selected problems solution - summer 1433-34h

King Saud University GE 404 Engineering Management

Civil Engineering Department Selected Problems Solution Summer semester 1433/ 1434H

1

Question 1 A small project activities and related information are given in table

(1).

a) Develop the AOA, AON, TSN

b) Calculate ES,LS,LS, AND LF and Find the critical path using

AON network

c) Find Gantt chart for each

d) Consider calendar date, the project starts Saturday, first of the

month and Friday is holiday. Draw the bar chart and

determine the project finish date and needed working days

e) Considering the constraints of ZZBD 3,2, FSCF 2, SSFG 4, SFEG 3; Determine:

i. Network data of forward and backward calculations and Total Float (TF)

ii. Critical path of the network. How it is differ from question 3b?

f) Develop load diagram of the resource, and find

i. average daily requirement;

ii. Critical Index

iii. Effectiveness value

iv. use Burgess Algorithm to level resource

Solution

a) AOA AON

a) TSN

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

A,2

B,5 D,10 E,3 G,8

C,1 F,6

Table (1)

Activity Immediate

predecessor(s)

Time,

days

Man-

power Equipment

A - 2 2 E1

B - 5 3 E1

C - 1 1 E2

D B 10 3 E2

E A, D 3 4 E3

F C 6 5 E4

G E, F 8 3 E5

1

2

3

4

6

5

7 8

A, 2

B, 5

C, 1

E, 3

D, 10

F, 6 G, 8

A

,

START B

,

C

,

D

,

E

,

F,

6

G

,



2

b) AON and critical path [carry calculation]

c) Gantt chart T 01 02 03 04 05 06 07 08 09 10 11 12 13 13 15 16 17 18 19 20 21 22 23 24 24 26

B B

D D

E E

G G

A A A

C C C

F F F

d) Calendar schedule July

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31

F S S M T W T F S S M T W T F S S M T W T F S S M T W T F S S

B

D D D

E

G G

A

C

F F

Activity A B C D E F G

Free Float 13 0 11 0 0 11 0

Total Float 13 0 11 0 0 11 0

A

0 2

13 2 15

C

0 1

11 1 12

B

0 5

0 5 5

D

5 1

5 1 1

G

18 26

18 8 26

E

15 18

15 3 18

F

1 7

12 6 18

0 0

0 0 0

START

L

S

E

S

CRITICAL



3

e) Solution with constraint (lag times)

Forward pass:----------------------------------------------------------------------------------------------

Activity A: ESA=0; EFA= ESA+DA=0+2=2; -------------------------------------------------------------------------------

Activity B: ESB=0; EFB= ESB+DB=0+5=5; -------------------------------------------------------------------------------

Activity C: ESC=0; EFC= ESC+DC=0+1=1; -------------------------------------------------------------------------------

Activity D: ESD=ESB+SSBD=0+3=3 OR EFB+FFBD-DD=5+2-10=0 Then ESD=3; EFD= ESD+DD=3+10=13--

Activity E: ESE=13; EFE= ESE+DE=13+3=16----------------------------------------------------------------------------

Activity F: ESF=EFC+FSCF=1+2=3; EFF= ESF+DF=3+6=9-----------------------------------------------------------

Activity G: ESG=ESE+SFEG-DG=13+4-8=9 OR ESF+SSFG=3+4=7 Then ESG=9; EFG= ESG+DG=9+8=17----

Backward pass: --------------------------------------------------------------------------------------------

Activity G: LFG= EFG =17; LSG= LFG - DG =17-8=9; ------------------------------------------------------------------

Activity F: LFF=LSG-SSFG+DF =9-4+6=11; LSF= LFF – DF =11-6=5------------------------------------------------

Activity E: LFE=LFG-SFEG+DE =17-4+3=16; LSE= LFE – DE =16-3=13--------------------------------------------

Activity D: LFD=13; LSD= LFD – DD =13-10=3; -------------------------------------------------------------------------

Activity C: LFC= LFF-FSCF = 10-2=8; LSC= LFC – DC=8-1=7--------------------------------------------------------

Activity B: LFB=LSD-SSBD+DB=3-3+5=5 OR LFD-FFBD=13-2=11 Then LFB=5; LSB= LFB – DB =5-5=0-----

Activity A: LFA=13; LSA= LFA – DA =13-2=11; --------------------------------------------------------------------

