the management problem of enterprise combi-
nation. Suppose a farmer involved in milk production knew the
combinations of grain and hay that would yield the most profitable
amount of milk for him to product. In other words suppose he
knew the isoproduct curve which represented the most profitable
amount of milk for him to produce. The decision problem might
be as pictured in Figure 5.
The price line in Figure 5 is an isomoney line. It is called an
isorevenue line when Yj and Y2 are considered as products and an
isocost line when Y1 and Y2 are considered as inputs.
The economic principles discussed thus far would indicate that
the dairyman should produce 20 tons of hay and 18 of grain. He
should feed 13.8 tons of hay and 27 of grain. Therefore, he
would need to buy 9 tons of grain and sell 6.2 tons of hay.
V. Linear Programming
Linear programming provides an alternative method of solving real
world minimization or maximization decision problems. The approach
is particularly suited to solutions with the aid of modern computers.
The cost minimization problem discussed in Section II and the profit
maximization problem discussed in Section III could be solved with.
the use of linear programming. However, in problems like these
Figure 5. Graphic Illustration of a Solution to a Problem of Vertical
18 - - -
SI rice line
S12 Production Posii ie
6 I I
5 10 15 20 25
where the decision-maker is faced with only a few alternatives
it is easier to solve them with budgeting.
In order for a problem to be solved with linear programming it
must have the following three parts.
1. A minimization or maximization objective. The decision-maker
must be trying to minimize or maximize something.
2. There must be certain restrictions on the attainment of the
3. There must be alternative courses of action or alternative
ways to attempt the attainment of the objective.
A dairy farmer who is attempting to minimize the cost of his ration
has a problem that contains these three parts. His objective is to
minimize cost. The restrictions on his decision are the nutritive
requirements which the ration should meet. It should have a certain
minimum level of protein, a minimum level of fat, a maximum level
of fiber, etc. The alternative ways he has of attempting to attain
the objective is the various feed stuffs available to him -- such
as soybean meal, corn, molasses, etc.
A farmer attempting to determine the most profitable combination
of enterprises provides an example of a profit maximization problem.
Here the objective would be to maximize profits. The restrictions
would be the amount of land, labor, capital, etc., available. The
alternatives would be the crop and perhaps livestock enterprises
the farmer could engage in.
Let us consider a profit maximization problem like that in Section
III where the problem was one of determining the most profitable
combination of palay and corn with a given amount of resources.
Example 1: Assume that a Filipino farmer has 1.5 hectares of
land and 1800 hours of labor to use on crops. He is
fairly certain that upland rice is his best choice
of crop to plant during the rainy (first crop) season.
However, during the dry or second season he wonders
if he should grow corn alone YI, or corn with
vegetables intercropped, Y2 He has half his
labor for use during the second season. Corn grown
alone uses 400 hours of labor per hectare while
corn intercropped uses 800. The farmer can earn a
return to his land and labor of ?1000 per hectare
with corn and P1500 with corn intercropped. What
should he do to maximize returns to his land and
With the land available the farmer can grow 1.5 hectares of corn,
1.5 hectares of corn intercropped, or some combination of the two
given by the relationship, Y1 + Y2 = 1.5, where Y1 is hectares of
corn and Y2 is hectares of corn intercropped with vegetables.
With the labor available he can produce 2 1/4 hectares of corn,
1 1/8 hectares of corn interplanted, or some combination given
by the relationship, 400Yi + 800Y2 = 900. Each hectare of corn
returns to land and labor y1000 and each hectare of corn inter-
planted P1500. His total returns will be equal to 1000Y + 1500Y2.
In summary the linear programming problem may be stated as follows:
Find that value for Y1 and Y2 which will maximize total returns =
1000Y1 + 1500Y2 subject to the conditions,
Y1 + Y2 1.5,land restriction
400Y1 + 800Y2 < 900,labor restriction
Y1 1 0 and Y2 > 0
The inequalities are introduced in the functions so that the farmer
will not have to use up all of his land or labor if it is most
profitable for him to leave some idle. Since it is mathematically
possible for Y1 and Y2 to be negative we must also restrict them
to zero or positive values. The problem is graphed in Figure 6.
