CM10250 2011
Week 6 Solution to questions
011.12
a. The cost figure as the basis for depreciation :
$11,000 + $100 + $200 + $400 : $11,700
011.13
(600)
Workings
Depreciation
20X9 20% x ($1,000 + $1 ,600)= $520 (Lorry A $200, Lorry B $320)
20x0 20% x ($1,6
CM10250
Financial Accounting 2011
Week 3 Solution to Questions
Q3.2 refer text/notes
Q 4.2
Date
1 Mar
2 Mar
4 Mar
6 Mar
9 Mar
11 Mar
13 Mar
16 Mar
18 Mar
20 Mar
22 Mar
24 Mar
26 Mar
28 Mar
30 Mar
31 Mar
Dr
Lease
Office Equipment
Purchases
Postage
Purchase
CM10250 2011/12
Week 8 Solutions to questions
Q 10.3(a) & (b) Refer text P 102
Q10.4 Refer text P 102 to 104
Q10.16
Appropriate for Nesales and not for Cadberry.
Q 10.17
The $10 million spent on staff recruitment, training and development should be recogn
axis([0 L+1 0 N+1]);
hold on; % keeps the current graphics
for j = 1:L
plot(j*ones(1,N),1:N,o);
end
plot(1:L, Q,-); % Q is a vector whose components are the states
% at time l = 1, ., L
% Hence a solid line segment drawn from the previous node to the new
This equation is between two polynomials in the variable ejw . Setting the constant terms
2
2
equal, we get k 1 + p |ak |2 = W or
k=1
2
U =
2
W
1 + |ak |2
2
2
where U < W if any ak = 0, k = 1, . . . , p.
11.20. Set D(k) [n]
X[n] X (k+1) [n]. Then from the
24
(b) Mercers Theorem:
( ) =
X
() ()
=0
Hence, 0 () = 1 () = 1 cos 2 2 () = 2 cos 4
R
functions by 0 | ()|2 = 1 we get
=
r
1
1 =
r
Normalizing these orthogonal
2
= 2
Then, we can compute the lambdas as: 0 = 3 1 = and 2 = 2 These are the
variances of
estimate of correlation function
4
3
R[m]
2
1
0
1
2
0
50
100
m
150
200
Figure 1: Estimate of correlation function for N = 100.
r(1,1)=z(N+1);
r(2,1)=z(N+2);
r(3,1)=z(N+3);
r
pause(5);
a=inv(R)*r
disp(May compare a vector values to ar(3) coefficients.);
b1
11.11. (a) Since Y [N ] Y [N + 1] . . . Y [N ]. Using Theorem 11.1-4 property (b), we have
E X[n]|Y [N ], Y [N + 1], . . . , Y [N ] = E[X[n]|Y [N ]+E X[n]|Y [N + 1], . . . , Y [N ] .
By the same procedure
N
E X[n]|Y [N ], Y [N + 1], . . . , Y [N ] =
E[X[
(a) W [n] is the innovation sequence for X[n] because
W [n] is a white (uncorrelated) sequence.
It is dened as a causal invertible linear transformation on X[n].
(b) A simple substitution will show the result. From the defn. of X[n],
n
W [n] = X[n] +
k=
18
Then, if [ ] = [()] = a constant, and if
2 , [( )2 ] 0
then () is ergodic in the mean.
(b) Using the sucient condition, i.e.
Z +
( )
or
( )
Z +
in the case of zero mean, we have
Z +
2 | | cos(2 )
=
=
=
=
2 + 2
2
2
0
"
#
(+2 )
(+2 )
2
+ 2
27
Secondly, we are asked to show using Chebyshevs inequality that for such processes, we
actually have () = ( + ) (pr. 1). Now, by Chebyshev
[|( + ) ()|2 ]
2
0
= 2
= 0
for all 0
[|( + ) ()| ]
Then, dene the events
1
, |( + ) ()|
1
Now, consider the
Solutions to Chapter 11
11.1. For minimum variance, we want to minimize diagonal terms of
2
(Y Y )(Y Y )T ,
where Y = AX.
