SIMPLE ECONOMIC APPLICATIONS OF MATRICES

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1 SIMPLE ECONOMIC APPLICATIONS OF MATRICES Eva Ulrychová Vysoká škola finanční a správní, Fakulta ekonomických studií, Estonská 5, Praha Astract: This article deals with simple examples pointing out to the use of mathematical models, especially of linear algera tools (matrix operations, eigenvectors, Markov chains, systems of linear equations, least squares approximation), in economic applications In some cases, solved prolems and data given in examples may have een simplified compared to real values as the main aim is to show possiilities of applications rather than to otain accurate results The presented examples can help to introduce students of economics-oriented universities into the matrix theory and illustrate the connection etween theory and practice Keywords: application, eigenvalues, matrix, matrix operations, system of linear equations JEL classification: A2, A22 Doručeno redakci: 222; Recenzováno: 222; 822; Schváleno k pulikování: 962 Introduction In the process of education it is often advisale to show the connection etween theory and practice In this article several economic applications of matrices are presented from quite simple to more complicated These examples are simple so that difficult economic theories do not have to e explained While the solved prolems and data given in examples can e simplified in comparison with real values for easier computation, the economic significance is not affected It is important to show possiilities of applications rather than to otain an accurate result On the other hand, it is interesting to include some real prolems if possile Matrix operations Scalar multiplication Formulation of the prolem A store discounts commodities c, c 2, c, c 4 y percent at the end of the year The values of stocks in its three ranches B, B 2, B prior to the discount are given in Tale Using matrices, find the value of stock in each of B, B 2, B after the discount Tale : Values of stocks in B, B 2, B c c 2 c c 4 B B B Source: Illustrative data Solution A percent reduction means that the commodities are eing sold for 7 percent of their original value If we organize the information given in Tale into a matrix V, then the matrix V 2 = 7 V expresses the value of stock after the discount: 9

2 65 V 2 = 7 V = = Distriutive law for matrix multiplication Formulation of the prolem A store sells commodities c, c 2, c in two its ranches B, B 2 The quantities of the commodities sold in B, B 2 in a week are given in Tale 2, the individual prices of the commodities are given in Tale, the costs to the store are given in Tale 4 Find the store s profit for a week, using a) total concepts ) per-unit analysis to show that matrix multiplication is distriutive Tale 2: Quantities of commodities Tale : Selling prices Tale 4: Costs c c 2 c c 2 c 5 B 2 5 B Source: (Tale 2,, 4) Illustrative data c 2 4 c 5 c 2 c 4 Solution The quantities (Q) of the commodities, the selling prices (P) and the costs (C) of the commodities can e represented in matrix form: Q = a) Using total concepts: The total revenue in B is: 2 P = C = and in B 2 : These calculations can e written using a product of two matrices: the total revenue (TR) is given y the matrix QP: TR = QP = 4 = Similarly, the total cost (TC) is given y the matrix QC: TC = QC = = Profits (Π) are Π = TR TC = QP QC = = ) Using per-unit analysis: The per-unit profit (U) is 9

3 2 5 5 U = P C = 4 = 5 4 The total profit (Π) is given y the matrix QU: Π = QU = = From a) and ) ve have Π = QP QC and Π = QU = Q(P C) Thus QP QC = Q(P C) Product of matrices Formulation of the prolem Let us consider five factories, each of them needs to supply each other Determine the numer of possiilities to transport commodities from one factory to the other one with one change at most, if there are truck and train connections as given thereinafter Solution Let us denote the factories as, 2,, 4, 5 and define the 5 5 matrix A: a ii = and for i j we put a i j = if there is a truck connection etween i and j, and a ij = otherwise Because of two way connection etween i and j the matrix A is symmetric For example the matrix A expresses that there is a truck connection etween factory and factory 2 (as a 2 = a 2 = ), whereas there is no connection etween factory and factory (as a = a = ) etc In a similar way we define the matrix T representing a train connection: T Then the matrix 2 AT 2 2 gives the numer of possiilities how to get from one of the factories to the other one in two steps y truck at first and then y train If we denote the elements of the matrix AT as (at) ij, we can see that for example (at) 45 = 2 there are two possiilities how to get from the 92

