# report

```Department of Material Science and Engineering
University of Ghana, Legon
First Semester 2018
Numerical Methods
FAEN 301
Course Project
Submitted by
ID
10607444
10616366
10620620
10625002
Students
Jacqueline Abeasi
Jonathan Selorm Degbedzui
Gabriel Opoku
Okempa Philip
Contents
1 Introduction
1
2 Solution Approach and Algorithm
2.1 General Problems . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Specific Questions . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.1 Question 1 - Investment In An Entrepreneurial Venture
2.2.2 Question 2 - Clock Production Company . . . . . . . .
2.2.3 Question 3 - Is Aid an Invariably Fatal Disease? . . . .
2
2
24
24
25
27
3 Conclusion
30
4 Appendix
31
4.1 Question 1 - Investment In An Entrepreneurial Venture . . . . 31
4.2 Question 2 - Clock Production Company . . . . . . . . . . . . 32
References
36
i
Chapter 1
Introduction
Numerical analysis is the study of algorithms that use numerical approximation (as opposed to general symbolic manipulations) for the problems of
mathematical analysis (as distinguished from discrete mathematics). Numerical analysis naturally
finds applications in all
fields of engineering and the physical sciences, but in the 21st century,
the life sciences and even the arts have adopted elements of scientific computations. Ordinary differential equations appear in the movement of heavenly
bodies (planets, stars and galaxies); optimization occurs in portfolio management; numerical linear algebra is important for data analysis; stochastic
differential equations and Markov chains are essential in simulating living
cells for medicine and biology. The overall goal of the
field of numerical analysis is the design and analysis of techniques to give
approximate but accurate solutions to hard problems. This project employees
the use of software such as Microsoft Excel, Visual Basic and MATLAB.
Solutions obtained can be used as models to solve similar problems in many
fields of study
1
Chapter 2
Solution Approach and
Algorithm
2.1
General Problems
2
Question 1

8
4
A=
2
1
5
9
4
5
7
8
3
7






2
4
1 2 3 4
1 5 8
7
6
 , B = 7 8 5 9 , C = 3 4 9 , D = 
1
7
5 6 1 4
2 7 6
9
4
5
2
8
2

6
9

3
8
(a) Matrix multiplication between AB will not be possible as the their dimensions do not agree.
A has a dimension of (4 x 4) and B (3 x 4). Two matrices can be multiplied if and only if the
column size of the first matrix is equal to the row size of the second matrix.
(b) Matrix A has 4 columns while second Matrix C has 3 rows. Hence matrix multiplication
between AB will not be possible as the their dimensions do not agree.Two matrices can be
multiplied if and only if the column size of the first matrix is equal to the row size of the second
matrix.
(c)

8
4
2
1
5
9
4
5
7
8
3
7
 
2
4


6 7
∗
7 1
9
4
5
2
8
2



6
82 110 130


9
 = 111 114 177
 67 56 113
3
8
82 89 144
(d) Matrix multiplication between BC will not be possible as the their dimensions do not
agree. B has a dimension of (3 x 4) and C (3 x 3). Two matrices can be multiplied if and only
if the column size of the first matrix is equal to the row size of the second matrix.
(e)


4
1 2 3 4
7
BD = 7 8 5 9 ∗ 
1
5 6 1 4
4

5
2
8
2



6
37 41 65

9
= 125 109 201
3
79 53 119
8
(f) Matrix C has 3 columns while second Matrix D has 4 rows. Hence matrix multiplication
between CD will not be possible as the their dimensions do not agree.Two matrices can be
multiplied if and only if the column size of the first matrix is equal to the row size of the second
matrix.
