Chapter 7. Project Scheduling and Resource Levelling – Construction Project Management: Theory and Practice


Project Scheduling and Resource Levelling

Introduction, resource levelling, resource allocation, importance of project scheduling, other schedules derived from project schedules, network crashing and cost-time trade-off


After the network calculations are made, it is important to present their results in a form that is easily understood by all the stakeholders involved with the project. Most of the times, the results of network techniques such as PERT/Critical Path Method/Precedence Diagram are plotted in the form of a Bar chart or a Gantt chart for its obvious advantages. A Gantt chart is plotted using either the early start time of activities or the late start time of activities. In the former case the chart is known as early Gantt chart, and in the latter case, late Gantt chart. The bar charts themselves are dependent on the ideal condition of availability of efficient resources as and when needed by an activity in a project. Due to the competition between activities for a particular resource, the demand may at times exceed the planned availability.

Through these charts, a range of information can be derived—for example, the total floats available in an activity, the critical activity, etc. As discussed earlier, these charts can also be used for plotting and comparing the planned progress versus the actual progress achieved. Further, these charts are useful for plotting the resources required on a day as well as for giving useful information on the cost aspects.

It may be recalled that while preparing the network diagram we concentrated mainly on the technological constraints (for instance, one activity cannot start until the other is over), and assumed that the resources are unlimited. Further, the resources can be mobilized on demand anytime we require them. Although such ideal conditions may prevail in some exceptional projects, in most of the real-life projects we face resource constraints. In real-life situations, the activity start times not only have to deal with technological constraints in the form of their precedence relationships, but also face the challenge of resource availability constraints. That is, we have limited quantity of resources at our disposal.

The planning for scheduling of activities, thus, has to account for the different constraints that may be imposed on the availability of any of these resources, and ensure that sudden changes in the requirement of these are avoided. For example, using an equipment at the outset and then at the end of the project should be avoided, as even idle (or idling) equipment carry a certain cost. Similarly, sudden increase (or retrenchment) of labour should also be avoided!! Thus, effort has to be made to ensure that resources are allocated to activities in a manner that any changes in demands are gradual to the extent possible, and that the available resources are optimally used.

Resource constraints pose several operational problems in network-based scheduling. The concepts of float and criticality of activities start losing their normal meanings. For example, now the early start of an activity is governed not only by technological constraints but resource constraints as well. Hence, there cannot now be a single early start schedule as in the case without resource constraints. Similarly, the value of slack or float is no longer unique since it is dependent on early start and late start schedule.

The problem of scheduling activities so that none of the precedence relationships are violated and none of the resource availabilities are exceeded is a difficult task, and it could result into a number of combinations of possible early start and late start schedules. In literature, scheduling problems with limited resources is classified as a large combinatorial problem. A number of programmes based on some rules of thumb (also called heuristic) have been developed to solve such problems. These heuristic programmes for resource scheduling can broadly be classified under two categories: (1) resource levelling and (2) resource allocation.


In resource levelling, the constraint is the fixed project duration. That is, the project must get completed by a fixed date. The attempt of such heuristic is to reduce peak requirement of resources and to smooth out period-to-period assignments. Such problems are also referred to as ‘time limited resource considerations’ problems. The assessment of resources is done using a resource-loading or resource-aggregation chart. We take an example to illustrate the concept of resource levelling.

Let us assume that there are a total of seven activities, A to G, in the example network (Figure 7.1). The duration of each of the activities is written below the arrow, while the resource requirement is shown in the bracket adjacent to the activity name. For example, the duration for activity A is three days and the resource required is two units. The early start and late start time of events or nodes are also shown in the network, from which the float available in a particular activity can be calculated and critical activities identified. The critical path of the network is 1-2-4-5-6 and it consists of activities A, C, E and G. The critical path is shown by bold arrows in the network.

Resource levelling can, in principle, be carried out through the following steps:

  1. The project network is prepared based on the data provided for each activity. Event times and activity times are computed as illustrated earlier; thus, total float is also computed for each of the activities.
  2. The activities are ranked in order of their earliest start date (refer to Table 7.1). Activity (1, 2) has ‘0’ as its earliest start time. Activities (2, 3), (2, 4) and (2, 5) all have their earliest start times on day 3, while activities (4, 5), (3, 5) and (5, 6) have their earliest start times on days 6, 7 and 11, respectively. The resources required on daily basis for each of the activities are summed up and shown in the form of a chart called resource-aggregation or resource-loading chart. Figure 7.2 shows a resource-loading chart based on the earliest start time of all the activities. The project takes a total of 87 man-days to complete, and the daily requirement varies from a minimum of 2 resources on days 1 to 3, to a maximum of 11 resources on days 4, 5 and 6.

    Figure 7.1 Network for resource leveling illustration

    Table 7.1 Resource-loading table showing daily requirements of workers based on early-start order

    Figure 7.2 Resource-loading chart based on early start

  3. Now, the activities are ranked in order of their latest start date (refer to Table 7.2). It may be noted that the latest start date of an activity is the latest time of the finish event less the duration. Thus, latest start times of activities (1, 2), (2, 4), (2, 3), (4, 5), (2, 5), (3, 5) and (5, 6) are on days 0, 3, 5, 6, 7, 9 and 11, respectively, in the ascending order. The resource-loading chart shown in Figure 7.3 is prepared based on the ascending order of latest start times of each activity. It can be noticed that the requirement of resources varies from a minimum of 2 to a maximum of 10 resources.