The area within which the farmer has enough resources to produce
is enclosed by the lines OABC. This area is called the area of
feasible solutions in linear programming jargon. Values of Y1 and
Y2 beyond the line CE are not attainable because of insufficient
land. Values of Y1 and Y2 beyond line AD are not attainable because
of insufficient labor. There is enough land but not enough labor
to produce those amounts of Y1 and Y2 within the triangle ABE.
Similarly there is enough labor but insufficient land to produce
within the triangle CBD. Land and labor are both restrictive
beyond the two line segments EB and BD. Production at values
of Y1 and Y2 along the line segment BC exhausts land but results
in surplus labor (except at the point B). Production at values of
Y1 and Y2 along the line segment AB exhausts labor but results in
some idle land (except at the point B). By leaving some land and
labor idle it would be possible to produce any nonnegative values
Figure 6. Graphic Presentation of Filipino Farmers Enterprise
Combination Problem during the Dry Season
Point B is the optimal
14 1 This is the point Y1 =
S / B Returns to land and lab<
o / constraint 1
w / / 1875 isoreven
''. f V -i i \
.75, Y2 = .75
or = Y1875.
Y1 (corn in hectares)
of Y1 and Y2 within the line segments AB and BC. The solution to
the problem is that value of Y1 and Y2 on the boundary or within
the area defined by OABC which maximize 1000Y1 + 1500Y2.
The solution will be at one of the corners of this area because if
it pays to substitute one of the products for the other along any
one of the straight line segments it will pay to go all the way
to the end of it. The revenue at each corner may be budgeted as
Corner Revenue in Pesos
0 (1000)(0) + (1500)(0) = 0
A (1000) (0) + (1500) (1.125) = 1687.50
B (1000) (.75) + (1500) (.75) = 1875.00
C (1000) (1.5) + (1500) (0) = 1500.00
Point B yields the greatest profit. The isorevenue function for
Y1875.00 just touches the production possibilities curve at this
point. The linear programming solution does not budget the returns
at every corner of the feasible area as we have done here. Rather
it provides a precise systematic procedure of going from one corner
to the next until it gets to the most profitable one. In going
from one corner to the next the procedure never leads to a corner
representing lower returns. This precise systematic procedure
is called the simplex method. The corners of the area of feasible
solutions are called basic feasible solutions.5/
5/ A basic theorem of linear programming establishes that one
only has to look at the basic feasible solutions and when one finds
the best of those, there will be no better solution in the set of all
feasible solutions. The simplex method is an efficient method of
finding a best basic feasible solution.
The reasoning behind the simplex method goes something like this.
(1) Start at 0 where all resources are idle and profits are zero.
(2) At 0, ask the question, "Will it pay to increase Yj or Y2 and
put some of the resources to work"? The answer is clearly
yes since each crop gives a money return. Let us go the Y1
route and move to point C.
(3) At C, ask the question, "Will it pay to give up a part of the
corn planted alone and add corn interplanted along the island
Along the island curve one hectare of corn must be given up
for each hectare of corn interplanted which is added. Corn
is worth 41000 per hectare and corn interplanted is worth
P1500. Consequently, for each hectare of corn interplanted
which is added there is a gain of P500 -- so the answer to
the question is yes. Therefore, corner B is more profitable
(4) At B, ask the question, "Will it pay to give up more corn for
additional corn interplanted along the isolabor curve"?
Along this curve each hectare of corn given up frees 400 hours
of labor' however, each hectare of corn interplanted requires
800 hours. Therefore, giving up an acre of corn will only
free enough labor for 1/2 hectare of corn interplanted. To
figure whether or not it will be profitable to move along
the isolabor curve one must determine if 1/2 hectare of
corn interplanted is worth more than a hectare of corn
planted alone. One-half of 1500 is less than 1000.
