Numerical Methods
Midterm 1 March 18, 2014
• Please give details of your calculation. A direct answer without explanation is not counted.
• Your answers must be in English.
Problem 1 (15%)
Give the double binary representation of the number −32.6
Soluiton
Because
32 = (100000)2, 0.6 = (1001)2, we can know that
32.6 = 100000.1001 = 1.000001001 × 25. (a) The sign bit: 1.
(b) The exponent part: 5 + 1023 = 1028 ⇒ 10000000100.
(c) The mantissa part: 0000010011001100110011001100110011001100110011001100.
Problem 2 (15%)
Consider the following 4 × 4 matrix
3 1 3 5 6 4 2 1 6 1 9 2 3 4 5 6
.
Calculate the LU factorization.
Soluiton
Because we want to do a LU factorization with no pivoting, we can assume that
3 1 3 5 6 4 2 1 6 1 9 2 3 4 5 6
=
1 0 0 0
l21 1 0 0 l31 l32 1 0 l41 l42 l43 1
u11 u12 u13 u14 0 u22 u23 u24 0 0 u33 u34 0 0 0 u44
.
(a) We can know that
u11 = 3, u12= 1, u13 = 3, u14= 5, l21= 2, l31 = 2, l41 = 1.
(b)
4 2 1 1 9 2 4 5 6
−
2 2 1
1 3 5 =
1 0 0
l32 1 0 l42 l43 1
u22 u23 u24 0 u33 u34
0 0 u44
⇒
2 −4 −9
−1 3 −8
3 2 1
=
1 0 0
l32 1 0 l42 l43 1
u22 u23 u24 0 u33 u34
0 0 u44
Then, we can get
u22= 2, u23 = −4, u24= −9, l32= −1/2, l42= 3/2.
(c)
3 −8 2 1
−−1/2 3/2
−4 −9 = 1 0 l43 1
u33 u34 0 u44
1 −25/2 8 29/2
= 1 0 l43 1
u33 u34 0 u44
Then, we can get
u33= 1, u34= −25/2, l43= 8, u44 = 229/2.
Finally, the conclusion is
3 1 3 5 6 4 2 1 6 1 9 2 3 4 5 6
=
1 0 0 0
2 1 0 0
2 −1/2 1 0 1 3/2 8 1
3 1 3 5
0 2 −4 −9
0 0 1 −25/2 0 0 0 229/2
.
Problem 3 (15%)
Conduct LU factorization with pivoting to solve the following linear system
2 4 1
4 −10 2
1 2 4
x1 x2 x3
=
5
−8 13
.
Soluiton
Since we want to use a LU decomposition with pivoting, the left hand side need to first transfer to the form P Ax = LU where
A =
2 4 1
4 −10 2
1 2 4
and we define
b =
5
−8 13
. (a) Let us first find U , L and P .
P1A =
0 1 0 1 0 0 0 0 1
2 4 1
4 −10 2
1 2 4
=
4 −10 2
2 4 1
1 2 4
L1(P1A) =
1 0 0
−1/2 1 0
−1/4 0 1
4 −10 2
2 4 1
1 2 4
=
4 −10 2
0 9 0
0 9/2 7/2
P2(L1P1A) =
1 0 0 0 1 0 0 0 1
4 −10 2
0 9 0
0 9/2 7/2
=
4 −10 2
0 9 0
0 9/2 7/2
L2(P2L1P1A)
1 0 0
0 1 0
0 −1/2 1
4 −10 2
0 9 0
0 9/2 7/2
=
4 −10 2
0 9 0
0 0 7/2
So, we can get
U =
4 −10 2
0 9 0
0 0 7/2
,
L = L−11 L−12 =
1 0 0 1/2 1 0 1/4 0 1
1 0 0
0 1 0
0 1/2 1
=
1 0 0
1/2 1 0 1/4 1/2 1
and
P = P2P1 =
1 0 0 0 1 0 0 0 1
0 1 0 1 0 0 0 0 1
=
0 1 0 1 0 0 0 0 1
. (b) Then, we can solve this linear system.
(a) Solve Ly = P b.
y = L2L1P b
=
1 0 0
0 1 0
0 −1/2 1
1 0 0
−1/2 1 0
−1/4 0 1
0 1 0 1 0 0 0 0 1
5
−8 13
=
−8 9 21/2
.
(b) Solve U x = y using back substitution. The linear system is as follows:
4 −10 2
0 9 0
0 0 7/2
x1 x2 x3
=
−8 9 21/2
.
Finally, we can easily get
x3 = 3 ⇒ x2 = 1 ⇒ x1 = −1.
