This document provides details about a proposed housing scheme project to construct houses for low-income families in a village. It includes information on the project scope, applicant selection criteria, construction plans, cost minimization analyses that will be performed, and a sample data collection and analysis on building materials. The project will use critical path analysis, linear programming, the Hungarian algorithm, and hypothesis testing to optimize the construction process and costs.

1. CARIBBEAN EXAMINATION COUNCIL
CARIBBEAN ADVANCED PROFICIENCY EXAMINATION
(C.A.P.E)
2012
APPLIED MATHEMATICS
Unit 2 Mathematical Applications
Project
Construction of Housing Scheme for Low Income
Families
Name of Candidate: ANEIZA KAYUM
Centre: Queen’s College
Centre Number: 090041
Candidate Number: 0900410493
Territory Guyana
©2016 Aneiza Kayum. All Rights Reserved.

2. INTRODUCTION
The population of Meten-Meer-Zorg, a village on the West Coast Demerara, Guyana, is
approximately four-hundred and sixty persons. Twenty-Five percent of these persons have a
family and do not own a home of their own. These persons are either living with family members
or renting homes. This community is one of the many depressed areas in Guyana in which
‘breadwinners’ can barely be able to cover daily living expenses.
This situation has influenced Mr. Bradley, a prominent businessman and a past community
member, to give back to this community. He intends to finance the construction of a housing
scheme for the low income earners in this community.
HOUSES
The houses to be constructed will be both wooden and concrete. Each house will be one storied
with two bedrooms, one bathroom, living room and kitchen. Electricity and water will be
provided, however the cost is to be covered by the ocupants.
APPLICANTS
Interested persons will have to fill out forms that requests information, both personal and work
related. All applicants will be required to provide proof that the information on provided on the
forms is true. The criteria for selecting persons to receive houses will be based on family size
and income levels.

3. CONSTRUCTION
Houses - This project will be undertaken by; Building Construction Co.
Fences – The construction of the fences will be done by local community masons.
INSPECTIONS
Inspections of the materials being used and the houses will be done by the Central Housing and
Planning Authorities.
STATEMENT OF TASK
Purpose/ Aim: To provide an analysis of the process of minimization of cost and time in the
construction of wood and concrete houses, using Critical Path Analysis, Linear Programming,
Hungarian Algorithm, and Normal Distribution with Hypotheses tests.
CRITICAL PATH ANALYSIS
This will be used to give earliest starting time and latest start time of each activity in the construction
process of each house (wood and concrete). It will also be used to give an overall time to finish each
house.
LINEAR PROGRAMMING
This will be used to aid Mr. Bradley in deciding how many houses to construct in order to
maximize land space. In total Mr. Bradley has 6 000 m2
of land available for this housing
scheme; he expects at most 16 wood houses (denoted as x) and at most 8 concrete houses
(denoted as y). He has already estimated that each plot of land to be allocated will be 300 m2
for
wooden houses and 200 m2
for concrete houses. The total number of wooden houses will not
exceed 16 and the total number of concrete houses will not exceed 8.

4. HUNGARIAN ALGORITHM
Four different groups of masons from the community were asked to quote the time it would take
them to complete each of the following tasks:
1) Digging of the foundation for the fences
2) Laying the Blocks
3) Plastering the Blocks
4) Installing Gates
These groups of persons will be responsible for the construction of the fences surrounding each
house. Each group will be given a different task 1,2,3,4, based on the least time it would take
them to complete the task. The assignment of the task will be done using the Hungarian
Algorithm.
NORMAL DISTRIBUTION AND HYPOTHESES TEST
Before the start of the project, inspections were carried out by the Central Housing and Planning
Authorities. The contractor claims that the mean mass of the blocks is 25 kg. The inspector
claims that the density of the blocks is weak and not suitable for this project.
Samples of 40 blocks were collected and their mass was recorded. The mean and variance were
calculated from this sample. A hypothesis test at five percent level of significance was used to
determine whether to accept the contractor’s claims and carry on construction or to reject his
claim and replace the blocks being used.

5. METHODOLOGY
The data for this project were collected through the use of questionnaires and sight surveys. A
questionnaire was used to gather data from the Manager of the construction company. The
information collected was based on costs of construction, number of workers to be employed, the
categories of workers and the time it would take them to finish different sections of each of the
houses.
Sight surveys were used to collect random sample of the weight of 40 blocks to be used in the
construction of the concrete houses.
Aims:
 To use critical path analysis to show the total time taken to complete one house and one
concrete house and to identify the critical activities.
 To use linear programming to derive the number of houses to be constructed in order to
maximize land space.
 To use the Hungarian Algorithm to allocate the four groups of workers to different tasks
in the construction of the fences so that the total time taken is minimized.
 To use the normal distribution and hypothesis test to decide whether to accept or reject
the Inspectors’ claim that the blocks being used are weak.

