computer-nec-license
  • NEC-Computer
  • 1. Concept of Basic Electrical and Electronics Engineering
    • 1.1 Basic Concepts
    • 1.2 Network Theorems
    • 1.3 Alternating Current Fundamentals
    • 1.4 Semiconductor Device
    • 1.5 Signal Generator
    • 1.6 Amplifiers
    • MCQs
      • MCQs On Basic Electrical
        • set-1
        • set-2
      • MCQs On Basic Electronics
        • set-1
        • set-2
  • 2. Digital Logic and Microprocessor
    • 2.1 Digital Logic
    • 2.2 Combinational & Arithmetic Circuit
    • 2.3 Sequential Logic Circuits
    • 2.4 Microprocessor
    • 2.5 Microprocessor System
    • 2.6 Interrupt Operations
    • MCQs
      • MCQs On Digital Logic
        • set-1
        • set-2
        • set-3
        • set-4
        • set-5
        • set-6
        • set-7
        • set-8
        • set-9
        • set-10
        • set-11
        • set-12
      • MCQs On Microprocessor
        • set-1
        • set-2
        • set-3
        • set-4
        • set-5
        • set-6
        • set-7
        • set-8
        • set-9
  • 3. Programming Language and Its Applications
    • 3.1 Introduction to C Programming
    • 3.2 Pointers, Structures, and Data Files
    • 3.3 C++ Language Constructs with Objects and Classes
    • 3.4 Features of Object-Oriented Programming
    • 3.5 Pure Virtual Functions and File Handling
    • 3.6 Generic Programming and Exception Handling
    • MCQs
      • set-1
      • set-2
      • set-3
      • set-4
      • set-5
  • 4. Computer Organization and Embedded System
    • 4.1 Control and CPU
    • 4.2 Computer Arithmetic and Memory System
    • 4.3 I/O Organization and Multiprocessor
    • 4.4 Embedded System Design
    • 4.5 Real-Time Operating and Control Systems
    • 4.6 Hardware Description Language (VHDL) and IC Technology
    • MCQs
      • set-1
      • set-2
      • set-3
      • set-4
      • set-5
      • set-6
      • set-7
      • set-8
      • set-9
      • set-10
      • set-11
  • 5. Concept of Computer Network and Network Security System
    • 5.1 Introduction to Computer Networks
    • 5.2 Data Link Layer
    • 5.3 Network Layer
    • 5.4 Transport Layer
    • 5.5 Application Layer
    • 5.6 Network Security
    • MCQs
      • Basic Networking
        • set-1
        • set-2
      • Advanced Networking
        • set-1
        • set-2
        • set-3
        • set-4
        • set-5
        • set-6
  • 6. Theory of Computation and Computer Graphics
    • 6.1 Introduction to Finite Automata
    • 6.2 Introduction to Context-Free Languages (CFL)
    • 6.3 Turing Machines (TM)
    • 6.4 Introduction to Computer Graphics
    • 6.5 Two-Dimensional Transformation
    • 6.6 Three-Dimensional Transformation
    • MCQs
      • MCQs on Theory of Computation
        • set-1
        • set-2
        • set-3
      • MCQs On Computer Graphics
        • set-1
        • set-2
        • set-3
        • set-4
        • set-5
        • set-6
  • 7. Data Structures and Algorithm, Database System and Operating System
    • 7.1 Introduction to Data Structures, Lists, Linked Lists, and Trees
    • 7.2 Sorting, Searching, Hashing and Graphs