Therefore, there would be a loss if one substituted corn inter-
planted for corn along the isolabor curve. And the answer to
question (4) is no, so corner B is the most profitable. The
value 1500-1000 = 500 under question (3) and the value
1/2(1500)-1000 = -250 under question (4) are the simplex criteria
which determined whether there is a more profitable corner than the
one being considered. One can see from Example 1 that the linear
programming approach to the problem is similar to the approach
studied in Section III. Linear programming provides the same
answer as budgeting if the same information is used and both tech-
niques are used correctly. However, for large problems it becomes
quite difficult with the budgeting approach for a person to take
into account all production alternatives.
In this small problem there are only two constraints (land and
labor) and two alternatives (grow corn alone and/or grow corn
interplanted with vegetables). Modern computers permit the
solution of problems that may have a few thousand constraints and
several thousand alternatives. However, irregardless of how large
the problem is there will be the three parts -- a linear objective
function to be maximized (or minimized), a set of constraints on
the attainment of the objective, and alternative ways (variables)
of attempting to maximize (or minimize) the objective function.
For every linear programming problem (called the primal problem)
there is another problem called its dual. For a resource alloca-
tion problem like that given in Example 1, the dual problem is a
resource valuation problem. When one solves a linear programming
problem the dual problem is also solved so there is never any reason
to numerically solve both the primal and the dual problems as
separate problems. The solution to either one is sufficient.
In the resource allocation problem of Example 1, our interest was
in determining what crops (alternatives) the farmer should produce
with his land and labor (constraints on production) in order to
maximize returns to land and labor (objective). The dual of this
problem asks what wage should be awarded to the farmers labor and
what rent to his land if we want to keep the cost of these inputs
as low as possible and yet pay them at least as much as they are
worth. Since the dual is also a linear programming problem it has
the same three parts -- i.e. objective, alternatives, and constraints.
Here the objective is to minimize the cost of resources used in
production. The alternatives are to award the receipts from
production to either land or labor. The constraints are that we
must pay the resources at least as much as they are worth.
The prices assigned to the resources are called shadow prices and
these prices provide valuable economic information about the
production process because those resources which have a relatively
large shadow price are those which place the greatest constraint on
the production process.
Mathematically the resource valuation problem -- the dual to the
resource allocation problem of Example 1 -- may be defined as
follows: Minimize, the cost of land and labor = P Xi + P 2X2
= 1.5PX1 + 900PX2
subject to the following constraints
PX + 400Px > 1000, corn alone
P + 800P2 > 1500, corn interplanted
X1 X2 =
PX, 1P 2 > 0
where P = rent per hectare, X1 = land measured in hectares,
PX = wages per hour, and X2 = labor measured in hours.
Since there are 1.5 hectares of land and 900 hours of labor, the
objective function simply represents the cost of land and labor.
The first constraint states that the rent per hectare PX1 plus
the amount awarded to labor, 400PX2, must be at least as much as
these resources earn in the production of corn alone. The second
constraint states that the rent plus the amount awarded to labor
in the production of a hectare of corn interplanted must be at
least as much as these resources earn in the production of corn
interplanted with vegetable (P1500 per hectare). The last con-
straints simply indicate that only nonnegative prices can be
assigned.to land and labor.
A graphic presentation of the dual problem is presented in Figure
7. The constraints on the problem require that only prices on or
Figure 7. Graphic Solution to the Resource Valuation (Dual)
Solution: PX = Y500/ha.
Cost = '1875
isocost for a cost
PX1, Rent per hectare in pesos
beyond the line segments AB and BC (the shaded area, i.e. the area
of feasible solutions) can be awarded to land and labor. One
cannot award land and labor prices in the area OEBD because the
prices are less than the amounts the resources earn in production.
Even though prices within the area DBC award land and labor what
they are worth in the production of corn alone, they do not award
them what they are worth in the prodcution of corn interplanted.
Therefore, prices in this region do not satisfy all the constraints.
Similarly, prices in the area ABE award land and labor what they
are worth in the production of corn interplanted, but fail to award
them what they are worth in the production of corn alone. Therefore,
prices within this area cannot be candidates for the solution.
The objective is of course to minimize cost. Costs are minimized
when the isocost line is as small as possible and yet has at least
one point in the area of feasible solutions. The smallest value
for cost which will do this job is "l875 and the point representing
an optimal solution is at point B.
The cost (objective function) for the problem is:
cost = 1.5PX + 900PX2.
