Construction Planning and Scheduling

Time-cost trade-off (TCTO) is the technique of shortening a project's duration by crashing (speeding up) selected activities, while looking at the total cost. Direct cost (labour, materials, machines) rises as an activity is crashed, while indirect cost (overhead, supervision, penalties) falls as the project becomes shorter. The total of the two is lowest at an optimum duration.

 Cost
  |\                         /  total cost
  | \  indirect          __/
  |  \                 _/
  |   \__    _____---/     direct cost
  |      \__/--  *  <- minimum total
  |_________|________|_____ Duration
          crash   normal

Use in determining project duration

• Cost slope of an activity = (crash cost - normal cost) / (normal time - crash time).
• Only critical activities are crashed, cheapest slope first, until the slope exceeds the indirect cost saved per unit time, or the crash limit is reached.
• The duration with minimum total cost is the optimum duration; the shortest feasible duration is the crash duration.

Example

Two activities in series: A (normal 6 days, Rs 3,000; crash 4 days, Rs 4,000) and B (normal 5 days, Rs 2,000; crash 4 days, Rs 2,300). Indirect cost Rs 400/day.

Slope of A = (4000 - 3000)/(6 - 4) = Rs 500/day; slope of B = (2300 - 2000)/(5 - 4) = Rs 300/day.

Crashing B costs Rs 300 and saves Rs 400, so it is worthwhile. Crashing A (Rs 500) would cost more than the Rs 400 saved. The optimum duration is 10 days with a total cost of Rs 9,300 (against Rs 9,400 at normal duration).

Preferred tool for building construction: the bar (Gantt) chart, supported by CPM for complex or large projects.

Comparison

Why the bar chart for buildings

1. Building works use well-known activities and standard durations, so PERT's three-time estimates are unnecessary.
2. Site engineers, foremen and the client understand it readily, so it works as a communication tool at the site.
3. Progress can be marked directly (planned vs actual) with little effort.
4. It can be linked to resource and material schedules, and cash flow.
5. It is cheap and quick to prepare without software.
6. Its limitation (no dependency shown) can be removed by linked bar charts or by a CPM network behind it for large buildings.

For a small or medium building a bar chart is enough; for a high-rise or a large complex a CPM network with a bar chart output is best.

Client

A client (owner or employer) is the person, organisation or government agency that needs the project, finances it, and awards the contract for its design and construction.

Role of the client in pre-tender stage planning

The pre-tender stage is the period from the idea of the project until tenders are invited. The client:

1. Defines the need and objectives, scope and requirements of the project.
2. Feasibility studies: technical, economic, financial, social and environmental studies (IEE/EIA).
3. Arranges finance and budget approval; fixes a cost ceiling.
4. Acquires land and clears legal issues, permits and right of way.
5. Appoints the consultant for survey, design, drawings, specifications, estimate and bill of quantities.
6. Decides the contract strategy: type of contract, procurement method, packaging of work, and the duration.
7. Prepares or approves tender documents: instructions to bidders, conditions of contract, technical specifications, drawings, BOQ, and the bid security amounts.
8. Fixes qualification criteria and decides pre-qualification if needed.
9. Approves the master programme and the cost estimate; appoints the tender evaluation committee.
10. Publishes the tender notice and arranges the site visit and pre-bid meeting.

Given data and method

Indirect cost = 2,500 Rs/week. Time-cost trade-off (TCTO) crashes the activities on the critical path, one time-unit at a time, always choosing the cheapest cost slope, as long as the saving in indirect cost is more than the extra direct cost.

Assumption: for P, Q, R, S and T the table lists the crash cost below the normal cost, which is impossible for a crash. These columns are taken as interchanged, i.e. the smaller value is the normal cost and the larger is the crash cost. U cannot be crashed (crash duration equals normal duration).

$$\text{Cost slope}=\frac{\text{Crash cost}-\text{Normal cost}}{\text{Normal duration}-\text{Crash duration}}$$

Paths at normal duration

• P-R = 12 weeks
• P-S-V = 19 weeks
• Q-T-V = 22 weeks
• Q-U = 21 weeks

Critical activities (on the longest path or paths): Q, T, V; normal duration = 22 weeks; normal direct cost = 24,000 Rs.

