Unit 23 Mathematics for Software Development Assignment

LO1: Understand core mathematical skills for software engineers

A.1 Solve the following linear and quadratic equations:

1. 2(3 – 5x) = 15

             (2 × 3) – (2 × 5x)=15

6-10x=15

6-15=10x

-9=10x

x= - 9/10

x= - 0.9

2. x² + x -20 = 0

            x² + 5x -4x -20 = 0

x(x + 5) – 4(x + 5) =0

(x – 4) (x + 5) =0

x – 4 = 0 or x + 5 =0

x = 4 or x = -5

A. 2 Solve the following sets of simultaneous equations by

(a) algebraic method (b) graphical method.

1. y=2x; y= -2x+1

          Algebraic method:

y=2x and y= -2x+1

2x = -2x+1

4x=1

x= 1/4 = 0.25 and y=2x

y= 2 × 0.25

y=0.50

        Graphical method:

                    (1) Y= 2x

                    (2) Y= -2x+1

1. y = 5x + 1; y =−5x + 1

Algebraic method:

y = 5x + 1 and y =−5x + 1

-5x+1=5x+1

1-1=5x+5x

10x=0

x= 0/10 =0 and y=5x+1

y=5 × 0 + 1

y=0+1=1

y=1 and x=0

           Graphical method:

                    (1)  y= 5x+1

                    (2)  y= -5x+1

1. −6y = 3x − 4; 2y = x + 5

Algebraic method:

3x+6y=4 and x=2y-5

3(2y-5)+6y=4

6y-15+6y=4

12y=4+15

12y=19

y= 12/19   = 1.58 and x=2y-5

x=2 × 19/12  – 5

x= 19/6 -5

  Graphical method:

                       (1) -6y = 3x - 4

                       (2) 2y = x + 5

3 Find the volume of the following shapes to three significant figures by showing your work step by step

 

1. a cube with a length of one side 27 meters

Volume of a cube = a³

= (27)³ = 27 × 27 × 27 = 19683 m³

Hence volume of a cube to three significant figures is 197m³.

A.4 Using Pythagoras’ theorem, proof that triangle ?ABC (9:12:15) is a right-angled triangle

1. Calculate sine, cosine and tangent for each angles of ?ABC.

 

1. Using an appropriate Excel function, demonstrate on a spreadsheet that ?ABC is a right-angled triangle.

Pythagoras theorem states that Square of hypotenuse=Square of 1^st side+ Square of 2^nd side for a Right Angled Triangle

h² = a² + b²

15 × 15 = 9 × 9 + 12 × 12

225 = 81 + 144

225 = 225

Hence it is proofed that triangle ABC is a right angle triangle.

A.5 Two robots, Alice and Bob are pulling a box as shown on the figure:

          

A           Calculate vector c = a + b.

C= (13,20) + (5,-6)

C=(13+5,20-6)

            C=(18,14)

B          Calculate magnitude of vector c.

Magnitude of vector c = <x, y>

c=<18,14>

C          Write a Pseudocode for calculating magnitude of vector c.

Algorithm: magnitudeOfVector(x, y)

Purpose: magnitude of a vector whose 2 components are given
Pre: Given both components of vector
Post: None
Return: Magnitude
{

            magnitude ← sqrt(x*x + y*y)

            Return magnitude

}

LO2: Understand the application of algebraic concepts

B.1  A certain British company has three departments. Following sets are showing departments, surnames and annual salaries of employees of this company:

A={ Martin, Marriott, Boast, Preston, Kans}

B= {24k, 25k, 26k, 27k, 30k}

C= {Production, Sales, Finance}

Mr Martin and Mrs Marriott are working at production department, Mrs Boast and Mrs Preston working at sales department and Mr Kans works at Finance department.

Find the Cartesian product of set A and set B. (R=A×B)

Cartesian of a product (R= A × B)

={ Martin, Marriott, Boast, Preston, Kans} × {24k, 25k, 26k, 27k, 30k}, ={Martin , 24k},{Martin , 25k},{Martin , 26k},{Martin , 27k},{Martin , 30k}, {Marriott , 24k},{Marriott , 25k},{Marriott , 26k},{Marriott , 27k},{Marriott , 30k}, {Boast , 24k}, {Boast , 25k}, {Boast , 26k}, {Boast , 27k}, {Boast , 30k}, {Preston , 24k}, {Preston , 25k}, {Preston , 26k}, {Preston , 27k}, {Preston , 30k}, {Kans , 24k}, {Kans , 25k}, {Kans , 26k},{Kans , 27k},{Kans , 30k}

Find the Natural join of R and C.