6. Linear Programming Graph
3x + 2y ≤ 60
Y
X ≤ 16
Y ≤ 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
30
28
26
2
4
6
8
10
12
14
16
18
20
22
24
Feasibility Region

7. The graph above shows the feasibility region (shaded in pink) to maximize the land space used in
the construction of the housing scheme.
The above diagram was derived from the following constraints:
1. 300x + 200y ≤ 6000 (divide throughout by 100,for simplification purposes, we get;
3x + 2y ≤ 60 )
2. x ≤ 16
3. y ≤ 8
4. y ≥ 0
5. x ≥ 0
The coordinates of the points of intersections of the feasibility are as follows:
x y
0 8
14.67 7.99
16 6
0 0
16 0
The number and type of houses that should be constructed to maximize land space are 16
wooden houses and 6 concrete houses.

8. Abbreviation Activities
A Gathering materials and Preparation of land.
B
Digging and casting of foundation.
This may also include setting beans (upright and
flooring)
C Roofing
D
Installing wood planks on walls and floor.
E Installing windows and doors.
F Interior walls such as room and bathroom walls.
G Wiring, plumbing, and installation of other amenities.

9. CRITICAL PATH NETWORK FOR THE WOOD HOUSES
2
12 5
.
0 2 1
12 1
2
7 2
7
Start A B
D
C
E
F
G
End
0 0
14 18
2 2
14 14
21 22
21 21
23 23
24 24

10. Critical Activities Activities Duration (in
days )
Predecessors Earliest Start
Time (EST)
Latest Start
Time (LST)
Float Time
 A 2 -
0 0 0

B 7 A 2 2 0
C 5 B 14 18 4

D 4 A, B 14 14 0
E 3 D 21 22 1
 F 5 D 21 21 0

G 3 C, E, F 23 23 0

11. Network Activity for Critical Path Analysis of Wood House
A A
B B
C C
D D
E E
F F
G G
A A
B B
C C
D D
E E
F F
G G
A A
B B
C C
D D
E E
F F
G G
A A
B B
C C
D D
E E
F F
G G
A A
B B
C C
D D
E E
F F
G G
A A
B B
C C
D D
E E
F F
G G
A A
B B
C C
D D
E E
F F
G G
A A
B B
C C
D D
E E
F F
G G

12. Abbreviation Activities
A Gather all materials needed
B Grading and preparation on of land
C Foundation digging- this also includes the casting of the
main columns and beans.
D Roofing
E Siding, laying blocks for walls, plastering walls.
F Installing windows and doors
G Installing bathroom, bedroom walls and other
amenities.

13. CRITICAL PATH ANALSYS FOR THE CONCRETE HOUSES
2 5 5
7
0 2 7 5 4 3
7
3
2
Start A B C D
F
G
E
End
14 14
0 0 2 2 9 9 14 14
19 19
22 22
9 16

14. Critical Activities Activities Duration (in days ) Predecessors Earliest Start Time Latest Start Time Float

A 2
_ 0 0 0

B 7 A 2 2 0

C 5 B 9 9 0

D 4 C 14 14 0
E 3 A, C 9 16 7

F 5
A, B, C
14 14 0

G 3 D, E, F 19 19 0

15. Network Activity for Critical Path Analysis of Concrete House
A A
B B
C C
D D
E E
F F
G G
A A
B B
C C
D D
E E
F F
G G
A A
B B
C C
D D
E E
F F
G G
A A
B B
C C
D D
E E
F F
G G
A A
B B
C C
D D
E E
F F
G G
A A
B B
C C
D D
E E
F F
G G
A A
B B
C C
D D
E E
F F
G G

16. HUNGARIAN ALGORITHM
1. Collect information based on the number of days it would take each group to complete
each activity.
Table above shows the time it would take each of group to complete each of the four activities.
2. Identify the minimum value from each row
1 2 3 4 Minimum Value
A 2 9 3 4 2
B 3 8 3 1 1
C 2 8 2 9 2
D 2 6 4 3 2
The diagram above shows the minimum value in each row
1 2 3 4
A 2 9 3 4
B 3 8 3 1
C 2 8 2 9
D 2 6 4 3

17. 3. Subtract each minimum value from each of the row values
1 2 3 4 Minimum Value
A 0 6 1 2 2
B 2 7 1 0 1
C 0 6 0 7 2
D 2 4 2 1 2
The values in the table above shows the minimum row value minus the individual values in
each row.
4. Identify the minimum value in each column
1 2 3 4
A 0 6 1 2
B 2 7 1 0
C 0 6 0 7
D 2 4 2 1
Minimum Value 0 4 0 1
The table above shows the minimum column value.

