Role of optimization techniques in agriculture jaslam

Department of Agricultural Statistics, COA vellayani 1
MUHAMMED JASLAM P K
2015-19-005
IInd year MSc. Agricultural Statistics

Department of Agricultural Statistics, COA vellayani
2

The only way to meet increasing demand of food,
fibre and fuel for the ever increasing population is by
increasing production per unit area which is possible by
more scientific utilization of the resources and their
optimal allocation to achieve maximum returns
(Hassan et al. 2015).

4

A schematic view of
modeling/optimization process
assumptions, abstraction,data,simplifications
optimizationalgorithm
interpretation
makessense?
changethemodel,
assumptions?
Real-world
problem

Decision making???
Decision making always involves making a choice between
various possible alternatives
production scheduling
vehicle routing and scheduling
feed mix
product mixes
fertilizer mix
Examples:
6

What is a model?
Model: A schematic description of a system, theory, or
phenomenon that accounts for its known or inferred
properties and maybe used for further study of its
characteristics.
Mathematical models
– abstract models
– describe the mathematical relationships
among elements in a system
Mathematical models are cheaper, faster, and safer than constructing
and manipulating real systems.
7

What do we optimize?
A real function of n variables
with or without constrains
),,,( 21 n
xxxf 
8

Optimization is the act of obtaining the best result under given
circumstances.
Optimization can be defined as the process of finding the
conditions that give the maximum or minimum of a function.
The optimum seeking methods are also known as mathematical-
programming techniques
Optimization
9

An Optimization Problem
Objective Function
Variables
Constraints
components
Once the design variables, constraints, objectives and the relationship between
them have been chosen,the optimization problem can be defined.

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mi ,....,3,2,1
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mj ,....,3,2,1
Statement of an optimization problem
An optimization problem can be stated as follows:
To find
Subject to the constraints
which minimizes or maximizes
11

Optimization
methods
constraints
Optimization
without
constraints
Optimization
with constraints
Type of solved
problem
Linear
programming
Non Linear
Optimization
Classification of Optimization methods

Classical optimization techniques
Single variable optimization
• Useful in finding the optimum solutions of continuous and
differentiable functions
• These methods are analytical and make use of the techniques
of differential calculus in locating the optimum points.
• Since some of the practical problems involve objective
functions that are not continuous and/or differentiable, the
classical optimization techniques have limited scope in
practical applications.

Necessary condition
a point x* at which f’(x*)=0 is called a stationary point.
If a function f (x) is defined in the interval a ≤ x ≤ b and has a
relative minimum at x = x*, where a < x* < b, and if the
derivative df (x) / dx = f’(x) exists as a finite number at x = x*,
then f’ (x*)=0

Sufficient condition
• Let f’(x*)=f’’(x*)=…=f (n-1)(x*)=0, but f(n)(x*) ≠ 0. Then f(x*) is
– A minimum value of f (x) if f (n)(x*) > 0 and n is even
– A maximum value of f (x) if f (n)(x*) < 0 and n is even
– Neither a minimum nor a maximum if n is odd

16
A Citrus grower estimates that if 60 orange trees are
planted; the average yield per tree will be 400 oranges. The
average yield will decrease by 4 oranges per tree for each
additional tree planted on the same acreage. How many
trees should the grower plant to maximize the total yield?

To maximize! Let’s find the critical numbers:
Y’ (n) = 160 − 8n = 0 n = 160 / 8 = 20 is the only critical number.
Moreover, Y” (n) = −8 Y” (20) = −8 < 0.
By the second derivative test, Y has a local maximum at n = 20, which is an
absolute maximum since it is the only critical number.
Let n= the number of additional trees.
Y= the total yield = number of trees × the yield per tree.
Then:
Y (n) = (60trees + n · trees) (400oranges − n · 4oranges) = (60 + n) (400 − 4n)
= 24, 000 + 160n − 4n2
17
The grower should plant 60+20 = 80 trees to maximize the total yield.

A landscape architect plans to enclose a 3000 square foot rectangular
region in a botanical garden; she will use shrubs costing Rupees 15 per
foot along three sides and fencing costing Rupees 10 per foot along the
fourth side, find the minimum total cost.

