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Chapter 3 · 4 hours

Loading and Scheduling Technique

Practice questions

Practice questions and answers

4 exam-style questions on this chapter, written for this site from the official syllabus. We haven’t found past IOE papers for this subject yet; if you have some, share them in the community.

  • Practice · 5 marks

What is a Gantt chart? Explain how it is used for machine loading and progress recording, and state its advantages and limitations.

Answer

A Gantt chart (Henry Gantt, about 1910) is a horizontal bar chart in which the vertical axis lists jobs or machines and the horizontal axis is time. Each bar shows the planned start, duration and finish of an activity.

Uses

  1. Machine loading chart: each machine is a row; jobs are bars on the time scale. It shows the load on each machine and idle time, so jobs can be assigned and re-sequenced.
  2. Progress (schedule) chart: the planned bar is drawn lightly; a thick line below it shows work completed. A vertical "today" line shows if the job is ahead or behind schedule.
 Machine | Day: 1  2  3  4  5  6  7  8
 --------+-----------------------------
 Lathe   |      [Job A ][Job C   ]
 Milling |         [Job B   ][Job D ]
 Drill   |               [Job A2][..]
 Done:    ====>   (progress on A)
              ^ today line

Advantages

  • Simple to draw and understand, even by shop-floor staff.
  • Shows the whole schedule and the idle time of machines at a glance.
  • Useful for comparing planned and actual progress.
  • Low cost; easy to update.

Limitations

  • It does not show the dependencies between activities clearly, nor the critical path.
  • It becomes complex for large projects with many activities.
  • It cannot show uncertainty in activity times.
  • Rescheduling needs the chart to be redrawn.

For this reason CPM and PERT are used for large projects, while the Gantt chart is retained for shop loading and for presenting schedules.

  • Practice · 8 marks

A project has the following activities (times in days):
ActivityABCDEFGH
Duration43564352
Predecessors--AA, BBC, DD, EF, G
(a) Draw the network (describe it), (b) find the earliest and latest start and finish times of every activity, (c) find the total float, the critical path and the project duration.

Answer

(a) Network (activity-on-node)

        +--> C(5) ---------+
 A(4) --+                   +--> F(3) --+
        +--> D(6) -----+----+           +--> H(2)
 B(3) ------>----+     |                |
        +--> E(4) ---+ +--> G(5) -------+

A and B start the project. C follows A; D needs both A and B; E follows B; F needs C and D; G needs D and E; H needs F and G.

(b) Forward and backward pass

Forward pass: EF=ES+tEF = ES + t, where ES is the largest EF of predecessors. Backward pass: LS=LF−tLS = LF - t, where LF is the smallest LS of successors. Project duration = largest EF = 17.

ActivityttESEFLSLFFloat (LS - ES)
A404040
B303141
C5497123
D64104100
E4376103
F3101312152
G5101510150
H2151715170

Sample: ES of D = max(EF of A, EF of B) = max(4, 3) = 4. ES of F = max(9, 10) = 10. LF of D = min(LS of F, LS of G) = min(12, 10) = 10.

(c) Critical path

Activities with zero total float are critical: A, D, G, H.

Check: 4+6+5+2=174 + 6 + 5 + 2 = 17 days. Other paths are shorter (e.g. B-D-G-H = 16, A-D-F-H = 15, A-C-F-H = 14).

Answer: Critical path A-D-G-H; project duration 17 days; floats: B = 1, C = 3, E = 3, F = 2 days.

  • Practice · 8 marks

A project has the activities below with optimistic (a), most likely (m) and pessimistic (b) times in weeks:
ActivityambPredecessors
A246-
B3513-
C468A
D5817A, B
E234C
F357D
G123E, F
(a) Calculate the expected time and variance of each activity. (b) Find the critical path and expected project duration. (c) Find the probability that the project is completed within 24 weeks and within 20 weeks.

Answer

(a) Expected time and variance

PERT uses the beta distribution:

te=a+4m+b6,σ2=(b−a6)2t_e = \frac{a + 4m + b}{6}, \qquad \sigma^2 = \left(\frac{b-a}{6}\right)^2
Act.tet_eσ2\sigma^2ESEFLSLFFloat
A(2+16+6)/6 = 40.44404262
B(3+20+13)/6 = 62.77806060
C(4+24+8)/6 = 60.44441011177
D(5+32+17)/6 = 94.0006156150
E(2+12+4)/6 = 30.111101317207
F(3+20+7)/6 = 50.444152015200
G(1+8+3)/6 = 20.111202220220

(b) Critical path

Zero-float activities: B, D, F, G. Expected project duration TE=6+9+5+2=22T_E = 6 + 9 + 5 + 2 = 22 weeks.

(c) Probability

Variance of the project = sum of variances on the critical path only:

σT2=2.778+4.000+0.444+0.111=7.333,σT=2.708 weeks\sigma_T^2 = 2.778 + 4.000 + 0.444 + 0.111 = 7.333, \qquad \sigma_T = 2.708\ \text{weeks} Z=TS−TEσTZ = \frac{T_S - T_E}{\sigma_T}
  • For 24 weeks: Z=(24−22)/2.708=0.74Z = (24-22)/2.708 = 0.74, so P=0.770P = 0.770 (from the standard normal table).
  • For 20 weeks: Z=(20−22)/2.708=−0.74Z = (20-22)/2.708 = -0.74, so P=1−0.770=0.230P = 1 - 0.770 = 0.230.

Answer: Critical path B-D-F-G, expected duration 22 weeks, sigma = 2.71 weeks; P(finish within 24 weeks) = 77%; P(within 20 weeks) = 23%.

  • Practice · 5 marks

Differentiate between CPM and PERT. Explain the terms critical path, float and event slack.

Answer

Both CPM (Critical Path Method, DuPont 1957) and PERT (Program Evaluation and Review Technique, US Navy, Polaris project 1958) are network techniques for planning, scheduling and controlling projects.

BasisCPMPERT
OriginIndustrial/construction projectsResearch and development projects
Time estimateOne deterministic timeThree estimates: a, m, b
NatureDeterministicProbabilistic
OrientationActivity orientedEvent oriented
CostTime-cost trade-off (crashing) is consideredMainly time; cost not considered
UseRepetitive, well-known jobsNon-repetitive, uncertain jobs
ProbabilityNot computedProbability of meeting a date is found
DistributionNot usedBeta distribution

Terms

  • Critical path: the longest path through the network; it fixes the shortest possible project duration. Any delay in a critical activity delays the whole project.
  • Total float: the time by which an activity can be delayed without delaying the project completion: TF=LS−ES=LF−EFTF = LS - ES = LF - EF. Critical activities have zero float.
  • Free float: delay possible without affecting the earliest start of the next activity.
  • Event slack (PERT): for an event, slack=TL−TE\text{slack} = T_L - T_E, the difference between latest allowable and earliest expected event times. Events on the critical path have zero slack.

Written from the official syllabus. Questions and answers are written for this site; check them against your class notes.

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