    • 7.3 Introduction to Data Models, Normalization, and SQL
    • 7.4 Transaction Processing, Concurrency Control, and Crash Recovery
    • 7.5 Introduction to Operating System and Process Management
    • 7.6 Memory Management, File Systems, and System Administration
    • MCQs
      • MCQs ON DSA
        • set-1
        • set-2
        • set-3
        • set-4
        • set-5
        • set-6
      • MCQs On DBMS
        • set-1
        • set-2
      • MCQs On Operating System
        • set-1
        • set-2
        • set-3
        • set-4
        • set-5
        • set-6
        • set-7
        • set-8
        • set-9
        • set-10
        • set-11
        • set-12
  • 8. Software Engineering and Object-Oriented Analysis & Design
    • 8.1 Software Process and Requirements
    • 8.2 Software Design
    • 8.3 Software Testing, Cost Estimation, Quality Management, and Configuration Management
    • 8.4 Object-Oriented Fundamentals and Analysis
    • 8.5 Object-Oriented Design
    • 8.6 Object-Oriented Design Implementation
    • MCQs
      • set-1
      • set-2
      • set-3
      • set-4
      • set-5
      • set-6
      • set-7
      • set-8
      • set-9
  • 9. Artificial Intelligence and Neural Networks
    • 9.1 Introduction to AI and Intelligent Agents
    • 9.2 Problem Solving and Searching Techniques
    • 9.3 Knowledge Representation
    • 9.4 Expert System and Natural Language Processing
    • 9.5 Machine Learning
    • 9.6 Neural Networks
    • MCQs
      • set-1
      • set-2
      • set-3
      • set-4
      • set-5
      • set-6
      • set-7
      • set-8
      • set-9
  • 10. Project Planning, Design and Implementation
    • 10.1 Engineering Drawings and Its Concepts
    • 10.2 Engineering Economics
    • 10.3 Project Planning and Scheduling
    • 10.4 Project Management
    • 10.5 Engineering Professional Practice
    • 10.6 Engineering Regulatory Body
    • MCQs
      • MCQs On Engineering Drawing
        • set-1
        • set-2
      • MCQs On Engineering Economics
      • MCQs On Project Planning & Scheduling
      • MCQs On Project Mangement
      • MCQs On Engineering Professional Practice
      • MCQs On Engineering Regulatory Body
  • Questions Sets
    • Set 1 (Chaitra, 2080)
      • Short Questions (60*1=60 Marks)
      • Long Questions (20*2=40 Marks)
    • Set 2 (Aasadh, 2081)
      • Short Questions (60*1=60 Marks)
      • Long Questions (20*2=40 Marks)
    • Set 3 (Asojh, 2080)
      • Short Questions (60*1=60 Marks)
      • Long Questions (20*2=40 Marks)
    • Model Set - Computer Engineering By NEC
      • Short Questions (60*1=60 Marks)
      • Long Questions (20*2=40 Marks)
    • Model Set - Software Engineering By NEC
      • Short Questions (60*1=60 Marks)
      • Long Questions (20*2=40 Marks)
  • Tips & Tricks
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On this page
  • 1. Understanding Project Cash Flow
  • 2. Discount Rate, Interest, and Time Value of Money
  • 3. Basic Methodologies for Engineering Economics Analysis
  • 4. Comparison of Alternatives
  • 5. Depreciation System in Nepal
  • 6. Taxation System in Nepal
  • Conclusion
  1. 10. Project Planning, Design and Implementation

