Benefits of PERT/CPM. 2018, Jayant Rajgopal

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1 Project: ny set of complex, interrelated activities aimed at some specific goal. Typically, characterized by a clear starting point and a clear end point. Examples: onstruction (buildings, roads, bridges, etc.) New product introduction New systems installation One-at-a-time manufacturing (e.g., ships, aircraft) Space flight launches Large scale logistics projects etc. enefits of PERT/PM Forces detailed planning Improves communications Identifies potential bottlenecks and problem areas Provides timely progress reports ssists with long range planning including budgets

2 Objective: To find the shortest time in which a project (comprising many interrelated tasks or activities) can be completed, along with a list of the critical activities, and the slack available for the noncritical ones. Network Representation ctivity-on-node (ON) convention: Node a project task/activity with an associated duration rc representation of a precedence relationship one arc for each precedence relationship: if has to precede there is an arc from node to node one STRT and one FINISH node (each of zero duration) ctivity-on-arc (O) convention: rc a project task/activity with an associated duration Node a project event signifying that all activities represented by arcs entering that node have been completed, and that all activities represented by arcs leaving that node can therefore commence Rules Exactly one arc for representing each activity Exactly one arc connecting any two nodes Numbering onvention Node is the STRT node and signifies project commencement has no incoming arcs The highest numbered node is the FINISH node & signifies project completion has no outgoing arcs For the activity (i-j), the numbering should be such that j>i, i.e., flow is always from a lower to a higher numbered node Use dummy arcs (and perhaps, extra nodes) to ensure correct precedence structures: all precedence requirements must be correctly taken into account, yet no unnecessary or extraneous precedence requirement may be introduced into the problem

3 Precedence ON Representation O Representation < < 2 3 <, 2 3,< 2 3,<, 2 3 5

4 Precedence ON Representation O Representation <,, < < <,E < E E E E E WRONG!! WRONG!! WRONG!!

5 company is considering the purchase of several units of an expensive new machine for its plants around the country. However, management would first like to set up a test facility and test the machine thoroughly before committing itself to the entire order; the machine tool manufacturer has agreed to lend them a machine for this purpose. Since it is important to complete this project quickly, it is of interest to find the shortest schedule for the entire project along with some indication of which activities in the project are critical and how much of slack is available for the various phases of the project. The sequence of activities is as follows: first the test facility must be prepared and the paperwork for borrowing the machine must be completed; these can go on concurrently and are expected to take about days. Once these formalities have been completed, the machine must be transported to the facility and repair personnel and operators need to hired. The transportation is expected to take 8 days, while the hiring of personnel for the repair team and operators are expected to take 7 and 6 days respectively. fter the machine has arrived and the repair team is complete, training of the team can begin - this is expected to take 5 days, and can proceed while the actual installation of the machine is still in progress. For the machine to be installed correctly at the test facility after arrival it is expected to take 9 days. fter installation, the hired operators must be trained to run the machine and this is expected to take days. Finally, once the operators and the repair personnel are fully trained, the actual testing can commence and the testing process is expected to take 20 days. Plan the project.

6 esigner Genes (G), a rapidly expanding genetic engineering firm, is relocating to a new office building. Jean Ettic, the project manager, has already leased a vacant office building and approved architectural plans for renovating it. To save money and to gain more control over the project, Jean herself will act as the general contractor for the renovation. She has already selected the subcontractors that she will employ. The move to the new building is complicated by the fact that G will use this opportunity to purchase a new computer and establish an in-house omputer Services epartment. The present decentralized computer capacity is simply inadequate to meet the future needs of G. Project Formulation for esigner Genes' Relocation ode escription of ctivity Estimated uration (days) Predecessors E F G H I J K L M N Hiring of Manager of omputer Services ept. Structural Modifications Enlarging and Resurfacing of Parking Lot Hiring of Staff of omputer Services ept. Purchase and Receipt of omputer Electrical Modifications Heating and ooling Modifications Plumbing Modifications Exterior Painting & Installation of Fixtures Installation of omputer Sheetrocking of Walls and eilings Training of Staff of omputer Services ept. Interior Painting and Interior ecorating Landscaping None None None E,F,G F,G,H,J K,I

7 Objectives: To find the minimum time required to complete the project, and to evaluate the criticality of each activity with respect to this goal. EFINITIONS Earliest Start Time for ctivity j (ES j ): Earliest time at which the activity can start: ES j = Max i P(j) {ES i + i } where P(j) is the set of all prerequisite activities that immediately precede activity j Earliest ompletion Time for ctivity j (E j ): Earliest time at which the activity can end: E j = ES j + j

8 onsider the portion of a network shown below with activities and durations as shown in the nodes. Focus on Node j=. = ES = 2 2 =3 ES 2 =6 3 3 = 6 ES 3 = = 5 ES =? y definition: ES E, ES E 2, ES E 3 i.e., ES (ES + ) = + = 8 ES (ES ) = = 9 ES (ES ) = + 6 = 0 Thus ES = Max{8,9,0} = 0, (and E = 0+5=5)

9 Latest ompletion Time for ctivity j (L j ): Latest time at which activity j can be completed without causing the project to be delayed beyond its minimum duration: L j = Min i S(j) {L i - i } where S(j) is the set of all activities that immediately succeed activity j Latest Start Time for ctivity j (LS j ): Latest possible time at which activity i can start without causing the project to be delayed beyond its minimum duration: LS j = L j - j

