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Bar chart; expected time and variance

Short answer

A bar chart shows activities as horizontal bars on a time scale; it is simple and shows start, finish and overlap, but not logical dependence, critical path or float. For to=6, tm=8, tp=12, te=(6+4*8+12)/6=8.33 time units and variance=((12-6)/6)^2=1.

Draw a bar chart and explain its advantages and disadvantages. Also find te and variance when to, tm and tp are 6, 8, 12.

Bar chart

A bar chart shows each activity as a horizontal bar on a time scale; the length of the bar is the duration. Example: house project (weeks).

                 0   4   8   12  16  20  24  (wk)
                +------------------------
A Site clearing |#                       
B Excavation    | ##                     
C Foundation    |   ###                  
D Plinth beam   |      ##                
E Walls+columns |        #####           
F Roof slab     |             ###        
G Plumb/electric|             ###        
H Plastering    |                ###     
I Floor,doors   |                   ###  
J Painting      |                      ##
                +------------------------

Advantages

  • Simple to prepare, read and understand, even for non-technical persons.
  • Shows start, finish, duration and overlap of activities on a time scale.
  • Cheap and quick; needs no special training.
  • Good for presenting the summary to top management and site staff.
  • Progress can be marked on the bars and compared with the plan.
  • Useful for resource scheduling on a time scale.

Disadvantages

  • Does not show the logical dependence between activities, so the effect of a delay on other activities is not seen.
  • Critical activities and float are not shown, so the manager cannot say where attention is needed.
  • Not suitable for large and complex projects; the chart becomes crowded.
  • Difficult to update; any change in logic needs redrawing.
  • Durations are single-point values; uncertainty is not considered.
  • Does not show resource or cost relations.

Expected time and variance (PERT)

Given: optimistic time to=6t_o = 6, most likely time tm=8t_m = 8, pessimistic time tp=12t_p = 12.

te=to+4tm+tp6=6+4(8)+126=506=8.33σ=tp−to6=12−66=1Variance=σ2=12=1\begin{aligned} t_e &= \frac{t_o + 4t_m + t_p}{6} = \frac{6 + 4(8) + 12}{6} = \frac{50}{6} = 8.33 \\ \sigma &= \frac{t_p - t_o}{6} = \frac{12 - 6}{6} = 1 \\ \text{Variance} &= \sigma^2 = 1^2 = 1 \end{aligned}

Answer: te=8.33t_e = 8.33 and variance =1= 1 (standard deviation 1).

✓ Answer checked by Claude Opus 5.5 (high effort)

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