USING AND MANAGING DATA & INFORMATION
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Contents
Introduction ..................................................................................................................................... 3
Task 1 .............................................................................................................................................. 4
Task 2 .............................................................................................................................................. 6
Task 3 .............................................................................................................................................. 8
Task 4 ............................................................................................................................................ 10
Task 5 ............................................................................................................................................ 12
Conclusion .................................................................................................................................... 15
References ..................................................................................................................................... 16
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Introduction
This report focuses on the use and management of data and information. It encompasses varied
segments of data and information management. The first task includes the basic arithmetic
procedures consisting addition, subtraction, multiplication, division, the unitary method, and
percentage calculations. The second task focuses on statistical procedures like the average and also
requires the thorough understanding and application of the summation operator. The third task
deals with combination and permutation procedures. Task 4 brings the coordinate geometry in the
scene. The last task introduces the linear programming model formulation and solving for the
optimal output given the various constraints. It also requires to show the application of the
coordinate geometry through the plotting of the constraints on the Cartesian plane and drawing out
the feasible region from the graph.
Overall the five tasks combined, takes the in depth test of the understanding and application of
different data and information management techniques.
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Task 1
These problems require the use of the unitary method. It is an arithmetic procedure in which the
value of a single unit is calculated from the value of multiple units and then that single unit value
is used to calculate the value of some other multiple units (Sancheti and KAPOOR, 2008).
There are essentially two steps in unitary method. The first step is to get the single unit value and
the second step is to derive the ultimate multiple units’ value leveraging that single unit value.
a) The problem in discussion asks to find out the duration of the call that cost Jill £6.00.
Step 1:
The first piece of information given in the problem is that phone calls to NY costs £2.50 for a 5minute call. Here the duration of a call is expressed as in multiple units of Pound. First the duration
of call for a single unit of Pound needs to be calculated. A division operation is required for that.
A call that costs £2.50 lasts for 5 minutes
So, a call that costs £1.00 lasts for (5/2.5) minutes or 2 minutes
Step 2:
Now as the value of the single unit that is the duration of the call for a single unit of Pound has
been calculated, this piece of information can be used in order to calculate the desired duration of
call that Jill made for £6.00. A multiplication operation is required for that.
A call that costs £1.00 lasts for 2 minutes (from step 1)
So, a call that costs £6.00 lasts for (2×6) minutes or 12 minutes
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b) The second problem can be solved by using simple multiplication and division operations on
the given information.
Step 1:
The buyer of mortgage can make 360 monthly payments each of £680
So, the total payments need to be made to settle the mortgage is (360×680) or £244,800
Step 2:
The total payments need to be made to settle the mortgage is £244,800 (from step 1)
So, if each payment is of £600, the total number of monthly payments would be (244,800/600) or
408
This procedure is actually a pseudo unitary method itself. If the words were arranged a bit
differently, it would appear as a proper unitary procedure.
c) The third problem requires the understanding of the concept of percentages. At 20% discount
the car was being sold for £1,800. That is, if the original price was £100, then at 20% discount it
was being sold at £80.
Step 1:
If the after discount price is £80, the before discount original price is £100
So, if the after discount price is £1, the before discount original price is (100/80) or £1.25
Step 2:
If the after discount price is £1, the before discount original price is £1.25 (from step 1)
So, if the after discount price is £1,800, the before discount original price is (1.25×1,800) or
£2,250
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Task 2
The data of the height and weight of ten athletes are given here. Some statistical procedures are
required to be conducted. The height variable is expressed as x here and the weight variable is
expressed as y. The population size is 10 which will be denoted as n.