    Table 7.2 Resource-loading table showing daily requirements of workers based on late-start order

    Figure 7.3 Resource-loading chart based on late start

  4. The two resource-loading charts obtained from steps 2 and 3 are compared. The two charts provide the two extreme arrangements of resource requirements. In the case where peaks and valleys are seen in the utilization pattern for a resource, the activities are manipulated by visual inspection and an acceptable resource requirement is found between the two extremes. The bottom line is to ensure continuous deployment of resources and to avoid large variations in the utilization pattern. One such compromise solution is shown in Figure 7.4 (also refer to Table 7.3). This figure has been obtained by allocating the resources in the following manner.

    Figure 7.4 Resource-levelled chart (time constrained)

    Table 7.3 Resource-loading table showing daily requirements of workers after levelling

    1. The early start time of non-critical activity (2, 3) has been followed.
    2. The late start time of non-critical activities(2, 5), and (3, 5) have been followed.
    3. The critical activities (1, 2), (2, 4), (4, 5), and (5, 6) have not been disturbed.

The above manipulations have resulted in reducing the peak requirement (from 11 to 10) besides bringing a gradual change in resource deployment. One can also note that there is no change in the project duration of 16 days. This is also known as time-constrained leveling.

As can be understood from this example, it would be extremely difficult to employ this technique of visual examination for large problems. For simple problems the levelling exercise can be completed in one attempt, but for larger problems, the resource levelling cannot be carried out in a single step and is a largely iterative process. Computers can be employed to good advantage for levelling of resources under a time-constrained situation. A number of heuristics have been developed for this purpose and they are discussed elsewhere.


Here, availability of resources is a constraint. In other words, the resources have fixed limits. For illustrating resource allocation, we take the same problem (project network of Figure 7.1) that was used for resource levelling. The difference here is resource constraint. Let us take the maximum availability of resources as eight. In case it is not possible to resolve the resource over-allocation on a particular day, some activities may have to be delayed. The basic objective here is to find out which activities can be delayed and by how much, and finally to arrive at the shortest possible time to complete the project satisfying the resource constraints.

When the resource requirement on a particular day exceeds eight, we decide the priority of competing activities on a predefined set of rules. Let the activities with earliest start time get the first priority. In case there is a tie between two or more activities, the priorities are decided on the basis of float available in the activities. That is, the activity with the minimum float gets first priority. Further, non-critical activities may need to be rescheduled in order to free the resources for critical activities. Also, let’s not stop an activity in-between once it has started.

The network of Figure 7.1 can be redrawn to time scale. This is shown in Figure 7.5.

Figure 7.5 Network of Figure 7.1 redrawn to time scale

Day 1

Only activity (1, 2) is scheduled for day 1 and only two resources are required, which is less than the maximum. Thus on day 1, requirement—2, availability—8, remaining—6.

Day 2

Activity (1, 2) continues. Requirement—2, availability—8, remaining—6.

Day 3

Activity (1, 2) continues. Requirement—2, availability—8, remaining—6.

The time-scale network for days 1–3 is shown in Fig 7.6.

Day 4

The three activities (2, 3), (2, 4) and (2, 5) can be started on day 4 (see Fig 7.7). Thus, resource requirement would become 3 1 4 1 4 5 11, which is more than 8, the maximum limit. The decision rule ‘activity with least float gets priority’ for resource assignment comes into play here.

Figure 7.6 The time-scale network on day 1-3

Figure 7.7 The time-scale network on day 4

Activity (2, 4) lies on the critical path, i.e., float 5 0, and hence, resources need to be allocated to it first. The activity (2, 4) requires 4 units of resources, while availability is 8; thus, resources remaining after assigning to (2, 4) are 4 (8 2 4 5 4). Next in queue is activity (2, 3) with float 5 2 days. Thus, resources are assigned to (2, 3). The requirement of this activity is 3, while now the availability is 4; thus, 1 resource is left out after this allocation.

Activity (2, 5) is left out since it needs 4 resources but the available resource is just 1; hence, we need to postpone this activity. Also, the float of this activity is reduced to 3 now.

Day 5

Since according to our ‘decision rule’, the activities in progress cannot be stopped in-between, the resources are allocated to them first. Thus, the resource assignment on day 5 is as below:

Activity (2, 4): requirement—4, availability—8, remaining—4

Activity (2, 3): requirement—3, availability—4, remaining—1

Activity (2, 5), again, has to be postponed because of the unavailability of resources. Also, the float of this activity is reduced to 2 now (see Fig 7.8).

Day 6

Based on the previous argument that the activities in progress cannot be stopped in-between, resources must be allocated to them first. We have the following resource assignment for day 6:

Figure 7.8 The time-scale network on day 5

Activity (2, 4): requirement—4, availability—8, remaining—4

Activity (2, 3): requirement—3, availability—4, remaining—1

Activity (2, 5) has to be postponed because of unavailability of resources. The float remaining for this activity is 1 day (see Fig 7.9).

Day 7

Since activities in progress cannot be stopped in-between, resources must be allocated to them first. Thus, resource assignment for day 7 is as below:

Activity (2, 3): requirement—3, availability—8, remaining—5

Activities (4, 5) and (2, 5) both compete for the remaining resources. The activity (4, 5) requires 3 resources while activity (2, 5) requires 4 resources. Thus total requirement for these two activities is 7 as against the remaining 5 resources. According to the algorithm, the first preference would be given to activity (2, 5) since it has an earlier start time compared to activity (4, 5). This is despite activity (4, 5) being on the critical path.