= (Production, Martin, 24K),(Production, Marriott, 25K), (Sales, Boast, 26K),(Sales, Preston, 27K), (Finance, Kans, 30K)

Fill in the below table by using provided information:

Employee Name	Salary	Department
Martin	24,000	Production
Marriott	25,000	Production
Boast	26,000	Sales
Preston	27,000	Sales
Kans	30,000	Finance

B.2 A small ICT firm, has three branches in

1. Redbridge,

2. Enfield,

3. Barnet.

Five technicians with following details are working at this company:

Ali (Location: Barnet, age: 25, salary: £21,000),

Steve (Location: Redbridge, age: 45, salary: 23,000),

Mike (Location: Enfield, age: 50, salary: 19,000),

Linda (Location: Barnet, age: 55, salary: 24,000 ),

Carol (Location: Redbridge, age: 43, salary: 27,000)

A. Draw required number of tables and fit in the above information there.

Employee Name	Age	Location	Salary (pounds)
Ali	25	Barnet	21,000
Steve	45	Redbridge	23,000
Mike	50	Enfield	19,000
Linda	55	Barnet	24,000
Carol	43	Redbridge	27,000

B List individuals satisfying the conditions below:

1. (Age<46) AND (Salary> £ 23,000)

{Carol}

2. (Age>26 ) OR (Salary < £24,000)

Steve ,Mike,Linda,Carol,Ali

3. (Age< 53) AND (Salary>29) OR (Location=1)

Steve,Carol

4. (Age> 25) XOR (Salary>30) OR (Location=2)

Steve ,Mike,Linda,Carol

B3. Create a magic square by identifying values of p, q, r, s, t, u, x, y, z in matrix A. [P2.2]

      0101    15   14    p

A=                12     06    q     r

          08      s      t    u     

            X    03     y   z

Magic numbers 4X4 , the total should be equal to 34 from all the sides. All the number 1 to 16 should be used to solve the matrix using each number only one.As per the grid 01, 15, 14, 12, 06, 08 and 03 are already present. The rest numbers to be used. Left are 2, 4, 5, 7, 9, 10,11,13,16. These numbers when used in combination gives the total 34

01+15+14+p = 34 => p = 34 – 30 => p = 4

01+12+8+x = 34 => x = 34 – 21 => x = 13

15+06+s+3 = 34 +> s = 34 – 24 => s = 10

x + s + q + p = 34 => 13+10+q+4 = 34 => q = 34 – 27 => q = 7

12+06+q+r = 34 => 12+06+7+r = 34 => r = 34 – 25 => r = 9

We have left to find the values for t, u, y, z.

Also from the 16 numbers that we have in our Matrix, we have left 2, 5, 11, 16.

08 + s + t + u = 34 => 08 + 10 + t + u = 34 => 18 + t + u = 34 => t + u = 34 – 18 =>

t + u = 16 => t = 11, u = 05;

x + 03 + y + z = 34 => 13 + 03 + y + z = 34 => 16 +y + z = 34 => y + z = 34 – 16 =>

y + z = 18 => y = 02, z = 16.

                               0 01    15    14    04   

               A =             12    06    07    09

                                  08    10    11    05

                                  13    03    02    16

Hence p = 04, q = 07, r = 09, s = 10, t = 11, u = 05, x = 13, y = 02, z = 16

B4. Show that if  

            

Then P is the inverse of Q. [P2.2]

If P × Q = I and Q × P = I then P = Q^-1 and Q = P^-1 (P and Q are inverse of each other)

If P is inverse of Q then below statements should hold true: P×Q=I where I = as below

    11        0

      0        1    

           

    P × Q =       1     2                −2          1

                        3     4                 1.5     −0.5 

 

P × Q =           1×(−2)+2×1.5        1×1+2×(−0.5)

                        3×(−2)+4×1.5        3×1+4×(−0.5)

   

