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Unformatted text preview: um must be s D 0.
:6 C :4a :6 :6a
a1
k
k
1
7 .:5/ ! 0 gives A ! A ; any A D
with
:4 :4a :4 C :6a
:4 C :6a 0 8 If P D cyclic permutation and u0 D .1; 0; 0; 0/ then u1 D .0; 0; 1; 0/; u2 D .0; 1; 0; 0/;
9 10
11
12
13 u3 D .1; 0; 0; 0/; u4 D u0 . The eigenvalues 1; i; 1; i are all on the unit circle. This
Markov matrix contains zeros; a positive matrix has one largest eigenvalue D 1.
M 2 is still nonnegative; Œ 1 1 M D Œ 1 1 so multiply on the right by M to
ﬁnd Œ 1 1 M 2 D Œ 1 1 ) columns of M 2 add to 1.
D 1 and a C d 1 from the trace; steady state is a multiple of x 1 D .b; 1 a/.
Last row :2; :3; :5 makes A D AT ; rows also add to 1 so .1; : : : ; 1/ is also an eigenvector
of A.
B has D 0 and :5 with x 1 D .:3; :2/ and x 2 D . 1; 1/; A has D 1 so A I has
D 0. e :5t approaches zero and the solution approaches c1 e 0t x 1 D c1 x 1 .
x D .1; 1; 1/ is an eigenvector when the row sums are equal; Ax D .:9; :9; :9/ Solutions to Exercises 85 14 .I A/.I CACA2 C / D .I CACA2 C / .ACA2 CA3 C / D I . This says that
0 :5
I C A C A2 C is .I A/ 1 . When A D
; A2 D 1 I; A3 D 1 A; A4 D 1 I
2
2
4
10
1
1
1
1 C 2 C 2 C 4 C
21
D
D .I A/ 1 .
and the series adds to
22
1 C 1 C 1 C 1 C
2
2
8
130
:5 1
15 The ﬁrst two A’s have max < 1; p D
and
;I
has no inverse.
6
32
:5 0 16 D 1 (Markov), 0 (singular), :2 (from trace). Steady state .:3; :3; :4/ and .30; 30; 40/.
17 No, A has an eigenvalue D 1 and .I A/ 1 does not exist.
"
#
F1 F2 F3
P1
0 D 3 C
18 The Leslie matrix on page 435 has det.A I / D det
0
P2
2
F1 C F2 P1 C F3 P1 P2 . This is negative for large . It is positive at D 1
provided that F1 C F2 P1 C F3 P1 P2 > 1. Under this key condition, det.A I / must
be zero at some between 1 and 1. That eigenvalue means that the population grows
(under this condition connecting F ’s and P ’s reproduction and survival rates).
19 ƒ times S 1 S has the same diagonal as S 1 S times ƒ because ƒ is diagonal.
20 If B > A > 0 and Ax D max .A/x > 0 then B x > max .A/x and max .B/ > max .A/. Problem Set 8.4, page 446
1 Feasible set D line segment .6; 0/ to .0; 3/; minimum cost at .6; 0/, maximum at .0; 3/.
2 Feasible set has corners .0; 0/; .6; 0/; .2; 2/; .0; 6/. Minimum cost 2x
3
4 5 6 y at .6; 0/.
Only two corners .4; 0; 0/ and .0; 2; 0/; let xi ! 1, x2 D 0, and x3 D x1 4.
From .0; 0; 2/ move to x D .0; 1; 1:5/ with the constraint x1 C x2 C 2x3 D 4. The new
cost is 3.1/ C 8.1:5/ D $15 so r D 1 is the reduced cost. The simplex method also
checks x D .1; 0; 1:5/ with cost 5.1/ C 8.1:5/ D $17; r D 1 means more expensive.
Cost D 20 at start .4; 0; 0/; keeping x1 Cx2 C2x3 D 4 move to .3; 1; 0/ with cost 18 and
r D 2; or move to .2; 0; 1/ with cost 17 and r D 3. Choose x3 as entering variable
and move to .0; 0; 2/ with cost 14. Another step will reach .0; 4; 0/ with minimum cost
12.
If we reduce the Ph.D. cost to $1 or $2 (below the student cost of $3), the job will
go to the Ph.D. with cost vector...
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 Spring '12
 Minki
 Mass

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