MONTE CARLO NUMERICAL METHOD

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MONTE CARLO NUMERICAL METHOD

• BY
• A. Lecture KARRAR DH. MOHAMMED

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History of the Monte Carlo Method

• The Monte Carlo name is taken from the Monte
  Carlo, the capital of Monaco which was a center
  for gambling .

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• Monte Carlo methods are used in many different
  ways e.g. as a technique of integration of a
  function. One of this ways is
• The Monte Carlo method as a numerical
  technique for calculating the integral by using
  sequences of random numbers.

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• In mathematics, Monte Carlo integration is
  numerical integration using random numbers. That
  is, Monte Carlo integration methods are
  algorithms for the approximate evaluation of
  definite integrals, usually multidimensional
  ones.

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• The usual algorithms evaluate the integrand at a
  regular grid.
• Monte Carlo methods, however, randomly choose
  the points at which the integrand is evaluated,
  informally, to estimate the area of a domain D,

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• First, pick a simple domain E whose area is
  easily calculated and which contains D.
• Now pick a sequence of random points that fall
  within E.
• Some fraction of these points will also fall
  within D.

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• The area of D is then estimated as this fraction
  multiplied by the area of E.
• The traditional Monte Carlo algorithm distributes
  the evaluation points uniformly over the
  integration region .

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Why are Monte Carlo techniques used ?

• Two of the main reasons why we use Monte Carlo
  methods
• their ant-aliasing properties, and
• their ability to approximate quickly an answer
  that would be very time-consuming to find out.

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• if we want to use a method to determine the exact
  answer, Monte Carlo method can be used to
  simulate problems that are too difficult and
  time-consuming.

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• An example is in the use of Monte Carlo
  techniques in integrating very complex
  multidimensional integrals.
• This is a task that other processes can not
  handle well, but which Monte Carlo can.

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• The first point refers to the fact that since
  Monte Carlo methods involve a random component in
  the algorithm, then they go some way to avoid the
  problems of ant-aliasing (only for certain
  applications).

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• An example that was brought to my attention was
  that of finding the area of the black squares on
  a chess board. Now, if I was using an
  acceptance-rejection method to attack this
  problem,
• I should come out with a fair approximation,

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• Due to the fact that I would be going to random
  points on the chessboard, I will do the same
  process by using an algorithm that moved to a
  certain next point in a set of distance away and
  then continued to do this,

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• thus not having any random point selection, the
  potential problem is that this person may have a
  bad step size and may over evaluate or under
  evaluate the number of successful trials he has,
  thus inevitably giving a poor approximation.

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• These are two solid reasons why people use Monte
  Carlo techniques. Other possible reasons could
  include its ease in simulating complex physical
  systems in the fields of physics, engineering and
  chemistry

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Transformation Function

• T c,d 0,1
• is defined by -
• T (x) (x-c) / (d-c)

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Generate a random number x uniform in X min,X
max

• Step 1
• A random number generated by the following
  relation -
• ni (kni) mod m . such that
• ni integer number
• k multiplier number
• m modulus number
• no initial value

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• step 2
• ri ni /m in0,1 where
• ni n1,n2 ,n3,
• m modulus
• step 3
• Xi (random) X min ri (X max X min)

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Crude Monte Carlo

• We are going to use this technique to solve the
  integral I.
• I
• The method is that you take a number N for random
  number where c xi d then for all xi we find
  the function
• f (xi) .

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• For the f(x) we sum all of these values and
  divide by N to get the mean value for the random
  number. We then multiply this value by the
  interval (b-a) to get the integral by
• I

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EXAMPLE

• Solve the following integral by using Monte Carlo
  method
• I
• Solution- we use the previous steps to get the
  solution

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Step 1-let no1,k3,m7

• We compute the random numbers (n) as follows-
• nik (ni-1) mod 7
• n1(31) mod 7 3
• n2(33) mod 7 2
• n3(32) mod 7 6
• n4(36) mod 7 4

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• n5 (34) mod 7 5
• n6 (35) mod 7 1
• n7 (31) mod 7 3 ... Stop
• Number of random m-17-16

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Step 2-by the following we shall compute ri

• rini/m
• r13/70.42
• r22/70.28
• r36/70.85
• r44/70.57
• r55/70.71
• r61/70.14

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Step 3-now we shall compute xi as following

• Xi xminri(xmax-xmin)
• X110.42(2)1.84
• X210.28(2)1.56
• X310.85(2)2.7
• X410.57(2)2.14
• X510.71(2)2.42
• X610.44(2)1.28

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F(xi)sin(xi) xi ri Ni
0.032 1.84 0.42 1
0.027 1.56 0.28 2
0.047 2.7 0.85 3
0.0373 2.14 0.57 4
0.0422 2.42 0.71 5
0.0233 1.28 0.14 6

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• I
• I (1/6)(0.207)0.0345

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Example

• Solve the following integral by using Monte Carlo
  method
• I
• Solution- we use the same steps in the previous
  example to get the solution

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Step 1-let no1,k3,m7

• We compute the random numbers (n) as follows-
• nik (ni-1) mod 7
• n1(31) mod 7 3
• n2(33) mod 7 2
• n3(32) mod 7 6
• n4(36) mod 7 4

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• n5 (34) mod 7 5
• n6 (35) mod 7 1
• n7 (31) mod 7 3 ... Stop
• Number of random

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• n5 (34) mod 7 5
• n6 (35) mod 7 1
• n7 (31) mod 7 3 ... Stop
• Number of random m-17-16

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Step 2-by the following we shall compute ri

• rini/m
• r13/70.42
• r22/70.28
• r36/70.85
• r44/70.57
• r55/70.71
• r61/70.14

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Step 3-now we shall compute xi as follows

• Xi xminri(xmax-xmin)
• X110.42(2)1.84
• X210.28(2)1.56
• X310.85(2)2.7
• X410.57(2)2.14
• X510.71(2)2.42
• X610.44(2)1.28

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F(xi)exi xi ri Ni
6.296 1.84 0.42 1
4.7588 1.56 0.28 2
14.879 2.7 0.85 3
8.4994 2.14 0.57 4
11.2458 2.42 0.71 5
3.5966 1.28 0.14 6

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• I
• I (1/6)(49.277) 8.2128

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The Monte Carlo method program by MATLAB

• clc
• clear
• cinput('please enter the lower bound of interval
  i.e c ')
• dinput('please enter the upper bound of interval
  i.e d ')
• minput('please enter the modulus number i.e m
  ')
• nm-1
• trans_fun(c,d)
• a0
• b1
• s0
• z1
• h(b-a)/n
• k0

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• while kltn
• p3z
• qmod(p,m)
• zq
• rq/m
• xc(r(d-c))
• fsind(x)
• ssf
• yhs
• kk1
• end
• y

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The transformation function program

• function g trans_fun(a,b)
• if (a0)(b1)
• winput('please enter the number of intervals
  points you wont to convert it w ')
• for v 1w
• xinput('please enter the point of interval you
  want to convert it x ')
• g (x-a)/(b-a)
• end
• disp('the original interval convert to a0
  b1')
• else
• disp('already a0 b1')
• end

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Finally
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Thanks for being so patient