problem 3 - aon network times\na large southeast city is requesting federal funding for a park - and - ride…
problem 3 - aon network times\na large southeast city is requesting federal funding for a park - and - ride project. one of the requirements in the request application is a network plan for the design phase of the project. sophie kim, the chief engineer, wants you to develop a project network plan to meet this requirement. she has gathered the activity time estimates and their dependencies shown here. show your project network with the activity early, late and slack times. mark the critical path.\nid description predecessor time\na survey none 5\nb soils report a 20\nc traffic design a 30\nd lot layout a 5\ne approve design b,c,d 80\nf illumination e 15\ng drainage e 30\nh landscape e 25\ni signage e 15\nj bid proposal f,g,h,i 10
Answer
Explanation:
Step1: Calculate early start (ES) and early finish (EF)
ES of A = 0, EF of A=ES of A + Time of A = 0 + 5=5. For activities with single - predecessor like B, C, D: ES of B = EF of A = 5, EF of B=ES of B + Time of B = 5+20 = 25; ES of C = EF of A = 5, EF of C=ES of C + Time of C = 5 + 30=35; ES of D = EF of A = 5, EF of D=ES of D + Time of D = 5+5 = 10. For activity E with multiple predecessors, ES of E = max(EF of B, EF of C, EF of D)=max(25, 35, 10)=35, EF of E=ES of E + Time of E = 35+80 = 115. Similarly, for F, G, H, I: ES of F = EF of E = 115, EF of F=ES of F + Time of F = 115+15 = 130; ES of G = EF of E = 115, EF of G=ES of G + Time of G = 115+30 = 145; ES of H = EF of E = 115, EF of H=ES of H + Time of H = 115+25 = 140; ES of I = EF of E = 115, EF of I=ES of I + Time of I = 115+15 = 130. For J, ES of J = max(EF of F, EF of G, EF of H, EF of I)=max(130, 145, 140, 130)=145, EF of J=ES of J + Time of J = 145+10 = 155.
Step2: Calculate late start (LS) and late finish (LF)
LF of J = 155, LS of J = LF of J - Time of J = 155 - 10=145. For F, G, H, I: LF of F = LS of J = 145, LS of F = LF of F - Time of F = 145 - 15 = 130; LF of G = LS of J = 145, LS of G = LF of G - Time of G = 145 - 30 = 115; LF of H = LS of J = 145, LS of H = LF of H - Time of H = 145 - 25 = 120; LF of I = LS of J = 145, LS of I = LF of I - Time of I = 145 - 15 = 130. For E, LF of E = min(LS of F, LS of G, LS of H, LS of I)=115, LS of E = LF of E - Time of E = 115 - 80 = 35. For B, C, D: LF of B = LS of E = 35, LS of B = LF of B - Time of B = 35 - 20 = 15; LF of C = LS of E = 35, LS of C = LF of C - Time of C = 35 - 30 = 5; LF of D = LS of E = 35, LS of D = LF of D - Time of D = 35 - 5 = 30. For A, LF of A = min(LS of B, LS of C, LS of D)=5, LS of A = LF of A - Time of A = 5 - 5 = 0.
Step3: Calculate slack time
Slack time of an activity = LS - ES (or LF - EF). For A: Slack = LS of A - ES of A=0 - 0 = 0; For B: Slack = LS of B - ES of B=15 - 5 = 10; For C: Slack = LS of C - ES of C=5 - 5 = 0; For D: Slack = LS of D - ES of D=30 - 5 = 25; For E: Slack = LS of E - ES of E=35 - 35 = 0; For F: Slack = LS of F - ES of F=130 - 115 = 15; For G: Slack = LS of G - ES of G=115 - 115 = 0; For H: Slack = LS of H - ES of H=120 - 115 = 5; For I: Slack = LS of I - ES of I=130 - 115 = 15; For J: Slack = LS of J - ES of J=145 - 145 = 0.
Step4: Identify critical path
The critical path consists of activities with zero slack. The critical path is A - C - E - G - J.
Answer:
| Activity | Early Start | Early Finish | Late Start | Late Finish | Slack Time |
|---|---|---|---|---|---|
| A | 0 | 5 | 0 | 5 | 0 |
| B | 5 | 25 | 15 | 35 | 10 |
| C | 5 | 35 | 5 | 35 | 0 |
| D | 5 | 10 | 30 | 35 | 25 |
| E | 35 | 115 | 35 | 115 | 0 |
| F | 115 | 130 | 130 | 145 | 15 |
| G | 115 | 145 | 115 | 145 | 0 |
| H | 115 | 140 | 120 | 145 | 5 |
| I | 115 | 130 | 130 | 145 | 15 |
| J | 145 | 155 | 145 | 155 | 0 |
Critical path: A - C - E - G - J