Requirement for activity (2, 4)−4, availability—5, remaining—1

Activity (4, 5) has to be postponed because of the unavailability of resources. This will increase the project duration by one day (see Fig 7.10). Thus the project duration has become 17 days now. Also note the 1 day float available now for activity (3, 5).

Day 8

Activity (2, 5) continues. Requirement—4, availability—8, remaining—4

Activity (4, 5) can be taken up now. The resources required are 3 and the availability is 4. The remaining 1 resource remains unutilized.

Figure 7.9 The time-scale network on day 6

Figure 7.10 The time-scale network on day 7

Figure 7.11 The time-scale network on day 8–9

Day 9

Activities (2, 5) and (4, 5) continue.

Activity (2, 5): Requirement—4, availability—8, remaining—4

Activity (4, 5): Requirement—3, availability—4, remaining—1

The time-scale network for days 8-9 is shown in Fig 7.11.

Day 10

Since activities in progress cannot be stopped in-between, resources must be allocated to them first.

Activity (2, 5): Requirement—4, availability—8, remaining—4

Activity (4, 5): Requirement—3, availability—4, remaining—1

Activity (3, 5) cannot be taken up on this day since it requires 3 resources as against the remaining 1 resource.

Thus it has to be delayed which would exhaust the 1 day float available with this activity (see Fig 7.12).

Day 11

Activity (2, 5) is completed.

Activity (4, 5) continues. Requirement—3, availability—8, remaining—5

Activity (3, 5) can be taken up now. Requirement—3, availability—5, remaining—2

Day 12

Activities (4, 5) and (3, 5) continue.

Activity (4, 5) : Requirement—3, availability—8, remaining—5

Activity (3, 5): Requirement—3, availability—5, remaining—2

The time scale network for days 11–12 is shown in Fig 7.13.

Figure 7.12 The time-scale network on day 10

Figure 7.13 The time-scale network on day 11-12

Day 13

Only activity (5, 6) requires resource.

Activity (5, 6): Requirement—4, availability—8, remaining—4

Day 14

Activity (5, 6) continues.

Activity (5, 6): Requirement—4, availability—8, remaining —4

Day 15

Activity (5, 6) continues.

Activity (5, 6): Requirement—4, availability—8, remaining —4

Day 16

Activity (5, 6) continues.

Activity (5, 6): Requirement—4, availability—8, remaining —4

Day 17

Activity (5, 6) continues.

Activity (5, 6): Requirement—4, availability—8, remaining —4

The time-scale network for days 13–17 is shown in Fig 7.14.

Thus, we observe that the project duration has been increased by 1 day from the original 16 days, with no resource constraint to 17 days with resource limit of 8. The resource requirement varies from 2 to 7. The resource-loading chart on different days (day 1 to day 17) for the total project duration is shown in Figure 7.15.

Figure 7.14 The time-scale network on day 13-17

Figure 7.15 Resource-allocation chart on different days (non-time constrained/resource constrained leveling)

The steps illustrated above can be carried out manually only if the project is small. For large projects involving multiple resources, we need to look to computer programs that have been illustrated in Chapter 18.

The examples presented for resource levelling and resource allocation are that of a very simplistic nature. First of all, we have considered only a single type of resource used for all types of activities. Further, we have considered that the resource is interchangeable across different activities. Unfortunately, in real-life situations, both these assumptions do not hold true. In practice, we have to deal with different types of resources such as equipment, materials, labours, etc., and further, a single activity may require more than one type of resource. Such complicated problems are solved using computer programs and are discussed in detail in Chapter 18.


The success of a project heavily depends on how effective the scheduling is and how tightly the project can be controlled. Poor scheduling can easily result in completion delays and cost overruns. These, in turn, result in claims and counter-claims, disagreements and disputes. The appropriate method should be used in planning and scheduling a project; management decisions should not depend on experience and intuition alone.


Table 7.4 Project schedule preparation steps in general

Identification and classification of tasks

This is the first critical step wherein all individual tasks or activities required to complete the project are listed out.

Defining duration of a task

This is essentially the first iteration at estimating the time durations involved in completing the different tasks. All available information should be used to make an accurate estimate of the duration of each task, and depending upon the nature of the project and the activity, the estimated time may be given in hours, days, or months.

Task sequence

Activities in a construction project may appear to be independent, but often depend upon each other—for example, the painting in a house is done by an independent agency, but the work can commence only after the plastering of the walls is completed. Continuing with the example of house construction, there is obvious interdependence in electrical, plumbing and finishing works. Although it appears trivial, such issues of ‘inherent’ logic and ‘predecessor’ and ‘successor’ tasks should be sorted out at the outset to reduce the possibility of difficulties later on.
In larger projects, it also helps to identify tasks that can be completed or carried out at the same time. Such tasks, or others that require careful handling, may be ‘flagged’ at the outset and special attention paid to their progress and any possible slippage. To that extent, it may be noted that an easy-to-understand depiction of relationships between different activities is an important function of the project planner. Different methods of such representations, such as a network diagram, are available and discussed in greater detail elsewhere.

Creating a calendar of events

Once the list of activities, their interdependence and the time involved in completing them is known from the steps above, a calendar of events can be easily created highlighting the milestone events.