a) This part asks for computing the sum of all the x values. The calculation can be expressed as:
∑x = x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10
= 1.79 + 1.57 + 1.70 + 1.85 + 1.69 + 1.75 + 1.72 + 1.61 + 1.65 + 1.77
= 17.1
b) This part asks for computing the sum of all the y values. The calculation can be expressed as:
∑y = y1 + y2 + y3 + y4 + y5 + y6 + y7 + y8 + y9 + y10
= 87 + 68 + 76 + 92 + 71 + 83 + 77 + 71 + 76 + 82
= 783
c) This part asks for computing the sum of all the xy values. The calculation can be expressed as:
∑xy = xy1 + xy2 + xy3 + xy4 + xy5 + xy6 + xy7 + xy8 + xy9 + xy10
= 155.73 + 106.76 + 129.2 + 170.2 + 119.99 + 145.25 + 132.44 + 114.31 + 125.4 + 145.14
= 13,389.3
d) This part asks for computing the sum of all the x2 values. The calculation can be expressed as:
∑x2 = X21 + X22 + X23 + X24 + X25 + X26 + X27 + X28 + X29 + X210
= 3.2041 + 2.4649 + 2.89 + 3.4225 + 2.8561 + 3.0625 + 2.9584 + 2.5921 + 2.7225 + 3.1329
= 292.41
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e) Average height (arithmetic mean) of the students, 𝑥̅ =
∑𝑥
𝑛
= 17.1/10
= 1.71
f) Average weight (arithmetic mean) of the students, 𝑦̅ =
∑𝑦
𝑛
= 783/10
= 78.3
g) To be calculated: 𝑛 × (∑ 𝑥𝑦) − ∑ 𝑥 × ∑ 𝑦
= {10 × (xy1 + xy2 + xy3 + xy4 + xy5 + xy6 + xy7 + xy8 + xy9 + xy10)} - {(x1 + x2 + x3 + x4 + x5 +
x6 + x7 + x8 + x9 + x10) × (y1 + y2 + y3 + y4 + y5 + y6 + y7 + y8 + y9 + y10)}
= 10 × (155.73 + 106.76 + 129.2 + 170.2 + 119.99 + 145.25 + 132.44 + 114.31 + 125.4 + 145.14)
– (1.79 + 1.57 + 1.70 + 1.85 + 1.69 + 1.75 + 1.72 + 1.61 + 1.65 + 1.77) × (87 + 68 + 76 + 92 + 71
+ 83 + 77 + 71 + 76 + 82)
= 10 × 13,389.3 – 17.1 × 783
= 133,893 – 13,389.3
= 120,503.7
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Task 3
The problems under this task require the use of the concept of combination. The formula for
calculating the total number of possible combinations is:
𝑛𝐶𝑟 =
𝑛!
𝑟! (𝑛 − 𝑟)!
Where n is the population size and r is the desired combination size (Levine, Szabat, and Stephan,
2016).
a) There are 20 students and the question asks how many combinations of 3 students from the 20
are possible to be made. So the total number of possible combinations:
20!
20𝐶3 = 3! (20−3)!
20!
= 3! 17!
=
=
=
20×19×18×17!
3! 17!
20×19×18
3!
20×19×18
3×2×1
= 1,140
b) The lottery involves choosing 5 different number between 1 and 30. The order in which the
numbers are chosen does not matter. So, the number of possible combinations:
30!
30𝐶5 = 5! (30−5)!
30!
= 5! 25!
=
=
=
30×29×28×27×26×25!
5! 25!
30×29×28×27×26
5!
30×29×28×27×26
5×4×3×2×1
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= 142,506
So to explain the procedure, it follows the law of combinations. As the lottery draws the numbers
between 1 and 30 to form a 5 number combination, the total number of possible combinations can
be calculated simply using the formula of combination which is:
𝑛𝐶𝑟 =
𝑛!
𝑟! (𝑛 − 𝑟)!
The probability of winning the jackpot can be calculated as:
Number of favorable combination
Number of total combination
The number of favorable combination is 1, which is the combination of 5 numbers that has been
chosen. The total number of combinations is 142,506 as calculated above.
So the probability of winning the jackpot is (1/142,506) or 0.0007%
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Task 4
a) The equation of a line is given as y = 3x + 4. This equation is called a linear equation which is
here presented in the slope-intercept form. The slope-intercept form of a linear equation is denoted
as: y = mx + b where y and x are variables in discussion, m is the slope and b is the intercept of
the line.
The slope of a line equals to the change in y variable divided by the change in x variable. It can be
calculated as:
𝑚=
(𝑦2 − 𝑦1)
(𝑥2 − 𝑥1)
The slope of a line shows the rise of line in the y-axis of the Cartesian plane along with the run of
the line in the x-axis. The slope of the line is also called a gradient.
Intercept of a line is essentially, the value of y variable given that the x variable equals to zero.
