Logical Analysis And Problem Solving Assignment Sample in UK

Table of Contents

Introduction to Logical Analysis And Problem Solving Assignment

4 Pages 877 Words

Introduction to Logical Analysis And Problem Solving Assignment

Section 1:

1. Task 1

Stage 1 of the circuit involves three gates NR2, AN2 and OR2. The input of NR2 is A and B`. The output of NR2 would be (A+B`)` which is equivalent to (A`+B). The input of AN2 is A and B` and its output is A.B`. The input of OR2 is A and B` and its output would be A+B`. In the next stage of the circuit, NR2 is present and its input is (A`+B) and A.B`. The output of this gate would be ((A`+B)+(A.B`))`. In the third stage of the system, ND2 is present and the input is x and y where x is ((A`+B)+(A.B`))` and y is (A+B`). The final output of the concerned gate would be (x.y)` which is denoted by Z. Based on the calculations performed using the properties of a boolean expression, the table below provides the view of the truth table for the concerned circuit.

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A	B	A`	B`	*Z = (A`+B+A.B`)*
0	0	1	1	1
0	1	1	0	1
1	0	0	1	1
1	1	0	0	1

Truth Table for the above circuit

(Source: Self-Created)

2. Task 2

The input given to the first AND gate is A` and B and its output would be A`.B. The input given to the second AND gate is A, C` and D and therefore, its output would be A.C`.D. The input given to the third AND gate is A, B` and C which in turn would generate the output as A.B`.C. All these outputs are applied to the OR gate present in the circuit and therefore, the output of the gate is (A`.B) + (A.C`.D) + (A.B`.C). This is the required sum of products that can be generated on the basis of the inputs that have been applied.

3. Task 3

2x₁ + 6x₂+ 4x₃ = 18…….1

5x₁ + 6x₂ + 4x₃ = 24……..2

-2x₁+ x₂ + 3x₃ = 3……..3

From equation 1, the value of x3 can be calculated as

x3 = (18 - 2x1 - 6x2) /4 ……..4

Putting the value of x3 in equation 2 and 3, we get,

5x1 + 6x2 + 4*(18-2x1-6x2)/4 = 24

5x1 + 6x2 + 18-2x1-6x2 = 24

3x1 = 24-18 = 6

x1 = 6/2 = 3

Putting the value of x1 in equation 1 and 3, we get,

2*3 + 6x2 + 4x3 = 18 [Equation 1]

6x2 + 4x3 = 12………5

-2*3 + x2 +x3 = 3 [Equation 3]

x2 + x3 = 9……..6

By solving the equations 5 and 6, the values of x2 and x3 are -12 and 21 respectively. Therefore, the values of x1, x2 and x3 are

x1 = 3, x2 = -12 and x3 = 21

4. Task 4

The total price of the mobile phone and smartwatch = £ 1,250

Consider the cost of the smartwatch to be x, therefore, the cost of the mobile would be 3x.

a) The expression to show the total cost in terms of the cost of the smartwatch is given below.

x + 3x = 1250…….1

b) Solving equation 1, we get,

x = 1250/4 = 312.5

Therefore, the cost of the smartwatch is £ 312.5 and for the mobile, the cost is 312.5*3 which is equivalent to £ 937.5.

c) VAT rate for the mobile phone is 14%, therefore, VAT charged on the customer is

VAT = (14/100)*937.5 = £ 131.25

5. Task 5

The values of the variables given in the given equation are t1 = 2.6, t2 = 0.45, A = 15.6, a = 12.57/144 , g = 32.2, h1 = 36 and h2 = 25. Putting the values of the above variables in the equation, the value of k can be calculated as follows,

2.6 - 0.45 = (2*15.6)/ 0.0872 (√(1+k)/2*32.2)(√36 - √25)

2.15 = (31.2)/0.0872 (√(1+k)/64.4)(6-5)

2.15 = 357.79(√(1+k)/64.4)

√(1+k)/64.4 = 0.006

(1+k)/64.4 = 0.000036

1+k = 0.0023

k = 0.0023 - 1 = -0.9977

Therefore, the value of k comes out to be -0.9977.

6. Task 6

The equation given in the question is 1.2v + 0.64u = 0.85. The value of u ranges from 5 to 100. The table below provides the view of the values of v for corresponding values of u using the above equation.

u	*v = (0.85 - 0.64u)/1.2*
5	-1.95
10	-6.04
15	-7.29
20	-9.95
30	-15.29
40	-20.625
50	-25.95
60	-31.29
70	-36.625
80	-41.95
90	-47.29
100	-52.625

Table 2: Table for showing the values of v for corresponding values of u

(Source: Self-Created)

The figure below provides the view of the graph generated in Excel showing the relation between u and v.

Relation between u and v

7. Task 7

The equation given in the question is 2u + 7v = 52. The value of v ranges from 2 to 50. The table below provides the view of the values of u for corresponding values of v using the above equation.

v	*u = (52 - 7v)/2*
2	19
4	12
6	5
8	-2
10	-9
15	-26.5
20	-44
30	-79
35	-96.5
40	-114
45	-131.5
50	-149

Table 3: Value of u for corresponding values of v

(Source: Self-Created)

Relation between u and v

The figure above provides the view of the relationship existing between u and v as per the equation given in the question.

8. Task 8

As per the details in the questions, A indicates Returning Customer and B indicates Discount Code.

a) The truth table showing when the customer would be able to get the discount is shown below.

A	B	Y
0	0	0
0	1	1
1	0	0
1	1	1

Table 4: Truth table

(Source: Self-Created)

The above truth table provides the idea that whenever the value of B which stands for discount codes is True, then the output of the system is also True.

b) The expression for the given truth table is

Y = A`.B + A.B`

Section 2:

9. Task 9

Monthly rent for 24 students are 1500, 1350, 350, 1200, 850, 900, 1500, 1150, 1500, 900, 1400, 1100, 1250, 600,610, 960, 890, 1325, 900, 800, 2550, 495, 1200, 690.

Figure 3: Calculation of Relative Frequency for students' monthly rent

(Source: Self-Created)

Histogram

The table above provides the view of the relative frequency calculation performed in Excel for the given monthly rent of the students.

10. Task 10