10.2 Engineering Economics

Engineering economics is the application of economic principles to the evaluation of engineering projects and investments. It involves assessing the financial viability and risks associated with engineering decisions. This section covers key concepts such as project cash flow, time value of money, and various methodologies used to analyze and compare engineering projects.


1. Understanding Project Cash Flow

Project cash flow represents the inflow and outflow of money during the life cycle of a project. It helps in assessing the financial feasibility of a project by determining whether the revenues generated by the project outweigh the costs.

  • Inflow: The cash received from the project, such as sales, investments, or cost savings.

  • Outflow: The cash spent on the project, such as operational costs, equipment, labor, and other capital expenses.

  • Net Cash Flow: The difference between the inflows and outflows at any given point in time.

To evaluate the project’s viability, these cash flows are typically projected over a specific time period and discounted to their present value using an appropriate discount rate.


2. Discount Rate, Interest, and Time Value of Money

Time value of money is the concept that money available now is worth more than the same amount in the future due to its potential earning capacity. This principle underlies most financial calculations and decision-making.

  • Discount Rate: The rate at which future cash flows are adjusted to reflect their present value. It reflects the opportunity cost of capital, inflation, and risk.

  • Interest: The cost of borrowing money or the return on invested capital. Interest is calculated based on the principal amount and the interest rate.

  • Present Value (PV): The current value of a future sum of money, discounted at a certain rate over time.

  • Future Value (FV): The value of an amount of money at a future time, after accounting for interest or discounting.

Formula for calculating the present value (PV) of a future cash flow: PV=FV(1+r)tPV = \frac{FV}{(1 + r)^t}PV=(1+r)tFV​

Where:

  • FV=Future value,FV = \text{Future value,}FV=Future value,

  • r=Discount rate,r = \text{Discount rate,}r=Discount rate,

  • t=Time period (years).t = \text{Time period (years).}t=Time period (years).


Effective Interest Rate Calculations

Problem 1: Effective Interest Rate (EIR) Per Year Question: A bank has a nominal interest rate of 9% per annum, compounded quarterly. Calculate the effective interest rate per year. Given:

  • Nominal interest rate r=9%=0.09r = 9\% = 0.09r=9%=0.09

  • Compounding frequency n=4n = 4n=4 (quarterly)

Formula: EIR=(1+rn)n−1EIR = \left(1 + \frac{r}{n}\right)^n - 1EIR=(1+nr​)n−1

Solution: EIR=(1+0.094)4−1=0.09308EIR = \left(1 + \frac{0.09}{4}\right)^4 - 1 = 0.09308EIR=(1+40.09​)4−1=0.09308

Answer: The effective interest rate per year is 9.308%.


Problem 2: Effective Interest Rate Per Semi-Annual Question: A bank has a nominal interest rate of 9% per annum, compounded quarterly. Calculate the effective interest rate per semi-annual period. Given:

  • Nominal interest rate r=9%=0.09r = 9\% = 0.09r=9%=0.09

  • Compounding frequency n=4n = 4n=4 (quarterly)

  • Semi-annual compounding means n=2n = 2n=2 periods per year.

Formula: EIRsemi-annual=(1+rn)n/2−1EIR_{\text{semi-annual}} = \left(1 + \frac{r}{n}\right)^{n/2} - 1EIRsemi-annual​=(1+nr​)n/2−1

Solution: EIRsemi-annual=(1+0.094)2−1=0.04550625EIR_{\text{semi-annual}} = \left(1 + \frac{0.09}{4}\right)^2 - 1 = 0.04550625EIRsemi-annual​=(1+40.09​)2−1=0.04550625

Answer: The effective interest rate per semi-annual period is 4.551%.


Problem 3: Effective Interest Rate Per Month Question: A bank has a nominal interest rate of 9% per annum, compounded quarterly. Calculate the effective interest rate per month. Given:

  • Nominal interest rate r=9%=0.09r = 9\% = 0.09r=9%=0.09

  • Compounding frequency n=4n = 4n=4 (quarterly)

  • Monthly compounding means n=12n = 12n=12 periods per year.

Formula: EIRmonthly=(1+rn)n/12−1EIR_{\text{monthly}} = \left(1 + \frac{r}{n}\right)^{n/12} - 1EIRmonthly​=(1+nr​)n/12−1

Solution: EIRmonthly=(1+0.094)1/3−1=0.007448EIR_{\text{monthly}} = \left(1 + \frac{0.09}{4}\right)^{1/3} - 1 = 0.007448EIRmonthly​=(1+40.09​)1/3−1=0.007448

Answer: The effective interest rate per month is 0.7448%.