4 factory 4 to the factory 5 (ut not from the factory 5 to the factory 4 the order train-truck makes a difference) For example, since (at) 45 = a 4 t 4 + a 42 t 25 + a 4 t 5 + a 44 t 45 + a 45 t 55 = ++++, it is possile to get from factory 4 to factory 5 changing in 2 or in Similarly, the matrix 2 TA 2 2 gives the numer of possiilities how to get from one of the factories to another one y a train at first and then y a truck While there are for example two possiilities of connection etween factory 4 and factory 5 in the order truck-train, there is no connection etween 4 and 5 in the order train-truck The numer of all the connections etween the particular factories with one change at most is given y the matrix A T AT TA For example, we can see that there are three possiilities to get from the factory 2 to the factory 4 one possiility of a direct connection (as a 24 =) and two possiilities with a change one in the order truck-train ((at) 24 =) and one in the order train-truck ((ta) 24 =) 2 System of linear equations Formulation of the prolem Average salaries in Czech Repulic for the years 99, 995, 2 and 25 are susequently 286, 872, 29, 8 44 crowns The values, rounded in order to compute easily without a calculator, are given in Tale 5 Tale 5: Average salaries in Czech Repulic Thousands of crowns 8 8 Source: We portal Kurzycz [online] Availale: Taking into consideration these data, we estimate the average salary in 25 Solution We consider the time to e an independent variale and the salary a dependent variale we otain a function of one variale If we denote the year 99 as, 995 as, 2 as 2 and 25 as, then we have four points [,], [,8], [2,], [,8] in plane We are looking for a function whose graph approximately goes through these points If we draw the picture, we can see that the points seem to lie on a straight line (see Figure ) 9

5 Figure : A line passing through the given points Source: author s own processing We will try to find a linear function so that coordinates of the given points satisfy the equation y = ax+ of this function: a a a 2 a 8 We otained a system of linear equations represented y the augmented matrix The solution is a = 5, = It is really surprising that the linear system has a solution the given points lie exactly on a straight line! The equation of this line is y = 5x+ To estimate the average salary in 25 we sustitute x = 5 (the value corresponding to the year 25) into the equation y = 5x+ and otain y = 28 Let us note that if the system has no solution (the given points do not lie on a straight line) it is possile to find the least squares solution of this system - see elow Least squares approximation Formulation of the prolem The ratios of households in Czech Repulic having computers (in percentage) are given in Tale 6 We find the least squares approximating paraola for these data, compute the norm of the least squares error and estimate the ratio of households having computers in the year

6 95 Tale 6: Ratios of households with computers % Source: We portal of Czech Statistical Office: Česká repulika od roku 989 v číslech [online] Availale: Solution As in the previous prolem we denote the year 99 as, 995 as, 2 as 2 and 25 as In this case it is ovious that the points [,], [,7], [2,2], [,42] do not lie on a straight line (see Figure 2) Figure 2: A least squares approximating paraola Source: author s own processing We will try to find a paraola that gives the least squares approximation to the given four points Sustituting these points into the equation of a paraola y = ax 2 +x+c, we otain the system of linear equations c a c a c a c a represented y the augmented matrix This system is inconsistent the four given points do not lie on a paraola To find a paraola that fits the points as much as possile, we use the least squares approximation The least squares solution (Poole, 2, p 58) of the system A x = ( m n n m R R x A,, ) ()

7 is the vector x ~ such that Ax ~ Ax for all and can e otained as a solution of the equation n x R (2) A T A x = A T () In our case we have a 7 A,, x 9 42 c We compute products A T A and A T, put into the equation A T A x = A T and otain the equation a c 7 that is the system of linear equations The solution of this system is a = 425, = 5, c = 285 and x~ = (425, 5, 285) T The desired equation of paraola is y = 425x 2 +5x+285 To otain the least squares error e A~ x (4) we compute the product A ~ x at first: ~ 745 A x Then e = (5, 45, 45, 5) T = 45 We estimate the ratio of households having computers in the year 2 sustituting x = 4 into the equation y = 425x 2 +5x+285: y = Thus, using the data in the tale the ratio of households having computers in the year 2 is estimated at 72% 4 Eigenvector Formulation of the prolem Three producers (denoted y P, P 2, P ) organized in a simple closed society produce three commodities c, c 2, c Each of these producers sells and uys from each other, all their products are consumed y them and no other commodities enter the system (the closed model (Friederg, 2, p 76)) The proportions of the products consumed y each of P, P 2, P are given in Tale 7: 96