3
10
(a)

8
4
A=
2
1
5
9
4
5
7
8
3
7

2
6

7
9
|A − λI| = 0




8 5 7 2
λ 0 0 0
4 9 8 6


 − 0 λ 0 0 | = 0
|
2 4 3 7
0 0 λ 0
1 5 7 9
0 0 0 λ


8−λ
5
7
2
 4
9−λ
8
6 
| = 0
|
 2
4
3−λ
7 
1
5
7
9−λ


21.6363


7

λ=
 2.6114 
−2.2477
The correspondent eigenvectors are:
for λ = -2.2477

for λ = 7

−0.3614
−0.2611

V =
 0.8189 
−0.3614

for λ = 21.6363

−0.8566
−0.2109

V =
 0.1450 
0.4481

for λ = 2.6114

−0.4890
−0.6208

V =
−0.3694
−0.4890
4
11


−0.3609
 0.8329 

V =
−0.2139
−0.3609
Matrix C


1 5 8
C = 3 4 9
2 7 6
|C − λI| = 0
The eigenvalues are :

 

1 5 8
λ 0 0
| 3 4 9 −  0 λ 0  | = 0
2 7 6
0 0 λ


1−λ
5
8
4−λ
9 | = 0
| 3
2
7
6−λ

for λ = -1.6792

15.2221
λ = −1.6792
−2.5430
forλ = 15.2221


−0.8721
V = −0.2266
0.4337

for λ = -2.5430
(b) determinant of Matrix A

−0.5405
V = −0.6096
−0.5799


−0.5428
V = −0.5829
0.6047
5
12
det(A) = λ1 × λ2 × λ3 × λ4
det(A) = 21.6363 × 7 × 2.6114 × −2.2477
det(A) = −888.9816
.
Matrix C
det(C) = λ1 × λ2 × λ3
det(C) = 15.2221 × −1.6792 × −2.5430
det(C) = 65.0015
(c) The inverse of matrix C
using Gauss Jordan method


1
0
0

1
0
0


1 5 8
C = 3 4 9
2 7 6

1 5 8 1 0 0
3 4 9 0 1 0
2 7 6 0 0 1

5
8
1 0 0
−11 −15 −3 1 0
−3 −10 −2 0 1

5
8
1 0 0
−11 −15 3 1 0 
0
65 13 3 −11


−65 −325 0 39 24 −88
 0 −715 0 0 110 −165
0
0
65 13 3
−11

−46475
 0
0

1
0
0

0
0 27885 −18590 −9295
232375 0
0
−35750 5625 
0
65
13
3
−11

0 0 −0.6
0.4
0.2
1 0
0
−0.1538 0.2308 
0 1 0.2 0.04615 −0.1692
6
13
C −1
(d)

8
0

0
0


−0.6
0.4
0.2
−0.1538 0.2308 
= 0
0.2 0.04615 −0.1692


8 5 7 2
4 9 8 6

A=
2 4 3 7
1 5 7 9



8 5 7 2 1 0 0 0
4 9 8 6 0 1 0 0



2 4 3 7 0 0 1 0
1 5 7 9 0 0 0 1



8 5
7
2
1 0
0
0
0 −13 −9 −10 1 −2 0
0



0 −11 −5 −26 1 0 −4 0 
0 −35 −49 −70 1 0
0 −8



8 5
7
2
1
0
0
0
0 −13 −9

−10 
0 

  1 −2 0

0 0


34 −228
2
22 −52
0 
0 0 −322 −560 −22 70
0 −104


5
7
2
1
0
0
0

−13 −9
−10 
−2
0
0 
 1



0
34 −228
2
22
−52
0 
0
0 −92456 −104 9464 −16744 −3536



1 0.625 0.875
0.25
0.125
0
0
0
0 −1 −0.6923 −0.7692 

0.07692
−0.1538
0
0



0



0
1
−6.7059
0.05882
0.6471 −1.5294
0
−3
0
0
0
−1
−1.1249 × 10
0.1024 −0.1811 −0.03824



0.1253
0.0256 −0.04528 −9.56 × 10−3
1 0.625 0.875
0
0 −1 −0.6923 0  
0.07779
−0.2325
0.1393
0.02941 



0
0.06636
−0.03958 −0.315
0.2564 
0
1
0 
0.1024
−0.1811
−0.03824
0
0
0
−1 −1.1249 × 10−3



−0.06724
