Construction-materials-and-management.pdf

Construction Materials and Management Solutions for Volume : I Classroom Practice Questions Chapter- 3

Chapter- 5

Introduction to Project Management

07. Sol: a)

Ans: 13 weeks

13. Ans: (a) t  4t L  t P Sol: t E  o 6 8  4  9  13  = 9.5 6 2 2  t p  t o   13  82  2    Variance,      6   6  25 2  36

A B Activity C D E F G H 0 2 4 6

PERT

8 10 12 13 14

Time (weeks)

29. Sol: (a)

Total completion time of period = 13 weeks

Chapter- 4 Elements of Networks 12. Sol:

1

A B C

13. Sol:

1

2 D 4 3

A B C

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F

(b) 5

E 30. Sol: a)

3 D 2 4

5 E

Ans: 50% & 95.2%

TE = 24 month  = 3.6 T  TE Z= S  24  24 Z 0 3.6 From table probability = 50% 30  24 3. 6 Z = 1.67 From table probability = 95.2% Z

Ans: 65.76 & 58.25 weeks

TE = 60 weeks 

2 = 20.25

P (%) = 90 %  Z = 1.28

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:2:

Critical path:

variance  20.25

Std deviation =

1–2–5 = 1–2–4–5= 1–3–4–5= 1–3–5 =

 = 4.5

1.28 

TS  TE 4.5

TS = 65.76 weeks

10 + 7 10 + 9 + 11 9 + 8 + 11 9+5

= 17 = 30 = 28 = 14

Critical path = 1 – 2 – 4 – 5 TE = 30 days; TS = 36 days

b)

31.

Solutions for Additional Syllabus

P(%) = 35% For P(%) = 35 % From table, Z =  0.387 T  TE  0.387  s 4.5 TE = weeks = 58.25 weeks

TS  TE  To calculate  :-

Ans: 90.32%

36  30 = 1.27 ≃ 1.3 4.69 From the table Z = 1.3  probability = 90.32 %

Z=

 2 of critical path

 2 = 4 + 9 + 9 = 22  = Z=

t  4t L  t P tE = o 6

Sol:

6-9-18 10 1 9 5-8-17

2

 tp  to  =   6

  

2

4-7-10 7 4-10-22 4 11 5 2-5-8

2

9 4-7-22 4-7-16 8 3

22 = 4.69

Chapter- 6 CPM 5

27. Ans: (c) Sol: 1

TE =1

TE =4

1

3

4

4

TE =0

Activity 1-2 1-3 2-4 3-4 2-5 4-5 3-5

to tL  tP 6-9-18 5-8-17 4-7-22 4-7-16 4-7-10 4-10-22 2-5-8

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tE 10 9 9 8 7 11 5

2 4 4 9 4 1 9 1

6

5

1 4

TE =4

2

TE =11

2 4

5 TE =9

 Earliest start time for activity 5  6 = 9 days

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:3: 12. Ans: (c)

29. Ans: (c) TE =5 Sol: B

Sol: From the data given, the maximum time for

TL =8

5 TE =0 6

the project is 11 days and minimum time is

4

TE =6

A TL =0

Solutions for Additional Syllabus

TE =18

TE =12 6

C TL =6

D TL =12

6

TE =9 3

9 days.

F TL =18 7

E

For 11 days, the total direct cost is: 800 + 1200 + 500 = 2500 units.

TL =11

For 10 days, the total direct cost is: FT(AB) = LST EST EST = 0 LST = 8 5 = 3 FT(AB) = 3 FT(CE) = LST  EST = (11 3)  6 = 2 FFEF   TEj  TEi  t ij





= 18 9 7 =2

800 + 1350 + 500 = 2650 units. For 9 days, the total direct cost is: 900 + 1500 + 500 = 2900 units. The feasible range of total direct cost varies from 2500 to 2900.

Chapter- 7 CPM Cost Analysis

13. Ans: (a) Sol: From the given diagram, on the 21st & 22nd day three concurrent activities are there with a total resources of 6 + 7 + 9 = 22.

11. Ans: (a) Sol: 9th

Parallel Activities A

Total Resource Load 6

11th

A+B

6 + 4 = 10

13th

A + B+ D

6 + 4 + 7 = 17

A+B+C+D

6 + 4 + 3 + 7 = 20

Week

th

15

Minimum resource occurs when only one activity exists. In the present case it is 6 per day.  Maximum resources is 22 and minimum is 6

From the above, the maximum resource load per week is 20

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