Crashing, step by step

Result

• Optimum (minimum-cost) schedule: duration 19 weeks, direct cost 24,450 + indirect 47,500 = 71,950 Rs.
• Activities crashed at the optimum: Q by 2, V by 1. Durations: P=7, Q=7, R=5, S=8, T=9, U=12, V=3.
• Minimum possible duration: 19 weeks, with total cost 71,950 Rs (direct 24,450 + indirect 47,500).

Answer: optimal crashed project completion time = 19 weeks; minimum total cost = Rs 71,950.

Given data and method

Indirect cost = 3,000 Rs/week. Time-cost trade-off (TCTO) crashes the activities on the critical path, one time-unit at a time, always choosing the cheapest cost slope, as long as the saving in indirect cost is more than the extra direct cost.

F cannot be crashed (crash duration equals normal duration).

$$\text{Cost slope}=\frac{\text{Crash cost}-\text{Normal cost}}{\text{Normal duration}-\text{Crash duration}}$$

Paths at normal duration

• A-C = 14 weeks
• A-D-G = 22 weeks
• B-E-G = 25 weeks
• B-F = 23 weeks

Critical activities (on the longest path or paths): B, E, G; normal duration = 25 weeks; normal direct cost = 22,500 Rs.

Crashing, step by step

Result

• Optimum (minimum-cost) schedule: duration 21 weeks, direct cost 23,350 + indirect 63,000 = 86,350 Rs.
• Activities crashed at the optimum: B by 2, E by 1, G by 1. Durations: A=8, B=8, C=6, D=9, E=9, F=13, G=4.
• Minimum possible duration: 21 weeks, with total cost 86,350 Rs (direct 23,350 + indirect 63,000).

Answer: optimal crashed project completion time = 21 weeks; minimum total cost = Rs 86,350.

Time-cost trade-off is the study of the relation between project duration and cost, in order to find the duration with minimum total cost. Direct cost rises when activities are crashed; indirect cost falls when the project is shorter.

Steps in planning

1. Draw the network and find the critical path and normal duration.
2. Find the normal and crash time and cost of every activity.
3. Compute the cost slope = (crash cost - normal cost)/(normal time - crash time).
4. Crash the critical activity with the lowest slope by one unit (not below its crash time). If there are several critical paths, crash a combination that cuts all of them.
5. Recompute project duration, direct cost, indirect cost and total cost.
6. Repeat until the slope exceeds the indirect cost saved per unit time, or all critical activities are at crash limit.
7. Choose the duration with the lowest total cost as the optimum.

Example

Activities in series: P (normal 4 days, Rs 2,000; crash 2 days, Rs 2,800) and Q (normal 3 days, Rs 1,500; crash 2 days, Rs 1,800). Indirect cost Rs 500/day.

Slopes: P = (2800 - 2000)/2 = Rs 400/day; Q = 300/1 = Rs 300/day.

Answer: all slopes are below the indirect cost of Rs 500/day, so the project is crashed fully: optimum duration 4 days, total cost Rs 6,600.

Time-cost trade-off

It is the method of balancing project duration against cost. Total cost = direct cost + indirect cost. When an activity is speeded up, direct cost rises (overtime, extra labour, more equipment) while indirect cost (overheads, supervision, penalties) falls with shorter time. The minimum of the total cost curve gives the optimum duration.

 Cost
  |  \         total
  |   \      /
  |    \___*/   <- optimum
  |  direct^  indirect falls
  |_________________ Time
   crash          normal

Example: if activity X normal 10 days, Rs 20,000, crash 8 days, Rs 24,000, slope = (24000 - 20000)/2 = Rs 2,000/day. If the project overhead is Rs 3,000/day, crashing X pays, saving Rs 1,000 per day.

