problem 2 - aon network times\nthe project information for the custom order project of the air control…

problem 2 - aon network times\nthe project information for the custom order project of the air control company is presented here. draw a project network for this project. compute the early and late activity times and slack times. identify the critical path.\n\n| id | description | predecessor | time |\n|----|----|----|----|\n| a | order review | none | 2 |\n| b | order standard parts | a | 3 |\n| c | produce standard parts | a | 10 |\n| d | design custom parts | a | 13 |\n| e | software development | a | 18 |\n| f | manufacture custom hardware | c,d | 15 |\n| g | assemble | b,f | 10 |\n| h | test | e,g | 5 |

problem 2 - aon network times\nthe project information for the custom order project of the air control company is presented here. draw a project network for this project. compute the early and late activity times and slack times. identify the critical path.\n\n| id | description | predecessor | time |\n|----|----|----|----|\n| a | order review | none | 2 |\n| b | order standard parts | a | 3 |\n| c | produce standard parts | a | 10 |\n| d | design custom parts | a | 13 |\n| e | software development | a | 18 |\n| f | manufacture custom hardware | c,d | 15 |\n| g | assemble | b,f | 10 |\n| h | test | e,g | 5 |

Answer

Explanation:

Step1: Draw the project network

Based on the predecessor - successor relationships, nodes represent activities and arrows represent the sequence.

Step2: Compute early start (ES) and early finish (EF) times

ES of the first activity A is 0. EF = ES+Time. For subsequent activities, ES is the maximum EF of its predecessors.

  • For A: ES(A)=0, EF(A)=0 + 2=2
  • For B: ES(B)=EF(A)=2, EF(B)=2 + 3=5
  • For C: ES(C)=EF(A)=2, EF(C)=2 + 10=12
  • For D: ES(D)=EF(A)=2, EF(D)=2 + 13=15
  • For E: ES(E)=EF(A)=2, EF(E)=2 + 18=20
  • For F: ES(F)=max(EF(C),EF(D))=max(12,15)=15, EF(F)=15 + 15=30
  • For G: ES(G)=max(EF(B),EF(F))=max(5,30)=30, EF(G)=30 + 10=40
  • For H: ES(H)=max(EF(E),EF(G))=max(20,40)=40, EF(H)=40 + 5=45

Step3: Compute late finish (LF) and late start (LS) times

LF of the last activity H is set to EF(H)=45. LS = LF - Time. For previous activities, LF is the minimum LS of its successors.

  • For H: LF(H)=45, LS(H)=45 - 5=40
  • For G: LF(G)=LS(H)=40, LS(G)=40 - 10=30
  • For E: LF(E)=LS(H)=40, LS(E)=40 - 18=22
  • For F: LF(F)=LS(G)=30, LS(F)=30 - 15=15
  • For D: LF(D)=LS(F)=15, LS(D)=15 - 13=2
  • For C: LF(C)=LS(F)=15, LS(C)=15 - 10=5
  • For B: LF(B)=LS(G)=30, LS(B)=30 - 3=27
  • For A: LF(A)=min(LS(B),LS(C),LS(D),LS(E))=2, LS(A)=2 - 2=0

Step4: Compute slack times

Slack = LS - ES = LF - EF.

  • For A: Slack(A)=LS(A)-ES(A)=0 - 0=0
  • For B: Slack(B)=LS(B)-ES(B)=27 - 2=25
  • For C: Slack(C)=LS(C)-ES(C)=5 - 2=3
  • For D: Slack(D)=LS(D)-ES(D)=2 - 2=0
  • For E: Slack(E)=LS(E)-ES(E)=22 - 2=20
  • For F: Slack(F)=LS(F)-ES(F)=15 - 15=0
  • For G: Slack(G)=LS(G)-ES(G)=30 - 30=0
  • For H: Slack(H)=LS(H)-ES(H)=40 - 40=0

Step5: Identify the critical path

The critical path consists of activities with zero slack. The critical path is A - D - F - G - H.

Answer:

Project network is drawn based on the given data. Early and late activity times:

Activity ES EF LS LF Slack
A 0 2 0 2 0
B 2 5 27 30 25
C 2 12 5 15 3
D 2 15 2 15 0
E 2 20 22 40 20
F 15 30 15 30 0
G 30 40 30 40 0
H 40 45 40 45 0
Critical path: A - D - F - G - H