# Engineering Economics in FE Electrical Exam

Welcome to the world where the elegance of electrical engineering meets the pragmatism of economics. In Engineering Economics, every decision we make carries profound financial implications, playing a pivotal role in the success of our projects. As we dive into this exciting discipline, we uncover the hidden powerhouse that shapes the landscape of electrical engineering.

In the dynamic field of electrical engineering, technical skills and knowledge alone cannot guarantee success. Enter Engineering Economics, the most critical subject that balances innovation with financial viability. With its array of tools and techniques, Engineering Economics empowers electrical engineers to make informed decisions, identify cost-effective solutions, and navigate the complex terrain of project finance.

This blog will discuss the methodologies and techniques of Engineering Economics in the FE Electrical exam. From cost estimation techniques to the time value of money, we will explore all the critical concepts of the Engineering Economics section of the NCEES® FE Reference Handbook.

We will help you try some practical examples that empower you to optimize projects, achieve efficiency, and set your engineering career to unparalleled growth. Let’s discuss why engineering economics for the FE exam aspirants is a critical topic that cannot be overlooked at any cost.

## Importance of Engineering Economics in Electrical Engineering

The study of Engineering Economics in Electrical Engineering can be well-reflected in a range of various perspectives and areas, including:

**Cost-Effective Design**– Engineering Economics enables electrical engineers to assess the financial feasibility of design choices, ensuring that projects are optimized for cost efficiency without compromising performance or safety. It helps engineers strike a balance between innovative solutions and economic viability.**Investment Decision-Making**– Electrical engineering projects require substantial equipment, infrastructure, and technology investments. Engineering Economics equips engineers with tools to evaluate the financial viability of these investments, considering factors such as cash flows, return on investment, and payback periods. This enables informed decision-making and maximizes the value of assets.**Risk Assessment and Mitigation**– Engineering Economics provides methods to quantify and analyze risks associated with electrical engineering projects. Engineers can assess potential financial risks through sensitivity analysis and probabilistic modeling and develop mitigation strategies. This ensures that projects are financially resilient and withstand uncertainties.**Life Cycle Cost Analysis**– Electrical engineering projects span their entire life cycles, from design and construction to operation and maintenance. Engineering Economics allows engineers to consider the long-term costs of these stages, including energy consumption, maintenance expenses, and equipment replacements. Engineers can optimize project decisions over the lifespan by incorporating life cycle cost analysis.**Economic Sustainability**– Engineering Economics fosters economic sustainability in a rapidly evolving technological landscape. By evaluating the financial viability of renewable energy systems, energy-efficient designs, and innovative grid solutions, electrical engineers can drive the adoption of sustainable practices that benefit both the environment and the bottom line.

## Time Value of Money

In Engineering Economics, “time value of money” refers to the principle that money has a different value at different points in time. It recognizes that receiving a certain amount today is more valuable than receiving the same amount in the future.

This concept is crucial in decision-making processes involving cash flows over time.

Several factors influence the time value of money, such as:

**Interest and Inflation****Value of Money****Time Periods****Annuities and perpetuities****Internal Rate of Return (IRR)**

In detail, let’s uncover each of these topics in engineering economics for the FE electrical exam.

### Interest and Inflation

Money can earn interest or be subject to inflation over time. Interest rates represent the cost of borrowing or the return on investment, while inflation erodes the purchasing power of money. These factors affect the value of money over time and should be considered in financial calculations.

#### Simple Interest

Simple interest is a straightforward method of calculating interest on a principal amount based on a fixed interest rate and the length of time the amount is borrowed or invested. The interest is calculated solely on the initial principal amount without considering any accrued interest over time.

Mathematically, the formula for calculating simple interest is:

Interest = Principal x Interest Rate x Time

Let’s consider an example from electrical engineering. Suppose a company borrows $10,000 to purchase electrical equipment with a simple interest rate of 5% per year. The loan duration is 3 years.

Using the simple interest formula:

Interest = $10,000 x 0.05 x 3 = $1,500

So, the company will pay $1,500 in interest over the 3 years and repaying the original $10,000 principal.

