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Unformatted text preview: Solution Markov Chains (J WA A special type of difference equation arises in the study of arkov chains or Markov
processes. We cannot go into the interesting theory of Markov chains, but we will give
an example that illustrates some of the ideas. An automobile rental company has three locations, which we designate as P, Q, and R.
When an automobile is rented at one of the locations, it may be returned to any of the three locations.
Suppose, at some speciﬁc time, that there are p cars at location P, q cars at Q, and r cars at R. Experience has shown, in any given week, that the p cars at location P are
distributed as follows: 10% are rented and returned to Q, 30% are rented and returned
to R, and 60% remain at P (these either are not rented or are rented and returned to P).
Similar rental histories are known for locations Q and R, as summarized below. Weekly Distribution History Location P: 60% stay at P, 10% go to Q, 30% go to R. Location Q: 10% go to P, 80% stay at Q, 10% go to R. Location R: 10% go to P, 20% go to Q, 70% stay at R. Let x], represent the state of the rental ﬂeet at the beginning of week k: p(k)
q(k)
r(k) Xk For the state vector xk, p(k) denotes the number of cars at location P, q(k) the number
at Q, and r(k) the number at R. From the weekly distribution history, we see that
p(k +1) =.6p(k)+.lq(k)+.lr(k)
q(k +1) =.lp(k)+.8q(k)+.2r(k)
r(k + l) = .3p(k) + .lq(k) + .7r(k).
(For instance, the number of cars at P when week k + 1 begins is determined by the 60% that remain at P, the 10% that arrive from Q, and the 10% that arrive from R.)
To the extent that the weekly distribution percentages do not change, the rental ﬂeet is rearranged among locations P, Q, and R according to the rule xk.“ = Axk,
k = 0, 1,..., where A is the (3 x 3) matrix
.6 .1 .1
A = .1 .8 .2 . .ﬂ
.3 .1 .7 Example 5 represents a situation in which a ﬁxed p0pulation (the rental ﬂeet) is
rearranged in stages (week by week) among a ﬁxed number of categories (the locations
P, Q, and R). Moreover, in Example 5 the rules governing the rearrangement remain EXAMPLE (5 Solution ﬁxed from stage to stage (the weekly distribution percentages stay constant). In general,
such problems can be modeled by a difference equation of the form ‘ xk+x=Axk, k=0,l,.... For such problems the matrix A is often called a transition matrix. Such a matrix has
two special properties: The entries of A are all nonnegative. (8a) In each column of A, the sum of the entries has the value I. (8h) It turns out that a matrix having properties (8a) and (8b) always has an eigenvalue of
A = 1. This fact is established in Exercise 26 and illustrated in the next example. Suppose the automobile rental company described in Example 5 has a ﬂeet of 600
cars. Initially an equal number of cars is based at each location, so that p(0) = 200,
q(0) = 200, and r(0) = 200. As in Example 5, let the weekbyweek distribution of cars be governed by xk+1 = Axk, k = 0, I, . . . , where
p(k) .6 .l .1 200
xk = q(k) , A = .l .8 .2 , and x0 = 200
r(k) .3 .l .7 200 Find lim;Moo xk. Determine the number of cars at each location in the kth week, for
k = 1,5, and 10. If A is not defective, we can use Eq. (6) to express X], as
Xk = 01(A1)kui + €120»sz + 03(As)kl13. where {u}, uz, ml is a basis for R3, consisting of eigenvectors of A.
It can be shown that A has eigenvalues A, = 1, A2 = .6, and A3 = .5. Thus A has
three linearly independent eigenvectors: 4 0
l1=1, 111: 9 , A2=6 u2= 1 ,
7 —1
—1
A3=.5, u3= ~l
2 The initial vector, x0 = [200, 200, 200]T, can be written as
x0 = 30u1 — 150u2 — 80u3.
Thus the vector xk = [p(k), q(k), r(k)]T is given by
xk = Akxo
= A"(30u, — 150u2 — 80u3)
= 30(A])kul — 150(x2)’<u2 — 80(A3)ku3
= 30u, — 150(.6)"u2 — 80(.5)"u3. (9) From the expression above, we see that 120
‘lim X; = 30u = 270
k—)oo 210 Therefore, as the weeks proceed, the rental ﬂeet will tend to an equilibrium state with 120 cars at P, 270 cars at Q, and 210 cars at R. To the extent that the model is valid, location Q will require the largest facility for maintenance, parking, and the like.
Finally, using Eq. (9), we can calculate the state of the ﬂeet for the kth week: I60 122.500 120.078
x]: 220 , x5 = 260.836 , and x10: 269.171 . ﬂ
220 216.664 210.751 : i'T‘ii'IT": EXERCISES In Exercises 1—6, consider the vector sequence {xk}, 12. A and x0 in Exercise 6
where xk = AXk_], k = 1,2, . . . . For the given starting 3 _1 1 3
vector x0, calculate x1, x2, X3, and x4 by using direct _ _ _
multiplication, as in Example 1. 13' A _ 12 0 5 ’ x0 _ —l4
1 A 0 l 2 4 ——2 —1 8
. — 1 0 . X0 — 4 —6 l 3 l
14. A = —3 0 2 , x0 = 1
2 A ’5 '5 16 20 2 IO 1
' " .5 .5 ’ x° ’ 8
5 25 128 In Exercises 15—18, solve the initialvalue problem.
