SUGGESTED SOLUTION
A1 – QUANTITAVE TECHNIQUES
NOVEMBER 2021
ANSWER 1
(a)
(i)
(ii)
(iii)
(iv)
(V)
(vi)
(vii)
(viii)
(ix)
(x)
C
A
B
C
C
A
B
A
C
D
(b)
(i)
(ii)
(iii)
(iv)
(v)
TRUE
TRUE
FALSE
TRUE
FALSE
c)
(i)
(ii)
(iii)
(iv)
(v)
B
E
I
C
G
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ANSWER 2
i.
Since the value of X and Y are known from the given table, the scatter plot is a plot on
an X-Y plane indicated by dots spread through the graph for values of x and Y as
indicated in figure 1 below:
Figure 1: Distribution of number of cars sold versus number of salespeople on duty
for a firm.
25
20
15
10
5
0
ii.
2
1
3
4
5
6
Salespeople on duty
7
8
9
If one is to use a method of least squares, then the following results are to be obtained:
∑𝑦 = 75, ∑𝑥 = 31∑𝑥𝑦 = 505
∑𝑦2=1261;∑𝑥 2=205
Ῡ=15,x = 6.2 and n =5
A simple regression line is given by y = a + bx where “a” is the y-intercept and “b” is
the slope parameter, all are estimates.
∑𝑥∑𝑦
Slope of the line is b = ∑𝑥𝑦 − 𝑛
_____________
∑𝑋2 −
(∑𝑥)2
𝑛
505 −
=
31(75)
5
__________
205 −
=
505−465
205−192.2
(31)2
𝑛
= 40⁄12.8
b = 3.125 or 3.1 to 1 decimal place
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and a
=
=
=
y̅– bx̅
15 – 3.1 (6.2)
15 – 19.22
a = -4.22 which is the required slope parameter
iii.
The least square line plotted in a Scattergram is as shown hereunder:
Figure 2: Scatter gram for data on number of cars sold versus number of
salespeople
25
20
15
10
5
-4
-3
0
1
2
3
4
5
6
Sales people on duty
7
8
9
vi.
In accordance with part (ii) of the question, the least square line can now be
written as follows:
ŷ=𝑎 + 𝑏𝑥 = −4.22 + 3.1𝑥
One number of cars sold= -4.22+3.1 (Number of salespeople on duty)
Now, if 5 sales people are kept on the show room floor each day, the number of cars
sold shall be predicated using the obtained equation as:
Number of cars sold =-4.22+3.1(5)
Number of cars sold =-4.22+15.5=11.28
Thus, the approximately number of cars sold shall be Eleven (11).
ANSWER 3
(a)
Tabular presentation of statistic data
•
It is the method of presenting data from masses of statistical data using complete
and comprehensive statistical table that are systematically organized in columns
and rows
•
That is, it facilitates a large amount of raw data to be sorted and organized in a
neat format. In doing so, the most relevant data will be included, facilitated
statistical and logical conclusions, and facilitates further statistical analyses.
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Graphical presentation of statistical data:
•
It is the use of charts and graphs to visually display numerical data.
•
Unlike the tabular method, the graphical method facilities presentation of data in
an attractive manner. In doing so, main features, comparisons, trends, and
fluctuations, can be easily understood.
(b)
Considering the given linear programming problem:
Formulating linear programming model
Let X be the number of radio ads and let Y be the number of TV ads
The objective function: 300,000X+900,000Y
Total number of ads: X+Y≥60
Radio ads:
X≥20
TV ads:
Y≥30
Thus, the LP model is
Minimize Cost=300,000X +900,000Y
Subject to
Y ≥ 60
≥ 20
Y ≥ 30
Y≥0
X+
X
X,
Graphical approach
Plot the constraints for the problem on XY-plane:
When X=0,Y=60 we have (0,60) and
When Y = 0, X=60 we have (60,0)
For X + 0Y =20 the line will cross X axis at the point (20,0)
For 0X+Y =30,the line will cross Y axis at the point (0,30)
(1mark)
Thus, the graph is
X = 10
60
Y = 30
30
X + Y = 60
10
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60
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The value of the objective function
For corner point A (10,50) =300,000(10) +900,000(50)= 48,000,000
For corner point B(30,30)=300,000(30)+900,000(30)=36,000,000
Therefore, the marketing manager should run 30 radio ads and 30 TV ads
to realize the minimum advertising cost of TZS36,000,000.