Activity A B C D E F G

Free Float 11 0 7 0 0 2 0

Total Float 11 0 7 0 0 2 0

A

0 2

11 2 13

C

0 1

7 1 8

B

0 5

0 5 5

D

3 13

3 10 13

G

9 17

9 8 17

E

13 16

13 3 16

F

3 9

5 6 11

0 0

0 0 0

START

jjj

jiji

jiji

iji

iji

ij

DESEF

DSFES

DFFEF

SSES

FSEF

allMaxES

SSFORWARD PA

]2[

Time Initial

)(]1[

iii

iijj

iijj

ijj

ijj

ji

DLFLS

DSFLF

DSSLS

FFLF

FSLS

e

allMinLF

ASSBACKWARD P

]4[

Tim Terminal

)(]3[



4

f) Resource and leveling

TI 01 02 03 04 05 06 07 08 09 10 11 12 13 13 15 16 17 18 19 20 21 22 23 24 24 26

B- B

D- D

E- E

G- G

A- A

C- C

F- F

T 01 02 03 04 05 06 07 08 09 10 11 12 13 13 15 16 17 18 19 20 21 22 23 24 24 26

B B

D D

E E

G G

A A

C C

F F

TI 01

02 03

04

05

06 07 08 09 10 11 12 13 13 15 16 17 18 19 20 21 22 23 24 24 26

R 6 10 8 8 8 8 8 3 3 3 3 3 3 3 3 4 4 4 3 3 3 3 3 3 3 3

ΣR

6 16 24

32

40

48 56 59 62 65 68 71 74 77 80 84 88 92 95 98 101

104

107

110

113

116

R2 36

100

64

64

64

64 64 9 9 9 9 9 9 9 9 16 16 16 9 9 9 9 9 9 9 9

Σ R2 648

TI 01

02

03

04

05

06

07

08

09

10

11

12

13

13 15 16

17

18

19

20

21 22 23 24 24 26

R 3 3 3 3 3 3 3 3 3 3 3 4 8 10 10 9 9 9 3 3 3 3 3 3 3 3

ΣR

3 6 9 12

15

18

21

24

27

30

33

37

45

55 65 74

83

92

95

98

101

104

107

110

113

116

R2 9 9 9 9 9 9 9 9 9 9 9 16

64

100

100

81

81

81

9 9 9 9 9 9 9 9

Σ R2 694

0

2

4

6

8

10

12

0 5 10 15 20 25 30

No

. of

Re

sou

rce

s (M

en

)

Time, Day

0

20

40

60

80

100

120

140

0 10 20 30

Acc

um

alat

ive

Re

sou

rce

(m

an)

Time, Days

E

S

CRITICAL

L

S

CRITICAL

ES

T

LS

T

ES

T

LS

T

Figure 2 man power requirement Figure 1 Man Load histogram (Earliest & Latest

Start



5

Analysis

𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑑𝑎𝑖𝑙𝑦 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑚𝑒𝑛𝑡, 𝐷𝑟 =𝑇𝑜𝑡𝑎𝑙 𝑚𝑒𝑛 (∑ 𝑅)

𝑝𝑟𝑜𝑗𝑒𝑐𝑡 𝑑𝑢𝑟𝑎𝑡𝑖𝑜𝑛 (𝐷)=

116

26= 𝟒. 𝟒𝟔𝟐

𝐸𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒𝑛𝑒𝑠𝑠 = 𝐷𝑟2 ∗ 𝐷 = (4.462)2 ∗ 26 = 𝟓𝟏𝟕. 𝟓𝟒

𝐸𝑆𝑇 𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙𝑖𝑡𝑦 𝐼𝑛𝑑𝑒𝑥 =√6482

26= 0.98

𝐿𝑆𝑇 𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙𝑖𝑡𝑦 𝑖𝑛𝑑𝑒𝑥 =√6942

26= 1.013

Load leveled

TI 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 24 26

B- B

D- D

E- E

G- G

A- A

C- C

F- F

𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙𝑖𝑡𝑦 𝑖𝑛𝑑𝑒𝑥 = √624

26

2

= 0.961

TI 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 24 26

R 3 3 3 3 3 3 5 5 4 8 8 8 8 8 8 4 4 4 3 3 3 3 3 3 3 3

ΣR 3 6 9 12 15 18 23 28 32 40 48 56 64 72 80 84 88 92 95 98 101 104 107 110 113 116

R2 9 9 9 9 9 9 25 25 16 64 64 64 64 64 64 16 16 16 9 9 9 9 9 9 9 9

Σ R2 624

0

2

4

6

8

10

12

0 5 10 15 20 25 30

No

. of

Re

sou

rce

s (M

en

)

Time, Day

CRITICAL

ES

T

LS

T

LEVELING

Figure 3 Load Histograms



6

Question 2 A small project activities and related information are given in table.

1. Develop the AOA, AON, TSN

2. Calculate network times and Find the critical path using AON network

3. Find Gantt charts

4. Considering the constraints of ZZAD 3,2, SFDF 2, FSFH 2, SSGI 2; ZZHI 2,1Determine:

a. Network data of forward and backward calculations and Total Float (TF)

b. Critical path of the network.