If one solves this for PX2 in terms of PX1, one obtains:
cost 1 .5
P c0 1 5 P
X2 900 900 X1
If one inserts 41875 for cost and plots the line on Figure 7 the
isocost line is obtained.
Insert a value of 1200 for cost in the equation
immediately above and plot the resulting line on
Figure 7 to show that this value for cost does not
satisfy all the constraints for the dual problem.
Budget the cost of resources at the prices corre-
sponding to points A, B, and C and show that point
B is indeed the solution.
Solution of Example 1 by the Simplex Method
A thorough discussion of linear programming would require a book
and would certainly be beyond the scope of this paper. References
3 5, and 6 will give the reader additional assistance with
the solution process. Our purpose here is to simply give the
reader an introductory feel for the solution process.
Again the problem is as follows:
Maximize, net returns to land and labor = 1000Y1 + 1500Y2,
subject to Y1 + Y2 = 1.5
400Yi + 800Y2 < 900
Y1, Y2 1 0
We need to allow for the possibility that it may be most profitable
to leave some land and/or labor idle. Therefore, we add two vari-
ables as follows:
Y3 = hectares of idle land, and
Y4 = hours of idle labor.
The constraints can now be written as:
Y1 + Y2 + Y3 = 1.5
400Y1 + 800Yz + Y4 = 900, and
Yi, Y2r Y3, Y4 a 0.
The first of these constraints simply says that the hectares of
corn plus the hectares of corn interplanted plus the hectares of
idle land must be equal to the amount available. The second con-
straint states that the labor used in the production of Y1 and Y2
plus idle labor must be equal to the amount available. And, the
last constraints say that all variables must be nonnegative. We
assume here that land and labor have zero coefficients in the
objective function so that it remains the same.
Y3 and Y4 are called slack variables because they allow for slack
in resource use if that is most profitable. Variables like Y3
and Y4 that are associated with a particular row (here Y3 is
associated with the land row and Y4 with labor) are also called
"logical" variables. With the addition of these two variables
the land and labor constraints represent two equations in four
unknowns. Since there is no unique solution to a system of
equations with more variables than equations, a unique solution
requires that two variables be set to zero and the two equations
solved for the remaining two variables.
The simplex method proceeds as follows:
1) It begins with a very easy basic feasible solution in which
the constraint equations are solved with a set of variables
equal to the number of constraints in the solution with
other variables at zero level.
2) At the end of this beginning solution, the simplex criterion
is applied to see if there is a better basic feasible solution.
If there is none, the optimal solution has been obtained.
3) If there is a better solution available, one of the variables
that was set at zero in the previous solution is permitted
to enter the solution to a level that drives a variable in
the solution to zero.
4) A new solution is obtained with the new variable.
5) Apply the simplex criterion and if all the (ZC-Cj) are positive
for a maximization problem (negative for a minimization problem)
the solution has been reached. If not repeat steps 3 and 4.
The completion of a new solution (through step 4) is called
To go about the numerical calculations required to carry out these
steps with a hand solution, it is quite helpful to put the problem
into a special format called a simplex tableau. Table 6 represents
such a tableau. Fractions have been used in the solution to avoid
any rounding errors. For a larger problem one would have to go to
decimal fractions and carry the work out a sufficient number of
places to maintain an acceptable level of accuracy.