Some Important Financial Formulas

  • Single-Payment Compound Amount: F=P(1+i)nF = P(1 + i)^nF=P(1+i)n

  • Single-Payment Present Worth Amount: P=F⋅(P/F,i,n)=F(1+i)nP = F \cdot (P/F, i, n) = \frac{F}{(1 + i)^n}P=F⋅(P/F,i,n)=(1+i)nF​

  • Equal-Payment Series Compound Amount: F=A⋅(F/A,i,n)=A⋅(1+i)n−1iF = A \cdot (F/A, i, n) = A \cdot \frac{(1 + i)^n - 1}{i}F=A⋅(F/A,i,n)=A⋅i(1+i)n−1​

  • Equal-Payment Series Sinking Fund: A=F⋅(A/F,i,n)=F⋅i(1+i)n−1A = F \cdot (A/F, i, n) = F \cdot \frac{i}{(1 + i)^n - 1}A=F⋅(A/F,i,n)=F⋅(1+i)n−1i​

  • Equal-Payment Series Present Worth Amount: P=A⋅(P/A,i,n)=A⋅(1+i)n−1i(1+i)nP = A \cdot (P/A, i, n) = A \cdot \frac{(1 + i)^n - 1}{i(1 + i)^n}P=A⋅(P/A,i,n)=A⋅i(1+i)n(1+i)n−1​

  • Equal-Payment Series Capital Recovery Amount: A=P⋅(A/P,i,n)=P⋅i(1+i)n(1+i)n−1A = P \cdot (A/P, i, n) = P \cdot \frac{i(1 + i)^n}{(1 + i)^n - 1}A=P⋅(A/P,i,n)=P⋅(1+i)n−1i(1+i)n​

  • Uniform Gradient Series Annual Equivalent Amount: A=A1+G⋅(A/G,i,n)=A1+G⋅(1+i)n−i⋅n−1i(1+i)n−iA = A_1 + G \cdot (A/G, i, n) = A_1 + G \cdot \frac{(1 + i)^n - i \cdot n - 1}{i(1 + i)^n - i}A=A1​+G⋅(A/G,i,n)=A1​+G⋅i(1+i)n−i(1+i)n−i⋅n−1​


3. Basic Methodologies for Engineering Economics Analysis

Engineering economic analysis is a critical aspect of evaluating the feasibility and financial viability of engineering projects. It combines economic and financial principles with engineering knowledge to assess costs, benefits, and risks. Below are some commonly used methodologies in this field:

  1. Payback Period

The payback period is the time required to recover the initial investment from the cash inflows generated by a project.

Simple Payback Period: Simple Payback Period=Initial InvestmentAnnual Cash Inflow\text{Simple Payback Period} = \frac{\text{Initial Investment}}{\text{Annual Cash Inflow}}Simple Payback Period=Annual Cash InflowInitial Investment​

Decision Rule:

  • Accept: If the payback period is less than or equal to the required period.

  • Reject: If the payback period is greater than the required period.


  1. Discounted Payback Period:

Time taken for the sum of discounted cash inflows to equal the initial investment.

Formula: Discounted Cash Flow=Ct(1+i)t\text{Discounted Cash Flow} = \frac{C_t}{(1 + i)^t}Discounted Cash Flow=(1+i)tCt​​

Decision Rule:

  • Accept: If the discounted payback period is less than or equal to the required period.

  • Reject: If the discounted payback period is greater than the required period.


  1. Equivalent Worth Method

This method involves converting cash flows to a common point in time using the time value of money. Sub-methods include:

  • Present Worth (PW):

    PW=∑Ct(1+i)t\text{PW} = \sum \frac{C_t}{(1 + i)^t}PW=∑(1+i)tCt​​

Decision Rule:

  • Accept: If the present worth is greater than or equal to zero (NPV ≥ 0).

  • Reject: If the present worth is less than zero (NPV < 0).


  • Future Worth (FW):

    FW=PW×(1+i)n\text{FW} = \text{PW} \times (1 + i)^nFW=PW×(1+i)n

Decision Rule:

  • Accept: If the future worth is greater than or equal to the initial investment.

  • Reject: If the future worth is less than the initial investment.


  • Annual Worth (AW):

    AW=PW×i(1+i)n(1+i)n−1\text{AW} = \text{PW} \times \frac{i(1 + i)^n}{(1 + i)^n - 1}AW=PW×(1+i)n−1i(1+i)n​

Decision Rule:

  • Accept: If the annual worth is greater than or equal to the required minimum annual return.

  • Reject: If the annual worth is less than the required minimum annual return.


