Pumps A and B are being considered for purchase in a chemical plant. Cost details are: Pump A — installed cost \(16000\), uniform end-of-year maintenance \(2400\), salvage value \(1000\), service life \(1\) year; Pump B — installed cost \(32000\), uniform end-of-year maintenance \(1600\), salvage value \(S\) (unknown), service life \(2\) years. The interest rate is \(10%\) per annum, compounded annually. For both pumps to have the same capitalized cost, find the salvage value \(S\) of pump B (in Rs., rounded to the nearest integer).
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For comparing alternatives with different lives, either use capitalized cost (perpetual service) or equivalent annual cost; both give the same decision at a fixed interest rate.
Capitalized cost equals the present worth of one full cycle divided by \(1-(1+i)^{-n}\).
Step 1: Capitalized cost (perpetual replacement)
For a unit with life \(n\), installed cost \(P\), uniform end-of-year maintenance \(A\), salvage \(F\) at year \(n\), and interest \(i\), the present worth of one cycle is
\[
PW_{\text{cycle}} = P + A\sum_{k=1}^{n}\frac{1}{(1+i)^k} - \frac{F}{(1+i)^n}.
\]
Capitalized cost (perpetual repetitions every \(n\) years) is
\[
K=\frac{PW_{\text{cycle}}}{1-(1+i)^{-n}}.
\]
Step 2: Pump A (\(n=1\))
\[
PW_A = 16000+\frac{2400}{1.1}-\frac{1000}{1.1}=17272.7273,
\quad
K_A=\frac{PW_A}{1-1/1.1}=190000.
\]
Step 3: Pump B (\(n=2\)) and equality of capitalized costs
\[
PW_B = 32000+\frac{1600}{1.1}+\frac{1600}{1.1^2}-\frac{S}{1.1^2},
\qquad
K_B=\frac{PW_B}{1-1/1.1^2}.
\]
Set \(K_B=K_A\) and solve for \(S\):
\[
\frac{32000+\dfrac{1600}{1.1}+\dfrac{1600}{1.1^2}-\dfrac{S}{1.1^2}}
{1-1/1.1^2}
=190000
\;\Rightarrow\;
S \approx 2179.999 \;\text{Rs}.
\]
Step 4: Rounded answer
\[
S \approx \boxed{2180\ \text{Rs}}.
\]