5. Develop load diagram of the resource, and find

a. average daily requirement;

b. Critical Index

c. Effectiveness value

d. use Burgess Algorithm to level resource

Solution Q1 -1-

A,4

C,4

D,7

B,4

K,10

F, 6

E, 7

END H, 14

G, 7

J, 15

I, 14

ZZAD 3,2

SFDF 2

FSFH 2

SSGI 2

ZZHI 2,1

AON network

5 9

6

A, 4 2 4

3 7

8 D, 7

B, 4

1

C, 4

E, 7

F, 6

K, 10

G, 7

J, 15

1

0

1

1

H, 14 1

2

I, 14

AOA network



7

1 2 3 4 5 6 7 8 9 1

0

1

1

1

2

1

3

1

4

1

5

1

6

1

7

1

8

1

9

2

0

2

1

2

2

2

3

2

4

2

5

2

6

2

7

2

8

2

9

3

0

3

1

3

2

3

3

3

4

3

5

3

6

2

7

3

8

3

9

4

0

4

1

4

2

4

3

4

4

4

5

Solution Q1 -2-

Critical Path activities: A-D-F-H-I

Project duration Time =45 weeks

A

0 0 4

0 4 4

B

4 3 8

7 4 11

C

4 27 8

31 4 35

D

4 0 11

4 7 11

K

8 17 18

35 10 45

E

11 6 18

17 7 24

G

18 6 25

24 7 31

J

8 22 23

30 15 45

H

17 0 31

17 14 31

F

11 0 17

11 6 17

I

31 0 45

31 14 45

Act

ES TF LS

LS D LF

CR@CS

END

45 0 45

45 0 45

TSN network

B, 4

A, 4

C, 4

D, 7

J, 15

F, 6 H, 14 I, 14

E, 7 G, 7

K, 10



8

Solution Q1 -3-

1 2 3 4 5 6 7 8 9 10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

27

38

39

40

41

42

43

44

45

A

D

F

H

I

B

J

E

G

C

K

1 2 3 4 5 6 7 8 9 10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

27

38

39

40

41

42

43

44

45

A

D

F

H

I

B

J

E

G

C

K

Solution Q1 -4 Solution with constraint Forward pass: Activity A: ESA=0; EFA= ESA+DA=0+4=4;

Activity B: ESB=4; EFB= ESB+DB=4+4=8;

Activity C: ESC=4; EFC= ESC+DC=4+4=8;

Activity D: ESD=ESA+SSAD=0+3=3 OR EFA+FFAD-DD=4+2-7=0

Then ESD=3; EFD= ESD+DD=3+7=10;

Activity E: ESE= EFD =10; EFE= ESE+DE=10+7=17;

Activity F: ESF=ESD+SFDF-DF=3+2-6=0 OR EFB= 8;

Then ESF=8; EFF= ESF+DF=8+6=14;

Activity J: ESJ= EFB= 8; EFJ= ESJ+DJ=8+15=23;

Activity G: ESG= EFE=17; EFG= ESG+DG=17+7=24;

Activity K: ESK= EFC=8; EFK= ESK+DK=8+10=18;

Activity H: ESH=EFF+FSFH=14+2=16; EFF= ESF+DF=16+14=30;

Activity I: ESI=ESH+SSHI=16+2=18 OR EFH+FFHI-DI=30+1-14=17 OR ESG+SSGI=17+2=19

Then ESI=19; EFI= ESI+DI=19+14=33;

EST GANTT CHART

LST GANTT CHART

jjj

jiji

jiji

iji

iji

ij

DESEF

DSFES

DFFEF

SSES

FSEF

allMaxES

SSFORWARD PA

]2[

Time Initial

)(]1[



9

Backward pass:

Activity I: LFI= EFI =33; LSI= LFI – DI =33-14=19;

Activity J: LFJ= EFI =33; LSJ= LFJ – DJ =33-15=18;

Activity K: LFK= EFK=33; LSK= LFK- DK =33-10=23;

Activity H: LFH=LSI-SSHI+DH=19-2+14=31 OR LFI-FFHI=33-1=32

Then LFH=31; LSH= LFH– DH =31-14=17

Activity G: LFG=LSI-SSGI+DG =19-2+7=24; LSG= LFG– DG =24-7=17

Activity F: LFF= LSH-FSFH = 17-2=15; LSF= LFF– DF=15-6=9

Activity E: LFE= LSG =17; LSE= LFE- DE=17-7=10;

Activity C: LFC= LSG =17; LSC= LFC- DC=17-4=13;

Activity B: LFB= LSF =9; LSB= LFB- DB=9-4=5;

Activity E: LFD=LFE-SFDE+DD=17-2+7=22; OR LFD = LSE=10;