−0.06023 −0.2303 0.23391
−1 −0.625 0 0
0

−1
0 0
0.1237
−0.2599 −0.0788 0.2069 



0


0
1 0
0.06636
−0.03958 −0.315
0.2564 
0
0
0 −1 −1.1249 × 10−3
0.1024 −0.1811 −0.03824
7
14



−1 0 0 0
−0.1440
0.1022 −0.1811 0.1046
 0 −1 0 0  
0.1237
−0.2599 −0.0788 0.2069 



0


0 1 0
0.06636
−0.03958 −0.315
0.2564 
0
0 0 −1 −1.1249 × 10−3
0.1024 −0.1811 −0.03824



1 0 0 0
0.1440
−0.1022 0.1811 −0.1046
0 1 0 0  −0.1237
0.2599
0.0788 −0.2069



0 0 1 0  0.06636
−0.03958 −0.315 0.2564 
0 0 0 1 1.1249 × 10−3 −0.1024 0.1811 0.03824
The inverse of the matrix A is therefore

0.1440
−0.1022

−0.1237
0.2599
A−1 = 
 0.06636
−0.03958
1.1249 × 10−3 −0.1024
(e) LU decomposition of matrix C

5 8
4 9
7 6

0 0
1 0
L32 1

U12 U13
U22 0 
U32 U33

U11
U12
U13

L21 U12 + U22
L21 U13 + U23
LU = L21 U11
L31 U11 L31 U12 + L32 U22 L31 U13 + L32 U23 + U33

.

1

C= 3
2

1
L = L21
L31

U11
U = U21
U31

0.1811 −0.1046
0.0788 −0.2069

−0.315 0.2564 
0.1811 0.03824
U11 = 1; U12 = 5; U13 ; U14 = 9
L21 U11 = 3 ⇒ L21 = 3
L21 + U12 + U22 = 4 ⇒ U22 = −11
L21 U13 + U13 = 9
U23 = −15
L31 U11 = 2
8
15
L31 = 2
L31 U12 + L32 U22 = 7
3
L32 =
11
L31 U13 + L32 U32 + U33 = 6
−65
U33 =
11
.

1
L = 3
2
.
0
1
3
11

0
0
1


1 5
8
U = 0 −11 −15
−65
0 0
11
 
14
b = 16
15
let the vector X be the solution

1
3
2

1
3
2
   
5 8 X1
14
4 9 X2  = 16
7 6 X3
15
   
Z1
14
0 0
1 0 Z2  = 16
3
1 Z3
15
11
Z1 = 14,
Z2 = −26,
−65
11

  

1 5
8
X1
14
0 −11 −15 X2  = −26
−65
−65
0 0
X3
11
11
Z3 =
9
16
−65
−65
X3 =
11
11
X3 = 1
−11X2 − 15X3 = −26, −11X2 = 11
X2 = 1
X1 + 5X2 + 8X3 = 14
X1 = 14 − 13
X1 = 1
Therefore the solution is
 
1

X = 1
1
10
17
Question 2
Forward Elimination of Unknowns
Since there are four equations, there will be three steps of forward elimination of unknowns.
First step
Divide Row 1 by 4.2857  10 7 and then multiply it by 4.2857  10 7 , that is, multiply Row 1 by
4.2857 107 4.2857 107 = 1 .

Row 1 (1) = 4.2857 107
− 9.2307 105

0 0
− 7.88710 
3
Subtract the result from Row 2 to get
4.2857  107

0


− 6.5

0

− 9.2307  105
3.7688 105
− 0.15384
0
0
− 4.2857  107
6.5
4.2857  107
  c1  − 7.887  103 
0
  

5.4619 105  c 2   7.887  103 
=
0.15384  c3   0.007 
  

− 3.6057  105  c 4  
0

Divide Row 1 by 4.2857  10 7 and then multiply it by − 6.5 , that is, multiply Row 1 by
− 6.5 4.2857  107 = −1.5167  10 −7 .
