Problem analysis and scheduling tools: Gantt, PERT, critical path

This week has two parts. The first is problem analysis: finding the real cause under the user’s complaint. The second is scheduling tools: turning the task list (WBS) you produced in week 3 into a calendar and a network. The exam calculation is here: a task table is given, and you build the network, do the forward and backward passes, and find the critical path and the slacks. Whoever solves this question by building a table does not miss it.

Golden rules:

1. Don’t solve the symptom, solve the cause. “The system is slow” is a symptom.
2. Critical path = longest path. The project duration is the length of the critical path.
3. Max on the forward pass, min on the backward pass.
4. Slack = LS − ES. It is 0 for critical tasks.

Problem analysis

At the beginning of the analysis phase the current system (as-is) is examined and problems are listed. Users usually state a symptom: “Borrowing a book takes too long.” The analyst’s job is to find its cause. If the cause is not solved, the new system will carry the same problem.

Root cause analysis

A simple technique: ask “Why?” several times.

The lending process takes long.       Why?
 -> The clerk writes each book by hand. Why?
 -> There is no barcode reader.          Why?
 -> The old system does not support barcodes.
Root cause: the system does not support barcodes.

Fishbone (Ishikawa) diagram

The problem is written at the head of the fish, and the bones are cause categories.

 People        Process
   \             \
    \ single clerk \ paper form
     \             \
------+-------------+-----> [Borrowing a laptop
     /             /         takes 20 min]
    / reader      / ID
   /  broken     /  photocopy
 Technology    Policy

Specific causes are written on each bone, and then which ones are really effective is checked with data (observation, measurement).

The PIECES framework

A checklist for classifying problems:

Problems are prioritized (impact, urgency) and improvement objectives are written: “Reduce the lending time from 20 minutes to 3 minutes.” A measurable objective makes it possible to test success in later stages.

Gantt chart

A Gantt chart shows the tasks as horizontal bars on a calendar. It is easy to read and ideal for presenting to managers. Progress is shown by filling the inside of the bar, and milestones with diamonds.

Day           0    5    10   15 18
A Requirements ####
B DB design       ###
C UI design       #####
D Backend            ######
E Frontend             ###
H Manual               ##
F Integration              ###
G Test                        ##
Delivery                          <>

Its weakness: dependencies (which task is waiting for which) are hard to see. A PERT network is used for that.

PERT and CPM

A PERT chart (program evaluation and review technique) shows the tasks as nodes and the dependencies as arrows. CPM (critical path method) finds the longest path on this network.

Three-point estimate

If the duration is uncertain, three estimates are taken for each task: optimistic ($o$), most likely ($m$), pessimistic ($p$).

$t_e = \frac{o + 4m + p}{6}$

Example: $o = 4$, $m = 6$, $p = 14$ days → $t_e = (4 + 24 + 14)/6 = 7$ days. The standard deviation is about $(p - o)/6 = 1.67$ days.

Four values

Slack (float) = LS − ES. How much the task can slip without delaying the project.

Worked example: library reservation system

Network

        +-->[B 3]-->[D 6]--+
        |                  v
[A 4]---+-->[C 5]-->[E 3]->[F 3]-->[G 2]
              |                     ^
              +------>[H 2]---------+

Forward pass

Left to right. ES = the largest of the EFs of the predecessors.

A: ES 0               EF 0+4  = 4
B: ES 4  (A)          EF 4+3  = 7
C: ES 4  (A)          EF 4+5  = 9
D: ES 7  (B)          EF 7+6  = 13
E: ES 9  (C)          EF 9+3  = 12
H: ES 9  (C)          EF 9+2  = 11
F: ES max(13,12)=13   EF 13+3 = 16
G: ES max(16,11)=16   EF 16+2 = 18
Project duration = 18 days

Backward pass

Right to left. The LF of the last task is the project duration. LF = the smallest of the LSs of the successors.

G: LF 18              LS 18-2 = 16
F: LF 16 (G)          LS 16-3 = 13
H: LF 16 (G)          LS 16-2 = 14
D: LF 13 (F)          LS 13-6 = 7
E: LF 13 (F)          LS 13-3 = 10
C: LF min(10,14)=10   LS 10-5 = 5
B: LF 7  (D)          LS 7-3  = 4
A: LF min(4,5)=4      LS 4-4  = 0

Table and slack

Critical path: A-B-D-F-G, 4 + 3 + 6 + 3 + 2 = 18 days. To check, the other paths: A-C-E-F-G = 17, A-C-H-G = 13. The longest is 18, consistent.

Interpretation

• If D slips by one day, the project goes to 19 days. If D slips by three days, 21 days.
• C or E can slip by one day without affecting the project. But if they slip, there will be two critical paths and the risk rises.
• H can slip by five days. If resources get tight, it makes sense to temporarily move the person on H to D.

PITFALL · using max in the backward pass. C has two successors: E (LS 10) and H (LS 14). LF takes the smallest: 10. If you take 14, you give C a false slack and delay E.

TIP · cross-check. At the end of the calculation, check that the tasks with slack 0 form a chain and that the length of this chain equals the project duration. If not, there is an error somewhere.

Common mistakes