  1. Rate of Return Method

This method calculates the rate of return that a project earns over its life. Here the project cashflow reinvested at the IRR. It includes:

  • Internal Rate of Return (IRR):

    NPV=∑Ct(1+IRR)t−Initial Investment=0\text{NPV} = \sum \frac{C_t}{(1 + \text{IRR})^t} - \text{Initial Investment} = 0NPV=∑(1+IRR)tCt​​−Initial Investment=0

Decision Rule:

  • Accept: If the IRR is greater than or equal to the required rate of return.

  • Reject: If the IRR is less than the required rate of return.


  • Minimum Attractive Rate of Return (MARR):

The MARR is the minimum return that an investor or organization requires from a project, typically based on the cost of capital or opportunity cost of funds. It serves as a benchmark to compare with the IRR to determine the project’s acceptability.

Decision Rule:

  • Accept: If IRR ≥ MARR.

  • Reject: If IRR < MARR.


  • External Rate of Return (ERR):

Also known as modified internal rate of return. On ERR, here the project cashflow reinvested at the cost of capital not the IRR. Here the slight modification makes the MARR more effective than that of IRR.

Decision Rule:

  • Accept: If the ERR exceeds the cost of capital or required return rate.

  • Reject: If the ERR is less than the cost of capital or required return rate.


  1. Benefit-Cost Ratio (BCR):

The Benefit-Cost Ratio compares the present value of benefits to the present value of costs.

Formula:

BCR=PV of BenefitsPV of Costs\text{BCR} = \frac{\text{PV of Benefits}}{\text{PV of Costs}}BCR=PV of CostsPV of Benefits​


  • Conventional B/C Ratio:

Where:

  • PV of Benefits is the present value of the benefits over the project's life.

  • PV of O&M Costs is the present value of the operating and maintenance costs.

  • PV of Salvage Value is the present value of the value at the end of the project's life.

  • PV of Costs is the present value of the total project costs (including initial investment and operating costs).

Decision Rule:

  • Accept: If B/C ≥ 1

  • Reject: If B/C < 1


  • Modified B/C Ratio:

Decision Rule:

  • Accept: If B/C ≥ 1

  • Reject: If B/C < 1


4. Comparison of Alternatives

When comparing multiple project alternatives, various factors are considered, such as:

  • Cost-benefit analysis: Evaluating the costs and benefits of each alternative.

  • NPV Comparison: Projects with higher NPVs are generally preferred as they add more value.

  • IRR Comparison: Projects with higher IRRs are typically more attractive, but the MARR must also be considered.

  • Payback Period: A shorter payback period is typically more desirable, especially for companies with liquidity concerns.

The comparison helps in choosing the alternative that offers the best financial returns and risk profile.


5. Depreciation System in Nepal

Depreciation is the process by which a company allocates the cost of a long-term asset over its useful life. This reflects the decrease in the asset's value as it is used in the business. In Nepal, the rules regarding depreciation are governed by the Income Tax Act, and companies are allowed to deduct depreciation expenses from their taxable income, which helps reduce their tax burden.

Methods of Depreciation:

  1. Straight-Line Method of Depreciation:

  • Depreciation Charge (Dn) per Year:

    Dn=P−SND_n = \frac{P - S}{N}Dn​=NP−S​

  • Where:

    • P=Cost of the asset, including installation expenses.P = \text{Cost of the asset, including installation expenses.}P=Cost of the asset, including installation expenses.

    • S=Salvage value at the end of the asset’s useful life.S = \text{Salvage value at the end of the asset’s useful life.}S=Salvage value at the end of the asset’s useful life.

    • N=Useful life of the asset (in years).N = \text{Useful life of the asset (in years).}N=Useful life of the asset (in years).

    • Dn=Depreciation charge for each year (uniformly distributed).D_n = \text{Depreciation charge for each year (uniformly distributed).}Dn​=Depreciation charge for each year (uniformly distributed).

  • Book Value after n Years (Bn):

    Bn=P−(D1+D2+⋯+Dn)B_n = P - (D_1 + D_2 + \dots + D_n)Bn​=P−(D1​+D2​+⋯+Dn​)

    Where:

    • Bn=Book value of the asset after n years.B_n = \text{Book value of the asset after n years.}Bn​=Book value of the asset after n years.