Then LFD=10; LSD=LFE – DE =10-7=3

Activity H: LFA=LSD-SSAD+DA=3-3+4=4 OR LFD-FFAD=10-2=8 OR LSJ=18 OR LSC=13

Then LFA=4; LSA= LFA– DA =4-4=0

Critical Path activities: A-D-E-G-I


A

0 0 4

0 4 4

B

4 1 8

5 4 9

C

4 9 8

13 4 17

D

3 0 10

3 7 10

K

8 15 18

23 10 33

E

10 0 17

10 7 17

G

17 0 24

17 7 24

J

8 10 23

18 15 33

H

16 1 30

17 14 31

F

8 0 14

9 6 15

I

19 0 33

19 14 33

Act

ES TF LS

LS D LF

CR@CS

END

33 0 33

33 0 33

ZZAD 3, 2 SFDF 2

FSFH 2

SSGI 2

ZZHI 2, 1

iii

iijj

iijj

ijj

ijj

ji

DLFLS

DSFLF

DSSLS

FFLF

FSLS

e

allMinLF

ASSBACKWARD P

]4[

Tim Terminal

)(]3[



10

Solution Q1 -5-

1 2 3 4 5 6 7 8 9 1

0

1

1

1

2

1

3

1

4

1

5

1

6

1

7

1

8

1

9

2

0

2

1

2

2

2

3

2

4

2

5

2

6

2

7

2

8

2

9

3

0

3

1

3

2

3

3

3

4

3

5

3

6

2

7

3

8

3

9

4

0

4

1

4

2

4

3

4

4

4

5

A 4 4 4 4

D 3 3 3 3 3 3 3

F 4 4 4 4 4 4

H 5 5 5 5 5 5 5 5 5 5 5 5 5 5

I 5 5 5 5 5 5 5 5 5 5 5 5 5 5

B 2 2 2 2

J 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3

E 5 5 5 5 5 5 5

G 4 4 4 4 4 4 4

C 3 3 3 3

K 2 2 2 2 2 2 2 2 2 2

1 2 3 4 5 6 7 8 9 1

0

1

1

1

2

1

3

1

4

1

5

1

6

1

7

1

8

1

9

2

0

2

1

2

2

2

3

2

4

2

5

2

6

2

7

2

8

2

9

3

0

3

1

3

2

3

3

3

4

3

5

3

6

2

7

3

8

3

9

4

0

4

1

4

2

4

3

4

4

4

5

A 4 4 4 4

D 3 3 3 3 3 3 3

F 4 4 4 4 4 4

H 5 5 5 5 5 5 5 5 5 5 5 5 5 5

I 5 5 5 5 5 5 5 5 5 5 5 5 5 5

B 2 2 2 2

J 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3

E 5 5 5 5 5 5 5

G 4 4 4 4 4 4 4

C 3 3 3 3

K 2 2 2 2 2 2 2 2 2 2

∑ R2 3295

∑ R2 3101

EST GANTT CHART

LST GANTT CHART

R 4 4 4 4 8 8 8 8 8 8 8 14 14 14 14 14 14 15 12 12 12 12 12 9 9 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5

∑R 4 8 12 16 24 32 40 48 56 64 72 86 100 114 128 142 156 171 183 195 207 219 231 240 249 254 259 264 269 274 279 284 289 294 299 304 309 314 319 324 329 334 339 344 349

R2 16 16 16 16 64 64 64 64 64 64 64 196 196 196 196 196 196 225 144 144 144 144 144 81 81 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25

∑ R2

16 32 48 64 128 192 256 320 384 448 512 708 904 1100 1296 1492 1688 1913 2057 2201 2345 2489 2633 2714 2795 2820 2845 2870 2895 2920 2945 2970 2995 3020 3045 3070 3095 3120 3145 3170 3195 3220 3245 3270 3295

R 4 4 4 4 3 3 3 5 5 5 5 4 4 4 4 4 4 10 10 10 10 10 10 10 9 9 9 9 9 9 12 11 11 11 11 10 10 10 10 10 10 10 10 10 10

∑R 4 8 12 16 19 22 25 30 35 40 45 49 53 57 61 65 69 79 89 99 109 119 129 139 148 157 166 175 184 193 205 216 227 238 249 259 269 279 289 299 309 319 329 339 349

R2 16 16 16 16 9 9 9 25 25 25 25 16 16 16 16 16 16 100 100 100 100 100 100 100 81 81 81 81 81 81 144 121 121 121 121 100 100 100 100 100 100 100 100 100 100

∑ R2

16 32 48 64 73 82 91 116 141 166 191 207 223 239 255 271 287 387 487 587 687 787 887 987 1068 1149 1230 1311 1392 1473 1617 1738 1859 1980 2101 2201 2301 2401 2501 2601 2701 2801 2901 3001 3101



11

Question 8 Consider the time-scaled network given below.

a) Draw the corresponding AON network showing each activity four timings and total float.

b) For Developed load diagram of the resource for EST below, find

i. Average daily requirement

ii. Effectiveness value

iii. Critical Index

c) Perform the resource leveling in order to keep the resource uniform with resource constraint of 5 and find

i. Average daily requirement

ii. Effectiveness value

iii. Critical Index

d) Comment on the budget costing before and after leveling of resources, Period, week 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23