Table 6. -- A Simplex Tableau with
Solution to the Problem of
C. 0 0 1000 1500
+ Activity Level of Slack Structural Row
Activity Activities Activities Check Ratio
3 Y4 Y y2 columns)
0 Y3 1 1/2 1 0 1 1 4 1/2 1 1/2
0 -Y4 900 0 1 400 002101 1 1/8
Z 0 0 0 -1000 -1500 -2500
column/ -900 1/2 0 0 600 700
0 +Y3 3/8 1 -1/800 /2) 0 1499/800 3/4
1500 +Y2 1 1/8 0 1/800 1/2 1 2101/800 9/4
Z 1687 1/2 0 1 7/8 -250 0 11,515/8
column -1688 0 -7/8 250 0
a/ The column check is equal to one minus the sum of the numbers in
If we write the objective function as Z = 1000YI + 1500Y2 and then
put it in implicit form (i.e. -10 Y1 1500Y2 + Z = 0) as the third
equation after the land and labor constraints we can write the
Y1 + Y2+ Y3 = 1.5
A) 400Yi + 800Y2 + Y4 = 900
-1000YI -1500Y2 + Z = 0
If Y1 and Y2 are set to zero we have a very easy first solution
to this system of three equations. The solution is:
Y3 = 1.5
Y4 = 900
This is precisely the solution given in the first section of Table
6. The column titled "activity" represents the variable in the
solution at the respective iterations. In the initial solution
Y3, Y4, and Z are the solution variables. Since we are interested
in,maximizing Z, that variable always remains in the solution and
is never a candidate to be removed in the solution process.
The column headed "level of activity" represents the solution, i.e.,
the values the variables take on at this iteration. The columns
designated "slack activities", Y3 and Y4, and "structural activities",
Y1 and Y2, give the coefficients of the variables in the initial
problem. If you look at the three equation system as it was written
above, you will see that the coefficients of Y, are 1, 400, and
-1000 in the land, labor, and objective function rows, respectively.
This is precisely what the coefficients are in the first section
of the tableau.
The row labeled C.+ at the top simply gives the coefficients of
the respective variables in the original objective function,
i.e. Z = 1000Y1 + 1500Y2 + 0Y3 + 0Y4. The column at the extreme
left labeled C. simply repeats the coefficients of the variable in
the objective function for the corresponding variable in the solution.
The "row check" column and the "column check" row have been introduced
to provide a check on the numerical accuracy. The numbers appearing
in the row check column are simply a summation of all the numbers
appearing in that row excluding the C. and ratio column. For
examples, the 4 1/2 = 1 1/2 + 1 + 1 + 1, and 2101 = 900 + 1 + 400
+ 800. The numbers appearing in the column check row are simply
one minus the sum of the numbers in the corresponding column
excluding the C.+ row. For examples, -900 1/2 = 1 (1 1/2 + 900),
and 600 = 1 (1 + 400 1000). Note that the row check and
g' column check do not work on one another.
Once the initial section of the tableau is set up with row and
column checks we are ready to begin step 2 of the solution process.
We ask the question, "Is there a better solution than Y3 = 1 1/2,
Y4 = 900, and Z = 0?" The simplex criterion answers this question..
This criterion is based on the sign of the coefficients in the
objective function of variables not in the solution. By scanning
the Z row we see that Y1 and Y2 both have negative coefficients,
i.e. -1000 and -1500, respectively. Therefore, if we let Y1 or
Y2 rise above zero the value of Z will increase. One can see
this easily by looking at the third equation of set A on page 47.
If, for example, Y1 were 1 and Y2 remained at zero, Z would increase
from zero to 1000. The most common rule for selecting a variable
to let rise above zero is to choose that one which has the most
negative coefficient. In our problem this is the -1500 for Y2.
The coefficient is enclosed with a rectangle to represent that the
corresponding variable is the one to enter the solution at the next
iteration. The coefficient for a variable, say the jth one, in the
objective function row is called Zj-C. in linear programming jargon.
Therefore Z2-C2 = -1500 is the most negative Z.-C..
We now proceed to step three. Since Y2 has been selected to enter
the solution we must determine which variable will be removed from
the solution, i.e., driven to zero. The variable to be removed is
determined by finding what level of Y2 will drive each variable in
the solution to zero. The variable driven to zero first, i.e.,
the one corresponding to the lowest level of Y2 is the variable
removed. The column titled "ratio" aids this determination. For
each unit or hectare that Y2, is increased, idle land Y3 must be
decreased by a hectare. Therefore, 1 1/2, the solution value of
Y3, divided by one, the coefficient in the Y3 row and Y2 column
is equal to 1 1/2, the value in the ratio column. Similarly, each
hectare of corn interplanted requires 800 hours of labor so
900/800 = 1 1/8 hectares can be grown before all the idle
labor is used up. Since 1 1/8 < 1 1/2, the farmer will run
out of labor before he runs out of land. As a result, if Y2
is increased from 0 in the initial solution to 1 1/8 in the
first iteration, Y4 will be driven to zero. The minimum
ratio, therefore determines which variable will be removed
from the solution in the next iteration. The number appear-
ing at the intersection of the Y2 column and the Y4 row is
called the pivot point. In the first section of the tableau
the pivot point is 800, and to set it off, it has been en-
closed in a circle. This is a very important number in the
calculations which are required for the second section of the
tableau -- the first iteration.
