Applied Statistics In Business And Economics,Doane,
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All about Convex Optimization
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Course
Applied Statistics In Business And Economics,Doane,
Institution
Harvard University
This tutorial will introduce various concepts involved in non-linear optimization. Linear programming problems are very easy to solve but most of the real world applications involve non-linear boundaries. So, the scope of linear programming is very limited. Hence, it is an attempt to introduce the ...
Applied Statistics In Business And Economics,Doane,
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Convex Optimization
In this Notes, the students will learn to solve the optimization problems like minf(x)
subject to some constraints.
These problems are easily solvable if the function f(x)
is a linear function and if the constraints are linear. Then it is called a linear programming problem
LPP.
But if the constraints are non-linear, then it is difficult to solve the above problem. Unless we can
plot the functions in a graph, then try to analyse the optimization can be one way, but we can’t plot a
function if it’s beyond three dimensions. Hence there comes the techniques of non-linear
programming or convex programming to solve such problems. In these tutorial, we will focus on
learning such techniques and in the end, a few algorithms to solve such problems. first we will bring
the notion of convex sets which is the base of the convex programming problems. Then with the
introduction of convex functions, we will some important theorems to solve these problems and some
algorithms based on these theorems.
Terminologies
The space Rn
− It is an n-dimensional vector with real numbers, defined as follows −
Rn={(x1,x2,…,xn)τ:x1,x2,….,xn∈R}
The space RmXn
− It is a set of all real values matrices of order mXn.
, Convex Optimization – Linear
Programming
Methodology
Linear Programming also called Linear Optimization, is a technique which is used to solve
mathematical problems in which the relationships are linear in nature. the basic nature of Linear
Programming is to maximize or minimize an objective function with subject to some constraints.
The objective function is a linear function which is obtained from the mathematical model of the
problem. The constraints are the conditions which are imposed on the model and are also linear.
From the given question, find the objective function.
find the constraints.
Draw the constraints on a graph.
find the feasible region, which is formed by the intersection of all the constraints.
find the vertices of the feasible region.
find the value of the objective function at these vertices.
The vertice which either maximizes or minimizes the objective
function accordingtothequestion������������ℎ��������� is the answer.
Examples
Step 1 − Maximize 5x+3y5�+3� subject to
x+y≤2�+�≤2,
3x+y≤33�+�≤3,
x≥0andy≥0�≥0����≥0
Solution −
The first step is to find the feasible region on a graph.
,Clearly from the graph, the vertices of the feasible region are
(0,0)(0,2)(1,0)(12,32)(0,0)(0,2)(1,0)(12,32)
Let f(x,y)=5x+3y�(�,�)=5�+3�
Putting these values in the objective function, we get −
f(0,0)�(0,0)=0
f(0,2)�(0,2)=6
f(1,0)�(1,0)=5
f(12,32)�(12,32)=7
Therefore, the function maximizes at (12,32)(12,32)
Step 2 − A watch company produces a digital and a mechanical watch. Long-term projections
indicate an expected demand of at least 100 digital and 80 mechanical watches each day. Because of
limitations on production capacity, no more than 200 digital and 170 mechanical watches can be
made daily. To satisfy a shipping contract, a total of at least 200 watches much be shipped each day.
, If each digital watch sold results in a $2$2 loss, but each mechanical watch produces a $5$5 profit,
how many of each type should be made daily to maximize net profits?
Solution −
Let x� be the number of digital watches produced
y� be the number of mechanical watches produced
According to the question, at least 100 digital watches are to be made daily and maximaum 200
digital watches can be made.
⇒100≤x≤200⇒100≤�≤200
Similarly, at least 80 mechanical watches are to be made daily and maximum 170 mechanical
watches can be made.
⇒80≤y≤170⇒80≤�≤170
Since at least 200 watches are to be produced each day.
⇒x+y≤200⇒�+�≤200
Since each digital watch sold results in a $2$2 loss, but each mechanical watch produces
a $5$5 profit,
Total profit can be calculated as
Profit=−2x+5y������=−2�+5�
And we have to maximize the profit, Therefore, the question can be formulated as −
Maximize −2x+5y−2�+5� subject to
100≤x≤200100≤�≤200
80≤y≤17080≤�≤170
x+y≤200�+�≤200
Plotting the above equations in a graph, we get,
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