#### Compound Interest

Compound interest considers the initial principal and any accumulated interest over time. The interest is reinvested or added to the principal at specific intervals, resulting in interest being earned on the interest itself.

Mathematically, the formula for calculating compound interest is:

A = P(1 + r/n)^{^}(nt)

Where:

A = the future value (including interest)

P = the principal amount

r = the annual interest rate (expressed as a decimal)

n = the number of compounding periods per year

t = the number of years

Consider an electrical engineering project that requires an investment of $5,000. The project offers a compound interest rate of 8% annually, compounded semi-annually (n = 2), over 5 years.

Using the compound interest formula:

A = $5,000(1 + 0.08/2)^{^}(2*5) = $7,267.08

So, the investment will grow to approximately $7,267.08 over the 5 years, with interest compounding semi-annually.

#### The difference in Simple and Compound Interest – How do they impact the final value of the investment?

Consider an electrical engineer who wants to save for a future project. He/She invests **$20,000** at an annual interest rate of **6%** using simple interest for 4 years. How much interest will they earn?

**Using the simple interest formula:**

Interest = $20,000 x 0.06 x 4 = $4,800

Therefore, the engineer will earn $4,800 in interest over the 4 years.

Now, let’s calculate the amount using compound interest instead.

**Using the compound interest formula:**

A = $20,000(1 + 0.06)^{^}(4) = $24,974.40

The investment will grow to approximately **$24,974.40** over the 4 years, with the interest compounding annually.

Comparing the two scenarios, we see that the simple interest method yields a fixed interest amount of **$4,800**, while compound interest generates a higher final value of **$24,974.40**. This highlights the potential advantage of compound interest in maximizing Return on Investment (ROI) over time.

### Value of Money

#### Present Value (PV)

Present value refers to the current worth of a future cash flow. It involves discounting future cash flows by an appropriate interest rate to determine their value in today’s terms. By calculating the present value, engineers can assess the economic viability of projects, investments, or cost estimation over the project’s lifespan.

Mathematically,

PV = CFt / (1 + r)^{^}t

Where:

PV = Present Value

CFt = Cash flow at time t

r = Discount rate

t = Time period

**Future Value (FV)**

The future value represents the worth of an investment or cash flow at a specific point in the future. It involves compounding the initial amount with earned interest over time. Engineers use future value calculations to assess the growth or accumulation of investments, considering factors such as compounding periods and interest rates.

CV = Initial Investment + ∑CFt

Where:

CV = Current Value

Initial Investment = Initial investment amount

Difference between Current and Present Value

Consider an electrical manufacturing firm evaluating an investment opportunity for a renewable energy project. The project requires an initial investment of $500,000. The projected cash inflows for the next five years are as follows:

**Year 1:**$100,000**Year 2:**$150,000**Year 3:**$200,000**Year 4:**$250,000**Year 5:**$300,000

The company’s discount rate, representing the minimum acceptable rate of return, is determined to be **8%**.

Let’s calculate the present value (PV) and future value (FV) using cash inflows, discount rates, and periods.

**Present Value (PV)**

Using the present value formula:

PV = CFt / (1 + r)^{^}t

For this problem, the discount rate (r) is **8%**, and each year’s cash inflows (CFt) are given.

Year 1: PV = $100,000 / (1 + 0.08)^{^}1

Year 2: PV = $150,000 / (1 + 0.08)^{^}2

Year 3: PV = $200,000 / (1 + 0.08)^{^}3

Year 4: PV = $250,000 / (1 + 0.08)^{^}4

Year 5: PV = $300,000 / (1 + 0.08)^{^}5

Calculating the present value for each year and summing them up:

PV = $100,000 / 1.08 + $150,000 / 1.08^{^}2 + $200,000 / 1.08^{^}3 + $250,000 / 1.08^{^}4 + $300,000 / 1.08^{^}5

Simplifying the equation, we get the following:

PV ≈ $92,592.59 + $129,629.63 + $162,037.04 + $184,027.78 + $195,808.83

Summing up the present values:

PV ≈ $764,095.87

The calculated present value of **$764,095.87** indicates the current worth of the projected cash inflows, considering the time value of money.

