Program Evaluation And Review Technique (PERT) Critical Path Method (CPM) www.labsistemtmip.wordpress.com Program Evaluation And Review Technique Untuk sebanyak mungkin mengurangi adanya penundaan, maupun gangguan produksi Mengkoordinasikan berbagai bagian suatu pekerjaan secara menyeluruh dan mempercepat selesainya proyek. Suatu pekerjaan yang terkendali dan teratur, karena jadwal dan anggaran dari suatu pekerjaan telah ditentukan terlebih dahulu sebelum dilaksanakan. Pencapaian suatu taraf tertentu dimana waktu merupakan dasar penting dari PERT dalam penyelesaian kegiatan-kegiatan bagi suatu proyek. 1
Critical Path Method (metode jalur kritis) Diselesaikan secara tepat waktu serta tepat biaya. Metode perencanaan dan pengendalian proyekproyek Prinsip pembentukan jaringan. Jumlah waktu yang dibutuhkan dalam setiap tahap suatu proyek dianggap diketahui dengan pasti, Hubungan antara sumber yang digunakan dan waktu yang diperlukan untuk menyelesaikan proyek. 2
NOMOR PENGERJAAN LAMA PENGERJAAN TANGGAL SELESAI TANGGAL MULAI 3
Pekerjaan Kelangsungan Proyek Data, Waktu, Biaya Informasi Sasaran Arti Panah PERT Perencanaan Dan Pengendalian Proyek Belum Pernah Dkerjakan, Belum Diketahui Waktu Pengerjaan Tercepat, Terlama Terlayak Tepat Waktu, Sebab Dengan Penyingkatan Waktu Maka Biaya Proyek Turut Mengecil, Anak Panah Menunjukkan Tata Urutan (Hubungan Presidentil) CPM Menjadwalkan Dan Mengendalikan Aktivitas Sudah Pernah Dikerjakan k Telah Diketahui Oleh Evaluator Waktu Pengerjaan Waktu Yang Paling Tepat Dan Layak Untuk Tepat Biaya Tanda Panah Adalah Kegiatan A Project Suatu set pekerjaan yang dilakukan secara sekuensial Tujuan (Goals) Menjamin suatu project mencapai tujuannya Selesai tepat waktu Sesuai Anggaran Sesuai dengan sumber daya Menyediakan mekasnisme monitoring 8 4
Jalur Kritis / Critical Path: Suatu aktivitas sekuensial yang menuju pada penyelesaian project. Slack: Jumlah fleksibitas dalam menjadwalan aktivitas yang tidak kritis. 9 An Activity On Node (AON) Network Representation of the Klonepalm 2000 Computer Project A 90 B 15 F 25 I 30 Immediate Estimated C E Activity 5 Predecessor Completion Time 21 A None 90 B A 15 C B 5 D G D G H 20 14 E 20 D 28 21 F A 25 G C,F 14 H D 28 I A J 30 J D,I 45 45 5
Seberapa cepat Turnamen dapat Disesaikan? Aktivitas manakah Yang kritis? Activity Description Immediate Predecessor Time Estimate (days) A Select teams 3 B Mail out invitations A 5 C Arrange accommodations 10 D Plan promotion B, C 3 E Print tickets B, C 5 F Sell tickets E 10 G Complete arrangements C 8 H Develop schedules G 3 I Practice D, H 2 J Conduct tournament F, I 3 11 Activity Expected Duration (weeks) A 2 B 2 Immediate Predecessors C 3 A D 2 B E 1 C,D Activities are represented by nodes: A,2 C,3 E,1 B,2 D,2 12 6
Forward Pass: Calculate Earliest Start Times, Earliest Finish Times Backward Pass: Calculate Latest Start Times, Latest Finish Times Slack Latest Start Time Earliest Start Time 13 A,2 C,3 Activit Expecte Immediate Earliest Earliest Latest Latest Slack y d Duration Predecess ors Start Time Finish Time Finish Time Start Time (weeks) A 2 E,1 B 2 B,2 D,2 C 3 A D 2 B E 1 C,D 14 7
A,2 C,3 Activit Expecte Immediate Earliest Earliest Latest Latest Slack y d Duration Predecess ors Start Time Finish Time Finish Time Start Time (weeks) A 2 0 2 2 0 0 E,1 B 2 0 2 3 1 1 C 3 A 2 5 5 2 0 B,2 D,2 D 2 B 2 4 5 3 1 E 1 C,D 5 6 6 5 0 Activities with 0 slack are on the critical path: Activity A Activity C Act. E Slack Activity B Activity D Act. E Time 0 1 2 3 4 5 6 15 Activity Duration (weeks) A 5 B 4 C 3 Imm Pred ES EF LF LS SLACK D 2 A E 6 B, C F 3 D, E G 7 E H 5 F I 4 F J 2 G 16 8