(
)
Row 1 − 1.5167 10−7 = − 6.5 0.14000 0 0
1.196210 
−3
Subtract the result from Row 3 to get
4.2857  107

0


0

0

− 9.2307  105
3.7688 105
− 0.29384
0
0
− 4.2857  107
6.5
4.2857  107
  c1   − 7.887  103 
0
  

5.4619 105  c 2   7.887  103 
=
0.15384  c3  5.8038 10−3 
  

− 3.6057  105  c 4  
0

Divide Row 1 by 4.2857  10 7 and then multiply it by 0, that is, multiply Row 1 by 0 4.2857 107 = 0 .
Row 1 (0) = 0 0 0 0
0
Subtract the result from Row 4 to get
4.2857  107

0


0

0

− 9.2307  105
3.7688 105
− 0.29384
0
0
− 4.2857  107
6.5
4.2857  107
11
  c1   − 7.887  103 
0
  

5.4619 105  c 2   7.887  103 
=
0.15384  c3  5.8038 10−3 
  

− 3.6057  105  c 4  
0

Second step
We now divide Row 2 by 3.7688 105 and then multiply it by − 0.29384, that is, multiply Row 2 by
− 0.29384 3.7688 105 = −7.7966 10−7 .
Row 2  (− 7.7966 10−7 ) =
0
− 0.29384 33.414 − 0.42584
− 6.1492 10 
−3
Subtract the result from Row 3 to get
4.2857  107

0


0

0

− 9.2307  105
3.7688 105
0
0
0
− 4.2857  107
− 26.914
4.2857  107
  c1  − 7.887  103 
0
  

5.4619 105  c 2   7.887  103 
=
0.57968  c3  1.1953 10 − 2 
  

− 3.6057  105  c 4  
0

Divide Row 2 by 3.7688 105 and then multiply by 0 , that is, multiply Row 2 by 0 3.7688 105 = 0 .
Row 2  (0) = 0 0 0 0
0
Subtract the result from Row 4 to get
4.2857  107

0


0

0

− 9.2307  105
3.7688 105
0
0
0
− 4.2857  107
− 26.914
4.2857  107
  c1  − 7.887  103 
0
  

5.4619 105  c 2   7.887  103 
=
0.57968  c3  1.1953 10 − 2 
  

− 3.6057  105  c 4  
0

Third step
We now divide Row 3 by − 26.914 and then multiply by 4.2857  10 7 , that is, multiply Row 2 by
4.2857  107 − 26.914 = −1.5924 106 .
) 
(
Row 3  − 1.5924 106 = 0 0 4.2857 107
− 9.2307 105

− 1.903410 
4
Subtract the result from Row 4 to get
4.2857  107

0


0

0

− 9.2307  105
3.7688 105
0
0
0
− 4.2857  107
− 26.914
0
Back substitution
From the fourth equation,
12
  c1  − 7.887  103 
0
  

5.4619 105  c 2   7.887  103 
=
0.57968  c3  1.1953 10− 2 
  

5.625  105  c 4   1.9034 104 
5.625 105 c4 = 1.9034 104
1.9034 104
5.625  105
c4 =
= 3.3837  10−2
Substituting the value of c4 in the third equation,
− 26.914c3 + (0.57968)c4 = 1.195310−2
1.1953 10 −2 − (0.57968)c 4
− 26.914
c3 =
=
1.1953 10 −2 − (0.57968)  3.3837  10 −2
− 26.914
= 2.8469 10 −4
Substituting the value of c3 and c 4 in the second equation,
(
)
3.7688105 c2 + − 4.2857107 c3 + 5.4619105 c4 = 7.887 103
c2 =
(
)
(
) (
7.887  103 − − 4.2857  107 c3 − 5.4619 105 c 4
3.7688 105
) (
) (
7.887  103 − − 4.2857  107  2.8469 10−4 − 5.4619 105  3.3837  10−2
=
3.7688 105
= 4.2615 10−3
Substituting the values of c2 , c3 and c 4 in the first equation,
(
)
4.2857107 c1 + − 9.2307105 c2 + (0)c3 + (0)c4 = −7.887 103
c1 =
=
(
)
(
) (
− 7.887  103 − − 9.2307  105 c 2 − (0)c3 − (0)c 4
4.2857  107
− 7.887  103 − − 9.2307  105  4.2615 10−3
4.2857  107
= −9.2244 10 −5
Hence the solution vector is
13
)
)
 c1  − 9.2244 10−5 
c  
−3 
 2  =  4.2615 10 
c3   2.8469 10− 4 
  
−2 
c4   3.3837 10 
The stress on the inside radius of the outer cylinder is then given by

 1 −  
c3 (1 +  ) + c4  r 2 



30  106 
 1 − 0.3 
=
2.8469 10− 4 (1 + 0.3) + 3.3837  10− 2 
2 
2 
1 − 0.3 
 6.5 
 =
E
1 − 2
= 30683psi
14
Question 3
1.