**Future Value (FV)**

Using the future value formula:

FV = PV * (1 + r)^{^}t

For this problem, the initial investment is **$500,000**, the discount rate (r) is **8%**, and the time period (t) is 5 years.

Calculating the future value:

FV = $500,000 * (1 + 0.08)^{^}5

Simplifying the equation, we get:

FV ≈ $500,000 * 1.46933

FV ≈ $734,665.08

The calculated future value of **$734,665.08** represents the accumulated worth of the investment and the projected cash inflows at the end of the five years without considering the time value of money.

**Comparing the present value and future value of the investment:**

The present value of **$764,095.87** indicates that the current worth of the projected cash inflows when discounted at a rate of **8%** over the respective time periods, exceeds the initial investment of **$500,000**. This suggests that the investment is financially attractive, as the present value is higher than the initial investment.

On the other hand, the future value of **$734,665.08** indicates the accumulated worth of the investment and the projected cash inflows at the end of the five years without considering the time value of money. The future value is lower than the initial investment, indicating that the investment does not generate a positive return when considering the future value alone.

The investment appears financially feasible based on the present value analysis. However, considering the future value alone, the investment does not generate a positive return. Therefore, the company needs to consider the time value of money and assess the investment’s financial attractiveness based on the present value.

CFt = Cash flow at time t

#### Net Present Value

Net Present Value (NPV) is a financial metric used in Engineering Economics to assess a project or investment’s profitability and economic feasibility. It considers the time value of money by discounting future cash flows to their present value and compares them to the initial investment.

Mathematically, the formula for calculating Net Present Value is:

NPV = ∑(CFt / (1 + r)^{^}t) – Initial Investment

Where:

NPV = Net Present Value

CFt = Cash flow at time t

r = Discount rate (required rate of return or cost of capital)

t = Time period

Let’s consider an example problem related to an electrical engineering project. Suppose a company is considering investing in a solar power plant. The initial investment required is **$1,000,000**. The projected cash flows for the next five years are as follows:

**Year 1:**$200,000**Year 2:**$300,000**Year 3:**$400,000**Year 4:**$400,000**Year 5:**$500,000

The company’s required rate of return or discount rate is **8%**.

To calculate the Net Present Value, we must discount each cash flow to its present value and subtract the initial investment.

Using the formula, the NPV can be calculated as follows:

NPV = (200,000 / (1 + 0.08)^{^}1) + (300,000 / (1 + 0.08)^{^}2) + (400,000 / (1 + 0.08)^{^}3) + (400,000 / (1 + 0.08)^{^}4) + (500,000 / (1 + 0.08)^{^}5) – 1,000,000

Simplifying the equation, we get the following:

NPV = 200,000 / 1.08 + 300,000 / 1.08^{^}2 + 400,000 / 1.08^{^}3 + 400,000 / 1.08^{^}4 + 500,000 / 1.08^{^}5 – 1,000,000

Calculating the values and summing them up, we find:

NPV ≈ 185,185.19 + 259,259.26 + 324,074.07 + 360,493.83 + 425,925.93 – 1,000,000

NPV ≈ 555,927.28

The calculated Net Present Value of $555,927.28 indicates that the solar power plant investment is economically feasible and potentially profitable. A positive NPV suggests that the project’s anticipated cash inflows outweigh the initial investment and the required rate of return.

By considering the Net Present Value, electrical engineers can make informed decisions regarding project feasibility, investment opportunities, and the allocation of resources, ensuring the long-term financial viability of electrical engineering endeavors.

### Time Periods

The time duration between cash flows also influences the time value of money. The longer the time period, the more significant the impact of compounding or discounting on the value of money. Engineers consider the time periods involved in projects to evaluate the financial implications of delaying or accelerating cash flows.

