The University of South Wisconsin has had steady enrollments over the past five

Question

The University of South Wisconsin has had steady enrollments over the past five years. The school has its own bookstore, called University Book Store, but there are also three private bookstores in town: Bill's Book Store, College Book Store, and Battle's Book Store. The university is concerned about the large number of students who are switching to one of the private stores. As a result, South Wisconsin's president, Andy Lange, has decided to give a student three hours of university credit to look into the problem. The following matrix of transition probabilities was obtained.

University Bill's College Battle's
University 0.6 0.2 0.1 0.1
Bill's 0 0.7 0.2 0.1
College 0.1 0.1 0.8 0
Battle's 0.05 0.05 0.1 0.8

At the present time, each of the four bookstores has an equal share of the market. What will the market shares be for the next period?



B----
Andy Lange, president of the University of South Wisconsin, is concerned with the declining business at the University Book Store. (see problem above). The students tell him that the prices are simply too high. Andy, however, has decided not to lower the prices. If the same conditions exist, what long-run market shares can Andy expect for the four bookstores?

Explanation / Answer

In mathematical terms we want to solve a system of the form: (P - I) x = 0 subject to the constraint: x1 + x2 + ... + xn = 1 This last simply adds an additional row to our "matrix equation", so an appropriate Matlab expression can be constructed using the "" operator to request the solution of a matrix. Let's assume that P is a square matrix and already has the coefficients (transition probabilities) for the given Markov chain. Then we can run: > [P - eye(size(P)); ones(1,columns(P))] [zeros(rows(P),1); 1] which will solve the system of equations above and provide the components of x as a "column" result. A brief explanation of this command is as follows: eye(size(P)) is an identity matrix of the same "shape" as P. After subtracting off those diagonal entries from P, we have (within the limits of numerical rounding) a singular matrix P - I. The "" operator solves a linear system. For example: > A b would ask for the solution x of a matrix equation Ax = b. The solution is obtained without "inverting" the matrix A, so it can applied (as we did above) to cases in which the matrix is not invertible. Here the matrix equation we are solving can be described in block form as: [ x_1 ] [ 0 ] [ (P - I) ] [ x_2 ] = [ 0 ] [ 1 1 ... 1 ] [ ... ] [...] [ x_n ] [ 0 ] [ 1 ] and the "matrix construction" expression given above reproduces this faithfully with the use of utility functions zeros(j,k) and ones(j,k) for submatrices of j rows and k columns that contain all zero or all one entries, respectively. If rounding errors are a problem, then a slightly fancier matrix formulation can be tried. Add a "slack variable" y to the "solution" vector x, and consider: [ (P - I) | U' ] [ x ] = [ 0 ] [ U | 0 ] [ y ] [ 1 ] where U = [1,1,...,1] is a row of all ones, the same length as rows of P, and U' is written for transpose of U. We would construct the "bordered" matrix shown here as follows: > Q = [P - eye(size(P)), ones(rows(P),1); ones(1,columns(P)), 0] Solve the new system like this: > Q [zeros(rows(P),1); 1] Ideally the extra value y turns out to be exactly zero, but in practice for larger systems the value y will only be "nearly zero" and gives a sense of the influence of rounding errors on the solution x. Readers who would like to duplicate these steps but don't have the funds with which to purchase Matlab may be interested in Octave, a similar computing environment which is freely available under the GNU general public license:

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.