If the rectangular region has dimensions x and y, then its area is A = x*y = 3000ft2.
So y = 3000/x.
If y is the side with fencing costing Rupees 10 per foot, then the cost for this side is
Rupees 10 y.
The cost for the three other sides, where shrubs costing Rupees 15 is used, is then
Rupees 15 (2x+y).
Therefore the total cost is: C(x) = 10y + 15(2x + y) = 30x + 25y.
Since y = 3000/x,
Then C(x) = 30x + 25 * (3000/x) that we wish to minimize.

20
Since C’(x) = 30 – 25(3000/x2), then C’(x) = 0 for x2 = (25 * 3000)/30 =
2500.
Therefore, since x is positive, we have only one critical number in the
domain which is x = 50ft.
Since C”(x) = 25*(6000/x3), we have C” (50) > 0. Thus, by the 2nd
derivative test, C has a local minimum
At x = 50, and therefore an absolute minimum because we have only
one critical number in the domain.
Hence, the minimum cost is C (50) = 3000, with the dimensions x = 50 ft
and y = 3000/50 = 60 ft.

Necessary condition
If f(X) has an extreme point (maximum or minimum) at X=X* and
if the first partial derivatives of f (X) exist at X*, then
Sufficient condition
A sufficient condition for a stationary point X* to be an extreme
point is that the matrix of second partial derivatives (Hessian matrix)
of f (X*) evaluated at X* is
Positive definite when X* is a relative minimum point
Negative definite when X* is a relative maximum point
0*)(*)(*)(
21
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x
f


Example (Discriminating Monopolist)
23
 A monopolist producing a single output has two types of
customers. If it produces q1 units for type 1, then these customers
are willing to pay a price of 50-5q1 per unit. If it produces q2 units
for type 2, then these customers are willing to pay a price of 100-
10q2 per unit.
 The monopolist’s cost of manufacturing q units of output is
90+20q.
 In order to maximize profits, how much should the monopolist
produce for each market?

24
 Profit is:
plan.supplymaximizing-profittheis(3,4)definitenegativeis
.0,20,10
.402020100
q
f
,30201050
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arepointscriticalThe
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Multivariable optimization with equality
constraints
• Problem statement:
Minimize f = f (X) subject to gj(X)=0, j=1,2,…..,m where
Here m is less than or equal to n, otherwise the problem becomes
overdefined and, in general, there will be no solution.
• Solution:
– Solution by direct substitution
– Solution by the method of Lagrange multipliers
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Solution by direct substitution
• Simple in theory
• Not convenient from a practical point of view as the
constraint equations will be nonlinear for most of the
problems
• Suitable only for simple problems

Necessary conditions for a general problem:
Minimize f(X)
subject to
gj (X)= 0, j=1, 2,….,m
The Lagrange function, L, in this case is defined by introducing one
Lagrange multiplier j for each constraint gj(X) as
Solution by Lagrange multipliers
)()()()(),,,,,,,( 22112121 XXXX mmmn gggfxxxL   

By treating L as a function of the n+m unknowns, x1, x2,…,xn,1, 2,…, m,
the necessary conditions for the extremum of L, which also corresponds
to the solution of the original problem are given by:
The above equations represent n+m equations in terms of the n+m
unknowns, xi and j
mjg
L
ni
x
g
x
f
x
L
j
j
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j
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The solution:
The vector X* corresponds to the relative constrained
minimum of f(X) (sufficient conditions are to be verified)
while the vector * provides the sensitivity information.
29
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Sufficient Condition
A sufficient condition for f(X) to have a constrained relative minimum at X* is
that the quadratic Q defined by
evaluated at X=X* must be positive definite for all values of dX for which the
constraints are satisfied.
If
is negative for all choices of the admissable variations dxi, X* will be a constrained
maximum of f(X)
ji
n
j ji
n
i
dxdx
xx
L
Q   
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1
2
1
ji
n
j ji
n
i
dxdx
xx
L
Q )(
1
2
1
*X*,  
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