By incorporating the time value of money into engineering economic analyses, engineers can make informed decisions regarding project viability, investment returns, cost estimation, and budgeting. Understanding this concept allows for proper evaluation of cash flows, risk assessment, and comparison of alternative financing options.

### Annuities and perpetuities

#### Annuities

Annuities are a series of equal cash flows that occur regularly over a specified period at specific intervals. These cash flows can be incoming (received) or outgoing (paid). Annuities are commonly used to calculate loan payments, retirement savings, and other financial scenarios involving regular payments.

Mathematically, the present value (PV) and future value (FV) of an annuity can be calculated using the following formulas:

PV of Annuity = C * [(1 – (1 + r)^{^}(-n)) / r]

FV of Annuity = C * [((1 + r)^{^}n – 1) / r]

Where:

C = Cash flow per period

r = Interest rate per period

n = Number of periods

Consider an electrical engineering company that is considering financing a project that requires an initial investment of **$500,000**. The company plans to make equal annual contributions over 5 years to repay the investment with an interest rate of **6%**. Calculate the annual payment required to repay the investment.

Given:

Initial investment (PV) = $500,000

Number of periods (n) = 5

Interest rate per period (r) = 6%

Using the formula for the PV of an annuity, we can calculate the annual payment (C):

PV = C * [(1 – (1 + r)^{^}(-n)) / r]

$500,000 = C * [(1 – (1 + 0.06)^{^}(-5)) / 0.06]

Solving for C:

C ≈ $133,164.97

Therefore, the company needs to make an annual payment of approximately **$133,164.97** to repay the investment over the 5 years.

#### Perpetuities

Perpetuities are a special type of annuity in which the cash flows continue indefinitely. In other words, perpetuities have an infinite time horizon. Examples of perpetuities include certain types of bonds or stocks that provide fixed payments indefinitely.

Mathematically, the present value (PV) of perpetuity can be calculated using the following formula:

PV of Perpetuity = C / r

Where:

C = Cash flow per period

r = Interest rate per period

Consider an electrical engineering company that invests in a renewable energy project that generates a fixed annual revenue of **$50,000**. The company’s required rate of return is **8%**. Calculate the present value of the perpetual revenue stream.

Given:

Cash flow per period (C) = $50,000

Interest rate per period (r) = 8%

Using the formula for the PV of perpetuity, we can calculate the present value:

PV = C / r

PV = $50,000 / 0.08

PV = $625,000

Therefore, the present value of the perpetual revenue stream from the renewable energy project is **$625,000**.

### Internal Rate of Return (IRR)

Internal Rate of Return (IRR) is a financial metric used in engineering economics to assess the profitability of an investment or project. It represents the discount rate at which cash flows’ net present value (NPV) becomes zero.

Since the Internal Rate of Return (IRR) is based on the concept of setting the net present value (NPV) of cash flows to zero, Mathematically, it can be represented as:

NPV = CF0 + CF1 / (1 + IRR) + CF2 / (1 + IRR)^{^}2 + … + CFn / (1 + IRR)^{^}n = 0

Where:

CF0 represents the initial investment or cash outflow at time 0.

CF1, CF2, …, CFn represent each period’s cash inflows or outflows.

IRR represents the discount rate or Internal Rate of Return. It considers the timing and magnitude of cash inflows and outflows associated with the investment. It provides insight into the project’s potential return by considering the time value of money.

It is different from Return on Investment (ROI). While ROI provides a percentage value representing the profitability of an investment, IRR goes a step further by determining the actual discount rate at which the project breaks even (NPV = 0). It considers the time value of money and provides a more precise measure of the project’s financial viability.

In essence, ROI indicates the efficiency of an investment, while IRR determines the rate of return at which the investment becomes financially viable.

## Decision-Making Tools

Decision-making methodologies are also a crucial aspect of engineering economics in the FE electrical exam. The following decision-making tools provide valuable insights to electrical engineers in assessing the financial viability, risk levels, and profitability of projects or investments.