Preparing project schedule is extremely important due to a number of reasons. Project schedule is the basis of extracting a number of other schedules and information. Even at the tendering stage, project schedules need to be prepared although there may not be required information for preparing project schedules. Project schedules are helpful not only for a contractor, but for the owners as well. In order to draw up a comprehensive schedule, the steps mentioned in Table 7.4 must be systematically followed.

For the contractors, project schedules help in timely mobilization of the required resources and help in identification of bottlenecks in the early stages of project duration. They also act as reference for comparison with actual progress, cost of construction, and profit margin at any given time during the course of monitoring. For the owner also, project schedule helps in understanding the requirement of finance to be mobilized. Also, with the project schedule in place, it is easy for the owner/client to monitor the project accordingly. The detailed steps carried out to prepare a project schedule by a contractor are listed below:

  • Identify and study the scope of work from the contract document (tender drawings, bill of quantities, terms and conditions, etc.). Contractors normally do not have access to all the details needed to prepare an exhaustive schedule at the beginning, but that should not be an excuse for non-preparation of schedule for the project. In fact, contractors should try to extract as much information as possible, even by unofficially meeting the owner’s representative and architects for the project. Owners try not to give information based on incomplete documentation on their part, for the fear of getting penalized at a later date.
  • Decide on the construction methodology for completing the project within the given time schedule. The time schedule specified in the tender document mostly governs the construction methodology adopted by the contractor.
  • The total project scope is finally brought down into work packages and activities. This breakdown into smaller activities helps the planner in estimating resource requirement and duration for completing the activities. The productivity norms as well as the experience of the planner are utilized to arrive at the estimates of duration and resource requirement for these activities.
  • Finally, using the relationships existing among different activities in the form of a network for the project is prepared. The network can be prepared by using either PERT/CPM or the precedence network.
  • From the network, a week-wise or month-wise project schedule is derived.
  • In large projects involving a number of subprojects, a master project schedule is prepared in addition to preparing schedules for each of the subprojects.
  • Finally, it should be kept in mind that these schedules need to be revised as and when any new information is available or at a fixed interval such as fortnightly or monthly.

From a project schedule, the following schedules can be prepared:

  1. Invoice schedule
  2. Cash inflow and cash outflow schedule
  3. Staff schedule
  4. Labour schedule
  5. Material schedule
  6. Specialized subcontractor schedule
  7. Plant and equipment schedule
  8. Working capital schedule
  9. Estimation of direct and indirect costs

These are explained with the help of an example. Let us assume that a contractor has been awarded a contract for the expansion of an automobile factory. The contract value is five crore rupees and the contract duration is 10 months from the date of receipt of letter of intent (explained in later chapters). The bill of quantities for the project is given in Table 7.5.

The schedule of the project (in bar chart form) from the project network prepared by the contractor is given in Figure 7.16. The schedule shows the thirteen items from 1 to 13. The durations for each of the items are given in months. For example, earthwork is scheduled to take 6 months, from month 1 to month 6.

The figure also gives the break-up of quantities in percentage terms on a monthly basis. For example, the quantities of earthwork planned to be executed from month 1 to month 6 are 5 per cent, 15 per cent, 20 per cent, 25 per cent, 25 per cent and 10 per cent, respectively. That is, 5 per cent of 5,000 m3 is planned for month 1, 15 per cent of 5,000 m3 is planned for month 2, and so on. The percentage break-up is given for all the items, which can be interpreted in a similar manner.

7.5.1 Preparing Invoice Schedule

Invoice is the amount of money realized by executing a particular activity or work. Accordingly, invoice value is the product of the quantities of work done each month for a particular activity and the corresponding rates. The rates are nothing but the quoted rates by the contractor. It must be kept in mind that the invoice schedule is the estimated billing amount on a monthly or weekly basis. The actual invoice may be exactly as per the planned one or it may be ahead or behind the estimated figure. This schedule also acts as the basis for monitoring progress of invoicing. There is a limitation, however, in using this schedule to monitor physical progress of a project, as completion dates of individual activities/milestones cannot be indicated herein.


Table 7.5 Bill of quantity for the expansion of automobile factory

Figure 7.16 The bar chart and break-up of planned quantities of different items in percent

The steps involved in the preparation of invoice schedule are as follows:

  1. From the quantities of items/activities taken from bill of quantities, a month-wise or week-wise quantities to be executed every month is prepared corresponding to the project schedule. In Figure 7.16, the month-wise break-up of quantities (in percent terms) for each activity is shown above the bar corresponding to that activity. For example, for earthwork—all soils activity, the monthly break-up is5 per cent, 15 per cent, 20 per cent, 25 per cent, 25 per cent and 10 per cent corresponding to months 1, 2, 3, 4, 5 and 6, respectively. One can also assume a uniform distribution of quantities for simplicity, although it may not be applicable in practice. For example, 833.33 m3 (16.67 per cent of 5,000 m3) of excavation may be assumed to get executed every month to complete the total quantity of 5,000 m3 in 6 months.
  2. Based on the quantities and unit rates (to be obtained from quoted rates) for each month, calculate invoicing amounts for all the activities. For large projects involving a number of items or activities, only major items need to be taken into account for this exercise. The remaining items can be clubbed together as other items. However, quantities for such items need not be mentioned. For projects having subprojects, a consolidated invoice schedule for the project can be prepared by combining the invoice schedules of subprojects.
  3. As discussed elsewhere, one of the major problems that may be faced in the preparation of an invoice schedule is insufficient details in the drawings and bills of quantities available at the time of scheduling. For example, the item-wise split-up of quantities (in terms of concrete, shuttering, steel, etc., in columns, footings, floors) may not be available. In such cases, reasonable assumptions and past experience are utilized to arrive at the quantities.
  4. The above steps are easy to carry out for an item rate contract. But for other types of contract such as lump-sum contracts, split-up of lump-sum price for each stage of payment will be taken into account since payment may not be based on quantities of work done.
  5. The total invoicing figure will also include escalations as per accepted terms in the contract. Further, a distinction should be made between the invoice figure and the actual payment due. The entire bill amount is not payable to the contractor. There are only certain percentages of the total work done that are paid to the contractor. The rest of the money is kept either as retention money or as certain deductions in lieu of the advances drawn earlier by the contractor and certain taxes. The percentage payment and deduction applicable are as per the agreed terms of payment and should be taken into account in arriving at invoicing figures.

