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 vellayani2

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).Department of Agricultural Statistics, COA vellayani 3

Role of optimization techniques in agriculture


A schematic view of modeling/optimization

process

Solution to model

To the real-world problem

assumptions, abstraction,data,simplifications

optimization algorithm

interpretation

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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:


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.


What do we optimize?

A real function of n variables

with or without constrains

),,,( 21 nxxxf


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


An Optimization Problem

Objective Function

Variables

Constraintscom

pone

nts

Once the design variables, constraints, objectives and the relationship between them have been chosen,the optimization problem can be defined.


nx

xx

x..2

1

xf

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Statement of an optimization problem

An optimization problem can be stated as follows:

To find

Subject to the constraints

which minimizes or maximizes


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


Department of Agricultural Statistics, COA vellayani

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.


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 pointNegative definite when X* is a relative maximum point

Multivariable optimization with no constraints

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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?



Profit is:

plan.supply maximizing -profit theis (3,4) definite negative is

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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

nx

xx

2

1

X



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



mjgL

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jj

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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.



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X*

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)



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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 have2y = 3λ ; λ = (2/3)y2x = 4λ ; λ = (1/2)x3x + 4y = 90 (2/3)y = (1/2)x4y = 3x Y = (3/4) x


3x + 4 * (3/4) x = 906x = 90

x = 15 units these are the values or x and y at which the Lagrange function is optimized.

45 + 4y = 904y = 90-454y = 45y= 11.25 units λ = 7.5

33


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LINEAR PROGRAMMING


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

Formulation

Solution

Interpretation and What-if Analysis

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.38Department of Agricultural Statistics, COA vellayani


• only two decision variables• • provide visualization

• large, real world LPP• most efficient and popular

method

GRAPHICAL METHOD

LIN

EAR

PRO

GRAM

MIN

GSEN

SITIVE ANALYSIS

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.


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.


What is Linear Programming?


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 16B 1 2 12C 4 2 28

Profit per unit 5 8


44

2

22

2

1

4Product 1

Product 2

Decision Variables:• x 1: Product 1 Units to be produced weekly• x 2: Product 2 Units to be produced weekly

Objective Function: Maximize 21 85 xx

Subjected to :

02824

12211622

2,1

21

21

21

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02,1 xx


01622

2,1

21

xxxx


012211622

2,1

21

21

xxxxxx


02824

12211622

2,1

21

21

21

xxxxxxxx

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 3nutrient A 20 30 70 110nutrient B 10 10 0 18nutrient C 50 30 0 90nutrient D 6 2.5 10 14

cost in Rs/- 41 36 96solver


321 963641 xxxz

110703020 321 xxx






SENSITIVITY ANALYSIS


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.


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.


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 in creased 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


66

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


1 UNIT

1 UNIT

A

B


Minimize 21 46 xxz

subjectd to

0601051005153002510

3,2,1

21

21

21

xxxxxxxxx

Education

Role of optimization techniques in agriculture jaslam