8 Tale 7: Proportions of products c, c 2, c consumed y P, P 2, P c c 2 c P 6 2 P P 5 Source: Illustrative data For example, the first column lists that 6% of the produced commodity c are consumed y P, % y P 2 and % y P As we can see, the sum of elements in each column is Let us denote x, x 2, x the incomes of the producers P, P 2, P Then the amount spent y P on c, c 2, c is 6 x +2 x 2 + x The assumption that the consumption of each person equals his or her income leads to the equation 6 x +2 x 2 + x = x, similarly for producers P 2, P We otain the system of linear equations 6x 2x x x x 7x 2x x x 5x x This system can e rewritten as the equation A x = x, where 6 2 A and x x x, x, T 2 Moreover, we assume the income to e nonnegative, ie x i, i=, 2, (we denote x o ) We can rewrite the equation A x = x into the equivalent form (A I) x = o, represented y the augmented matrix It means that we are looking for an eigenvector of A corresponding to the eigenvalue The general solution of the system has the form x = t(,,) T ; the condition x o is satisfied for t Thus, to ensure that this society operates, the incomes of the producers P, P 2, P have to e in the proportions :: 5 Markov chains Eigenvector Formulation of the prolem Suppose a market research monitoring a group of people, 2 of them use a product A and use a product B In any month 8% of product A users continue to use it and 2% switch to the product B and 9% of product B users continue to use it and % switch to the product A The percentages of users loyal to the original product are assumed to e constant in next months it means the proaility of changing from one product to the other is always the same That is a simple example of so called Markov chains Following this research we determine, how many people will e using each product one, two and k months, respectively, later, and estimate the state in the long run x 97

9 Solution The numer of product A users after one month is given y the following formula 8 2 7, since 8% of 2 A users (that is 8 2 ) stay with A and in addition % of B users (that is ) convert to A Similarly, the numer of product B users is given y the following formula 2 2 9, We can rewrite these two formulas using the matrix (transition matrix) 8 T 2 9 (an entry t ij represents the proaility of moving from state corresponding to i to state corresponding to j): If we denote x o = (2,) T (initial vector) and x = (7,) T, we can write x = T x Numers of each of A and B users after one month are given y the components of the vector x (let us note that these components are not necessarily integers they are only approximations of numers of people) Similarly to computing numers of users after one month we determine numers of users after two months (represented y the vector x 2 ): x 2 Tx (we can also write x 2 = T x = T(T x ) = T 2 x ) It is ovious that numers of A and B users after k months are determined y k x or x T x k Txk k (5) (an entry (t k ) ij of this matrix T k represents the proaility of moving from state corresponding to i to state corresponding to j in k transitions) It is possile to show (Poole, 2, p 2) that if the transition matrix is a matrix, which has some power positive, then vectors x k (state vectors) converge for large k to an unique vector x (steady state vector); when this vector is reached, it will not change y multiplying y T: x Tx (6) (it means that T has as an eigenvalue and the steady state vector is one of eigenvectors corresponding to this eigenvalue) Moreover, steady state vector does not depend on the choice of the initial vector x To determine numers of A and B users after a long time we compute the steady state vector (I is a unit matrix): x Tx Tx x o ( T I) x o (7) Since 2 T I, 2 we otain the following homogeneous linear system

10 the general solution of which is x = (t, 2t) T Since components of each of state vectors represent numers of A resp B users, the sum of these components must e equal to the gloal numer of users; in our case components of the steady stage vector must satisfy t+2t =, from which it follows that t= and x = (, 2) T After a long time, people will e using the product A and 2 people will e using the product B (and this result does not depend on the initial distriution of A and B users) Conclusion The aim of this article is to show easy examples, which point out to the use of matrices in economic applications To use the tools of linear algera for solving some real prolems it is usually necessary to have deeper knowledge of linear algera or other ranches of science In this article a few examples motivated y tasks in economy are provided Their solutions illustrate the use of linear algera tools such as matrix operations, eigenvectors, Markov chains, system of linear equations and least squares approximation These examples can e used in mathematical courses taught at economics-oriented universities References [] FRIEDBERG, S H, A J INSEL and L E SPENCE, 2 Linear Algera 4th edition, Prentice Hall, ISBN [2] POOLE, D, 2 Linear Algera: A Modern Introduction st edition, Brooks/Cole, ISBN

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