Acti vity Description Imm Pred A Select teams 3 B Mail out invitations A 5 C Arrange 10 accommodations D Plan promotion B, C 3 E Print tickets B, C 5 F Sell tickets E 10 G Complete C 8 arrangements H Develop schedules G 3 I Practice D, H 2 J Conduct tournament F, I 3 Dur ES EF LS LF SLCK 17 Decision Variables: Objective Function: A,2 C,3 E,1 Constraints: B2 B,2 D2 D,2 18 9
The terminal activity The single activity that identifies when the project is completed. If there is no natural terminal activity, add a dummy node with 0 duration: A, 1 B, 3 C, 1 D, 4 E, 2 19 Aktivitas digambarkan melelui tanda panah A,2 B,2 C,3 D,2 E,1 Source: 1 Source: 1 Find the maximum cost flow Interpretation: an arc has a flow of 1 if it is on the critical path A,2 B,2 C,3 D,2 E,1 Demand: 1 20 10
Often there are penalties and bonuses for late or early completion of a project. 21 Sebuah kontrak untuk menyelesaikan pekerjaan dalam waktu 16 minggu. Terdapat bonus sebesar $12,000 untuk setiap pekerjaan yang lebih awal per minggunya dari jadwal. Penalti sebesar $15,000 per minggu keterlambatan. Kapankah waktu ideal dalam menyelesaikan proyek tsb? Aktivitas manakah yang harus diakselerasi dan seberapa banyak? 22 11
Activity Standard Minimum Extra Cost at Imm Maximum Incremental Duration Duration Minimum Time Pred Reduction Cost (weeks) (weeks) ($000) A 5 3 8 B 4 2 14 C 3 1 16 D 2 1 7 A E 6 3 21 B, C F 3 2 4 D, E G 7 3 8 E H 5 3 8 F I 4 3 8 F J 2 2 N/A G 23 The critical path method (CPM) is a deterministic approach to project planning. Completion time depends only on the amount of money allocated to the activity. Reducing an activity s completion time is called crashing. 12
There are two crucial time durations to consider for each activity. Normal completion time (NT) Crash completion time (CT) NT is achieved when a usual or normal cost (NC) is spent to complete the activity. CT is achieved when a maximum crash cost (CC) is spent to complete the activity. The Linearity Assumption [Normal Time - Crash Time] [Normal Time] = [Crash Cost - Normal Cost] [Normal Cost] 13
Time 20 18 16 14 12 10 8 6 4 2 Normal NC = $2000 NT = 20 days and save on and completion save more time on completion time Add more Add to to the the normal normal cost... cost... Save 25% on completion time Marginal Cost A demonstration of the Linearity assumption Total Cost = $2600 Job time = 18 days Add 25% to the normal cost Crashing CC = $4400 CT = 12 days Additional Cost to get Max. Time Reduction = Maximum Time reduction = (4400-2000)/(20-12) = $300 per day 5 10 15 20 25 30 35 40 45 Cost ($100) Meetings a Deadline at Minimum Cost Let D be the deadline date to complete a project. If D cannot be met using normal times, additional resources must be spent on crashing activities. The objective is to meet the deadline D at minimal additional cost. Tom Larkin s political campaign problem illustrates the concept. 14
TOM LARKIN S POLITICAL CAMPAIGN Tom Larkin has 26 weeks of mayoral election campaign to plan. The campaign consists of the following activities Immediate Normal Schedule Reduced Schedule Activity Predecessor Time Cost Time Cost A. Hire campain staff None 4 2.0K 2 5.0K B. Prepare position paper None 6 3.0 3 9 C. Recruit volunteers A 4 4.5 2 10 D. Raise funds A,B 6 2.5 4 10 E. File candidacy papers D 2 0.5 1 1 F. Prepare campaign material E 13 13.0 8 25 G. Locate/staff headquarters E 1 1.5 1 1.5 H. Run personal campaign C,G 20 6.0 10 23.5 I. Run media campaign F 9 7.0 5 16 NETWORK PRESENTATION A C To meet the deadline H date of 26 weeks some activities must G be crashed. FINISH B D E F WINQSB CPM schedule with normal times. Project completion (normal) time = 36 weeks I 15