Resources, R 5 5 5 5 7 7 5 6 6 8 5 5 5 5 5 5 5 2 2 5 5 3 3

Budget, Cs,

1000SR 2 2 2 3 4 4 3.5 3 3 4 3 3 3 3 4 4 4 2.5 2.5 2.5 2.5 2 2

a) AON

Critical Path activities: A-C-F-I-K


D

2@1000SR

K

3@2000SR

H

3@1500SR

I

2@2500SR

F

3@2000SR

E

3@1000SR

C

3@2500SR

B

2@500SR

A

5@2000SR

G

2@1000SR

J

2@500SR

E

7 8 10

15 3 18

0

C

3 0 7

3 4 7

B

3 9 6

12 3 15

D

3 3 6

6 3 9

F

7 0 14

7 7 14

I

14 0 19

14 5 19

0 0 3

0 3 3

A

G

6 3 11

9 5 14

K

19 0 23

19 4 23

H

14 4 17

18 3 21

2

J

19 2 21

21 2 23

END

23 0 23

23 0 23



12

b) Load diagrams

Period, week 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23

Resources, R 5 5 5 7 7 7 5 8 8 8 5 3 3 3 5 5 5 2 2 5 5 3 3

Budget, Cs,

1000SR 2 2 2 4 4 4 4.5 4 4 4 3 2 2 2 4 4 4 2.5 2.5 2.5 2.5 2 2

week 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23

R 5 5 5 5 7 7 5 6 6 8 5 5 5 5 5 5 5 2 2 5 5 3 3

ΣR 5 10 15 22 29 36 41 49 57 65 70 73 76 79 84 89 94 96 98 103 108 111 114

R2 25 25 25 49 49 49 25 64 64 64 25 9 9 9 25 25 25 4 4 25 25 9 9

ΣR2 642

Analysis 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑑𝑎𝑖𝑙𝑦 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑚𝑒𝑛𝑡, 𝐷𝑟

=𝑇𝑜𝑡𝑎𝑙 𝑚𝑒𝑛 (∑ 𝑅)


114

23= 𝟒. 𝟗𝟓𝟕

𝐸𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒𝑛𝑒𝑠𝑠 = 𝐷𝑟2 ∗ 𝐷 = (4.957)2 ∗ 23

= 𝟓𝟔𝟓. 𝟏𝟓𝟑 = ∑ 𝑹𝑬𝟐

𝐸𝑆𝑇 𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙𝑖𝑡𝑦 𝐼𝑛𝑑𝑒𝑥 =∑ 𝑹𝑨

𝟐

∑ 𝑹𝑬𝟐

=642

565.153= 𝟏. 𝟏𝟑𝟔

Comments

1- Load histogram shows resource variations

through the time.

2- The average resources required to be leveled is

>= 5

3- available resources only 5 at any time, then

Hence, when available resource is 5; Then,

𝐸𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒𝑛𝑒𝑠𝑠 = 𝐷𝑟2 ∗ 𝐷 = (5)2 ∗ 23 = 𝟓𝟕𝟓 = ∑ 𝑹𝑬𝟐

𝐸𝑆𝑇 𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙𝑖𝑡𝑦 𝐼𝑛𝑑𝑒𝑥 =𝟒. 𝟗𝟓𝟕

𝟓= 𝟎. 𝟗𝟗𝟏

D

2@1000SR

K

3@2000SR

H

3@1500SR

I

2@2500SR

F

3@2000SR

E

3@1000SR

C

3@2500S

B

2@500SR

A

5@2000

G

2@1000SR

J

2@500SR

0

2

4

6

8

10

0 10 20 30

No

. of

Rec

ou

rses

Project Duration



13

c) Resource leveling

Arrange according LST (A, C, D, B, F, G, E , I, H, K, J)

EAS

OSS

A 5 5 5

C 3 3 3 3

D 2 2 2

B 2 2 2

F 3 3 3 3 3 3 3

G 2 2 2 2 2

E 3 3 3

I 2 2 2 2 2

H 3 3 3

K 3 3 3 3

J 2 2

WEEK 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5

- - - 2 2 2 2 3 3 2 2 2 2 2 2 2 2 3 3 2 2 2 2 2

R 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 3 5 5 3 3

ΣR 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 98 103 108 111 114

R2 5 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 9 25 25 9 9

ΣR2= 532

Analysis

𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑑𝑎𝑖𝑙𝑦 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑚𝑒𝑛𝑡, 𝐷𝑟 =𝑇𝑜𝑡𝑎𝑙 𝑚𝑒𝑛 (∑ 𝑅)


114

24= 𝟒. 𝟕𝟓

𝐸𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒𝑛𝑒𝑠𝑠 = 𝐷𝑟2 ∗ 𝐷 = (4.75)2 ∗ 24 = 𝟓𝟒𝟏. 𝟓 = ∑ 𝑹𝑬𝟐

