Question

a n b b Activity C, E E A Activity А 6 B 2 с 8 D 5 E 6 8 8 12 6 3 9 E Immediate Predecessor(s) G 3 3 5 Н 2 2 2 I 5 5 5 J 4 7 12 K 3 3 10 24 16 10 4 21 A A D, G, I, 1 B C, E

a.) Create project network for activites A thru G

b.) Then compute on the network above, the correct Activity on Arrow format (AOA), the earliest start, earliest finish, latest start, and latest finish times for project.

Answer 1

B (9:67 3 E(2.84) A(1015) I(5) 1 A(8) C(12) 2 4 7 Kiz 8 D(6.57 (71(3.33 H29 KG F135) 6

A B C D E F H J K L M 1 2 a m b Mean time EF LS LF Slack Activity Predecessor A ES 0 3 6 8 10 8 0 4 B A 2 8 24 8 9.666666667 12 8 0 17.66667 20.5 5 C A 8 12 16 8 0.5 6 D A 5 6 10 33 18.5 8 0 17.66667 8 20 8.5 14.5 26.5 20.5 17.66667 31 20.5 23.83333 29.66667 33 31 7 E B 1 3 4 20.5 6.5 8 2.833333333 17.66667 10.5 20.5 3.333333333 20.5 0 F C,E 6 9 21 31 0 8 9 G C,E 3 3 5 33 9.166667 10 H F 2 2 2 2 31 33 0 11 F 5 5 5 5 31 36 35.3333340.333334.333333 33 40.33333 0 12 4 7 12 40.33333 ا D,G,H 1,J 7.333333333 33 3.166666667|40.33333 13 K 3 3 4 43.5 40.33333 43.5 0 14 15

Formulas used are shown in the below table,

J 1 =K3-G3 B -KE.CE K6-16 Line COR. CONC А B D E F H 1 K L 2 Activity Predecessor m b Mean time ES EF LS LF Slack 3 А 6 8 10 =(D3+(4*E3)+F3)/6 8 =MIN(14,15,16) =K3-13 4 A A 2 8 24 =(D4+(46E4)+F4)/6 8 =H4+G4 =K4-G4 -J7 =K4-14 5 C А 8 12 16 =(D5+(4*E5)+F5)/6 8 =H5+G5 =K5-G5 -MIN(J8J9) =K5-15 6 D A 5 6 10 =(D6+(4+E6)+F6)/6 8 =H6+G6 =J12 7 1 Pinz...E E 3 ME B 4 LV =(D7+(4+E7)+F7)/6 =14 =H7+G7 =K7-G7 =MINJ8,19) =K7-17 24 8 F CE 6 9 21 LMINI =MAX(17,15) -ve.co =H8+G8 =K8-G8 =MIN(J11,J10) =K8-18 9 G CE 3 3 5 =(D9+(4*E9)+F9)/6 =H8 =H9+G9 Lvo vo.co =K9-G9 =J12 =K9-19 10 10 H F 190 2 2 2 =(D10+(4*E10)+F10)/6 =18 =H10+G10 =K10-G10 =J12 -10.14 =K10-110 44 11 5 5 5 =(D11+(4*E11)+F11)/6 =18 =H11+G11 =K11-G11 =J13 =K11-111 40 12 D,G,H 4 7 12 =(D12+(4*E12)+F12)/6 =MAX(16,19,110) =H12+G12 =K12-G12 =J13 =K12-112 12 13 к II 3 3 4 = (D13+(4*E13)+F13)/6 =MAX(111,112) =H13+G13 =K13-G13 43.5 =K13-113 14 15 16 17 18 19 F - -

Formulas used to calculate ES,EF,LS and LF are shown below:

Forward pass -

ES = Maximum(EF of the predecessor activities)

EF = ES + Mean time

Backward pass -

LF = Minimum(LS of all succeeding activities)

LS = LF - Mean time

Note,

ES for Activity A = 0

LF for activity K = EF for activity K

Critical path is the path with the activities with zero slack

Hence,

critical path is A-B-E-F-H-J-K

Time required for critical path = 43.5 units of time.