Utility maximization from consumption of two goods x and y with
the constraint of total income available (I) - 90 rupees and prices of
these goods (p1)- 3 rupees for good x and (p2)- 2 rupees for good y
with parallel solution.
Max x,y f(x,y) subject to g(x,y) = I

Utility (objective function) = U = u(x,y) = 2xy
Budget (constraint) = I = g(x,y) , I = p1x + p2y , 90 = 3x + 4y
L = UL = U(x,y) +λ ( I – g(x,y)) L =2xy +λ (90 – 3x -4y)
Lx = dL / dx = df(x,y)/dx – λ[dg(x,y)/dx] = 0 Lx = dL / dx = 2y – 3λ = 0
Ly = df(x,y)/dy – λ[dg(x,y)/dx] = 0 Ly = dL / dy = 2x – 4λ = 0
Lλ = dL/dλ = 1 – g(x,y) = 0 Lλ = dL / dλ = 90 – 3x – 4y = 0

from these three equations we have
2y = 3λ ; λ = (2/3)y
2x = 4λ ; λ = (1/2)x
3x + 4y = 90
(2/3)y = (1/2)x
4y = 3x Y = (3/4) x
3x + 4 * (3/4) x = 90
6x = 90
x = 15 units
these are the values or x and y at which the Lagrange function
is optimized.
45 + 4y = 90
4y = 90-45
4y = 45
y= 11.25 units λ = 7.5
33

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=48 >0

36
Mathematical programming is used to find the best or
optimal solution to a problem that requires a decision
or set of decisions about how best to use a set of
limited resources to achieve a state goal of objectives.
A mathematical tool for maximizing or
minimizing a quantity (usually profit or cost of
production), subject to certain constraints.
Of all computations and decisions made by
management in business, 50-90% of those
involve linear programming.

– Conversion of stated problem into a mathematical model that abstracts all
the essential elements of the problem.
– Exploration of different solutions of the problem.
– Finding out the most suitable or optimum solution.
Steps involved in mathematical programming
37Department of Agricultural Statistics, COA vellayani

The Linear Programming Model (1)
Let: X1, X2, X3, ………, Xn = decision variables
Z = Objective function or linear function
Requirement:- Maximization of the linear function Z.
Z = c1X1 + c2X2 + c3X3 + ………+ cnXn
subject to the following constraints:
where aij, bi, and cj are given constants.

• only two decision variables
•
• provide visualization
• large, real world LPP
• most efficient and popular
method

Examples of LP Problems (1)
1. A Product Mix Problem
• The decision maker wishes to produce the
combination of products that will maximize total
income.
40
E.g. Optimum crop-mix.
2. A Blending Problem
• The problem is to determine how much of each commodity
should be purchased and blended with the rest so that the
characteristics of the mixture lie within specified bounds and the
total cost is minimized.
. E.g. Fertilizer mix, feed mix or diet problem.

Examples of LP Problems (3)
3. A Production Scheduling Problem
The problem is to determine the production schedule that
minimizes the sum of production and storage costs.
4. A Transportation Problem
The problem is to determine the amount to be shipped
from each origin to each destination such that the total cost
of transportation is a minimum.
5. A Flow Capacity Problem
The problem is to determine the maximum flow, or capacity
of the network.

42
What is Linear Programming?

43
A workshop has three (3) types of machines A, B and C; it can
manufacture two (2) products 1 and 2, and all products have to go to
each machine and each one goes in the same order; First to the
machine A, then to B and then to C. The following table shows:
Formulate and solve using the graphical method a Linear Programming model for
the situation that allows the workshop to obtain maximum gains.
Type of Machine Product 1 Product 2 Available hours per week
A 2 2 16
B 1 2 12
C 4 2 28
Profit per unit 5 8