𝐸𝑆𝑇 𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙𝑖𝑡𝑦 𝐼𝑛𝑑𝑒𝑥 =𝟒. 𝟕𝟓

𝟓= 𝟎. 𝟗𝟒

0

2

4

6

8

10

0 5 10 15 20 25

No

. of

Res

ou

rces

Project Duration, week

R (without resource limit) R (with resource limit=5)



14

Comments

1- Load histogram shows resource variations reduced by leveling through the time.

d) Budgeted Cost

Week 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Budget, Cs 1000SR (Without resource level)

2 2 2 4 4 4 4.5 4 4 4 3 2 2 2 4 4 4 2.5 2.5 2.5 2.5 2 2

Cumulative Cs 2 4 6 9 13 17 20.5 23.5 26.5 30.5 33.5 36.5 39.5 42.5 46.5 50.5 54.5 57 59.5 62 64.5 66.5 68.5

Budget, Cs 1000SR (With resource level=5)

2 2 2 4 4 4 4.5 4 4 4 3 2 2 2 4 4 4 2.5 2.5 2.5 2.5 2 2 2

Cumulative Cs 2 4 6 9.5 13 16.5 19.5 22 24.5 27.5 30.5 33.5 36.5 39.5 43 46.5 50 54 58 59.5 62 64.5 66.5 68.5

Comment:

Since the duration is extended when resources is limited to 5, this is resulted in small reduction of budgeted cost

over periods.

Figure 4 Cumulative Budgeted Cost

0

10

20

30

40

50

60

70

0 5 10 15 20 25

Cu

mu

lati

ve b

ud

get

ed c

ost

, 1o

oo

SR

Project Duration, week

CumulativeCs1000SR(Withoutresource level)

Cumulative Cs1000SR(Withresource level=5)



15

Question 12 (Crushing)

Development of a new version of software is

considered by a software company. The activities

necessary for completion of the project are given

in the following table (1)

(a) What is the project completion date?

(b) What is the total cost required to completing

this project on normal time?

(c) If it is required to reduce the completion time

by one week, which activity should be

crashed? , and how much will this increase the total cost increase?

(d) What is the maximum time that can be crashed? How much would cost increase?

Solution (a) Project completion time 16

(Critical Path Activities A–D–G)

(b) Total cost on normal duration time =Σ normal cost of activities = $12,300

(c) Crash one week

Find cost slope Crash activity (D) by one week at extra cost of $75 (d) Maximum time to be crashed

Activities Predecessor(s) Tome, weeks Cost, $

Normal Crash Normal Crash

A - 4 3 2000 2600

B - 2 1 2200 2800

C - 3 3 500 500

D A 8 4 2300 2600

E B 6 3 900 1200

F C 3 2 3000 4200

G D, E 4 2 1400 2000

Activity

(X) Normal Time – Crash Time

(Y) Crash $–Normal $

Cost Slope=(Y/X);

$/time

A 1 600 600 B 1 600 600 C 0 0 — D 4 300 75 E 3 300 100 F 1 1,200 1,200 G 2 600 300

Activity Crash Cost

D 4 $300

G 2 600

A 1 600

E 1 100

7 weeks $1,600

A,

4

START B,

2

C,

3

E,

6

D,

8

F,

3

G,

4

END



16

Question 13 (financing) Consider data for a small project given in Table below. Revenue is paid to the contractor every 6 months

according to completed quantities of work. Assume that markup is 10% of total project cost and is uniformly

distributed over the work.

Assume also that work of each activity is uniformly distributed over its duration.

a. Draw on a scale the project cumulative cost and revenue curves according to ES timings.

b. Draw on the diagram produced in a) above effect of receipt of 10% of total revenue as an advance

payment. This payment will be deducted from each periodic revenue by the same percentage.

Solution

1-

Project completion time 30 ; (Critical Path Activities A–C–D)

2- Activity Cost distribution

TIME 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30

A-20 C-60 D-50

B-60

ACC. COST

0040 0080 0120 0160 0200 0240 0480 0720 0960 1080 1200 1320 1420 1520 1620

3- Revenue calculation

4- Total payment = 1620*(1+0.1) x 1000=(1620+162) x 1000 =1782 x 1000

Advanced payment = 0.1*1782 x 1000 =178.2 x 1000

Revenue every 6 months =

Activity Predecessor(s) Duration, Months Monthly cost, SR1000

A - 12 20

B A 6 60

C A 12 60

D B,C 6 50

Month R= Revenue =(Cost*1.1),

1000 SR Adj. Revenue = 0.9 R, 1000

SR Cumulative Revenue,

1000 SR

0 178.2

6 120(1.1)=132 118.8 118.8+178.2=297.0

12 240(1.1)=264 237.6 237.6+178.2=415.8

18 960(1.1)=1056 950.4 950.4+178.2=1128.6

24 1320(1.1)=1452 1306.8 1306.8+178.2=1485.0

30 1620(1.1)=1782 1603.8 1603.8+178.2=1782.0

A,

12

B,

6

C,

12

D,

6



17

5- Graph presentations

Figure 5 Cumulative Cash Flow

0

200

400

600

800

1000

1200

1400

1600

1800

0 5 10 15 20 25 30

Pro

ject

Cas

h, 1

00

0 S

R

Project Duration, weeks

Cash out= Cost

Adj. Revenue

Revenue

-700

-500

-300

-100

100

300

0 10 20 30

Cas

h, 1

00

0 S

R

Project Duration, weeks

Adj. Revenue

Revenue

Figure 6 Net Cash Flow



18

Question17 (project control) A small engineering project has budgeted cost of each activity as shown