3, A = [ ' ' ], x0 = [ ] 15. u’(l) = 5u(t) — 6v(t), u(0) = 4
.5 .75 64 v’(t) = 314(2) —4v(t). W) = 1
2 —1 3 16. u’(t) = 140) +2v(r), u(0) = l
4 A = 1 2 . x0 = 1 v’(t) = 2u(t) + v(t), v(0) = 5
l7. u’(t) = u(!) + v(t) + w(t), 14(0) = 3
5 A = ‘ 4 x = " v'(:) = ave) +3wm. um) = 3
' 1 1 ’ ° 2 mm = —2u(:) + v(:) + wrr). mm) = 1
3 1 2 18. u’(t) = —2u(t) + 2v(t) — 3w(t), u(0) = 3
6. A = , x0 = [ ] v'(t) = 2u(t) + v(t) — 6w(t), 12(0) = —l
4 3 0 w’(t) = —u(t) — 2w), w(0) = 3
In Exercises 714, let xk = Axk_;, k = l, 2, . . . , for 19 Consider the matrix A given by
the given A and x0. Find an expression for xk by using
Eq. (6), as in Example 3. With a calculator, compute x; A = 1 2 .
and x10 from the expression. Comment on limb“, xk. 0 1 7. . .‘
A and x0 1n Exercrse 1 Note that A = 1 is the only eigenvalue of A. 8. A and x ' Exerc'se 2
0 m l a) Verify that A is defective. . _ b) Consider the sequence {39,} determined by
10. A and x0 m Exercrse 4 Xi = Axk—hk = 1’ 2’ _ ‘ _, where 11. A and x0 in Exercise 5 x0 = [1, HT. Use induction to show that 9. A and x0 in Exercise 3 xk = [2k + 1, HT. (This exercise gives an
example of a sequence xk = Axk_1, where
lim,H00 lekll = 00, even though A has no
eigenvalue larger than 1 in magnitude.) In Exercises 20 and 21, choose a value a so that the
matrix A has an eignevalue of A = 1. Then, for
x0 = [1, HT, calculate limkam xk, where xk = Axk_,,
k = l, 2, . . . . .5 .5
20. A =[ :l
.5 1+0: [0 .3 ]
21.A=
.6 1+0: 22. Suppose that {uk} and {vk} are sequences satisfyv ing uk = Auk“ k = 1,2,..., and vk = Avk_.,
k =1,2,....Showthatifuo = v0, then u, = v,
foralli. 23. Let B = (b;,) be an (n x n) matrix. Matrix B is
called a stochastic matrix if B contains only non
ncgative entries and if b” + big +   + b;,, = 1,
l 5 i 5 n. (That is, B is a stochastic matrix if 87
satisﬁes conditions 8a and 8b.) Show that A = 1
is an eigenvalue of B. [Him: Consider the vector w=[1,l,...,l]T.] 24. Suppose that B is a stochastic matrix whose entries
are all positive. By Exercise 23, A = l is an eigen
value of B. Show that if Bu = u, u are 0, then u
is a multiple of the vector w deﬁned in Exercise 23. [Hint Deﬁne v = all so that v. = l and [va 5 1,
l 5 j 5 n. Consider the ith equations in Bw = w
and Bv = v.] 25. Let B be a stochastic matrix, and let A by any eigen
value of B. Show that lAl 5 1. For simplicity, as
sume that A is real. [Hinn Suppose that Bu 2 Au,
u ,4: 0. Deﬁne a vector v as in Exercise 24.] 26. Let A be an (n x n) matrix satisfying conditions
(8a) and (8b). Show that A = l is an eigenvalue of
A and that ifAu = Bu, 1! ye 0, then lﬁl 5 l. [Hint
Matrix AT is stochastic.] 27. Suppose that (A — Al)u = 0, 11 9E 0, and there
is a vector v such that (A — A1)v = n. Then v is
called a generalized eigenvector. Show that {u, v}
is a linearly independent set. [Hint Note that
Av = Av + u. Suppose that an + bv = 0, and
multiply this equation by A.] 28. Let A, u, and v be as in Exercise 27. Show that
Akv = A"v+kA""u, k = 1, 2. 29. Consider matrix A in Exercise 19.
a) Find an eigenvector u and a generalized
eigenvector v for A.
b) Express x0 = [1, HT as x0 = au + bv.
c) Using the result of Exercise 28, ﬁnd an
expression for Akxo = A"(au + bv). d) Verify that Akxo = [2k + 1, HT as was shown
by other means in Exercise 19. SUPPLEMENTARY EXERCISES 1. Find all values x such that A is singular, where x 12
A: 3 x 0
011 2. For what values x does A have only real eigenvalues,
where 3. Let A = ,
c d wherea+b = 2andc+d = 2. ShowthatA = 2i5
an eigenvalue for A. [Hint: Guess an eigenvector.] 4. Let A and B be (3 x 3) matrices such that det(A) = 2
and det(B) = 9. Find the values of each of the
following. a) det(A“Bz)
b) det(3A)
c) det(ABzA“) ...
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 Fall '11
 OSU

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