### 1. Cost-Benefit Analysis

Cost-Benefit Analysis (CBA) is a technique used to assess a project or investment’s potential benefits and costs. It involves quantifying and comparing different alternatives’ positive (advantages) and negative (downsides) impacts.

The goal is to determine whether the benefits outweigh the costs and whether the project or investment is financially viable.

#### Use-case of Cost-Benefit in Electrical Engineering

Cost-Benefit Analysis can be applied in electrical engineering for evaluating projects such as implementing energy-efficient technologies, developing renewable energy sources, or upgrading electrical systems.

It helps engineers assess the financial feasibility of such projects by considering the costs involved (e.g., installation, maintenance) and the potential benefits (e.g., energy savings, reduced emissions) over the project’s lifespan.

### 2. Break-Even Analysis

Break-Even Analysis is a tool used to determine when a project or investment starts generating enough revenue to cover its costs, resulting in neither profit nor loss. It helps identify the minimum level of activity or sales volume required to reach the break-even point.

#### Use-case of Break-Even Analysis in Electrical Engineering

Break-Even Analysis can be valuable when manufacturing or selling electrical components or products in electrical engineering. Engineers can use this tool to determine the number of units they need to sell to cover the manufacturing and operational costs, including production costs, overhead expenses, and pricing.

It assists in making informed decisions about pricing strategies, production volumes, and profitability projections.

### 3. Sensitivity Analysis

Sensitivity Analysis involves assessing the impact of changes in critical variables or assumptions on the outcomes of a project or investment. It helps identify the most critical factors influencing the project’s financial performance and provides insights into its robustness and potential risks.

#### Use-case in Sensitivity Analysis Electrical Engineering

Sensitivity Analysis can be used in electrical engineering when evaluating complex projects or systems involving various factors, such as costs, performance parameters, market conditions, or technological variables.

For instance, when designing an electrical power distribution network, engineers can analyze the sensitivity of the project’s profitability to variables like equipment costs, energy prices, or demand fluctuations.

This analysis helps identify the variables that have the most significant impact on project outcomes and allows engineers to assess the project’s sensitivity to changes in those variables.

## Depreciation and Taxes

### Depreciation

Depreciation in the study of engineering economics is the systematic allocation of an asset’s cost over its useful life. It represents the reduction in value due to factors like wear, obsolescence, or technological advancements. It impacts financial calculations and helps engineers assess asset value, tax deductions, and project feasibility.

Taxation is another crucial consideration in electrical engineering, affecting cash flows and profitability. Understanding depreciation methods and their impact on taxes help engineers make informed decisions about asset acquisition, project costs, and financial planning.

The three significant depreciation techniques used in engineering economics that hold key importance as must-to-cover topics in engineering economics in the FE electrical exam are outlined below:

Depreciation Method | Description | Use-Cases in Electrical Engineering |

Straight Line Depreciation | Evenly spreads asset cost over its useful life | Depreciation of electrical equipment and infrastructure |

Calculation of the depreciation expense for power generation systems | ||

Determining the decrease in value of research and development equipment over time | ||

Declining Balance Depreciation | Higher depreciation expenses in early years, decreasing over time | Assets experiencing rapid technological advancements in electrical engineering |

Equipment subject to higher wear and tear in the initial stages, such as specialized electrical systems | ||

Assets with higher productivity and value in their early years, like advanced technology components and systems | ||

Modified Accelerated Cost Recovery System (MACRS) | Depreciation method for tax purposes in the US | Depreciation of electrical assets used in projects, such as power generation equipment and transmission systems |

Calculation of depreciation deductions for research and development equipment | ||

Optimizing tax benefits and cash flow for electrical engineering investments |

### Conclusion

Now you have rich insights into the importance of engineering economics in the FE electrical exam and the career, specifically regarding decision-making and subject matter expert (SME) roles. The key topics we discussed in the blog are crucial for success.

Therefore, preparing this topic with all the technical depth in engineering economics for the FE exam preparation is recommended. For dedicated guidance from PE-licensed experts, explore Study for FE – a recognized online study platform and your go-to place for all FE and engineering subject-related queries.

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