Case study
It is known that the software development activity table is.
| KEGIATAN | | PENDAHULU | DURASI |
|----------|------------------------------------|-----------|--------|
| A | Analisis sistem yang berjalan | - | 10 |
| B | Studi kelayakan | A | 12 |
| C | Analisis kelemahan sistem yang ada | - | 15 |
| D | Analisis kebutuhan sistem | C | 5 |
| E | Desain data | A | 14 |
| F | Desain user interface | B, D | 8 |
| G | Desain proses | E | 15 |
| H | Instalasi program | F, G | 10 |Based on the table, then determine!
- Network diagram image
- Critical path diagram and table image
- The probability value, if the time limit is known (n) = 90 days and the estimated time table is as follows.
| KEGIATAN | | PENDAHULU | DURASI |
|----------|------------------------------------|-----------|--------|
| A | Analisis sistem yang berjalan | - | 10 |
| B | Studi kelayakan | A | 12 |
| C | Analisis kelemahan sistem yang ada | - | 15 |
| D | Analisis kebutuhan sistem | C | 5 |
| E | Desain data | A | 14 |
| F | Desain user interface | B, D | 8 |
| G | Desain proses | E | 15 |
| H | Instalasi program | F, G | 10 |Completion
1 ]. Network Diagram
2 ]. a ]. Critical Path Diagram
2 ]. b ]. Critical Path Table
| Kegiatan | Mulai Terdahulu ES | Selesai Terdahulu EF | Mulai Terakhir LS | Selesai Terakhir LF | Slack LS – ES | Jalur Kritis |
|----------|--------------------|----------------------|-------------------|---------------------|---------------|--------------|
| A | 0 | 10 | 5 | 13 | 5 | Tidak |
| B | 15 | 27 | 15 | 27 | 0 | Ya |
| C | 10 | 15 | 10 | 15 | 0 | Ya |
| D | 15 | 27 | 15 | 27 | 0 | Ya |
| E | 10 | 24 | 13 | 27 | 3 | Tidak |
| F | 27 | 42 | 27 | 42 | 0 | Ya |
| G | 27 | 42 | 27 | 42 | 0 | Ya |
| H | 42 | 52 | 42 | 52 | 0 | Ya |3 ]. Probability
Estimated Time Table
| Kegiatan | Waktu Optimis (a) | Waktu Realistis (m) | Waktu Pesimis (h) | Waktu Yg Diharapkan t = (a+4m+b)/6 | Varians [(b – a) / 6]2 |
|----------|-------------------|---------------------|-------------------|------------------------------------|------------------------|
| A | 0 | 10 | 5 | 13 | 0,25 |
| B | 15 | 27 | 15 | 27 | 0,11 |
| C | 10 | 15 | 10 | 15 | 0,69 |
| D | 15 | 27 | 15 | 27 | 0,25 |
| E | 10 | 24 | 13 | 27 | 0,45 |
| F | 27 | 42 | 27 | 42 | 0,25 |
| G | 27 | 42 | 27 | 42 | 0,45 |
| H | 42 | 52 | 42 | 52 | 0,25 |S2 = B variance + C variance + D variance + F variance + G variance + H variance
S2 = 0.11 + 0.69 + 0.25 + 0.25 + 0.45 + 0.25
S2 = 2
S = ? 2
S = 1.41
Z = (deadline -- completion time) / 1.41
Z = (90 -- 88.5) / 1.41
Z = 1.06
Z table = 0.3554
P = Z table + 0.5000
P = 0.3554 + 0.5000
P = 0.8554
So the probability of completing the project in 90 days is 0.8554