18. 5. Subtract the value from the individual values in the columns
The values in the table above show the minimum column value minus the individual values in
each column.
6. Shade rows and columns to cover as many zeros as possible.
The table above shows all zeros (whether in rows or in columns) shaded.
1 2 3 4
A 0 2 1 1
B 2 3 1 0
C 0 2 0 6
D 2 0 2 0
Minimum Value 0 4 0 1
1 2 3 4
A
0
2 1
1
B 2 3 1 0
C 0 2 0 6
D 2 0 2 0

19. 7. Pick the smallest number not shaded (in this case it is =1) and subtract this from each
unshaded value and add double it to each value on the shaded intersection.
1 2 3 4
A 0 1 0 1
B 2 2 0 0
C 4 2 0 8
D 6 0 2 2
The table above shows all the values after 1 has been subtract from each unshaded value and
double this value (2) is added to each value on the shaded intersection.
8. The following persons were chosen for the specific activity
1 2 3 4
A 0 1 0 1
B 2 2 0 0
C 4 2 0 8
D 6 0 2 2
Hence, group A has to complete activity one, group B has to complete activity four, group C
has to complete activity three and group D has to complete activity two.

20. 22.4 1 22.4 501.76
23.7 2 47.4 1123.38
23.9 6 143.4 3427.26
24.2 12 290.4 7027.68
24.5 4 98 2401
25.2 6 151.2 3810.24
25.3 2 50.6 1280.18
26.5 3 79.5 2106.75
27.3 4 109.2 2981.16
∑= 223 ∑= 40 ∑=992.1 ∑=24659.41
Assignment of Activities
Group Activity
A Digging of the foundation for the fences
B Laying the Blocks
C Plastering the Blocks
D Installing Gates
The table above shows the name of the groups and the activities they are assigned to.
Normal Distribution
Table showing the mass of 40 blocks in kilograms collected from the construction site.
x represents the mass of each block.
f represents the frequency of the mass of each block.

21. The mean (𝑥̅) =
∑(𝑓𝑥)
∑(x)
= 992.1/40 = 24.802 kg
The Variance = (
∑ (𝑓𝑥2)
∑(f)
) − ( 𝑥̅2) = (
24659.41
40
) − (24.802)2
= 1.346046 kg
X ~ N 





40
346.1
,25
It is assumed that the variance of the sample is a good estimator of the population variance.
Hypothesis test
Using the hypothesis test, we can derive the null and alternative hypotheses as follows:
H0: The mean mass of the blocks used in the construction is 25 kg, (µ = 25 kg).
H1: The mean mass of the blocks used in the construction is not kg, ( µ ≠ 25 kg).
𝑋̅~N (25,
1.346
40
)
Using a two tailed test at a 5% level of significance we get the following:
95%
ACCEPTANCE REGION
-a a
-1.96 1.96
REJECTION
REGION 2.5%
REJECTION
REGION 2.5%

22. P (Z<a) = (95+2.5) %
P (Z<a) =0 .975
ɸ a = 0.975
a = ɸ-1
0.975
Hence, a = 1.96 , and –a = -1.96
This implies that we reject the H0 if Z statistic < -1.96 or if Z statistic is >1.96.
Using the formula:
Z statistic =
24.802−25
√(
1.346
40
).
. = 1.079 > 1.96
The Z statistic value falls within the acceptance region, so we can conclude at a 5% level of
significance that there is supporting statistical evidence that the claim made by the contractor is
true. Hence, we continue the construction process after the blocks have been replaced with more
dense ones.

23. CONCLUSION
 The total time for the completion of a wood houses is 24 days. During the construction
process Roofing, Installing windows and doors are not critical activities and can be
delayed for 4 days and 1 day respectively, without affecting the construction process in
any way. The critical activities are Gathering materials and Preparation of land, Digging
and casting of foundation (this may also include setting beams (upright and
flooring)),Wiring, plumbing, and installation of other amenities, Interior walls such as
room and bathroom walls, installing wood planks on walls and floor.
 Through this critical path analysis it can be noted that the total time it would take to
complete building a concrete house is 22 days. It can also be noted that the siding,
laying blocks for walls, plastering walls has a float time of seven days; hence this activity
can be delayed for seven days without affecting any other activity. The Critical Activities
are installing windows and doors, Installing bathroom, bedroom walls and other
amenities, gather all materials needed, Grading and preparation of land, Foundation
digging- this also includes the casting of the main columns and beans, Roofing.
 To maximize land space 16 wooden houses and 6 concrete houses should be
constructed.
 To minimize the time for the construction of the fence, group A has to complete activity
one, group B has to complete activity four, group C has to complete activity three and
group D has to complete activity two.
 The claim made by the inspector was true, so the blocks being used for the construction
should be replaced with stronger more dense blocks.

24. REFERENCES
Bloomfield, I & Stevens, J. (2002). Discrete & Decision. United Kingdom: Nelson Thrones.
Mahadeo, Rudolph D. Statistical Analysis, A Comprehensive Text (First Edition). Trinidad, W. I.:
Caribbean Educational Publishers 2003 Ltd.

25. ©2016 Aneiza Kayum. All Rights Reserved.