in Table (15a). The budgeted cost of each activity is assumed to be

uniformly distributed over its duration. Project status at the end of the 15th

day, is given the table (15b). The actual cost of work performed (ACWP)

at the end of day 15 = 20200 SR

Calculate the CV, SV, CPI, SPI, and % overrun (underrun); BAC, EAC;

comment on progress of works.

Solution

Using Equations

BCWS (SR) =20,200

BCWP (SR) = 4000 + 6000 (0.6) + 2000 + 8000(0.8) + 7000 + 6000 (0.8) = 27,800

ACWP (SR) = 4200 + 3000 + 1600 + 5000 + 7200 + 4200 = 25,200

CV (SR) = BCWP – ACWP = 27800-25200=+2600 Over Budget

SV (SR) = BCWP – BCWS = 27,800-20,200=+7,600 Ahead of the schedule

BAC (SR) = 4000 + 6000 + 2000 + 8000 + 7000 + 6000 +9000 + 4000 = 46,000

EAC (SR) = (ACWP/ BCWP) × BAC = (25,200/27,800)*46,000= 41,697.84

ETC (SR) = EAC – ACWP =41,697.84-25,200= 16,497.84

% overrun or underrun = (ACWP-BCWP)/BCWP =(25,200-27,800)/ 27,800= -9.35% underrun

Comment: 1- Expenses are less than estimated and project is under control.

Table (1b) Project status

Activity % completed Expense to date,

1000 SR

A 100 4.2

B 60 3.0

C 100 1.6

D 80 5.0

E 100 7.2

F 80 4.2

G 0 0

H 0 0

Total 25.2

Table (1a). Activity budgeted cost

Activity A B C D E F G H

Budgeted cost (SR) 4000 6000 2000 8000 7000 6000 9000 4000

Project Cost Control

Activity Budgeted (scheduled) Cost,

1000 SR

To date value (Over /Under) Run

Cost, 1000 SR % completed Estimated Cost,

1000 SR Expense to date,

1000 SR

A 4 100 4*1.0=4 4.2 +0.2

B 6 60 6*0.6=3.6 3.0 -0.6

C 2 100 2 1.6 -0.4

D 8 80 8*0.8= 6.4 5.0 -1.4

E 7 100 7 7.2 +0.2

F 6 80 6*0.8=4.8 4.2 -0.6

G 9 0 0 0 0

H 4 0 0 0 0

Total 46 BCWP = 27.8 ACWP = 25.2 -2.4 (under-Run)



19

Question18 (Crashing & project control) Table (15) gives the work activities of a project and its related data.

Table (16a) project activity data

Activity Predecessor (s)

Normal Schedule Crashed Schedule

Time, week

Cost, 10,000 SR

Time, week

Cost, 10,000 SR

A - 4 2.0 2 5.0

B - 6 3.0 3 9.0

C A 4 4.5 2 10.0

D A, B 6 2.5 4 10.0

E D 2 0.5 1 1.0

F E 13 13.0 8 25.0

G E 1 1.5 1 1.5

H C, G 20 6.0 10 23.5

I F 9 7.0 5 16.0

Total 40.0 101.0

a) Determine EST, LST, TF schedule for each work activity at normal activity duration.

b) Determine a minimum schedule to complete the project within 26 weeks, what will be the required budget?

c) Consider that the project is scheduled according earliest time schedule for normal activity duration, only

246,000 SR has been spent and the project status as shown in table (3b). Determine the project is ahead or

behind schedule and the amount it is over or under budget. Suggest corrective actions in case that the project

is behind schedule or over budget.

Solution (a)

Critical Path activities: B-D-E-F-I

Project duration Time =36 Weeks

Table (16b) Project status

Activity % completed Expense to date,

10,000 SR

A 100 2.6

B 100 5.0

C 100 3.0

D 100 5.0

E 100 0.7

F 40 5.6

G 100 0.7

H 25 2.0

I 0 0

Total 24.6

B, 6 F, 13 D, 6

A, 4 C, 4

E, 2

FINIS

H, 20

G, 1

AON network

I, 9

STAR



20

Solution (b) Crashing

Activity Predecessor (s) X=(Normal –Crash)

Time, week

Y=(Normal –crash)