2
-1
A= 
0

0.5
-1
2
-1
0
0
-1
2
-1
0.5
0 
-1; 

1 
0.5
5 
b=  
 −5 
 
1.5 
a.


U= 



1.4142 -0.7071
0
0.3536


0
1.2247 -0.8165 0.2041


0
0
1.1547 -0.7217

0
0
0
0.5590 
0
0
 1.4142
 -0.7071 1.2247
0
U' = 

0
-0.8165 1.1547

 0.3536 0.2041 -0.7217
U  U  x  = b
T
U  x    y 
U   y  = b
T
b.
0
0
0   y1  0.5
 1.4142
 -0.7071 1.2247
0
0   y 2  5 


=

0
-0.8165 1.1547
0   y3   −5 

    
 0.3536 0.2041 -0.7217 0.5590   y 4  1.5 
 -0.3536 
 3.8784 

y= 
 -1.5877 


 -0.5590 
c.
U  x =  y 
15
0 
0 
0 

0.5590 






1.4142 -0.7071
0
0.3536
  x1   -0.3536 
  x 2  3.8784 
0 1.2247 -0.8165 0.2041
  = 

  x3   -1.5877 
0
0
1.1547 -0.7217
   

0
0
0
0.5590   x 4  -0.5590 
 1
 2

x=
 -2 


 -1 
2.


D=



2.0000
0
0
0
0
1.5000
0
0
0
0
1.3333
0
0 
0 
0 

0.3125
16
Question 4
(a)
00
mu (t) + ku(t) = F0 coswt
w0 =
u(t) =
r
k
m
F0
[coswt − cosw0 t]
− w2 )
m(w02
Figure 6: Graph of u(t), m= 1, k = 9;F0 =1; w = 2 and t [0,2π]
1
u(t) = [cos2t − cos3t]
5
The integral, I
I=
Z2π
u(t)dt
0
Exact value,
Z2π
u(t)dt = 0
0
Let the n = 20;
i. Trapezoidal Rule i.e. Approximating integral using Trapezoidal rule
I = −8.7197 × 10−17
ii. Simpson’s 1/3 i.e. Approximating integral using Simpson’s 1/3
I = −8.4290 × 10−17
n = 18
17
6
iii. Simpson’s 3/8 i.e. Approximating integral using Simpson’s 3/8
I = −8.7646 × 10−16
iv. Romberg n = 20 i.e. Approximating integral using Romberg
I = −2.9587 × 10−15
18
7
(b)
0
Adding the term cu (t) to the left side of the eqn in a.
u(t) = C1 er1 t + C2 er2 t +
c2 w 2
F0
[cwsin(wt) + m(w02 − w2 )cos(wt)] ...(1)
+ m2 (w02 − w2 )
p
p
−c − (c2 − 4w0 m2 )
(c2 − 4w0 m2 )
r1 =
andr2 =
2m
2m
for k =9; m = 1; F0 = 1; c = 10; w = 2;
calculating for r1 and r2 ;
−c +
r1 = -1
r2 = -9
Subtituting the value of r1 and r2 into eqn (1)
u(t) = C1 e−t + C2 e−9t +
1
[20sin(2t) + 5cos(3t)]
425
taking u(0) = 0 and u(1) = 0;
u(0) = C1 + C2 +
5
= 0 ...(1)
425
u(1) = C1 e−1 + C2 e−9 +
16.1052
= 0 ...(2)
425
solving (1) and (2) simultaneously yields:
C1 = -0.1030
C2 = 0.09127
Figure 7: Resulting Graph of solving the above
The integral, I
I=
Z
2π
u(t)dt
0
Taken n = 20,
19
8
i. Trapezoidal rule
I = - 0.0875
ii. Simpson’s 1/3 rule
I = - 0.0909
iii. Simpson’s 3/8 rule
let n = 18
I = - 0.09258
iv. Romberg
I = - 0.09266
(c)
i.