We have now determined that Y4 will leave the solution and
Y2 will enter Therefore, the second section of the tableau
has Y3, Y2, and Z as variables. We are now ready to proceed
with step 4 of the simplex method. This step involves obtain-
ing a basic feasible solution in terms of the new variables.
This solution process reduces to a couple of computational rules
which are rather straightforward to apply. These rules for
obtaining a new solution are as follows:
1. Each number or coefficient in the new (or incoming)
row associated with the incoming variable is equal to the
number in the same column of the old (or outgoing) row of the
outgoing variable divided by the pivot point.
This rule can be expressed by the following formula:
x = ,where
x = the number in the jth column of the row in the
new tableau associated with the incoming variable
(Y2 in the second section of the tableau for our
x = the number in the jth column of the row in the
old tableau associated with the outgoing vari-
able (Y4 in the first section of the tableau
for our example), and
xtk = pivot point (800 in the first section of the
tableau of our example).
Applying this rule, the coefficients in the Y2 row of the
second section of the tableau are computed as follows:
1 1/8 = 900/800, 0 = 0/800, 1/800 = 1/800, 1/2 = 400/800,
1 = 800/800, and 2101/800 = 2101/800.
You will note that the now "row chock" value is computed just
as. any other number. The check is made by reading the num-
bers in the row to see if they agree with the new row check
figure. If the same number is obtained by the two different
methods of computing it, one can be fairly certain that there
is no error. To check we verify that
1 1/8 + 0 + 1/800 + 1/2 + 1 = 2101/800.
2. Every other coefficient in the new section of the
tableau (not in the incoming row) is equal to the coefficient
in the incoming row and same column times the coefficient at
the intersection of the old (outgoing) column associated with
the incoming variable and the same row of the old section sub-
tracted from the coefficient in the same location in the old
section. This rule can be put as the following formula:
xi* xij xik J
Sij Xik xj where
x.. = number in a particular row, the ith, and column,
the jth,of the new section of the tableau,
x.. = number in the same location in the old section of
Xik = number at the intersection of the ith row and the
column in the old section of the tableau associated
with the incoming variable, and
x = as defined under rule 1.
As a demonstration of how to use rule 2 let us compute the
numbers in the Y3 row of the second section of the tableau.
The new value for Y3 is equal to the old value minus the
product of the new value of Y2 and the number in the Y3 row
and Y2 column of the old tableau, i.e. applying the formula:
3/8 = 1 1/2 1 *1 1/8.
The other values in the row follow directly from the formula
1 = 1 1 0
-1/800 = 0 1 1/800
1/2 = 1 1 1/2
0 = 1 1 1
1499/800 = 4 1/2 1 2101/800
The values for the Z row are computed with the same formula
as is the values for the column check row. The column check
row can also be computed as one minus the sum of the other
elements in the column. If both ways of computing the col-
umn check give the same answer, one can feel rather confident
that no error has been made.
When the second section of the tableau is completed using
rules 1) and 2), we again apply the simplex criterion (step
5). If all the coefficients of variables not in the solution
in the objective function row (the (Zj C.)'s) are positive,
we have an optimal solution. If not, at least one more itera-
tion is required. In the example problem in the second sec-
tion of the tableau, we see that (z1 C1), the coefficient
in the objective function, Z, row and the Y1 column is equal
to -250. Therefore, the simplex criterion indicates that we
are not at the optimum solution ahd we must go through at
least one more iteration. Yi will enter the solution in the
second iteration and Y3 will leave because the ratio for
Y3 = 3/4 < 9/4 = ratio for Y2 Therefore in the third
section of the tableau we have a solution with Y1, Y2, and Z.