Crashing

Crashing is the deliberate shortening of the duration of an activity by adding resources (more labour, shifts, overtime, better equipment, faster method) at an extra direct cost. Only critical activities, cheapest slope first, are worth crashing.

My suggestion if the boss wants a considerable reduction

1. Check the critical path and the available float of non-critical activities.
2. Compute cost slopes and crash the cheapest critical activities one step at a time, and compare the extra cost with the indirect saving.
3. Re-plan the logic: fast-track (overlap activities), change the sequence, use precast, readymade concrete, more formwork sets, and a larger crew.
4. Add shifts, overtime, or night work, and subcontract some work.
5. Ensure materials, equipment and finance are available early, and remove approval delays.
6. Explain to the boss the extra cost, risk to quality and safety, and give the revised cost-time curve to choose the final target.

Construction planning is deciding in advance what is to be done, how, by whom, with what resources and when. Scheduling assigns dates and sequence to the planned activities.

Importance

• Gives a clear path to complete the project on time and within budget.
• Allows proper use of men, materials, machines and money, avoiding idle time.
• Identifies the critical activities and allows control.
• Helps coordinate trades, subcontractors and suppliers.
• Forecasts cash flow and finance requirements.
• Provides a baseline for monitoring and for claims on delay.
• Reduces risks, uncertainty and conflict.

Steps

1. Define the scope and objectives of the project.
2. Break the project into work packages and activities (WBS).
3. Decide methods and technology of construction.
4. Establish the logical relationship (sequence) between activities.
5. Estimate quantities, resources and the duration of each activity.
6. Draw the network or bar chart and compute the project duration, float and critical path.
7. Allocate and level resources, and plan the materials, equipment and labour.
8. Prepare the budget and cash flow.
9. Review with stakeholders and approve the baseline schedule.
10. Monitor progress, compare with plan, update and take corrective action.

After the contract is awarded, the client's planning in the implementation stage ensures that the contractor and consultant can work without delay. Activities include:

1. Contract signing and notice to proceed: issuing the letter of acceptance, signing the contract, and obtaining performance security and insurance.
2. Handing over the site: possession of land, clearance of obstructions, access, and utility shifting (electricity, water, telephone).
3. Appointing the project manager/supervision team and defining the Engineer's duties and authority.
4. Mobilisation advance and finance planning: arranging funds, advance payment, and a payment schedule for interim bills.
5. Review and approval: the contractor's work programme, method statement, quality plan, material sources and key personnel.
6. Providing drawings and information on time, and issuing any client-supplied materials.
7. Monitoring and control: progress meetings, reports, inspection, quality testing, and cost and time tracking.
8. Change control: evaluating variations, extensions of time and claims, and approving them.
9. Obtaining permits and clearances from authorities, and managing community relations.
10. Preparing for commissioning, handover, training of operators, defects liability and maintenance.

The contractor's pre-tender planning decides whether to bid and at what price. Steps:

1. Tender notice study: reading the notice and deciding whether the work suits the contractor's experience, capacity, finance and workload.
2. Collecting documents and checking eligibility or pre-qualification requirements.
3. Studying the tender documents: drawings, conditions of contract, specifications, BOQ, bid security and time of completion. Risk clauses, such as liquidated damages and payment terms, are noted.
4. Site visit and investigation: access, ground, water, power, labour and material sources, local rates, weather, and nearby facilities. Attend the pre-bid meeting and ask clarifications.
5. Method statement and construction planning: select the construction methods and equipment, prepare a preliminary programme (bar chart/CPM) and check the duration.
6. Resource planning: plan labour, materials and plant, including quotations from suppliers and subcontractors.
7. Cost estimating: direct costs from quantities and rate analysis, site overheads, head office overhead, contingency, risk, and profit; decide the cash flow and financing cost.
8. Bid decision: a management review of risk and competitor position, then the final bid price.
9. Preparing the bid: completing forms, bid security, and required documents, and submitting before the deadline.