Table 7.6 Distribution of quantity planned to be executed activity-wise on a monthly basis

The month-wise break-up of quantities for all the activities are given in Table 7.6, and the invoice schedule is given in Table 7.7 for the case project.


Table 7.7 Estimated bill value or invoice value on weekly basis for the case project (values in Rs. lakh)

7.5.2 Schedule of Milestone Events

Milestone events are very important in the proposed schedule of a project. Top management is more concerned with the scheduled completion of a milestone event. Some examples of milestone events are—completion of foundation, completion of superstructure, and completion of sanitary and plumbing works. Thus, from the schedule of the project, the details of milestone events are separated and reported in the schedule of milestone events. The slippage, if any, and its financial implications are closely monitored, and the impact of slippage on other activities is also studied. The corrective actions to be taken to bring the slipped milestone event back on track are proposed. The person responsible for correcting the slippage is also identified in such reports. A typical format for preparing this report is given in Figure 7.17.

Figure 7.17 Schedule of milestone events

The schedule of milestone events has gained importance these days as any slippage from the scheduled completion date of a milestone event can become the cause for invoking ‘liquidated damages’ clause by the owner. Thus, from the perspective of a contractor, this schedule is very important.

7.5.3 Schedule of Plant and Equipment

In this schedule (see Table 7.8), the date of particular plant and equipment is also mentioned. The release date of a plant and equipment that is surplus at project site is mentioned as well. Finally, the time required for doing pre-dispatch maintenance is mentioned.


Table 7.8 Schedule of plant and equipment for the case project

7.5.4 Schedule of Project Staff

Depending on the invoicing to be done per month, a month-wise schedule for project staff is maintained (see table 7.9). In this schedule, the date by which a particular staff is required is also mentioned. The release date of a particular category of staff that is surplus at project site is to be mentioned as well.

The number of staff is dependent on the nature of activity, the quantum of work, the working hours adopted by the organization, the proportion of work which the main contractor is planning to subcontract, etc. As discussed in Chapter 1, a large number of project staff is needed for performing different activities. Some of the staffs are specialized in some areas such as planning, billing, supervision, and so on.

7.5.5 Schedule of Labour Requirement

A typical format of labour requirement schedule is shown in Figure 7.18. From the split-up of quantity, labour requirement is calculated. The requirement of labour is estimated for each activity. The number of labourers is dependent on the quantity involved in an activity and the productivity of labourers. While the productivity aspect has been dealt with separately elsewhere in the book, it has been referred to for illustration here. Thus, in our example, the quantity of earthwork involved in month 1 is 833.33 m3. If the productivity of labourers for this work is assumed to be 3 man hours per m3, the total man hours required for this activity alone would be 2,500 man hours. If it is assumed that each day consists of 10 working hours, and there are 25 working days, the number of workers required for this activity alone would be 10 (2500/10/25) for month 1. The requirement of workers for other months for the same activity is also computed in a similar manner.


Table 7.9 Schedule of project staff

Figure 7.18 Schedule of labour requirement

A similar exercise is carried out for other activities also. The monthly requirement of labourers is obtained by summing up the requirement for each activity planned for a month. Care should be taken that computations for different types of workers (for example, unskilled, semi-skilled and skilled) are carried out separately. This way, one can find out the requirement for workmen of different categories and present the same in the form of a schedule.

7.5.6 Schedule of Materials Requirement

Based on the quantity of a particular item, the type and quantity of materials to be purchased are established. The schedule helps the project manager to assess the timing at which a particular material is desired for the project. The material requirement as a function of time shown month-wise in the schedule (could be any other measurement of duration such as week) is predetermined based on the bar chart of the project. The sample schedule of materials requirement is provided in Table 7.10 for the case project.

For example, if the material for the item brickwork is required to be calculated, we find that the number of bricks required in months 3, 4, 5, 6, 7, 8 and 9 are 85,000, 127,500, 170,000, 170,000, 170,000, 85,000 and 42,500, respectively. This is on an assumption that 500 bricks are needed for 1 m3 of brickwork. Such predetermined schedules help a project manager in taking the procurement action well in time, thus realizing advantages such as discounts from the material supplier.

If, on the other hand, material is procured on an emergency basis, chances are that one may compromise on quality besides getting the material at higher prices.

7.5.7 Schedule of Specialized Agencies

In a project, a contractor may not be in a position to do all the activities in an economical manner. Further, for some of the activities in a project, he may not have the requisite expertise. In such cases, the contractor enters into a contractual arrangement with some specialized agencies called subcontractors or speciality contractors. As a first step, a list of activities for which the contractor wants to engage subcontractors is prepared. The scheduled dates of start and completion of such activities are noted. Let us assume that the contractor wants to engage subcontractors for activities ‘waterproofing’, ‘aluminium work’, ‘sanitary and plumbing works’, and ‘road works’. The scheduled starts for these works are shown in Table 7.11, which has been replicated from Figure 7.16.