The numbers in the third section of the tableau are then com-
puted using rules 1 and 2 for obtaining a new solution.
The Z. C. in the final tableau are all positive so the
simplex criterion indicates that we are at the optimal solu-
tion, which is:
Y1 = 3/4
Y2 = 3/4
Z = 1875
Earlier we indicated that the solution of the primal problem
also gave a solution to the dual problem. The coefficients
in the Z row of section three of the tableau give the solution
to the dual problem. That is
P = 500 = Z3-C3 for Y3 ,
P = 1 1/4 = Z-C4 for Y4 and
Z = 1875
Therefore, when we solve the resource allocation problem, we
also solve the resource valuation problem.
Besides the row and column checks there are a few other checks
one can apply during or after the solution process. These are
1) With each iteration the objective function must be
at least as good as it was before. The value of the objective
function can never deteriorate -- i.e., get smaller for a max-
imization problem or larger for a minimization problem.
2) At each iteration the solution obtained must satisfy
all constraints on the problem.
3) Since the values of the objective function for the
primal and dual problems are identical, one can check the value
of the objective function for the primal with the objective
function of the dual. That is the sum of the Zj Cj of re-
strictive resources in a resource allocation problem times
the level of the resources must equal the value of the objec-
tive function for the resource allocation problem. In the
PX P X1 + P X2 = 1875
(500) (1 1/2) + (1 1/4)(900) = 1875
Interpretation of Z* C.
1) As we have seen above, the Zj C. of restrictive
resources represent what resources are worth per unit at the
margin of production. They are linear programming measures
of the value of the marginal product. They are the shadow
prices of the resources. They represent the solution for
the structural variables of the dual problem.
If a resource is in surplus, i.e. its slack variable is in
the solution, it will have a zj C. = 0.
2) The Zj C. of a structural variable (productive
enterprise or activity) in the solution represents the amount
the objective function coefficient ("price") for the variable
can be increased before the variable would enter the solution.
These values provide useful information about the stability
of the solution. These .j C.) are also the values of the log-
ical variables (slack and surplus) of the dual problem.
If a structural variable is in the solution its Z. C. = 0.
Linear Programming and the Principles of Production
With the above very brief introduction to linear programming,
we return to the principles of production to more clearly
relate the two approaches. Earlier we had discussed the con-
cept of a production function relating elemental nitrogen
applications to rough rice yields (Figure 1). If we want to
represent different fertilizer rates in a linear programming
problem we need to introduce a separate variable for each
rate. For example, in a particular linear programming prob-
lem we may have several variables representing rice production
each at a different rate. Yi may be rice production with 25
kgs. of elemental N per hectare, Y2 50 kgs. per hectare, etc.
For any one of these variables the N and rough rice will
combine in fixed proportions but variable proportions can be
represented by adding the other variables. A variable in a
linear programming problem therefore represents one point on
a production function. More points can be represented by
adding more variables. Indeed one can approximate a produc-
tion function quite closely in a linear programming model by
adding more variables representing different rates of use of
the input. The situation is as depicted in Fiqure 8.
In a similar manner more than one variable can be introduced
to approximate an isoproduct curve. Again using the three
variables depicting three fertilization rates in Figure 8,
we have the following approximation to the production function
Elemental N (kgs. per ha.) Rough Rice (kgs./ha.)
If we depict these three rates of fertilization as a func-
tion of the two inputs land and elemental nitrogen we get the
situation depicted in figure 9. The isoproduct curve for
2000 kgs. of rice is approximated as ABC. If more variables
were introduced a closer approximation to a curvilinear iso-
product curve could be obtained.
The point we wish to make is that linear programming models
can be used to represent diminishing returns to an input and
the fact that inputs and outputs combine in fixed proportions
Figure 8. Relationship Between a Variable or Activity in a
Linear Programming Model and the Production Function
Y3, rice fertilize(
at 75 kg/ha.
19-- Y2, rice fertilized
at 50 kgs./ha
17t at 25 kg
~-the input-output relationship
assumed by variable Y3 in the
linear programming model
5 10 15 20 25 30 35
40 45 50 55 60 65 70 75
Input of Elemental Nitrogen (kgs./ha.)