Given data and method

Indirect cost = 2,000 Rs/day. Time-cost trade-off (TCTO) crashes the activities on the critical path, one time-unit at a time, always choosing the cheapest cost slope, as long as the saving in indirect cost is more than the extra direct cost.

$$\text{Cost slope}=\frac{\text{Crash cost}-\text{Normal cost}}{\text{Normal duration}-\text{Crash duration}}$$

Paths at normal duration

• A-B-E-G = 20 days
• A-C-G = 14 days
• A-D-F-G = 21 days

Critical activities (on the longest path or paths): A, D, F, G; normal duration = 21 days; normal direct cost = 46,000 Rs.

Crashing, step by step

(Rows with several activities show activities crashed together because two paths are critical.)

Result

• Optimum (minimum-cost) schedule: duration 13 days, direct cost 51,000 + indirect 26,000 = 77,000 Rs.
• Activities crashed at the optimum: A by 3, B by 1, D by 2, G by 3. Durations: A=2, B=3, C=3, D=4, E=5, F=4, G=3.
• Minimum possible duration: 11 days, with total cost 77,800 Rs (direct 55,800 + indirect 22,000).

Answer: minimum-cost schedule = 13 days, total cost Rs 77,000; minimum-duration schedule = 11 days, total cost Rs 77,800. Crashing beyond 13 days costs more (Rs 2,400 per day) than the Rs 2,000/day saved.

Given data and method

Indirect cost = 1,000 Rs/day. Time-cost trade-off (TCTO) crashes the activities on the critical path, one time-unit at a time, always choosing the cheapest cost slope, as long as the saving in indirect cost is more than the extra direct cost.

$$\text{Cost slope}=\frac{\text{Crash cost}-\text{Normal cost}}{\text{Normal duration}-\text{Crash duration}}$$

Paths at normal duration

• A-B-E-G = 20 days
• A-C-G = 14 days
• A-D-F-G = 21 days

Critical activities (on the longest path or paths): A, D, F, G; normal duration = 21 days; normal direct cost = 46,000 Rs.

Crashing, step by step

(Rows with several activities show activities crashed together because two paths are critical.)

Result

• Optimum (minimum-cost) schedule: duration 14 days, direct cost 49,800 + indirect 14,000 = 63,800 Rs.
• Activities crashed at the optimum: A by 3, D by 1, G by 3. Durations: A=2, B=4, C=3, D=5, E=5, F=4, G=3.
• Minimum possible duration: 11 days, with total cost 66,800 Rs (direct 55,800 + indirect 11,000).

Answer: minimum-cost schedule = 14 days, total cost Rs 63,800; minimum-duration schedule = 11 days, total cost Rs 66,800.

Given data and method

Indirect cost = 1.6 thousand Rs/week. Time-cost trade-off (TCTO) crashes the activities on the critical path, one time-unit at a time, always choosing the cheapest cost slope, as long as the saving in indirect cost is more than the extra direct cost.

$$\text{Cost slope}=\frac{\text{Crash cost}-\text{Normal cost}}{\text{Normal duration}-\text{Crash duration}}$$

Paths at normal duration

• A-C = 36 weeks
• A-D-E = 42 weeks
• B-E = 40 weeks

Critical activities (on the longest path or paths): A, D, E; normal duration = 42 weeks; normal direct cost = 26 thousand Rs.

Crashing, step by step

Result

• Optimum (minimum-cost) schedule: duration 40 weeks, direct cost 29 + indirect 64 = 93 thousand Rs.
• Activities crashed at the optimum: D by 2. Durations: A=26, B=28, C=10, D=2, E=12.

Further crashing beyond 40 weeks costs more than the Rs 1.6 thousand/week saved (E slope 1.83 > 1.6), so total cost rises. Crashing all activities fully gives about 28 weeks, but at a much higher cost.