Table 7.10 Schedule of materials requirement for the case project

1Cement is required for concreting, brickwork and plastering. For concrete the per-m3 requirement is assumed to be approximately 6.5 bags; for one m3 of brickwork (with 1:6 cement mortar) it is assumed to be 1.25 bags; and for plastering (1:4 cement mortar) it is assumed to be 0.109 bags per sqm. The exact requirement would come from the design mix. Wastage factor of 3 per cent is also added on top of this to get the total requirement.


Looking at the schedule, the contractor would know the latest time to start the process of finalization of contractor for a particular activity. For example, if the contractor takes one month to finalize the deal with his subcontractors, he would start the process of subcontractor selection for ‘waterproofing’ at the start of month 6, while the selection process for ‘aluminium work’, ‘sanitary and plumbing works’, and ‘road works’ would be initiated at the start of months 4, 2 and 5, respectively.

There are other types of schedules that can be prepared based on the project schedule. These are schedule of direct costs, schedule of overheads, schedule of cash inflows and outflows, etc. Typical formats for all these schedules have been given in figures 7.19 to 7.22. These have been described at appropriate places in detail.


Table 7.11 Schedule of specialized agencies

7.5.8 Schedule of Direct Costs

Figure 7.19 Schedule of direct costs

7.5.9 Schedule of Overheads

Figure 7.20 Schedule of overheads

The site contribution listed in Figure 7.31 is equal to invoicing minus overhead minus direct costs.

7.5.10 Schedule of Cash Inflow

Figure 7.21 Schedule of cash inflow

7.5.11 Schedule of Cash Outflow

Figure 7.22 Schedule of cash outflow


In the previous chapter, we discussed the computational procedures of critical path method. In this section, we discuss one of the application areas of this network technique. One of the assumptions in critical path method is that the duration of an activity can be reduced or crashed to a certain extent by increasing the resources assigned to it. As is known, the execution of an activity involves both direct costs and indirect costs. These terminologies are explained in detail elsewhere.

It is known that any reduction in duration (by increasing resources) of critical path activities can reduce the project duration and, thereby, enhance the possibility of reduction in project cost. However, as will be explained shortly, there is no point in attempting to crash all the activities by increasing the resources.

An activity can be performed at its normal or most efficient pace or it can be performed at higher speed. The duration associated with the former is called ‘normal duration’ and the duration associated with the latter is called ‘crash duration’.

Some activities along the critical path sometimes need to be shortened in order to reduce the overall duration of the project. This leads to a decrease in the indirect expenses (due to decrease in duration), as shown in Figure 7.23, and an increase in the direct expenses (due to more mobilization of resources), as shown in Figure 7.24. As is evident from Figure 7.23 and Figure 7.24, the relationship between the cost of the job and the duration has been assumed to be linear. The steeper the slope of the line, the higher the cost of expediting the job at an earlier date.

The expediting of an activity to an earlier time is referred to as crashing. There are three cases that normally arise:

  1. Line sloping down to the right—The steeper the slope, the higher the cost of crashing
  2. Horizontal line—There would not be any cost of crashing
  3. Vertical line—The activity cannot be shortened regardless of the extra resources applied to it

Figure 7.25 depicts the three possible cases.

Figure 7.23 Indirect cost vs time

Figure 7.24 Direct cost vs time

Figure 7.25 The three possible cases in ‘direct cost vs time’

Some important inferences drawn from Figure 7.25 are:

  1. There is a point beyond which there can be no decrease in direct costs.
  2. Slowing the project even further may only result in the increase in expenditure (last portion of the graph).

The point of the minimum cost of project is known as the optimum point. In order to find the optimum point, the project network is drawn based on the normal duration of the activities. This is the ‘maximum length’ schedule. The duration of the project is thus noted. It can be shortened by expediting jobs along the critical path. If the added cost of expediting the job is less than the saving in the indirect expenses which result from shortening the project, then a less expensive schedule can be found. New schedules are found as long as there is a reduction in the cost of the project.

At each step of the process, only the activities along the critical path are considered for crashing. The cost-time slope of each activity is examined and the activity with the least slope determined. This is the activity that can be shortened with the least increase in direct costs. If the increase in the direct costs is less than the savings in the indirect costs, then the activity is crashed up to the point where no further shortening of the schedule is possible (either because the activity duration cannot be reduced further or some other activity has become critical along a parallel path). The remaining critical activities are examined and the one with the flattest cost slope value is selected. The process is repeated until no further shortening of critical activities is possible or until the costs of such shortening would exceed the savings that result from reducing the project length.

We illustrate the process of crashing by the network shown in Figure 7.26. Suppose that the project consists of seven activities—A, B, C, D, E, F and G. As evident from the network, the critical path is 1-2-4-5-6 consisting of activities A, C, E and G, with project duration of 16 days. Please note that the 16 days duration is based on the normal duration.

Figure 7.26 Network used for illustrating crashing

The normal and crash duration of each activity is given under the arrow. The cost of crashing per day of all activities is given in the parentheses of all activities adjacent to the activity name. For example, the normal cost of activity A is 3 days while the crash duration is 2 days, and the crash cost per day is Rs 2,000. The crash cost per day is given by the cost slope of an activity, which is calculated from the information on normal duration, crash duration, normal cost and crash cost. The cost slope has been computed from the expression—

In other words, the cost slope indicates the extra cost incurred by reducing the activity duration by one day. The higher the value of the slope, the more is the cost of crashing the activities.