2 = (AX Y )(AX Y )T = AK1 AT + K2 AK12 K21 AT .
Now write A = A0 + ; is = 0 for minimum variance?
2 = (A0 + )K1 (A0 + )T + K2 (A0 + )
21
so we have
Z
+ 2
2
( )() () = () 2 )() ()
which means that both () and () have the same K-L orthonormal basis functions, i.e.
()
()
()
()
()
() = () and the eigenvalues of () are = 2 , or equivalently =
()
+ 2 We can then express the K-L expansion
9
So,
Z
2 (1 2 )
= ( )
() ( )
1 2
1 =2 =
Z
() ( ) + ( )
where
( ) =
Z
Z
(1 ) (2 ) (1 2 )1 2
as in part (a).
13. (a) We have that [] in the m.s. sense, and hence in probability and distribution.
Each [] is Gaussian distributed with mean and varianc
12
with appropriate regions of convergence for each of these terms. We nd
( + 2)
1
1
( + 2)
and =
=
=
=
( + 2)( + 2) =2 4
( + 2)( + 2) =+2 4
Now, for stability, the region of convergence (ROC) of the
1
4
1
4
+2
term is Re() 2, while
the ROC of the +2 ter
29
we can conclude that 0 is [20 20 + ] Then, since all the functions cos(20 +
20 ) are periodic with period 2 we can just as well integrate them over [ +] Thus
we have
#
"
#
"
\
\
cfw_cos(20 + 20 + ) 0 =
cfw_cos(20 + 0 ) 0
=1
=1
which is independent
6
Then
2
( )
2
!
2 ( )
( ) 2
= 4 ( )
+
2
( ) =
9. Let
1
0
() , [ ( + ) ()]
1
1
= ( + ) cos 20 ( + ) () cos 20
and then use the substitution cos 20 cos 2 0 sin 2 0 sin 20 for the rst cosine term,
to get
1
0
1
0
) cos 20 cos 2 ( + ) sin 2
sin 20
(
44
62. (a) The probability of leaving state 2 for the rst time at time is zero, since the waiting
time is an exponential RV, a continuous RV.
(b)
1 ( + ) = (1 1 )1 () + 2 2 () + 03 ()
2 ( + ) = +1 1 () (2 + 2 ) 2 () + 3 3 ()
3 ( + ) = 01 () + 2 2 () + 3 3
32
(c) Using above equation (1)
[Y( + )Y ()] = [AY( + )Y ()] + [B( + )Y ()]
= ARYY ( ) + BRY ( )
RYY ( )
=
(d) Taking the Fourier transforms of the results in part (b), we get
SY () = SY ()A + ()B
Solving for SY we then obtain
SY () = ()B (I A )1
From p
35
(b) For this part, we assume the input random process () is zero-mean.
[ () ( + )] = [ 2 () 2 ( + )]
= [()()( + )( + )]
= (0) (0) + ( ) ( ) + ( ) ( )
by 4th -order moment property for zero-mean Gaussian RVs (see problem 4.66), and so
2
2
( ) = (0) + 2
23
So now,
XX
=1 =1
(
) =
XX
=1 =1
Z
+
( + 0 ) ( + 0 )0
Z + X X
=
( + 0 ) ( + 0 ) 0
=1 =1
2
Z + X
0
=
( + ) 0
=1
0
since integral of non-negative function.
32. (a) The input process () has constant mean 128. So () = (0) = 128 1 = 128
(b)
8
For = 3 we have
(; 3) = (; 2) ()
= 2 () ()
Z
3
( )
=
()
0
Z
3
()
=
0
2
= 3 ()
2
Now we can guess the general result
1
()
( 1)!
To prove this result for general we appeal to mathematical induction, and assume the
2
result (; 1) = 1 (2)! () and
2
so that
(1 2 )
(2 ) = sinc
X
2
=
( )sinc
X
2
1 ( + )
sinc
=
sinc
3. The random sequence [] is Bernoulliand its values 1 occur with equal probabilities 12
We have () , sin 20 + [] where and 0 are given real numbers.
2
(a)
() , [()]
h
i
sin 20 + []