SME
Construction Engineering · Problem type
Crashing an activity
Use this for “An activity takes 10 days for $12,000, or 6 days at most for $15,200. What does cutting it by 3 days cost?”
Chart
C(7) = 14,400
Kept: 6 ≤ T ≤ 10; the rest of the curve is dashed
This is an example. Type your own values to start a new problem.
s ($/day)
=
$
−
$
Tₙ (days)
−
Tc (days)
$
ΔC
=
s ($/day)
×
Δt (days)
Δt is the days cut from the normal duration and ΔC what they add; Cc is the cost at the crash duration Tc.
T
New duration
calculated · days
C
New cost
calculated · $
Formulas
s
=
Cc − Cₙ
Tₙ − Tc
800
=
15,200 − 12,000
10 − 6
ΔC
=
s
×
Δt
2,400 = 800 × 3
T
=
Tₙ
−
Δt
7 = 10 − 3
C
=
Cₙ
+
ΔC
14,400 = 12,000 + 2,400
Assumptions
•
Direct cost rises in a straight line from the normal point to the crash point.
•
An activity cannot be cut below its crash duration.
•
Only critical activities shorten the project: cut the one with the smallest cost slope first.
Step-by-step
WE KNOW
Tₙ = 10 days, Cₙ = $12,000, Tc = 6 days, Cc = $15,200, Δt = 3 days
FIND
cost slope (s), added cost (ΔC), new duration (T), new cost (C)
Step 1 · Find cost slope (s)
Use s = (Cc − Cₙ) ÷ (Tₙ − Tc)
The cost slope is the extra cost over the days saved, from the normal point to the crash point.
s
=
15,200 − 12,000
10 − 6
s
=
3,200
4
s
=
800
$/day
Step 2 · Find added cost (ΔC)
Use ΔC = s × Δt
On the straight line each day cut costs the slope, so the added cost is the slope times the days cut.
ΔC = 800 × 3
ΔC = $2,400
Step 3 · Find new duration (T)
Use T = Tₙ − Δt
Take the days cut off the normal duration.
T
=
10
−
3
T
=
7
days
Step 4 · Find new cost (C)
Use C = Cₙ + ΔC
The new direct cost is the normal cost plus what the cut adds.
C
=
12,000
+
2,400
C
=
$14,400
Check
Put the numbers back in. Both sides should match.
800
=
15,200 − 12,000
10 − 6
✓
2,400 = 800 × 3 ✓
7 = 10 − 3 ✓
14,400 = 12,000 + 2,400 ✓
Topic check
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