The details of normal duration, normal cost, crash duration and crash cost of each of the activities of the project are given in Table 7.12.


Table 7.12 Details of activity cost and duration

As can be observed from this small example, it is possible to crash some activities such as (1, 2), (2, 3), (2, 4), (2, 5), (4, 5) and (3, 5), while it may not be possible for some activities such as (5, 6) (both normal and crash durations are same here).

Suppose that the indirect expenses are Rs 6,000/day. Thus, the total project cost if all the activities are executed at their normal pace is given by:

Total Project Cost



=[Normal cost of activities (1, 2), (2, 3), (2, 4), (2, 5), (4, 5), and (3, 5)]


    + (Indirect cost per day) 3 Duration of the project)


= (5,000 1 6,000 1 9,000 1 5,000 1 7,000 1 8,000 1 20,000) 1 (6,000 × 16)


= (60,000) 1 (96,000) 5 156,000.


It may be noted that the crashing exercise would have given the same result even if we had assumed the base cost to be equal to zero, as this was a constant in all the steps. Since every activity has to be performed to complete the project, a base cost of every activity has to be paid. These costs may be regarded as the ‘fixed’.

To shorten the project duration, we have to shorten the duration of activities along the critical path (2-4-5-6). We observe that the activity (1, 2) is on the critical path and has the least slope (Rs 2,000/day), and hence, can be crashed first. This activity can be crashed by one day; thus, the project duration reduces by a day. Project duration has become 15 days now.

The project cost to complete in 15 days



= cost to complete the project in 16 days 1 cost of crashing by a day


    – saving in indirect cost


= 156,000 + 2,000 – 6,000


= 152,000.


Thus, we have obtained a reduction of Rs 4,000 and a reduction in duration by 1 day.

The next higher cost slope (Rs 3,000 per day) on the critical path is for activity (4, 5) and we can crash this activity to a maximum of 3 days (from a normal duration of 5 days to a crash duration of 2 days). However, we will crash it in two steps of one day each. With one day of crashing, the project duration will become 14 days now.


The project cost to complete in 14 days


= the project cost to complete in 15 days 1 cost of crashing by a day


   – saving in indirect cost


= 152,000 + 3,000 – 6,000


= 149,000.


The project duration, thus, reduces by another day and the cost has also decreased by Rs 3,000 over the previous crash cost.

The activity (4, 5) is crashed again by a day. The project duration becomes 13 days now.


The project cost to complete in 13 days


= the project cost to complete in 14 days 1 cost of crashing by a day


  – saving in indirect cost


= 149,000 + 3,000 – 6,000


= 146,000.


The project duration, thus, reduces by another day and the cost has also decreased by Rs 3,000 over the previous crash cost.

There is still scope of crashing activity (4, 5). However, note that there are two critical paths 1-2-4-5-6 and 1-2-3-5-6, both of 13 days duration. The available options for crashing are given in Table 7.13. It may further be noted that the cost of crashing one day is the summation of individual cost slopes.

The lowest-cost slope option is given by activities E and F. The crash costs of these two activities combined together is Rs 4,500. Thus, we crash activities E and F by 1 day, which makes the project duration equal to 12 days.


Table 7.13 Options available for further crashing

Option Cost (Rs./day)

C and B

4,000 + 2,000 = 6,000

C and F

4,000 + 3,000 = 7,000

E and B

3,000 + 2,000 = 5,000

E and F

3,000 + 1,500 = 4,500


The project cost to complete in 12 days


= the project cost to complete in 13 days 1 cost of crashing by a day


   – saving in indirect cost


= 146,000 + 4,500 – 6,000


= 144,500.


The project duration, thus, reduces by another day and the cost has also decreased by Rs 1,500 over the previous crash cost.

Again, we have two critical paths 1-2-4-5-6 and 1-2-3-5-6, both with project duration of 12 days. The available option for crashing is activities C and B together. The crash cost would be equal to 4,000 1 2,000 5 Rs 6,000. The project duration reduces by another day and becomes equal to 11 days.


The project cost to complete in 11 days


= the project cost to complete in 12 days 1 cost of crashing by a day


   – saving in indirect cost


= 144,500 + 6,000 – 6,000


= 144,500.


The project duration reduces by another day but there is no increase or decrease in the cost over the previous crash cost.

Now, we observe that we have 3 critical paths, 1-2-4-5-6, 1-2-3-5-6 and 1-2-5-6, each of 11 days duration. Our only available option for crashing is to crash activities B, C and D together by one day, and the cost of crashing the three activities together is equal to the sum of cost slopes of activities (2, 3), (2, 4) and (2, 5)—that is, Rs 10,000. Thus, project duration becomes 10 days.


The project cost to complete in 10 days


= the project cost to complete in 11 days 1 cost of crashing by a day


   – saving in indirect cost


= 144,500 + 10,000 – 6,000


= 148,500.


The project duration reduces by another day but the cost has increased by Rs 4,000 over the previous crash cost.

Thus, we have reached a stage where the decrease in duration is accompanied by a significant increase in the direct cost, forcing us to stop further crashing. If we combine all our results in a graph showing how project length affects the schedule costs, we obtain the curve as in Figure 7.27, which shows the minimum project cost corresponding to project durations of 11 days and 12 days.

