The base case for the economic analysis presented in this example is the case where the boiler is not replaced. Moreover, the salvage value of the boiler is assumed insignificant after ten years. Therefore, the only annual cash flows (A) after the initial investment on a new boiler are the net savings due to higher boiler efficiency as calculated below:
A= Fuel-Use_{before} *(η_{befor}/η_{fter})*Fuel-cos/gallon
A=5000*(1-0.60/0.85)*$1.20=$1.765
The cost-effectiveness of replacing the boiler is evaluated as indicated below:
1. Net Present Worth. For this method CF_{0} = $10,000 and CF_{1} = … = CF_{10} = A, d = 0.05, and N= 10 years. Using Eq. (3.24) with USPW = 7.740 (see Example 3.4):
NPW = −CF_{0} + A*USPW(d,N) (3.24)
NPW = $3,682
Therefore, the investment in purchasing a new boiler is cost-effective.
2. Rate of return. For this method also CF_{0} = $10,000 and CF_{1} = …=CF_{10} = A, whereas SPPW(d′,k) is provided by Eq. (3.17). By trial and error, it can be shown that the solution for d′ is:
SPPW(d,N) = P/F = (1+d)^{−N} (3.17)
d′ = 12.5%
Inasmuch as d′ > d = 5%, the investment in replacing the boiler is cost- effective.
3. Benefit–Cost Ratio. In this case, B_{0} = 0 and B_{1}= … = B_{10} = A whereas C_{0} = $10,000 and C_{1} =… = C_{10} = 0.
Note that because the maintenance fee is applicable to both the old and new boiler, this cost is not accounted for in this evaluation method (only the benefits and costs relative to the base case are considered).
Using Eq. (3.26):
BCR=\frac {\sum\limits_{K=0}^{N}{B_{k}*SPPW(d,k)}}{\sum\limits_{K=0}^{N}{C_{k}*SPPW(d,k)}} (3.26)
BCR = 1.368
Thus, the benefit–cost ratio is greater than unity (BCR > 1) and the project of getting a new boiler is economically feasible.
4. Compounded Payback Period. For this method, CF_{0} = $10,000 and CF_{1}= … = CF_{10} = A. Using Eq. (3.27), Y can be solved: Y = 6.9 years.
Thus the compounded payback period is shorter than the lifetime of the project (Y > N = 10 years) and therefore replacing the boiler is cost-effective.
CF_{0}=\sum\limits_{K=1}^{Y}{CF_{k}*SPPW(d,k)} (3.27)
5. Simple Payback Period. For this method, CF_{0} = $10,000 and A =$1,765. Using Eq. (3.29), Y can be easily determined: Y = 5.7 years.
Thus, the simple payback period method indicates that the boiler retrofit project can be cost-effective.
Y=\frac {CF_{0}}{A} (3.29)