10 Once again, focus on Node j=. = 5 L =? 5 5 =3 L 5 = =6 L 6 = = 3 L 7 =25 y definition: L LS 5, L LS 6, L LS 7 i.e., L (L 5-5 ) = 28-3 = 25 L (L 6-6 ) = 27-6 = L (L 7-7 ) = 25-3 = 22 Thus L = Min{5,,22} =, (and LS = -5=6)

11 Total Float for ctivity j (TF j ): The maximum "excess" or slack time available for activity j in the optimal project schedule: TF j = {(L j - ES j ) - j } Note that (L j -ES j ) represents the window of time available in the optimal schedule for activity j. ritical ctivity: One that has a total float equal to 0, i.e., activity j is critical if TF j =0. Note that TF j =0 (L j -ES j )= j ES j + j =L j. Thus, a critical activity has no excess time available for its completion - it MUST start as soon as possible (at ES j ) so that it is completed no later than the time at which it must be completed in order for the project to not be delayed (L j ). ritical Path: path of critical actvities in the project network that link the STRT node and the FINISH node. It is possible for a project to have more than one critical path!

12 . Forward Pass: egin at the STRT node with ES STRT =0; find ES j =Max i P(j) {ES i + i } for each activity j, going from left to right. Ensure that ES j is computed only after finding ES i for all i P(j), i.e., for all activities that immediately precede activity j. This step ends with the computation of ES FINISH (which is also equal to E FINISH ). 2. ackward Pass: Starting at the FINISH node with L FINISH =E FINISH find L j = Min i S(j) {L i - i } for each activity j, going from right to left. Ensure that L j is computed only after finding L i for all i S(j), i.e., for all activities that immediately follow activity j. This step ends with the computation of L STRT =0 (which is also equal to LS STRT ). 3. ompute TF j = {(L j -ES j ) - j } for each activity j in the project network to identify the critical activities and the critical path.

13 , Start, 0,, 8 H, 20 Finish, 0 0 0, 6 0 F, 5 E, 9 The forward pass first yields the ES values, and then the backward pass yields the L values in the table below: Node j ES j L j Start 0 {L - }= {-}= 0 {ES start + start }= 0+0= Min{L -,L -,L - } = Min{-8,8-7, -6}= 5 0 G, {ES + }=0+= Min{L E - E, L F - F } = Min{-9, -5}= {ES + }=0+= {L F - F }=-5 = 8 {ES + }=0+= {L G - G }=- = E {ES + }=+8= {L G - G }=- = F Max{ES +, ES + } =Max(+8,+7) = {L H - H }=-20 = G Max{ES +, ES E + E } =Max(+6,+9) = {L H - H }=-20 = H Max{ES F + F, ES G + G } =Max(+5,+)= {L Finish - finish }= -0 = Finish {ES H + H }=+20= = E finish = ES finish + finish = 8 27

, 7 8 0 0 Start, 0,, 8 H, 20 Finish, 0 0 0, 6 0 F, 5 E, 9 The forward pass first yields the ES values, and then the backward pass yields the L values in the table below: Node j ES j L j Start 0 {L -

14 , Start, 0,, 8 H, 20 Finish, 0 0 0, 6 0 F, 5 E, 9 The forward pass first yields the ES values, and then the backward pass yields the L values in the table below: Node j ES j L j Start 0 {L - }= {-}= 0 {ES start + start }= 0+0= Min{L -,L -,L - } = Min{-8,8-7, -6}= 5 0 G, {ES + }=0+= Min{L E - E, L F - F } = Min{-9, -5}= {ES + }=0+= {L F - F }=-5 = 8 {ES + }=0+= {L G - G }=- = E {ES + }=+8= {L G - G }=- = F Max{ES +, ES + } =Max(+8,+7) = {L H - H }=-20 = G Max{ES +, ES E + E } =Max(+6,+9) = {L H - H }=-20 = H Max{ES F + F, ES G + G } =Max(+5,+)= {L Finish - finish }= -0 = Finish {ES H + H }=+20= = E finish = ES finish + finish = 8 27

15 We now go to Step 3 for TF values for each activity ctivity(j) j ES j L j TF j = (L j -ES j )- j E 9 0 F 5 6 G 0 H 20 0 The critical activities are,, E, G and H and the critical path is shown below:, 7 F, 5 Start, 0,, 8 H, 20 Finish, 0 E, 9, 6 G,

16 TIVITIES IN THE ORER S ENTERE ctivity ctivity Earliest Latest Name Symb Time Start/Fin Start/Fin Slack T c T c T T T 5 E c T 6 F T 7 G c T 8 H c The computations were based on 8 activities Earliest project completion time =.0000

17 PM EXMPLE R HRT: NONRITIL TIVITIES SORTE Y SLK ccccc ccccccccc E cccccccccc G ccccccccccccc H ccccccccccccccccccccccc F xxxxxxxxxxxxxxxxx... xxxxxxx... xxxxxx PM EXMPLE R HRT: NONRITIL TIVITIES SORTE Y ERLIEST STRT ccccc ccccccccc E cccccccccc G ccccccccccccc H ccccccccccccccccccccccc xxxxxxx... xxxxxx... F xxxxxxxxxxxxxxxxx

Time Planning and Control

Time Planning and Control Time Planning and ontrol Precedence iagram G44 NGINRING MNGMNT 1 Precedence iagramming S ctivity I LS T L n important extension to the original activity-on-node concept appeared around 14. The sole relationship