Solution

Given: $K = 600$ PY (Person-Years), $t_d = 3.5$ years = 42 months

Using Norden / Rayleigh / Putnam model (used in textbooks for this numerical):

Formulas:

• Peak Manning $m_0 = \\frac{K}{t_d \\sqrt{e}} \\approx 0.606 \\frac{K}{t_d}$
• Peak Time $t_0 \\approx \\frac{t_d}{\\sqrt{2}} \\approx 0.707 t_d$ (some books use $t_0 = t_d / 1.5$)
• Cumulative manpower up to time $t$: $y(t) = K(1 - e^{-a t^2})$ with $a = 1/(2t_0^2)$

(i) Peak Manning & Peak Time

Using standard SE approximation (Jalote/Pressman variant):

• $t_0 = 0.5 \\times t_d = 1.75$ years = 21 months (if assuming peak at mid-point, common simplified answer)
• Alternate Rayleigh: $t_0 = t_d / \\sqrt{2} = 2.47$ years = 29.7 months
• $m_0 = K/(t_d \sqrt{e}) = 600 / (3.5 \times 1.648) = 600 / 5.77 = 104 persons (approx)

If using Putnam $K = 600/3 =200$ PY per year? Some solutions state $m_0 \\approx 104-120$ persons, $t_0 \\approx 1.75-2.47$ years.

Exam expected answer (simplified): $t_0 = 1.75$ years, $m_0 \\approx 103-110$ persons. If using $t_d/\\sqrt{2}$ then $t_0=2.48$ years, $m_0 \\approx 92$ persons.

(ii) Manpower cost after 1 year 2 months ($t = 1.17$ years)

$y(t) = K[1 - e^{-(t/t_0)^2}]$ approximation
For $t=1.17, t_0=2.47$: $(t/t_0)^2 = 0.224$, $e^{-0.224}=0.799$
$y=600(1-0.799)=600\times0.201= 120.6 PY consumed

Effort remaining $= K - y \\approx 480$ PY
Cost ∝ manpower; if rate $c$ per PY, cost incurred $\\approx 120c$.

Using simplified $t_0=1.75$: $(1.17/1.75)^2=0.447$, $y=600(1-e^{-0.894})=600(1-0.409)=354$ PY? Variation shows model sensitivity - state assumption.

Note: Mention assumption clearly in exam; any of the standard Rayleigh/Norden derivations with formula is accepted.