Time-cost trade-off chart

 Cost (Rs '000)          total cost (direct + 1.6/week)
  93.7 |                          *  37 wk
  93.5 |                       *      38
  93.2 |                    *         39
  93.0 |                 *  <-- minimum, 40 weeks
  93.1 |              *               41
  93.2 |           *                  42 (normal)
       +------------------------------> weeks
         36 37 38 39 40 41 42

Answer: optimum duration 40 weeks; minimum total cost Rs 93.0 thousand (direct 29.0 + indirect 64.0).

Given data and method

Indirect cost = 225 Rs/day. Time-cost trade-off (TCTO) crashes the activities on the critical path, one time-unit at a time, always choosing the cheapest cost slope, as long as the saving in indirect cost is more than the extra direct cost.

$$\text{Cost slope}=\frac{\text{Crash cost}-\text{Normal cost}}{\text{Normal duration}-\text{Crash duration}}$$

Paths at normal duration

• A-C-E-G = 19 days
• A-D-F = 19 days
• B-E-G = 17 days

Critical activities (on the longest path or paths): A, C, D, E, F, G; normal duration = 19 days; normal direct cost = 9,200 Rs.

Crashing, step by step

(Rows with several activities show activities crashed together because two paths are critical.)

Result

• Optimum (minimum-cost) schedule: duration 19 days, direct cost 9,200 + indirect 4,275 = 13,475 Rs.
• Activities crashed at the optimum: none. Durations: A=6, B=8, C=4, D=3, E=6, F=10, G=3.
• Minimum possible duration: 12 days, with total cost 15,466.67 Rs (direct 12,766.67 + indirect 2,700).

Answer: minimum cost Rs 13,475 at the optimum (normal) duration of 19 days, since the cheapest critical slope (A, Rs 250/day) is above the Rs 225/day indirect cost; the minimum possible time to complete the project is 12 days (total cost Rs 15,466.67).

Given data and method

Indirect cost = 400 Rs/week. Time-cost trade-off (TCTO) crashes the activities on the critical path, one time-unit at a time, always choosing the cheapest cost slope, as long as the saving in indirect cost is more than the extra direct cost.

Assumption: for C the table shows crash cost (2,600) lower than normal cost (2,700). The two values are taken as interchanged, so normal cost = Rs 2,600 and crash cost = Rs 2,700. F cannot be crashed.

$$\text{Cost slope}=\frac{\text{Crash cost}-\text{Normal cost}}{\text{Normal duration}-\text{Crash duration}}$$

Paths at normal duration

• A-D = 17 weeks
• B-C-E = 26 weeks
• B-F = 22 weeks

Critical activities (on the longest path or paths): B, C, E; normal duration = 26 weeks; normal direct cost = 18,700 Rs.

Crashing, step by step

Result

• Optimum (minimum-cost) schedule: duration 22 weeks, direct cost 19,250 + indirect 8,800 = 28,050 Rs.
• Activities crashed at the optimum: B by 2, C by 1, E by 1. Durations: A=8, B=8, C=5, D=9, E=9, F=12.
• Minimum possible duration: 22 weeks, with total cost 28,050 Rs (direct 19,250 + indirect 8,800).

Answer: lowest-cost schedule = 22 weeks, total cost Rs 28,050.

Given data and method

Indirect cost = 300 Rs/week. Time-cost trade-off (TCTO) crashes the activities on the critical path, one time-unit at a time, always choosing the cheapest cost slope, as long as the saving in indirect cost is more than the extra direct cost.

F cannot be crashed.

$$\text{Cost slope}=\frac{\text{Crash cost}-\text{Normal cost}}{\text{Normal duration}-\text{Crash duration}}$$

Paths at normal duration

• A-D = 15 weeks
• B-C-E = 23 weeks
• B-F = 20 weeks

Critical activities (on the longest path or paths): B, C, E; normal duration = 23 weeks; normal direct cost = 17,800 Rs.