Mayoral Campaign Crash Schedule Activity NT NC($) CT CC T M($) M A 4 2000 2 5000 2 $1,500 B 6 3000 3 9000 3 2000 C 4 4500 2 10000 2 2750 D 6 2500 4 10000 2 3750 E 2 500 1 1000 1 500 F 13 13000 8 25000 5 2400 G 1 1500 1 1500 *** *** H 20 6000 10 23500 10 1750 I 9 7000 5 16000 4 2250 Heuristic Approach Three observations lead to the heuristic. The project time is reduced only by critical activities. The maximum time reduction for each activity is limited. The amount of time a critical activity can be reduced before another. path becomes critical is limited. Small crashing problems with small number of critical paths can be solved by this heuristic approach. Problems with large number of critical paths are better solved by a linear programming model. 16
Linear Programming Approach Variables X j = start time for activity j. Y j = the amount of crash in activity j. Objective Function Minimize the total additional funds spent on crashing activities. Constraints The project must be completed by the deadline date D No activity can be reduced more than its Max. time reduction Start time of an activity Finish time of immediate predecessor Min1500Y + 2000Y + 2750Y + 3750Y + 500Y + 2400Y + 17500Y + 2250Y A B Minimize total crashing costs C D E F H J Meet the deadline Maximum time-reduction constraints A C ST X (FIN ) 26 Y Y Y Y Y Y Y A B C D E F H G 2 3 2 2 1 5 10 H FINISH X(FIN) X + (9 Y ) I X(FIN) X + (20 Y ) X X X X X X X X X X + 1 I H X X + (13 Y ) X X X H H G F I E D D C F E X X G C D X B E A A + (4 Y + (2 Y ) + (2 Y ) E E ) + (6 Y ) D + (4 Y + (4 Y F C + (6 Y ) B A A ) ) Activity can start only after all the predecessors are completed. B D E F I 17
Most of the activities become critical!! Deadline WINQSB Crashing Optimal Solution Crashing costs Other Cases of Project Crashing Operating Optimally within a given budget When a budget is given, minimizing crashing costs is a constraint, not an objective. In this case the objective is to minimize the completion time. Incorporating Time-Dependent Overhead Costs When the project carries a cost per time unit during its duration, this cost is relevant and must be figured into the model. In this case the objective is to minimize the total crashing cost + total overhead cost 18
TOM LARKIN - Continued The budget is $75,000. The objective function becomes a constraint Minimize X(FIN) 1500 Y A + 2000 Y B + 2750 Y C + 3750 Y D + 500 Y E + 2400 Y F +1750 Y H + 2250 Y J This constraint becomes the objective function X(FIN) 26 1500 Y A + 2000 Y B + 2750 Y C + 3750 Y D + 500 Y E + 2400 Y F +1750 Y H + 2250 Y J 75,000-40,000 = 35,000 The rest of other crashing model constraints remain the same. Normal time is 13 weeks Project completion time Normal time is 17 weeks Overall crashing cost WINQSB Crashing Analysis with a Budget of $75000 19
Administrative Costs of $100 per week. The campaign must be completed within 26 weeks, but there are weekly operating expenses of $100. The Objective Function becomes Minimize 1500 Y A + 2000 Y B + 2750 Y C + 3750 Y D + 500 Y E + 2400 Y F +1750 Y H + 2250 Y J + 100X(FIN) The other crashing model constraints remain the same. 20