    • D1,D2,…,DnD_1, D_2, \dots, D_nD1​,D2​,…,Dn​ = Depreciation charges for the years 1,2,...,n1, 2, ..., n1,2,...,n (which are all the same using the straight-line method


  1. Declining Balance Method:

  • Depreciation Charge (D) per Year:

    D=C×(1−(VC)n)D = C \times \left( 1 - \left( \frac{V}{C} \right)^n \right)D=C×(1−(CV​)n)

    Where:

    • D=Depreciation charge for the year.D = \text{Depreciation charge for the year.}D=Depreciation charge for the year.

    • C=Original cost of the asset.C = \text{Original cost of the asset.}C=Original cost of the asset.

    • V=Scrap (or salvage) value of the asset.V = \text{Scrap (or salvage) value of the asset.}V=Scrap (or salvage) value of the asset.

    • n=Useful life of the asset (in years).n = \text{Useful life of the asset (in years).}n=Useful life of the asset (in years).


  1. Sum-of-Years-Digits (SOYD) Method:

  • Sum of the Years' Digits (SOYD):

    SOYD=1+2+⋯+N=N(N+1)2SOYD = 1 + 2 + \dots + N = \frac{N(N + 1)}{2}SOYD=1+2+⋯+N=2N(N+1)​

    Where:

    • N=Useful life of the asset (in years).N = \text{Useful life of the asset (in years).}N=Useful life of the asset (in years).

  • Depreciation Charge (Dn) per Year:

    Dn=(N−n+1)(P−S)SOYDD_n = \frac{(N - n + 1)(P - S)}{SOYD}Dn​=SOYD(N−n+1)(P−S)​

    Where:

    • Dn=Depreciation charge for the nth year.D_n = \text{Depreciation charge for the nth year.}Dn​=Depreciation charge for the nth year.

    • P=Cost of the asset, including installation expenses.P = \text{Cost of the asset, including installation expenses.}P=Cost of the asset, including installation expenses.

    • S=Salvage value at the end of the asset’s useful life.S = \text{Salvage value at the end of the asset’s useful life.}S=Salvage value at the end of the asset’s useful life.

    • N=Useful life of the asset (in years).N = \text{Useful life of the asset (in years).}N=Useful life of the asset (in years).

    • n=Year in which the depreciation is being calculated (e.g., for year 1, n = 1).n = \text{Year in which the depreciation is being calculated (e.g., for year 1, n = 1).}n=Year in which the depreciation is being calculated (e.g., for year 1, n = 1).

    • SOYD=Sum of the years’ digits.SOYD = \text{Sum of the years' digits.}SOYD=Sum of the years’ digits.


  1. Sinking Fund Method:

The Sinking Fund Method of depreciation is used to accumulate funds over time to replace an asset once its useful life has ended. It assumes that a fixed annual deposit is made into a sinking fund, and the interest earned on this fund will cover the depreciation of the asset.

  • Depreciation Charge (Dn) per Year:

    Dn=P−SPresent Value of Annuity Factor for N YearsD_n = \frac{P - S}{\text{Present Value of Annuity Factor for N Years}}Dn​=Present Value of Annuity Factor for N YearsP−S​

    Where:

    • P=Cost of the asset, including installation expenses.P = \text{Cost of the asset, including installation expenses.}P=Cost of the asset, including installation expenses.

    • S=Salvage value at the end of the asset’s useful life.S = \text{Salvage value at the end of the asset’s useful life.}S=Salvage value at the end of the asset’s useful life.

    • Dn=Depreciation charge for the nth year.D_n = \text{Depreciation charge for the nth year.}Dn​=Depreciation charge for the nth year.

    • N=Useful life of the asset (in years).N = \text{Useful life of the asset (in years).}N=Useful life of the asset (in years).