Z1.5
x2 lnxdx
1
For n = 2
- The exact integral is I = 0.192259
- The Gauss quadrature I = 0.192269
Relative error for Gauss Quadrature is very small hence gives the best estimate.
ii.
Z0.35
0
x2
2
dx, n = 3
−4
- The exact integral is I = - 0.176820
- The Gauss quadrature gives I = - 0.176820
The ralative error =
−0.176820 + 0.176820
=0
−0.176820
Both methods yield the same result and hence the accuracy is 100%
Z 3.5
x
√
dx, n = 4
x2 − 4
3
- The exact integral is I = 0.636213
- The Gauss quadrature yields I = 0.636213
The two answers are the same so the accuracy is 100%
20
9
Question 5
a)
f ( xi +1 ) − f ( xi )
x
f ( xi ) 
a = 0.24
xi = 0.1
x = 0.05
xi+1 = x + x
= 0.1 + 0.05
= 0.15
f (0.1) = 1 − e( −0.240.1)
= 0.023714
f (0.15) = 1 − e( −0.240.15)
= 0.035360
f (0.1) 
=
f (0.15) − f (0.1)
0.05
0.035360− 0.023714
0.05
= 0.23291
The exact value of f (0.1) can be calculated by differentiating
f (x ) = 1 − e − ax , x  0
as
f ( x ) =
d
 f (x )
dx
Knowing that
 
d − ax
e
= −ae − ax
dx
21
we find
f ( x ) =
d
(1 − e −ax )
dx
= ae− ax
= 0.24e −0.24 x
(
f (0.1) = (0.24) e − ( 0.240.1)
)
= 0.23431
The absolute relative true error is
True Value − Approximat e Value
100
True Value
t =
=
0.23431− 0.23291
100
0.23431
= 0.59761%
b)
a = 0.12
f (0.1) = 1 − e − ( 0.120.1)
= 0.011928
f (0.15) = 1 − e − ( 0.120.15)
= 0.017839
f (0.1) 
=
f (0.15) − f (0.1)
0.05
0.017839− 0.011928
0.05
= 0.11821
f ( x ) =
d
(1 − e −ax )
dx
= ae− ax
= 0.12e −0.12 x
f (0.1) = (0.12)(e −( 0.120.1)
= 0.11856
22
The absolute relative true error is
t =
=
True Value − Approximat e Value
100
True Value
0.11857− 0.11821
100
0.11857
= 0.29940%
The estimate of the derivative decreased.
23
2.2
2.2.1
Specific Questions
Question 1 - Investment In An Entrepreneurial
Venture
(a). With the Wyndor Glass Co problem, the optimal solution for two activities had to be found for a certain limited amount of resources. An optimal
combination of this two activities had to be found. Let x be the fraction of
purchase of the ownship in the first friends company Let y be the fraction of
purchase of the ownship in the second friends company
(b)
maximize 4500x + 4500y
Subject to : 5000x + 4000y ≤ 6000
400x + 500y ≤ 600
x ≤1
y ≤1
xG , x W ≥ 0
(c) Optimal Solution:(x,y) = (2/3, 2/3) and P = 6000
24
2.2.2
Question 2 - Clock Production Company
Three friends, David, LaDeana, and Lydia are the sole partners and workers
in a company which produces fine clocks. David and LaDeana work a maximum of 40 hours each per week while Lydia works a maximum of 20 hours
per week The company makes two different types of clocks, namely:
• A grandfather clock and a wall clock
• A wall clock
Each grandfather clock built and shipped yields a profit of 300 dollars,
while each wall clock yields a profit of 200 dollars.