Figure 9. Combinations of Land and Nitrogen Fertilizer Required
to Produce 2000 Kilograms of Rough Rice
Y3, rice fertilized at
,Y2, rice fertilized
at 50 kgs./ha.
isoproduct curve for
2000 kgs. of rough rice
Y1, rice fertilized
at 25 kgs./ha.
for a particular linear programming variable in no way re-
stricts us from adding more variables to depict other pro-
portions. 6-/ The linear programming model will select the
variable or variables that yields the greatest value of the
In a competitive environment the theory of the firm (principles
of production) takes prices as given and asks how much of the
variable input should be utilized in conjunction with the in-
puts held fixed and/or it asks how much of the product should
be produced given the production function, the prices of the
inputs and the prices of the outputs. For given prices of the
output(s),it asks how much of the inputs will be demanded at
various prices of the inputs, and for givenprices of the inputs
it asks how much of the outputs will be supplied at various
prices of the outputs. The approach is outward looking into
Linear programming uses the same prices and production func-
tion information but approaches the answer to the production
decision problem from the point of view of the firms fixed
6/ A word of caution. It is not possible to depict a
production function showing increasing returns to a variable
input in a linear programming model. without some special ac-
commodations. Technically the problem arises because the area
of feasible solution is not a convex set. The increasing re-
turns problem can be solved using integer programming or by
breaking the problem up into several linear programming prob-
lems--each with a convex set of feasible solutions.
resources. Given the market environment, it asks how best to
use the firms fixed resources to maximize profits. By para-
metric programming the supply of a resource or the price of
a product, a linear programming model can also get at the
question of the firm's demand for an input or the firm's
supply of a product. Though the two approaches attempt to
resolve the firm's decision problems from different perspec-
tives, they do arrive at essentially the same conclusions
given a well defined production situation.
Linear programming has an advantage for empirical applications
since the model is more efficient data wise (a model of sorts
can be defined with as few as one observation) and numerical
solutions with the aid of electronic computers are quite well
developed for rather large problems.
VI. A Concluding Remark and Further References
The principles of production provide considerable insight into
the economic behavior of the producing firm. An understanding
of these principles is essential for the serious student of
farm management. The intent of this chapter has been to pro-
vide a general overview of some of the basic principles
through introducing examples and exercises from the Philippines.
The references cited below will permit the interested student
to further develop his understanding of these basic production
1. Buse, Rueben C., and Daniel W. Bromley, Applied Economics,
Resource Allocation in Rural America. Ames: Iowa State
University Press, 1975.
Although this book carries a reference to rural America
in its subtitle, the principles and ideas discussed are
applicable to the rural sector of many Asian countries. The
book is clearly written.
2. Doll, John P., and Frank Orazem, Production Economics,
Theory with Applications. Columbus, Ohio: Grid, Inc.,
This book is written in a style which is somewhat similar
to this chapter. The book first develops the principles of
production and then introduces linear programming and some of
the ideas of modern decision theory. Appendices give a brief
introduction to some basic mathematics and to nonlinear pro-
3. Gass, Saul I., Linear Programming, 4th edition, New York:
McGraw-Hill Book Company, 1975.
This is a rather complete book on linear programming at
an intermediate to advanced level. It is written by a person
who has an interest in agricultural applications of the methods.
4. Heady, Earl 0., Economics of Agricultural Production and
Resource Use. Englewood Cliffs, N.T.: Prentice Hall, Inc.,
This is an internationally recognized book on the economics.
of agricultural production. The ideas are presented in a way
that is understandable to the beginning student and it is
thorough in its treatment of the topic.
5. Heady, Earl O., and Wilfred Candler. Linear Programming
Methods. Ames: The Iowa State University Press, 1958.
This is an old but good introduction to the methods of
linear programming. The examples used in presenting the
methods are from zjriculturc.
6. Sposito, V.A., Linear and Nonlinear Programming. Ames:
The Iowa State University Press, 1975.
This is another good book that focuses more on the theory
of linear programming than the methods. It is a good refer-
ence at the intermediate to advanced level.