G��Uniform gradient amount per interest period

Engineering Economics · FE Reference Handbook section

Engineering Economics

0 formulas

10 exam-style examples

~45 min

All Engineering Economics lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers G��Uniform gradient amount per interest period within Engineering Economics. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what g��uniform gradient amount per interest period describes physically and when it applies.
State every one of the 0 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: i per period must match n periods.

Lecture

Why this section exists. G��Uniform gradient amount per interest period is the part of Engineering Economics that lets you connect an alternative being compared over a study period to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as one cash-flow diagram converted with one factor. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. i per period must match n periods. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Three engineers in hard hats and safety vests reviewing drawings on a truck tailgate. — Photo 1. Where this shows up in practice: g��uniform gradient amount per interest period.
Capstone Studio instructional photograph

Engineering Economics — G��Uniform gradient amount per interest period: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes an alternative being compared over a study period. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 0 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

i ��Interest rate per interest period
ie ��Annual effective interest rate

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

This section is conceptual; there are no equations to memorise.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Present worth of a culvert alternative

A culvert costs $85,000 now and $4,000 per year to maintain for 20 years. At i = 6%, what is the present worth of the total cost?

Given

P₀ = $85,000
A = $4,000/yr
n = 20 yr
i = 6%

Find

Total present worth

Start with the thinking

Uniform series converts with (P/A, i, n).
Costs are all negative — keep one sign convention.

Step-by-step solution

Series factor
$(P / A) = [(1 + i)^{n} - 1] / [i (1 + i)^{n}]$
Growth term
$(1.06)^{20} = 3.2071$
Factor
$(P / A) = (3.2071 - 1) / [0.06 (3.2071)] = 2.2071/0.19243 = 11.47$
Maintenance PW — 4,000(11.47) = $45,880
Total — PW = 85,000 + 45,880 = $130,880

Answer: PW ≈ $131,000

Why the other options are there

$165,000 (annual costs summed undiscounted)
$45,900 (initial cost omitted)

Reference: FE Reference Handbook — Engineering Economics — Uniform series

Example 2

Benefit-cost ratio of a widening project

A widening saves $210,000 per year in delay, costs $1.6 M to build (20-yr life, i = 7%, no salvage) and $30,000 per year to maintain. Compute B/C.

Given

Benefits = $210,000/yr
First cost = $1,600,000
O&M = $30,000/yr
n = 20, i = 7%

Find

Benefit-cost ratio

Start with the thinking

Convert everything to the same time basis — annual is easiest here.
Maintenance belongs in the denominator, not as a negative benefit.

Step-by-step solution

Capital recovery
$(A / P) = i (1 + i)^{n} / [(1 + i)^{n} - 1]$
Growth term
$(1.07)^{20} = 3.8697$
Factor
$(A / P) = 0.07 (3.8697) /2.8697 = 0.09439$
Annual cost — 1,600,000(0.09439) + 30,000 = 151,024 + 30,000 = $181,024
Ratio
$B / C = 210, 000/181, 024 = 1.16$

Answer: B/C = 1.16, so the project is justified

Why the other options are there

1.31 (O&M left out)
0.86 (ratio inverted)

Reference: FE Reference Handbook — Engineering Economics — Benefit-cost analysis

Example 3

Future worth of a single payment — G��Uniform gradient amount per interest period

$26,000 is invested at 9.5% compounded annually for 10 years. What is the future worth?

Given

P = $26,000
i = 9.5%
n = 10 years

Find

Start with the thinking

Single sum moved forward — use (F/P, i, n).
i and n must be in the same period.

Step-by-step solution

Factor
$F = P (1 + i)^{n}$
Substituting — F = $26,000(1 + 0.095)^10
Factor value
$(1 + i)^{n} = 2.4782$
Result — F = $64,434

Answer: F ≈ $64,434

Why the other options are there

$50,700 (simple interest)
$10,491 (factor inverted)

Reference: FE Reference Handbook — Engineering Economics → G��Uniform gradient amount per interest period

Example 4

Present worth of an arithmetic gradient — G��Uniform gradient amount per interest period

Maintenance is $13,000 in year 1 and grows by $7,000 each year for 10 years. At 7.5%, what is the present worth?

Given

A₁ = $13,000
G = $7,000/yr
n = 10
i = 7.5%

Find

Total present worth

Start with the thinking

Split the cash flow into a uniform base plus a gradient.
The gradient factor starts in year 2 by definition.