Problem 4 (20%)
Recall that we prove by using p + 1 digits for x − y, then the relative rounding error < 2.
Consider β = 10 and p = 3.
(a) For Case 1 and Case 2 in the proof, give an example achieving the largest possible relative error. The purpose of this problem is to check if you understand the proof. Note that you need to give an example for Case 1 and an example for Case 2.
(b) For examples that you used in (a), what is the relative error if we use p + 2 digits?
(c) Same as (b) but using p + 3 digits.
(d) Same as (b) but using p + 4 digits.
Soluiton
(a) Case 1:
We set x = 1.01 and y = 0.00599. Then, ¯y = 0.005 and
x − y = 1.00401, x − ¯y + δ = 1.005 ≈ 1.01.
Therefore, the relative error is
|1.00401 − 1.01|
1.00401 ≈ 5.97 × 10−3.
The upper bound in the proof is (β
2 + 1) × β−p = 6 × 10−3. Case 2:
We set x = 1.00 and y = 0.00999. Then, ¯y = 0.009 and
x − y = 0.99001, x − ¯y + δ = 0.991.
Therefore, the relative error is
|0.99001 − 0.991|
0.99001 ≈ 10−3. The upper bound in the proof is
β−p = 10−3. (b) Case 1:
We set x = 1.01 and y = 0.00599. Then, ¯y = 0.0059 and
x − y = 1.00401, x − ¯y + δ = 1.0041 ≈ 1.00.
Therefore, the relative error is
|1.00401 − 1.00|
1.00401 ≈ 3.99 × 10−3. Case 2:
We set x = 1.00 and y = 0.00999. Then, ¯y = 0.0099 and
x − y = 0.99001, x − ¯y + δ = 0.9901 ≈ 0.990.
Therefore, the relative error is
|0.99001 − 0.990|
0.99001 ≈ 1.01 × 10−5. (c) Case 1:
We set x = 1.01 and y = 0.00599. Then, ¯y = 0.00599 and
x − y = 1.00401, x − ¯y + δ = 1.00401 ≈ 1.00.
Therefore, the relative error is
|1.00401 − 1.00|
1.00401 ≈ 3.99 × 10−3.
Case 2:
We set x = 1.00 and y = 0.00999. Then, ¯y = 0.00999 and
x − y = 0.99001, x − ¯y + δ = 0.99001 ≈ 0.990.
Therefore, the relative error is
|0.99001 − 0.990|
0.99001 ≈ 1.01 × 10−5. (d) The same as (c).
Problem 5 (20%)
Assume β = 10 and p = 3. Consider
x2− y2 versus (x + y)(x − y).
Give an example so that the relative error of the former is at least four times that of the latter.
We assume exactly rounded operations are conducted using the rounding even scheme. Note that do not find the relative error of the latter which can equal to zero.
Soluiton
We set x = 1 and y = 0.982. The exactly solution is
12− 0.9822 = (1 + 0.982)(1 − 0.982) = 0.035676.
(a) For x2− y2:
1 ⊗ 1 = 1.00, 0.982 ⊗ 0.982 = 0.964 ⇒ 1 0.964 = 0.036.
(b) For (x + y)(x − y):
1 ⊕ 0.982 = 1.98, 1 0.982 = 0.018 ⇒ 1.98 ⊗ 0.018 = 0.0356.
So, the ratio of these two relative errors is
|0.036 − 0.035676|
|0.0356 − 0.035676| = 0.000324 0.000076 > 4.
Problem 6 (15%)
Consider the following 4 × 4 symmetrix matrix
2 2 1 4 2 8 7 9 1 7 4 6 4 9 6 4
.
(a) Use the procedure taught in the class to do Cholesky factorization.
(b) Can you finish the procedure? If not, why?
Soluiton
(a) Let us define
L =
a 0 0 0 e b 0 0 f g c 0 h i j d
.
The procedure of Cholesky factorization is:
(1)
a =√
2, e = 2/√
2, f = 1/2, h = 4/√ 2.
(2)
8 7 9 7 4 6 9 6 4
−
2/√
2 1/√
2 4/√
2
2/√
2 1/√
2 4/√ 2
=
6 6 5
6 7/2 4
5 4 −4
So, we can derive b =√
6, g = 6/√
6, i = 5/√ 6.
(3)
7/2 4 4 −4
−6/√ 6 5/√
6
6/√
6 5/√ 6
=−5/2 −1
−1 −49/6
. Therefore, we failed at this step.
(b) The reason is that this matrix is a non-positive-definite matrix.