Cost, 10,000 SR

Cost Slope =Y/X,

10,000 SR

A - 2 3.0 1.5

B - 3 6.0 2.0

C A 2 5.5 2.75

D A, B 2 7.5 3.75

E D 1 0.5 0.5

F E 5 12.0 2.4

G E - - -

H C, G 10 17.5 1.75

I F 4 9 2.25

Cycle #

Activity to Shorten

Can Be Shortened

Week Shortened

Cost per week, 10,000 SR

Cost for Cycle, 10,000 SR

Project Duration

New activity duration time, week

0 36

1 E 1 1 0.5 0.5 35 1

2 B, A 3, 2 3, 1 2, 1.5 6+1.5=7.5 32 3, 3

3 I, H 4, 10 4, 3 2.25, 1.75 9+5.25=14.25 28 5, 17

4 D 2 2 3.75 7.5 4

Total extra cost paid, 10,000SR 29.75

Project budget,10,000 SR 40+29.75=69.75

Time Schedule analysis for Project normal activity duration

Activity Predecessor (s) Normal Time, week

Forward pass Backward pass Slack, (LS_ES)= TF

On Critical Path ES LS LS LF

A - 4 0 4 2 6 2

B - 6 0 6 0 6 0 yes

C A 4 4 8 12 16 8

D A, B 6 6 12 6 12 0 yes

E D 2 12 14 12 14 0 yes

F E 13 14 27 14 27 0 yes

G E 1 14 15 15 16 1

H C, G 20 15 35 16 36 1

I F 9 27 36 27 36 0 yes

Time Schedule analysis for Project after crashing

Activity Predecessor (s) Activity

Time, week

Forward pass Backward pass Slack, (LS_ES)= TF

On Critical Path ES LS LS LF

A - 3 0 3 0 3 0 yes

B - 3 0 3 0 3 0 yes

C A 4 3 7 5 9 2

D A, B 4 3 7 3 7 0 yes

E D 1 7 8 7 8 0 yes

F E 13 8 21 8 21 0 yes

G E 1 8 9 8 9 0 yes

H C, G 17 9 26 9 26 0 yes

I F 5 21 26 21 26 0 yes



21

Critical Activities; A, B, D, E, F, G, H, I

Project Duration =26 WEEKS –

Number OF Critical path = 4

Budget =697,500 SR

Solution (c) Project control project cost control Report

Activity Predecessor

(s)

Normal Schedule To date values

Time, week

Cost (scheduled), 10,000 SR

% completed

Cost (estimated), 10,000 SR

Expense to date, 10,000 SR

Cost overrun/underrun,10,000SR

A - 4 2.0 100 2.0 2.6 +0.6

B - 6 3.0 100 3.0 5.0 +2.0

C A 4 4.5 100 4.5 3.0 -1.5

D A, B 6 2.5 100 2.5 5.0 +2.5

E D 2 0.5 100 0.5 0.7 +0.2

F E 13 13.0 40 13*0.4=5.2 5.6 +.4

G E 1 1.5 100 1.5 0.7 -0.8

H C, G 20 6.0 25 6*0.25=1.5 2.0 +.5

I F 9 7.0 0 7*0=0 0 0

Total 40.0 BCWP = 20.7 ACWP = 24.6 Over-run = +3.9

OR BCWS (SR) =24.6 x 104

BCWP (SR) = (2 + 3 + 4.5 + 2.5 + 0.5 + 13*0.4+ 1.5 + 6*0.25) = 20.7 x 104

ACWP (SR) = (2.6 + 5 + 3 + 5 + 0.7 + 5.6 + 0.7 + 2) = 24.6 x 104

CV (SR) = BCWP – ACWP = 20.7 x 104-24.6 x 104=-3.9 x 104 under Budget

SV (SR) = BCWP – BCWS = 20.7 x 104-24.6 x 104 =-3.9 x 104 behind the schedule

BAC (SR) = (2 + 3 + 4.5 + 2.5 + 0.5 + 13+1.5 + 6 + 7) = 40 x 104

EAC (SR) = (ACWP/ BCWP) × BAC = (24.6 x 104 /20.7 x 104)* 40 x 104= 47.54 x 104

ETC (SR) = EAC – ACWP =47.54 x 104-24.6 x 104 = 22.94 x 104

% overrun or underrun = (ACWP-BCWP)/BCWP =(24.6 x 104 -20.7 x 104)/ 20.7 x 104= 18.84% Over-run

Comments:

1- The project is found to be behind schedule / experiencing cost over-run.

2- The experienced cost over-run or delay is lost causes due to not meeting project deadlines or budgets.

3- Action should be taken for cost saving, (a) in case the project behind schedule, expediting some activity

at extra cost may be possible; (b) in case over budget, non-critical activities are evaluated for cost saving.,

(c) in case of fund reduction causing delay, evaluating project duration time is evaluated and may require

to be extended.

selected problems solution - summer 1433-34h

Documents

c f f activity

b b d d e e g g

t w t f s s b d d d

b c d e f g free float

e3 f c

e2 d b

e4 g e

activity e