  • Present Value of Annuity Factor:

    Present Value of Annuity Factor=1−(1+r)−Nr\text{Present Value of Annuity Factor} = \frac{1 - (1 + r)^{-N}}{r}Present Value of Annuity Factor=r1−(1+r)−N​

    Where:

    • r=Interest rate per period (usually annual).r = \text{Interest rate per period (usually annual).}r=Interest rate per period (usually annual).

    • N=Useful life of the asset (in years).N = \text{Useful life of the asset (in years).}N=Useful life of the asset (in years).

  • Book Value after n Years (Bn):

    Bn=P−(D1+D2+⋯+Dn)B_n = P - (D_1 + D_2 + \dots + D_n)Bn​=P−(D1​+D2​+⋯+Dn​)

    Where:

    • Bn=Book value of the asset after n years.B_n = \text{Book value of the asset after n years.}Bn​=Book value of the asset after n years.

    • D1,D2,…,Dn=Depreciation charges for each year.D_1, D_2, \dots, D_n = \text{Depreciation charges for each year.}D1​,D2​,…,Dn​=Depreciation charges for each year.


6. Taxation System in Nepal

In Nepal, the taxation system significantly impacts engineering economics, particularly in project financial evaluations. Taxes affect the net cash flows, project costs, and pricing strategies, influencing the decision-making process for businesses and project managers. Below is a detailed breakdown of the main taxes that influence the financial landscape of projects:

1. Net Income

Net Income is the amount of money a company has left after subtracting all its expenses, taxes, interest, and depreciation from its total revenue. It represents the company's profitability during a given period.

Certainly! Here's the financial structure in a table format:

Item

Description

Formula

Revenue

Total income from sales and services.

-

Cost of Goods Sold (COGS)

Direct costs associated with the production of goods sold.

-

Gross Profit

Revenue minus the cost of goods sold.

Gross Profit = Revenue - COGS

Operating Expenses

Costs associated with running the business, excluding COGS.

-

Depreciation

Allocation of the cost of fixed assets over their useful life.

-

Taxable Income

Gross Profit minus Operating Expenses and Depreciation.

Taxable Income = Gross Profit - Operating Expenses - Depreciation

Income Tax

Tax rate applied to taxable income.

Income Tax = Tax rate × Taxable Income

Net Income

Taxable Income minus Income Tax.

Net Income = Taxable Income - Income Tax

Cash Flow

Net income adjusted for depreciation.

Cash Flow = Net Income + Depreciation

This table summarizes the income statement flow, from revenue through to the calculation of cash flow.


Example of Net Income

Item

Amount ($)

Total Revenue

100,000

Cost of Goods Sold (COGS)

(20,000)

Gross Profit

80,000

Operating Expenses

- Salaries

(10,000)

- Rent

(10,000)

- Utilities

(5,000)

- Depreciation

(5,000)

Total Operating Expenses

(30,000)

Taxes

(10,000)

Net Profit

30,000

Calculation Steps:

  1. Gross Profit = Total Revenue - COGS = 100,000 - 20,000 = 80,000

  2. Total Operating Expenses = Salaries + Rent + Utilities + Depreciation = 10,000 + 10,000 + 5,000 + 5,000 = 30,000

  3. Net Profit = Gross Profit - Total Operating Expenses - Taxes = 80,000 - 30,000 - 10,000 = 30,000


2. Corporate Income Tax

Corporate income tax is levied on the earnings of businesses operating in Nepal. Companies are required to pay a certain percentage of their profits as tax. The rates for corporate income tax may vary based on factors such as:

  • Company size: Larger corporations may face different tax rates compared to small and medium enterprises (SMEs).

  • Nature of the business: Certain types of businesses, such as banks or insurance companies, may be taxed differently compared to manufacturing or service industries.

This tax directly affects the profitability of a project, as businesses must account for tax liabilities when calculating their overall income and project costs.

3. Value Added Tax (VAT)

Value Added Tax (VAT) is a consumption tax imposed on the sale of goods and services. In Nepal, VAT is applicable to most products and services, and it affects both businesses and consumers. For project-based activities, VAT impacts:

  • Costs: VAT paid on purchases (e.g., raw materials, services, etc.) can be reclaimed if the business is VAT-registered. However, this affects the cash flow since the company must initially pay the VAT before reclaiming it.

