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Quantitative Models in Marketing - I

Quantitative Models in Marketing - I. 2 1 3 2 6 0. Mathematics for Modeling – Matrix Algebra. Used to derive values of m independent variables using m simultaneous equations Used in problems involving Markov chain approach to derive steady state probabilities

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Quantitative Models in Marketing - I

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  1. Quantitative Models in Marketing - I

  2. 2 1 • 3 2 • 6 0 Mathematics for Modeling – Matrix Algebra • Used to derive values of m independent variables using m simultaneous equations • Used in problems involving Markov chain approach to derive steady state probabilities • A Matrix is a mXn array of characters , mXn is called the order of the matrix • m rows = 3 • n columns = 3 • A is a 3x3 matrix • Is also called a square matrix because m=n • Otherwise it is a rectangular matrix A =

  3. Mathematics for Modeling – Matrix Algebra • Each matrix element is located by an address denoted by (a)ij which is an element in the ith row and jth column , hence (a)13 = 1 in the matrix A • Two matrices are equal only if they are of the same order and all their elements are equal • Transpose of a matrix is obtained by interchanging rows and columns and denoted by A′ • It is possible to add and subtract matrices only if they are of the same order , in this case the corresponding elements in the matrices are added or subtracted from each other • To multiply or divide a matrix by a scalar K , each element in the matrix has to be multiplied or divided by scalar K

  4. 2 1 • 3 2 • 6 0 1 0 1 3 1 2 4 0 0 13 2 7 21 3 10 23 6 17 Mathematics for Modeling – Matrix Algebra • Two matrices can be multiplied to each other only if the number of columns in the 1st matrix is equal to number of rows in the 2nd.The product will have number of rows equal to 1st matrix and number of columns equal to 2nd matrix • To find the value of the a(ij) element of the product matrix , the ith row elements of the 1st matrix are multiplied to the corresponding jth column elements of the 2nd matrix and the products are added =

  5. 2 1 • 3 2 • 6 0 Mathematics for Modeling – Matrix Algebra • Determinant of matrix A is denoted by │A│or det(A) • Det(A) = ∑ a(1j) c(1j) , where c(ij) is called the cofactor of element a(ij) of the matrix and is derived as c(ij)=(-1) │m(ij)│ , where m(ij) is the minor of the element a(ij) obtained by deleting the ith row and jth column from the matrix A • For example if we to find det(A) in the previous example i+j

  6. Finding determinant • The minors can be listed as m11= 3 2 m21= 2 1 m31= 2 1 6 0 6 0 3 2 m12= 4 2 m22= 3 1 m32= 3 1 5 0 5 0 4 2 m13= 4 3 m23= 3 2 m33= 3 2 5 6 5 6 4 3

  7. Finding Determinant • Finding the determinant of the minors of order 2x2 as follows Det(m11) = 3x0-6x2 = -12 , similarly we can find the determinants of all the other minors of order 2x2 • Using the expression c(ij)=(-1) │m(ij)│ we can derive the co-factor matrix as i+j -12 10 9 6 -5 -8 1 -2 1 C =

  8. Finding the Determinant • Now using the formula Det(A) = ∑ a(1j) c(1j) We have, Det(A)= 3(-12)+2(10)+1(9) = -7

  9. Mathematics for Modeling – Matrix Algebra -1 • Inverse of a matrix is denoted by A • A =( Adj (A)) / (det (A)) • Adj (A) is the adjoint of matrix A determined by taking a transpose of the matrix formed by the cofactors of all elements of matrix A -1 ′ -1 -12 10 9 6 -5 -8 1 -2 1 -12 6 1 10 -5 -2 9 -8 1 A = -1/7 = -1/7

  10. Mathematics for Modeling – Matrix Algebra • Consider a system of m simultaneous equations with n independent variables where m=n • The system can be described using matrix algebra as Where i runs from 1 to m and j runs from 1 to n , a (ij) is matrix of coefficients of independent variables x(j1) is matrix of independent variables and b(i1) is matrix of RHS constants a(ij) x(j1) = b(i1) This can also be written as AX=B To solve this system we have to find the elements of matrix X given as X = A B -1

  11. Mathematics for Modeling – Differential Calculus • Laws of differentiation • d/dx of a constant = 0 • dxⁿ/dx = nxn-1 • d(kxⁿ)/dx = knxn-1 , where k is a constant • f´(x) = u′(x) ± v′ (x) for f(x) = u(x) ± v(x) • f′(x) = u(x)v′(x) + u′(x)v(x) for f(x)=u(x)v(x) • f′(x) = {v(x)u′(x)-u(x)v′(x)}/(v(x))² for f(x)=u(x)/v(x) Caselet: P= -10.5x²+672x+205 where x is number of salesmen employed and P is profit find x for maximum profit and also find value of maximum profit

  12. Mathematics for Modeling – Differential Calculus • Y=f(x) • Derivative of curve f(x) denoted by dy/dx = cd • As cd approaches point ‘a’ dy/dx= slope at a a f(x) d c b b b x

  13. Mathematics for Modeling – Differential Calculus • Concept of Maxima and Minima of the curve y=f(x) • As curve rises slope is positive • As curve falls slope is negative • At Maxima and Minima slope=dy/dx=0 • Also gradient of the slope = d²y/dx² is negative at maxima and positive at minima • These are the conditions to find the value of the independent variable at which any function reaches its maximum and minimum value • The gradient of slope is called 2nd order derivative , by successive differentiation we can generate any nth order derivative till the point the function remains differentiable

  14. Optimisation Problem using differentiation with maxima-minima technique Caselet: P= -10.5x²+672x+205 where x is number of salesmen employed and P is profit find x for maximum profit and also find value of maximum profit

  15. Solution • Using Maxima-Minima , at the value of x where the function attains maximum or minimum value dP/dx=0 • d/dx(-10.5x²+672x+205 )=0 • -21x+672=0 , hence x=32 • With 32 employees Profit is either maximum or minimum • Also here d2P/dx2 = d/dx(-21x+672)=-21<0 • Hence we conclude that profit is maximum with 32 employees and the value of maximum profit is obtained as • P= -10.5(32)2 + 672(32) + 205 = -10752 + 21504 + 205 • P= 10957

  16. Quantitative Models Thank You

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