# Rate of Return

In this method, the first step is to determine the specific value of the discount rate *d* which reduces the NPW to zero. This specific discount rate is called the rate of return (ROR). Depending on the case, the expression of *NPW* provided in Eq. (3.23) or Eq. (3.24) can be used. For instance, in the general case of Eq. (3.23), the ROR *d' *is solution of the following equation:

To solve this equation accurately, any numerical method (such the Newton- Raphson iteration method) can be used. However, an approximate value of *d'* can be obtained by trial and error. This approximate value can be determined by finding the two «/-values for which the *NPW* is slightly negative and slightly positive, and then interpolating linearly between the two values. It is important to remember that a solution for ROR may not exist.

Once the ROR is obtained for a given alternative of the project, the actual market discount rate or the minimum acceptable ROR is compared to the ROR value. If the value of ROR is larger *id'* >*d),* the project is cost-effective.

# Benefit-Cost Ratio

The benefit-cost ratio (BCR) method is also called the savings-to-investment ratio (SIR) and provides a measure of the net benefits (or savings) of the project relative to its net cost. The net values of both benefits *(B _{k})* and costs

*(C*are computed relative to a base case. The present worth of all the cash flows is typically used in this method. Therefore, the

_{k})*BCR*is computed as follows:

The alternative option for the project is considered economically viable relative to the base case when the *BCR* is greater than one *(BCR >* 1.0).

# Payback Period

In this evaluation method, the period *Y* (typically expressed in years) required to recover an initial investment is determined. Using the cash flow diagram in Figure 3.1, the value of *Y* is the solution of the following equation:

If the payback period *Y* is less than the lifetime of the project *N (Y < N),* then the project is economically viable. The value of *Y* obtained using Eq. (3.27) is typically called the discounted payback period (DPB) because it includes the value of money.

In the vast majority of applications, the time value of money is neglected in the payback period method. In this case, *Y* is called the simple payback period (SPB) and is the solution to the following equation:

In the case where the annual net savings are constant *(CF _{k} =* A), the SPB can be easily calculated as the ratio of the initial investment over the annual net savings:

The values for the SPB are shorter than for the DPBs because the undiscounted net savings are greater than their discounted counterparts. Therefore, acceptable values for SPBs are typically significantly shorter than the lifetime of the project.

# Cost of Energy

For electrical distributed generation projects using cogeneration and renewable energy systems, the cost of energy is often used to determine their cost effectiveness compared to the baseline options such as buying power from the utility grid system. Specifically, the cost of energy is the cost of generating 1 kWh by the power generating facility or technology accounting for its capital costs, financing costs, fuel costs, and fixed and variable operating and maintenance (O&M) costs over an assumed lifetime. Typically, the term levelized cost of energy (LCOE) is utilized when the present worth of all costs are considered (Krarti, 2018). The specific calculations for *LCOE *depend on the generating technology, but a general method is provided by Eq. (3.30):

where *C _{k}* is the total cost incurred in year

*к*and includes investment expenditures as well as O&M and fuel costs, while

*E*is the total electrical energy produced at year

_{k}*k.*For a generating project or technology to be economically competitive,

*LCOE*should be lower than the baseline (typically from the grid) price,

*COE*

_{h}.For renewable energy systems, the *LCOE* can be determined based on the initial cost, /С; the operation and maintenance rate, *r _{a&M}* and the annual electricity generation,

*E*assuming that this generation does not degrade over time as noted by Eq. (3.31):

_{ge}„,# Summary of Economic Analysis Methods

Table 3.4 summarizes the basic characteristics of the economic analysis methods used to evaluate single alternatives of an energy retrofit project.

It is important to note that the economic evaluation methods described above provide an indication of whether a single alternative of a retrofit project is cost-effective. However, these methods cannot be used or relied upon to compare and rank various

TABLE 3.4

Summary of the Basic Criteria for the Various Economic Analysis Methods for Energy Conservation Projects

Evaluation Method |
Equation |
Criterion |

Net present worth |
||

Rate of return |
||

Benefit-cost ratio |
||

Discounted payback period ( |
||

Simple payback period ( |
||

Levelized Cost of Energy |

alternatives for a given retrofit project. Only the LCC analysis method is appropriate for such endeavor. The application of various single alternative analysis methods is provided by Example 3.5 to illustrate that they are consistent in assessing whether or not a project is cost-effective.

**Example 3.5**

After finding that the old boiler has an efficiency of only 60 percent, whereas a new boiler would have an efficiency of 85 percent, the building owner in Example 3.1 has decided to invest the $10,000 in getting a new boiler. Determine whether this investment is cost-effective if the lifetime of the boiler is ten years and the discount rate is 5 percent. The boiler consumes 5,000 gallons/year at a cost of $1.20/gallon. An annual maintenance fee of $150 is required for the boiler (independent of its age). Use all five methods summarized in Table 3.4 to perform the economic analysis.

SOLUTION

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 clue to higher boiler efficiency as calculated below:

Thus,

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, = ■■■ = CF*

_{]0}=*A, d =* 0.05, and *N =* 10 years. Using Eq. (3.24) with *USPW =* 7.740 (see Example 3.4):

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, = ■■■ = CF*

_{M}=*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:

Inasmuch as *d' > d =* 5%, the investment in replacing the boiler is cost-effective.

3. *Benefit-Cost Ratio.* In this case, *B _{o} = 0* and 6, - •" - 8,o =

*A,*whereas C

_{0 }= $10,000 and C, = — = C

_{10}= 0.

Note that because the maintenance fee is applicable to both the old and the 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):

Thus, BCR 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, = = Cf_{l0} *= 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.

5. *Simple Payback Period.* For this method, *CF _{0} =* $10,000 and

*A =*$1,765. Using Eq. (3.29),

*Y*can be easily determined: V = 5.7 years.

Thus, the *SPB* method indicates that the boiler retrofit project can be cost-effective.