  • Pricing strategies: Projects that involve the sale of goods or services must factor in VAT when determining their pricing structure. The inclusion of VAT in the project’s cost structure may affect its overall profitability and financial planning.

Example:

  • The company purchases raw materials worth NPR 100,000 with a VAT of 13% (NPR 13,000).

  • The total payment made upfront is NPR 113,000 (NPR 100,000 for raw materials + NPR 13,000 for VAT).

  • If the business sells its finished product for NPR 200,000 and charges VAT of 13% (NPR 26,000), it will collect VAT on its sales.

  • The business can then reclaim the NPR 13,000 VAT it paid on raw materials (input VAT) from the NPR 26,000 VAT it collected from the customer (output VAT).

  • The net VAT the company must remit to the tax authorities will be NPR 13,000 (NPR 26,000 collected - NPR 13,000 paid).

Cash Flow Impact:

  • Even though the VAT can be reclaimed, the company has to pay the VAT upfront when making the purchase. This means the business initially has to spend extra cash (the VAT amount), which affects its cash flow until the VAT is refunded or offset against future sales.

  • Where if the second transaction is smaller (selling for NPR 50,000 with a VAT of NPR 6,500), the business can reclaim the input VAT (NPR 13,000) up to the output VAT it collects (NPR 6,500).

  • The business can reclaim this carried-forward VAT against future sales where it collects more output VAT.

  • If the business cannot offset the remaining NPR 6,500 input VAT against future sales, it can apply for a VAT refund.

  • This typically happens when:

    • The business stops operations or completes its current cycle.

    • Input VAT exceeds output VAT due to circumstances like lower sales or no sales at all.

  • The refund process usually involves:

    1. Filing a VAT return showing the excess input VAT.

    2. Submitting the necessary supporting documents, such as purchase invoices and sales records.

    3. Awaiting verification and approval from the tax authorities.

4. Tax Incentives

In order to encourage investment in specific sectors, the government of Nepal offers various tax incentives and deductions. These incentives are particularly targeted at industries that are crucial for the nation’s economic development, such as:

  • Infrastructure: Projects related to roads, bridges, public utilities, and other infrastructure developments may be eligible for tax reductions or exemptions to incentivize investment.

  • Manufacturing: To promote industrial growth, manufacturing companies may receive tax holidays or reduced tax rates for a specific period.

These tax incentives reduce the financial burden on businesses, making it easier to undertake large-scale projects in these sectors. When evaluating projects in such industries, understanding available tax incentives is crucial for accurately calculating the project’s net cash flow and return on investment.

Importance of Understanding the Taxation System

Understanding the taxation system in Nepal is essential for making informed financial decisions in project planning. By considering the impact of corporate income tax, VAT, and available tax incentives, project managers can:

  • Accurately forecast net cash flow: Taxes influence both inflows (e.g., revenue generation) and outflows (e.g., costs), so it’s important to account for them when calculating projected profits and cash flows.

  • Adjust pricing strategies: For businesses involved in the sale of goods or services, VAT and other taxes can affect the final price. Thus, understanding the tax landscape helps in setting competitive yet profitable pricing.

  • Evaluate project viability: By accounting for taxes and incentives, businesses can better evaluate the financial feasibility of projects, ensure profitability, and minimize tax liabilities.

Here is the table with income tax slab for the resident of Nepal.

Source:

In conclusion, taxes play a key role in determining the financial outcome of projects in Nepal. A comprehensive understanding of the tax system ensures that all financial aspects are properly accounted for and helps businesses optimize their operations and investments.


Conclusion

Engineering economics helps in making informed decisions by providing quantitative methods to evaluate the financial viability of projects. By considering cash flow, the time value of money, and applying methodologies like NPV, IRR, and the discounted payback period, engineers can assess the profitability of projects. Additionally, understanding depreciation systems and taxation is crucial in optimizing the financial outcomes of projects, especially in specific regions like Nepal.

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