44
2
2
2
2
1
4
Product 2
Decision Variables:
• x1: Product 1 Units to be produced weekly
• x2: Product 2 Units to be produced weekly
Objective Function:
Maximize 21 85 xx 
Subjected to :
0
2824
1221
1622
2,1
21
21
21




xx
xx
xx
xx

02,1 xx

0
1622
2,1
21


xx
xx

0
1221
1622
2,1
21
21



xx
xx
xx

0
2824
1221
1622
2,1
21
21
21




xx
xx
xx
xx
21 85 xx Z=
@(0,6) z=40
@(7,0) z=35
@(6,2) z=46
@(4,4) z=52

International Wool Company operates a large farm on which sheep are raised. The
farm manager determined that for the sheep to grow in the desired fashion, they
need at least minimum amounts of four nutrients (the nutrients are nontoxic so the
sheep can consume more than the minimum without harm). The manager is
considering three different grains to feed the sheep. Table lists the number of units
of each nutrient in each kg of grain, the minimum daily requirements of each
nutrient for each sheep, and the cost of each grain. The manager believes that as
long as a sheep receives the minimum daily amount of each nutrient, it will be
healthy and produce a standard amount of wool. The manager wants to raise the
sheep at minimum cost.
Grain Min. Daily
req.1 2 3
nutrient A 20 30 70 110
nutrient B 10 10 0 18
nutrient C 50 30 0 90
nutrient D 6 2.5 10 14
cost in Rs/- 41 36 96

50
321 963641 xxxz 
110703020 321  xxx

 Optimal solutions to LP problems have been examined
under deterministic assumptions.
 Conditions in most real world situations are dynamic and
changing.
 After an optimal solution to a problem is found, input data
values are varied to assess optimal solution sensitivity.
 This process is also referred to as sensitivity analysis or post-
optimality analysis.
57

 Sensitivity analysis determines the effect on optimal solutions
of changes in parameter values of the objective function and
constraint equations
 Changes may be reactions to anticipated uncertainties in the
parameters or the new or changed information concerning the
model
58
Eg. OPTIMUM CROP MIX

Sonmez and Altin (2004) developed a linear programming model
for irrigation scheduling and cropping pattern with adequate and
deficit water supply in Harrran plain, Turkey. It was found that even
with very low water supply, it is possible to keep the farm income at
high levels.
John and Nair (1998) worked out an optimal
Integrated Farming System (IFS) model through linear
programming for small farmers (0.2 ha and less)
comprising of 43 enterprises with a cropping
intensity of 161 per cent and a cost benefit ratio of
1:2.5.

63
Subhadra (2007) conducted a study to identify the optimum
activity mix of dairy enterprise and crop production to enhance farm
income with the given resource use efficiency and technology in
Thrissur and Palakkad districts of Kerala. It was found that net income
of different farm size groups could be enhanced in between Rs.4,275
to Rs.15,252 by adding two animals to large and small farmers each
and three animals to marginal farmers.
Dey (2011) applied linear programming to study on optimum
allocation of vegetable crops in Kakdwip block of South Parganas
district in West Bengal. In the optimal crop plan, resources were
allocated in favour of brinjal and pointed gourd. Net return earned from
optimal crop plan increased by 49.79 percent over the net return
earned in existing crop.

64
Traditionally, judgment based on experience has been the basis for
planning in agriculture, but increased specialization and the adoption of
capital intensive production systems have stimulated the development of
more formal planning methods based on the construction and analysis of a
mathematical model.
Once a solution to the model has been derived and tested, the solution
can be implemented and its performance is monitored and controlled.
Mathematical modeling is quicker and less expensive than using the trial-
and-error approach or constructing and manipulating real systems.
CONCLUSION

A Diet is to contain 300 units of carbohydtrates ,100 units of fat 60 units of protien,
Two foods A&B are available
10 units of carbohydrate
20 units of Fat
15 units of protien
25 units of carbohydrate
10 units of fat
20 units of protien.
Formulate and solve LPP so as to find the minimized cost for diet that Consist of
mixture of these two foods and also meet the minimum nutrient requirements.
Cost of food A is Rs 6/unit and B is Rs 4/unit
66
1 UNIT
1 UNIT
A
B

67
Minimize 21 46 xxz 
subjectd to
0
60105
100515
3002510
3,2,1
21
21
21




xxx
xx
xx
xx

Role of optimization techniques in agriculture jaslam

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