Crashing, step by step

Result

• Optimum (minimum-cost) schedule: duration 20 weeks, direct cost 18,225 + indirect 6,000 = 24,225 Rs.
• Activities crashed at the optimum: B by 2, C by 1. Durations: A=7, B=7, C=4, D=8, E=9, F=11.
• Minimum possible duration: 19 weeks, with total cost 24,250 Rs (direct 18,550 + indirect 5,700).

Answer: optimal crashed result = 20 weeks; minimum total cost = Rs 24,225. (Crashing E to reach 19 weeks costs Rs 325 against a saving of Rs 300, so it is not worthwhile.)

Given data and method

Indirect cost = 30 Rs/day. Time-cost trade-off (TCTO) crashes the activities on the critical path, one time-unit at a time, always choosing the cheapest cost slope, as long as the saving in indirect cost is more than the extra direct cost.

The four activities lie in series on the critical path (normal duration 2 + 4 + 9 + 5 = 20 days). Crashing limits: 1-2, 2-3 and 5-6 by 1 day each; 3-5 by 3 days. No new critical path appears.

$$\text{Cost slope}=\frac{\text{Crash cost}-\text{Normal cost}}{\text{Normal duration}-\text{Crash duration}}$$

Paths at normal duration

• 1-2 + 2-3 + 3-5 + 5-6 = 20 days

Critical activities (on the longest path or paths): 1-2, 2-3, 3-5, 5-6; normal duration = 20 days; normal direct cost = 580 Rs.

Crashing, step by step

Result

• Optimum (minimum-cost) schedule: duration 16 days, direct cost 685 + indirect 480 = 1,165 Rs.
• Activities crashed at the optimum: 1-2 by 1, 3-5 by 3. Durations: 1-2=1, 2-3=4, 3-5=6, 5-6=5.
• Minimum possible duration: 14 days, with total cost 1,320 Rs (direct 900 + indirect 420).

Answer: the lowest-cost schedule is obtained by crashing 1-2 by 1 day and 3-5 by 3 days: duration 16 days, total cost Rs 1,165 (direct Rs 685 + overhead Rs 480), against Rs 1,180 at the normal duration. Crashing 2-3 (Rs 65/day) or 5-6 (Rs 150/day) costs more than the Rs 30/day saved.

Given data and method

Indirect cost = 2,000 Rs/week. Time-cost trade-off (TCTO) crashes the activities on the critical path, one time-unit at a time, always choosing the cheapest cost slope, as long as the saving in indirect cost is more than the extra direct cost.

Assumption: the costs of activity 2-3 (C) are blurred in the scan; they are taken as normal Rs 8,000 and crash Rs 10,000 (slope Rs 1,000/week). The result (optimum 8 weeks) holds for any crash slope of C below Rs 2,000 per week. Network: 1-2 (A) then 2-3 (C) in series, with 1-3 (B) in parallel.

$$\text{Cost slope}=\frac{\text{Crash cost}-\text{Normal cost}}{\text{Normal duration}-\text{Crash duration}}$$

Paths at normal duration

• A-C = 9 weeks
• B = 8 weeks

Critical activities (on the longest path or paths): A, C; normal duration = 9 weeks; normal direct cost = 17,000 Rs.

Crashing, step by step

(Rows with several activities show activities crashed together because two paths are critical.)

Result

• Optimum (minimum-cost) schedule: duration 8 weeks, direct cost 18,000 + indirect 16,000 = 34,000 Rs.
• Activities crashed at the optimum: C by 1. Durations: A=4, B=8, C=4.
• Minimum possible duration: 7 weeks, with total cost 36,000 Rs (direct 22,000 + indirect 14,000).

Answer: least cost schedule = 8 weeks (A = 4, B = 8, C = 4 weeks), total cost Rs 34,000 against Rs 35,000 for the normal 9-week schedule. A further week would need A or C together with B (Rs 4,000 > Rs 2,000 saved).