        −2+3        1+(−1)              P × Q =        1             0  

        −6+6         3+(−2)                                  0             1

         P × Q = I

              Q × P =       −2        1                   1        2

                                  1.5     −0.5                3        4       

              Q × P =        (−2)×1+1×3                (−2)×2+1×4                        

                                  1.5×1+(−0.5)×3          1.5×2+(−0.5)×4       

               Q × P =       −2+3               −4+4              Q × P =     1        0

                                 1.5+(−1.5)        3+(−2)                             0        1          

               Q × P = I

            P × Q = I and Q × P = I then P = Q^-1 and Q = P^-1 (P and Q are inverse of each other)              

            Hence P is inverse of Matrix Q

LO3: Be able to apply the fundamentals of formal methods

C1. Suppose that two sets are A and B, defined by

A = { g, e, r, m, a, n, i }

B = { p, o, l, a, n, d }

Identify the following statements as true or false:

• a ∈ A = True since a is an element of set A
• b ∈ B = False because b is not an element in set B
• d ∉ B = False since d is an element of set B
• u ∉ A = True since u is not an element of A
• a ∈ A?B = Since a is an element of A?B hence true
• |A| = |B| = False since the count of elements in set A is 7 which is not equal to element count in set B 6.
• { i, r, a, n} ⊂ A = True as { i, r, a, n} is a subset of set A
• |A?B| = 8 =  False [ g, e, r, m, a, n, i, p, o, l,d] = 11 ∉8

C2. Suppose we have a universal set

{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27} and consider two sets P and O defined as follows:

P = “all multiples of 3”

O = “the first ten even numbers”

Represent all of the elements in a Venn diagram and identify the elements in P?O, P?O

and PΔO.

P = {3, 6, 9, 12, 15, 18, 21, 24, 27}

O= {2, 4, 6, 8, 10, 12, 14, 16, 18, 20}

Those elements that they make the union of elements in P and O.

where P?O = { 2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 27}

Those elements that they are common in P and O.

and P?O = {6, 12, 18}

PΔO= P?O \ P?O

VENN DIAGRAM

C3. For all of the following sets defined in set−theoretic notation, list out all of the elements:

S1 = {x: x = 2n, where 1 ≤ n ≤ 6}

Values for n = (1, 2, 3, 4, 5, 6)

Replacing with the n values in x = 2n => x = 2 × (1, 2, 3, 4, 5, 6)

S1 = {2, 4, 6, 8, 10, 12}

S2 = {x: x = 3n², where 1 ≤ n ≤ 5}

Values for n = (1, 2, 3, 4, 5)

Replacing with the n values in x = 3n² => x = 3 × [(1²), (2²), (3²), (4²), (5²)] =>

x = 3 × (1, 4, 9, 16, 25)

S2 = {3, 12, 27, 48, 75}

S3 = {y: y = 5n³, where 1 ≤ n ≤ 4}

Values for n = (1, 2, 3, 4)

Replacing with the n values in x = 5n³ => x = 5 × [(1³), (2³), (3³), (4³)] =>

x = 5 × [(1×1×1), (2×2×2), (3×3×3), (4×4×4)]    →      x = 5 × (1, 8, 27, 64)

S3 = {5, 40, 135, 320}

S4 = {x: x = √n, where 3 < n < 5}    →    Values for n  = 4

Replacing with the values for n in x = √n => x = √4

S4 = {2}

C4. For the circuit shown below, construct a truth table for each intermediate function. Hence, find the output function X.

A	B	C	B·C	A·B·C	Not
0	0	0	0	0	1
0	0	1	0	0	1
0	1	0	0	0	1
0	1	1	1	0	1
1	0	0	0	0	1
1	0	1	0	0	1
1	1	0	0	0	1
1	1	1	1	1	0

C5. Suppose that a salesman has 4 differently-located customers.

Find the number of different ways that the salesman can leave home, visit two different customers and then return home.

Write a pseudocode for calculating the answer for the previous section.

Algorithm: PermutationOfTwo(x,y)

Purpose: No of ways of y out of x with repetition
Pre: Given both x and y
Post: None
Return: permutation
{

 

          b    What is the Mode value of number of children’s per household?

        Mode value = None

D2. A company has ten sales territories with approximately the same number of sales people working in each territory. Last month the sales orders achieved were as follows:

              So 5^th no will be median
               =140

X = Sales	mean	X - mean	(X-mean)2	|X – mean |
150	150	0	0	0
130	150	-20	400	20
140	150	-10	100	10
150	150	0	0	0
140	150	-10	100	10
300	150	150	22500	150
110	150	-40	1600	40
120	150	-30	900	30
140	150	-10	100	10
120	150	-30	900	30

D3. Identify a topic in one of the following areas and conduct a research on its application in software development.