In the given equation y = 3x + 4, 3 is the value of the slope and 4 is the value of the intercept. The
slope value of 3 indicates that for a single unit of change in x variable will cause the y variable to
change by 3 units. The intercept value of 4 indicates that even if the x variable is zero, the y variable
would have a value of 4.
b) The following graph is provided which contains a line with two coordinates: (-2, 0) and (0, 3).
Figure 1: Line with two coordinates (Source: Given)
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i) The gradient or slope of the line,
𝑚=
(𝑦2 − 𝑦1)
(𝑥2 − 𝑥1)
(3-0)/ {0 - (-2)}
= 3/2
= 1.5
ii) The intercept, b = 3 and the slope m = 1.5.
So the linear equation written in the form of y = mx + b:
y = 1.5x + 3
iii) When x = 6, the calculated value of y variable would be:
y = 1.5×6 + 3
y= 9 + 3
y = 12
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Task 5
Linear programming is defined as a mathematical technique that management follows to fulfill
firm-specific objectives given a set of constraints (Hillier, 2012). Linear program includes the
following elements:
I. Control Variables: These are variables under the decision maker’s control. These variables
directly appear in the mathematical models that are formulated.
II. Objective Function: Every linear program has a linear objective function which represents the
goal either to be minimized or maximized.
III. Constraints: A mathematical inequality (an inequality constraint) or equality (an equality
constraint) that must be satisfied by the variables in the model.
IV. Non-negative Constraints: These conditions are those in a model which stipulate that the
decision variables can have only nonnegative values (Taha, 2011).
To summarize the problem given in the task:
Particulars
Base Unit (B)
Cabinet Unit (C)
Available Time
Production Department
60 minutes
25 minutes
840 minutes
Assembly Department
30 minutes
20 minutes
480 minutes
£20
£30
Profit
Control Variables: Control variables in the stated problems are:
B = Number of units of base units to be produced
C = Number of units of cabinet units to be produced
Objective: The objective of the problem is to maximize the profit gained by the sales of the two
different products. The profit function for the problem is: 20B + 30C.
To max 20B + 30C
Constraints: There are two constraints related to the available production and assembly hours each
day.
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60B + 25C ≤ 840 [Production Department Constraint]
30B + 20C ≤ 480 [Assembly Department Constraint]
B, C ≥ 0 [Non-negativity Constraint]
Now for developing the feasible region in the graph, each of the production and assembly
constraints need to be solved.
Constraint 1: 60B + 25C ≤ 840
If C = 0, B = 14
If B = 0, C = 33
Constraint 2: 30B + 20C ≤ 480
If C = 0, B = 16
If B = 0, C = 24
Due to the non-negativity constraint, the above values should be plotted on the non-negative
quadrant of the Cartesian plane. Base unit (B) is shown on the x-axis and cabinet unit (C) is shown
on the y-axis.
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Figure 2: Drawing graph of the feasible region (Source: Prepared by author)
Both the constraints are less than constraints and these are satisfied below the line. The above
graph shows the feasible region ABCD (shaded area). The four extreme points of the feasible
region are:
A= (0, 24); B= (0, 0); C= (14, 0); D= (11, 8)
After that the optimality test is performed by putting each co-ordinate value into the objective
function:
A= 20 × 0 + 30 × 24 = 720
B= 20 × 0 + 30 × 0 = 0
C= 20 × 14 + 30 × 0 = 280
D= 20 × 11 + 30 × 8 = 460
So, the profit maximizing quantities are 0 units of B and 24 units of C. Also the maximum profit
to be achieved at this optimal solution is £720.
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Conclusion
This report focuses on the use and management of data and information. It encompasses varied
segments of data and information management. There were a total of five tasks to perform. Overall
the five tasks combined, takes the in depth test of the understanding and application of different
data and information management techniques.
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References
Hillier, F.S., 2012. Introduction to operations research. Tata McGraw-Hill Education.
Levine, D.M., Szabat, K.A. and Stephan, D., 2016. Business statistics: A first course. Pearson.
Sancheti, D.C. and KAPOOR, N., 2008. Business Mathematics. Sultan Chand.
Taha, H.A., 2011. Operations research: an introduction (Vol. 790). Upper Saddle River, NJ, USA:
Pearson/Prentice Hall.
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