Problem: The three partners now want to determine how many clocks of
each type should be produced per week to maximize the total profit
Solution Approach The Following is the linear programming model
for the Clock Production Company:
a
maximize 300xG + 200xW
Subject to : 6xG + 4xW ≤ 40
8xG + 4xW ≤ 40
3xG + 3xW ≤ 20
xG , x W ≥ 0
e C(G): \$300 -¿ \$375 % of allowable increase = 100(375-300/100) = 75%
The optimal solution will remain the same since \$75 is within the allowable increase.
f By the 100% rule for simultaneous chnages in the objective function,
the optimal solution may or may not chnage.
C(G): \$300 -¿ \$375 % of allowable increase = 100(375-300/100) = 75%
C(W): \$200 -¿ \$175 % of allowable decrease = 100(200-175/50) = 50%
25
g ”Graph” When the profit for grandfather clock increases to \$375, the
optimal solution is still (G,W) = (x1,x2) = (3.333.3.333) and P =
1666.67 if both profit estimates change thus grandfather clock increases
to \$375 and wall clock decreases to \$175, then the optimal solution
changes to: (G,W) = (x1,x2) = (5,0) and P = 1875
h Lydia should increase her hours slightly since she has the highest shadow
price connected to her
i The shadow price for David is zero because all of his available hours
are not being used and consequently an increase in his hours would not
impact total profit
j Yes, this increase is within the allowable increase. The increase in total
will be \$33.33 x 5 = \$166.65
k By the 100the optimal solution may or may not chnage. The shadow
prices may not be used to determine the effect on profit.
C(L): 20 -¿ 25 % of allowable increase = 100(25-20/100) = 25% C(D):
40 -¿ 35 % of allowable increase = 100(40-35/100) = 75% sum = 125%
26
2.2.3
Question 3 - Is Aid an Invariably Fatal Disease?
Solution Approach
QUESTION 1
s(t) = e−kt
show thats Tavg =
Tavg =
Z
∞
1
k
s(t) dt
0
Tavg =
Z
∞
e−kt dt
0
lim
b→∞
lim [
b→∞
QUESTION 2
Z
b
e−kt dt
0
e−kt
e−k∗0
+
]
−k
−k
Tavg = 6.4 months
s(t), where t = 5 years
Find the fraction remaining after 5 years
−t
s(t) = e Tavg
−60
s(t) = e 6.4
s(t) = 8.4818 ∗ 10−5
QUESTION 3a
s(t) = e−kt
show that S(t) = 2
27
−t
T1
2
0.5 = e
−kt 1
2
−kt 1
ln(0.5) = ln(e
2
)
ln(0.5) = −kt 1
2
ln(0.5)
−k
t1 =
2
t1 =
ln(2)
k
k =,
ln(2)
t1
2
2
−
0.5 = e
ln(2)
∗t
t1
2
0.5 = 2
−t
T1
2
QUESTION 3b
show thats t 1 = Tavg ln(2)
2
t1 =
2
k =
ln(2)
k
1
Tavg
t 1 = Tavg ln(2)
2
QUESTION 4
Incurable f raction that survives af ter 5 years = 0.14 − 0.10 = 0.04 = 4%
s(t) = 0.04 = e−k5
28
k = 0.644/yr
s(t) = e−0.644t
F raction surviving af ter 2 years = e−0.644∗2 = 0.2757 = 27.58
29
Chapter 3
Conclusion
Advanced numerical analysis are essential in making numerical weather predictions feasible, computing the trajectory of spacecraft using numerical solution of a solution of differential equations.
30
Chapter 4
Appendix
4.1
Question 1 - Investment In An Entrepreneurial
Venture
31
4.2
Question 2 - Clock Production Company
a
32
b
c
33
d
34
35
References
[1] W. L. Winston, et. al, Developing Spreadsheet- Based Decision
Support Systems Using Excel and VBA for Excel, Dynamic Ideas, 2007.
[2]Hillier and Lieberman,Introduction to Operations Research, 7th
edition.
[3]D. G. Zill, A First Course in Differential Equations with Modeling
Applications, 10th ed. New York: Brooks/Cole, 2013
36
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