It may be noted that the crashing exercise would have given the same result even if we had assumed the base cost to be equal to zero, as this was a constant in all the steps. Since every activity has to be performed to complete the project, a base cost of every activity has to be paid. These costs may be regarded as the ‘fixed’ or ‘sunk’ costs of the project. Neglecting the base cost of activities since these remain the same irrespective of crashing, base cost of activities has been assumed to be Re 0. This cost has to be paid for all activities and is independent of the crashing done.

Figure 7.27 Time vs cost for the example problem

The time-cost curve shown in Figure 7.27 helps the project manager to choose a suitable schedule for the project. The decision to select a particular duration does not depend on the cost alone, but is determined by a number of other criteria such as the considerations of safety of workers, the risks involved in a particular completion schedule, and so on. Thus, a project manager may not always go for the schedule associated with the least cost and he may choose to go for the moderate cost schedule after incorporating other factors.



1. Weist, J.D. and F.K. Levy, 2005, A Management Guide to PERT/CPM with GERT/PDM/DCPM and other Networks, 2nd ed., New Delhi: Prentice Hall.

  1. State whether True or False:
    1. During resource levelling exercise, the critical activities are given priority for resource allocation.
    2. Time-cost trade-off analysis in CPM tries to optimize the project duration to minimize the total project cost.
    3. An activity with the highest cost slope is to be considered first for crashing.
    4. When there is more than one critical path in a network, crashing an activity on any one path will reduce the project duration.
    5. Resource levelling exercise uses activity floats to reschedule activities, without delaying project completion.
  2. Short-answer type questions
    1. What do you understand by ‘crashing’ of an activity? How is it used for optimizing project cost? Explain with help of a diagram.
    2. Explain the step-by-step process of ‘resource scheduling’. What do you understand by ‘resource levelling’ and how do you achieve it?
    3. Define ‘crashing’ and explain the basic assumptions that enable crashing in CPM. Why crashing cannot be attempted in PERT method?
    4. What is the need of resource levelling in any construction project?
    5. Differentiate between resource smoothening and resource levelling.
    6. What do you mean by project scheduling and why is it important?
    7. Discuss resource levelling and allocation.
    8. What is meant by ‘project crashing’?
    9. Discuss time-cost trade-off.
    10. Discuss different types of schedules.
    11. Define direct cost and indirect cost.
    12. What are the different problems associated with project scheduling?
  3. Numerical problems
    1. A construction company has been awarded a contract to construct a flyover in a city with a completion period of 18 months. The major activities in the project and the relationships among them, the normal and crash durations, and the corresponding normal and crash costs are given in Table Q7.1.1.


      Table Q7.1.1 Data for Question 3a

      1. If the project attracts a penalty of Rs 10,000 per month for project completion beyond 18 months, find out the total project cost including penalty if the company does not ‘crash’ any activity.
      2. Which of the activities should be crashed first in order to reduce the penalty payable and reduce the total cost?
    2. The precedence requirements, normal and crash activity times, and normal and crash costs for a construction project are given in Table Q7.2.1. The overhead costs are Rs 4 lakh per week.


      Table Q7.2.1 Data for Question 3b

      1. Find the critical path and time for completion under normal working conditions.
      2. Find the optimum duration for the project to minimize the total project cost. What is the total project cost for the optimum duration?
    3. Construction of a school building is scheduled to start from May 1, 2008. The details of different activities of the project are given in Table Q7.3.1. The budgeted costs shown in the table for different activities are to be spent uniformly over their duration. The activities start on the first day of the start month and finishes on the last day of the finish month. Draw a Gantt chart for project schedule and work out the monthly and cumulative cash flows for the project.


      Table Q7.3.1 Data for Question 3c

    4. The details of activities for a small project are given in Table Q7.4.1. If the indirect cost for the project is Rs 11.5 lakh per week, (a) what is the minimum time in which the project can be completed considering maximum possible crashing, and (b) what would be the total project cost if the schedule is crashed to the maximum?


      Table Q7.4.1 Data for Question 3d

    5. A critical activity in a project is estimated to need 15 days to complete at a cost of Rs 15,000. The activity can be expedited to complete in 12 days by spending a total amount of Rs 27,000. If the indirect costs of the project are Rs 3,000 per day, is it economically advisable to complete the activity early by crashing?
    6. In a small project there are nine activities. The duration for each of these activities and the labour required to do them are given in Table Q7.6.1. The project must be completed within 27 days. Nevertheless, the contractor wishes to carry out some resource levelling/smoothing in order that there are no excessive peaks or troughs in his labour schedule. Prepare labour schedule based on early start and late start of activities, and by visual inspection indicate the adjustment you would make in activities in order to perform resource scheduling.


      Table Q7.6.1 Data for Question 3f

      Activities Duration (days) Labour

      (1, 2)


      (2, 3)


      (2, 5)


      (2, 4)


      (3, 4)


      (3, 7)


      (5, 6)


      (6, 7)


      (4, 7)

    7. For the data given in Table Q7.7.1, perform resource levelling within the minimum project duration for the smallest number of workers if activities cannot be split.


      Table Q7.7.1 Data for Question 3g

      Activities Duration (days) Labour

      (1, 2)


      (1, 3)


      (1, 5)


      (2, 3)


      (3, 4)


      (3, 5)


      (3, 6)


      (4, 6)


      (5, 6)