(c)
Considering the given information, we have
Arrival rate (λ) = 15 customers per hour
Service rate (µ) = 20 customers per hour
(i)
The probability that an arriving customer has to wait for service:
𝜆
Pw=µ
15
PW=20 = 0.75 or 75%
Therefore, the probability that an arriving customer has to wait for service is 75%.
(ii)
The average number of customers in the queue:
𝜆2
Nq = µ(µ−𝜆)
Nq=
(15)2
20920−15
225
=
= 2.25
100
Therefore, the average number of customers waiting in the queue is 2.25
customers.
(iii)
The average time a customer spends in the queue:
𝜆
Tq = µ(µ−𝜆
T𝑞 =
15
20(20−15)
=
15
100
= 0.15
Therefore, the average time a customer spends in the queue is 0.15 hours or 9 minutes.
QUESTION 4
Step 1: Reduce each row by its smallest element. The row wise reduced is as shown in the
table below:
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TASKS
OPERATORS
A
B
C
1
7
0
23
2
15
15
4
3
6
16
3
4
0
13
0
D
9
16
14
0
Step 2: Reduce the new matrix by its smallest element. The column wise reduced
is as shown in the table below:
TASKS
OPERATORS
A
B
C
1
7
0
23
2
11
11
0
3
3
13
0
4
0
13
0
D
9
12
11
0
Step3: Draw the minimum number of lines possible to cover all the zeroes in the
matrix given in the table
TASKS
OPERATORS
A
B
C
1
7
0
23
2
11
11
0
3
3
13
0
4
0
13
0
D
9
12
11
0
From this matrix; number of lines drawn Order of matrix
A further step has to be taken.
Step 4:
(a) Reduce all the uncrossed values by the lowest uncrossed value. In this example
the lowest uncrossed value is 3)
b)
Add this value to all value crossed by two lines leaving unaltered all values
which are crossed only by one line.
TASKS
OPERATORS
A
B
C
1
7
0
26
2
8
8
0
3
0
10
0
4
0
13
3
D
9
9
8
0
Step 5: Now, draw the minimum number of lines possible to cover all the zeroes.
Here the minimum number to cover all the zeroes is 4 which is equal to the order
of the matrix as shown in the following matrix:
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TASKS
OPERATORS
A
B
C
1
7
0
26
2
8
8
0
3
0
10
0
4
0
13
3
D
9
9
8
0
Step 6: Assign the Tasks to the operators:
TASKS OPERATORS
A
1
2
7
8
3
4
x
0
B
8
10
13
0
X
3
0
C
26
D
9
9
8
0
Therefore, the optimal assignment is:
Task
Operator
A
3
B
1
15
C
D
2
4
21
12
TOTAL
b)
Cost
19
67
Since we want to know whether the intelligence of students is above average, this is a
one tail test. Thus
The null hypothesis is
𝐻 0:µ=90
against
The alternative hypothesis 𝐻 1:µ ˃90
Assume 𝐻 O is true;
At the 5% level, the critical value is ± 1.64
At the 10% level, the critical value is ± 1.295
At the 1% level, the critical value is ± 2.325
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Acceptance
Region
4
3
2
1
1
0
2
4
3
Rejection Regions
The calculated value (Normal test) is Z =
𝑥− µ
𝜎/
√𝑛
=
115−90
15/
√40
=
25
2.37
= 10.54
Since 10.54 > 1.64 (or the calculated value falls in the rejection region), reject the null
hypothesis H0and accept H1at 5% level.
Thus, the principal’s claim that the students in his university are above average
intelligence is true.
c)
i.
2𝐷𝑆
Q =√
𝐻
Were:
D =
S =
H =
Stands for annual demand
Ordering cost
Holding cost
2𝐷𝑆
Thus, the economic order quantity Q = √
2 𝑥 5000 𝑥 15,000
=√
𝐻
4000
= 194 units
ii.
Total cost is given by Ordering cost plus Holding cost:
𝐷
That is C = 𝑄 𝑆 +
𝑄
2
ℎ=
5000
194
𝑥 15,000 +
194
2
𝑥 4000
386,598 + 388,000 = 774,598.
Thus, the total annual cost will be TZS.774,598/=.
________________  _______________
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