Using the solutions obtained in parts (a) and (b) above to explain why initial errors should need to be avoided as much as possible. There is greater variation of solutions obtained in parts (a) and (b) due to initial errors as from (x , y)=(2.415,0) to (1.1) respectively , therefore should be avoided as much as possible to reduce variation of solutions.

1. SOKOINE UNIVERSITY OF AGRICULTURE
Department of Biometry and Mathematics
Bsc.Irrigation and Water Resources Engineering
MTH 108
NUMERICAL METHODS 1
GROUP ASSIGNMENT-GROUP 4
INSTRUCTOR: Dr. G.K. KARUGILA
SUBMISSION DATE: Monday 30th
April 2018

2. List of members
S/N NAME OF MEMBER REGISTRATION NUMBER SIGNATURE
1 DOTO MUSA GESE IWR/D/2016/0011
2 HENRY PAULO B IWR/D/2016/0012
3 IDDI SHABANI IWR/D/2016/0067
4 UFINYU FREDRICK M IWR/D/2016/0057
5 JORAM GEORGE IWR/D/2016/0066
6 ZAMBI CHARLES IWR/D/2016/0064
7 MBANGO MARTIN HABAKUKI IWR/D/2016/0032
8 PHILIBERT FRANK IWR/D/2015/0028
9 METHOD JUSTINE IWR/D/2016/0035
10 KILINDO ABUBAKARI S IWR/D/2016/0065
11 WAMBURA NYAMBULI N IWR/D/2016/0068
12 KAPINGA FREDRICK IWR/D/2016/0017
13 ISAYA ALLY IWR/E/2016/0082

3. QUESTIONS
9. (a) Find the exact solution of the following linear system of equations
5x + 7y = 12.075
7x + 10y = 16.905
(b Round the values on the right hand side of each equation to two significant
digits and then find the exact solution of the resulting system of linear
equations.
(c) Use the solutions obtained in parts (a) and (b) above to explain why initial
errors should need to be avoided as much as possible.
10.Given the quantity £ = 0.333333333 + 0.272727273 – 0.15, perform the
following calculations
(a) Find the exact value of £ correct to five significant digits.
(b) Approximate the value of £ using three digit chopping arithmetic
(rounding without making adjustments).
(c) Approximate the value of £ using three digit rounding arithmetic.
(d) Calculate the absolute errors and percentage relative errors in the
approximations obtained in parts (b) and (c).
11. The number  =5.436563657 can be defined by 









0
!
1
2
k
k
 , where
  12)^2(1!  kkkk for 0k and .1!0  Compute the true absolute error and
true percentage relative error in the approximation of  by .
!
1
2
5
0








k
k
(Round-off your final results to four significant figures).

4. SOLUTIONS
9. (a) Finding the exact solution of the following linear system of equations.
5x + 7y = 12.075………..equation 1
7x + 10y = 16.905………..equation 2
Taking equation 1
5x + 7y = 12.075
5x = 12.075-7y
x = (12.075-7y)/5………………….equation 3
Substituting equation 3 into equation 2 to obtain
7((12.075-7y)/5) + 10y =16.905
16.095-9.8y+10y =16.095
0.2y =16.095-16.095
y = 0
Recalling equation 3 and substitute the value of y obtained above.
X = (12.075-7y)/5
x = (12.075-7(o))/5
x = (12.075)/5
x = 2.415
Therefore , the exact solution is (x , y) =(2.415 , 0)
1

5. (b) Rounding the values on the right hand side of each equation to two
significant digits and then find the exact solution of the resulting system of
linear equations.
5x + 7y = 12………..equation 1
7x + 10y = 17………..equation 2
Taking equation 1
5x + 7y = 12
5x = 12-7y
x = (12-7y)/5………………….equation 3
Substituting equation 3 into equation 2 to obtain
7((12-7y)/5) + 10y =17
16.8-9.8y+10y =17
0.2y =17-16.8
y = 1
Recalling equation 3 and substitute the value of y obtained above.
x = (12-7y)/5
x = (12-7(1))/5
x = (12-7)/5
x = 1
Therefore , the exact solution is (x , y) =(1 , 1)
(c) Using the solutions obtained in parts (a) and (b) above to explain why initial
errors should need to be avoided as much as possible.
There is greater variation of solutions obtained in parts (a) and (b) due to
initial errors as from (x , y)=(2.415,0) to (1.1) respectively , therefore
should be avoided as much as possible to reduce variation of solutions.
2

6. 10. (a) Find the exact value of £ correct to five significant digits.
Given that
£ = 0.333333333 + 0.272727273 – 0.15
£ = 0.606060606 – 0.15
£ = 0.456060606
£ ≈ 0.45606
(b) Approximating the value of £ using three digit chopping arithmetic
(rounding without making adjustments).
Given that
£ = 0.333333333 + 0.272727273 – 0.15
£ = 0.456060606
£ ≈ 0.45
(c) Approximating the value of £ using three digit rounding arithmetic.
Given that
£ = 0.333333333 + 0.272727273 – 0.15
£ = 0.456060606
£ ≈ 0.46
(d)Calculate the absolute errors and percentage relative errors in the
approximations obtained in parts (b) and (c).
Given that
£ (b) ≈ 0.45 and £(c) ≈ 0.46
For approximations obtained in parts (b)
| |=| |=| |
Absolute error =
| |x100% =| | x100%`
1.328903641%
Relative error =1.328903641%
3

7. Approximations obtained in parts (c)
| |=| |=| |
Absolute error =
| |x100% =| | x100%`
0.8637873888%
Relative error = 0.8637873888%

8. 11. Given The number  =5.436563657 can be defined by 









0
!
1
2
k
k
 , where
  12)^2(1!  kkkk for 0k and .1!0 
Finding the approximated value by
 * = .
!
1
2
5
0








k
k
For k = 0,1,2,3,4 and 5 obtain the following series
∑ = 2[ + + + + + ]
 * = 2[ + + + + ]
 * =2[1+1+0.5+0.166666666+0.041666666+0.008333333333]
 * =2[2.716666665]
 * 5.433333330
(i) True absolute error
| |=| |=| |
True absolute error
(ii) True percentage relative error
| |x100% =| | x100%
= 0.05941854458%
True relative error 0.05942%
4
5