Propositional logic: That branch of mathematics which is mainly concerned with the binary terms ‘true’ and ‘false’ is popularly known as Propositional Logic. But its scope is not just limited to mathematics, and extends to the field of English and Computer Science also. We will be discussing here the uses of Propositional Logic in the area of computers. For a sound understanding of logic in day to day life, English language can be used to well organize it. Logic can be divided as: and (conjunction), or (disjunction), not (negation), and if (material conditional). Propositional calculus is the basis for categorizing in the field of computer science.

Computers and Propositional Logic: With the introduction of automation of computer systems , logical mathematics has strongly impacted the concept of Artificial Intelligence. With the realisation of the fact that in solving conclusive operations on the computer, logical inferences can be highly effective, this process was initiated.

When FOL (first order logic) was accepted as a fine way of evaluating and analyzing information, a same kind of representation was expressed to the pinnacle in actual. Research projects like Knowledge Based Software Assistant also uses it. Propositional Logic can help in transforming computer specifications into a diversified range of codes. The use of Propositional Logic extends in the field of High Frame Languages where complex semantics of Frame languages can be simplified, ‘KL One’ for instance in this regard. A calculus form had been derived which helps in analyzing the sets, subsets etc. which is a characteristics of this. Propositional Logic is linked with the field of computers in many ways, where the most effective in all are as presented here.

Attributes of a Propositional Logic: The attributes of Propositional Logic in the field of computer science are as follows:

• Syntax: It is a formula which can be linked with the syntactical objects in a well organised form.
• Semantics: A meaningful formula that depicts the association of logic is called Semantics.
• Proof theory: The manipulation of formula according to different set of rules is Proof Theory.

The science of atomic propositions is the appropriate definition of Propositional Logic in the field of computer science. Some complex issues termed as connectives that are associated with Propositional Logic, are discussed below for a better understanding.

And: It can be termed as a conjunctive form of Propositional Logic. For example, if A and B are two propositions, then they can be expressed as:

A∧B

For the whole system to be true, both A and B need to be true.

The concept of Boolean algebra is used to present it like the binary system.

Or: It is termed as a disjunctive form of Propositional Logic and the relation between A and B here can be expressed as:

A∨B

For the overall logical system to be true, either one of them should may be true.

If and Then: Propositional Logic having this combination is termed as conditional logic. If A and B are two arbitrary elements, then they can be expressed as:

A ⇒B

This is true only if either A is false or B is true, that is, a conditional relationship between A and B is depicted here.

Not: Above two considerations are binary while this is exclusively Unary and it is a negation statement, that is, totally opposite for an element.

When we write, ¬B, it means that if A is false, then it can be true.

Applications and Analysis: To calculate logical consequences, this concept is effectively applicable where tools like Proof Assistants follow logic that is based on propositions. An error-free proof script van be developed effectively because the validity of proofs can be checked with the development of a proof-checker.

Industry applications can take the benefit of languages like HML, Scheme etc. that are logic based. The designing of Java Card specifications uses an effective method of Jakarta Toolset. The ad-hoc model transformations that are highly structure based can be easily passed into virtual machine operations with the help of this Jakarta Toolset. Thus proof obligations can be developed effectively.

References

The Pythagorean Theorem[Online] Available at [HYPERLINK http://www.mathsisfun.com/pythagoras.html] [Accessed on 15/11/2014]
Graphical drawing [online] Available at [HYPERLINK
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[Accessed on 08/11/2014]
Mean, Median, Mode, and Range [online] Available at [HYPERLINK http://www.cimt.plymouth.ac.uk/projects/mepres/book8/bk8i5/bk8_5i2.htm] [Accessed on
15/11/2014]
Standard Deviation and Variance [online] Available at [HYPERLINK  http://www.quickmba.com/stats/standard-deviation] [Accessed on 08/11/2014]
Calculating Magnitude with Vectors [online] Available at [HYPERLINK http://hotmath.com/hotmath_help/topics/magnitude-and-direction-of-vectors.html] [Accessed on
15/11/2014]
Brown, F. (2003), Boolean Reasoning: The Logic of Boolean Equations, 1st edition, Kluwer Academic Publishers, Norwell, MA. 2nd edition, Dover Publications, Mineola, NY.

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