(a) Forward/backward pass gives project duration 36 days and critical path 1–2–3–5–8. (b) Logarithmic differentiation gives $\dfrac{dy}{dx} = \dfrac{y}{2}\big[\tfrac{1}{x-1}+\tfrac{1}{x-2}-\tfrac{1}{x-3}-\tfrac{2x+1}{x^2+x+1}\big]$.

Part (a): CPM network. Activities (with durations): 1-2:7, 1-6:6, 2-3:14, 2-4:5, 3-5:11, 4-5:7, 6-7:11, 5-8:4, 7-8:18.

Forward pass (earliest event times $E$):

$E_1=0,\ E_2=7,\ E_3=21,\ E_4=12,\ E_6=6,$

$E_5=\max(E_3+11,\ E_4+7)=\max(32,19)=32,\quad E_7=E_6+11=17,$

$E_8=\max(E_5+4,\ E_7+18)=\max(36,35)=36.$

Project duration $= 36$ days.

Backward pass (latest event times $L$), $L_8=36$:

$L_5=32,\ L_7=18,\ L_3=21,\ L_4=25,\ L_6=7,$

$L_2=\min(L_3-14,\ L_4-5)=\min(7,20)=7,\quad L_1=\min(L_2-7,\ L_6-6)=0.$

Activity times (EST $=E_i$, EFT $=$ EST$+t$, LFT $=L_j$, LST $=$ LFT$-t$):

Zero-float (critical) activities: 1-2, 2-3, 3-5, 5-8.

Critical path: 1–2–3–5–8, duration $= 7+14+11+4 = 36$ days.

Part (b): Differentiate $y = \sqrt{\dfrac{(x-1)(x-2)}{(x-3)(x^2+x+1)}}$. Use logarithmic differentiation.

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