Step-by-step solution

Uniform part
$(P / A, i, n) = [(1 + i)^{n} - 1] / [i (1 + i)^{n}]$
Evaluate
$(P / A) = 6.8641$
Gradient part
$(P / G, i, n) = (1/ i) [(P / A) - n / (1 + i)^{n}]$
Evaluate
$(P / G) = 26.8286$
Combine — PW = $13,000(6.8641) + $7,000(26.8286) = $277,033

Answer: PW ≈ $277,033

Why the other options are there

$89,233 (gradient omitted)
$569,719 (gradient treated as uniform)

Reference: FE Reference Handbook — Engineering Economics → G��Uniform gradient amount per interest period

Example 5

Future worth of a single payment — G��Uniform gradient amount per interest period (2)

$89,000 is invested at 8.0% compounded annually for 18 years. What is the future worth?

Given

P = $89,000
i = 8.0%
n = 18 years

Find

Start with the thinking

Single sum moved forward — use (F/P, i, n).
i and n must be in the same period.

Step-by-step solution

Factor
$F = P (1 + i)^{n}$
Substituting — F = $89,000(1 + 0.080)^18
Factor value
$(1 + i)^{n} = 3.9960$
Result — F = $355,646

Answer: F ≈ $355,646

Why the other options are there

$217,160 (simple interest)
$22,272 (factor inverted)

Reference: FE Reference Handbook — Engineering Economics → G��Uniform gradient amount per interest period

Example 6

Present worth of an arithmetic gradient — G��Uniform gradient amount per interest period (2)

Maintenance is $18,000 in year 1 and grows by $5,000 each year for 14 years. At 7.0%, what is the present worth?

Given

A₁ = $18,000
G = $5,000/yr
n = 14
i = 7.0%

Find

Total present worth

Start with the thinking

Split the cash flow into a uniform base plus a gradient.
The gradient factor starts in year 2 by definition.

Step-by-step solution

Uniform part
$(P / A, i, n) = [(1 + i)^{n} - 1] / [i (1 + i)^{n}]$
Evaluate
$(P / A) = 8.7455$
Gradient part
$(P / G, i, n) = (1/ i) [(P / A) - n / (1 + i)^{n}]$
Evaluate
$(P / G) = 47.3718$
Combine — PW = $18,000(8.7455) + $5,000(47.3718) = $394,277

Answer: PW ≈ $394,277

Why the other options are there

$157,418 (gradient omitted)
$769,601 (gradient treated as uniform)

Reference: FE Reference Handbook — Engineering Economics → G��Uniform gradient amount per interest period

Example 7

Future worth of a single payment — G��Uniform gradient amount per interest period (3)

$122,000 is invested at 8.0% compounded annually for 17 years. What is the future worth?

Given

P = $122,000
i = 8.0%
n = 17 years

Find

Start with the thinking

Single sum moved forward — use (F/P, i, n).
i and n must be in the same period.

Step-by-step solution

Factor
$F = P (1 + i)^{n}$
Substituting — F = $122,000(1 + 0.080)^17
Factor value
$(1 + i)^{n} = 3.7000$
Result — F = $451,402

Answer: F ≈ $451,402

Why the other options are there

$287,920 (simple interest)
$32,973 (factor inverted)

Reference: FE Reference Handbook — Engineering Economics → G��Uniform gradient amount per interest period

Example 8

Present worth of an arithmetic gradient — G��Uniform gradient amount per interest period (3)

Maintenance is $14,000 in year 1 and grows by $2,000 each year for 10 years. At 6.5%, what is the present worth?

Given

A₁ = $14,000
G = $2,000/yr
n = 10
i = 6.5%

Find

Total present worth

Start with the thinking

Split the cash flow into a uniform base plus a gradient.
The gradient factor starts in year 2 by definition.

Step-by-step solution

Uniform part
$(P / A, i, n) = [(1 + i)^{n} - 1] / [i (1 + i)^{n}]$
Evaluate
$(P / A) = 7.1888$
Gradient part
$(P / G, i, n) = (1/ i) [(P / A) - n / (1 + i)^{n}]$
Evaluate
$(P / G) = 28.6395$
Combine — PW = $14,000(7.1888) + $2,000(28.6395) = $157,923

Answer: PW ≈ $157,923

Why the other options are there

$100,644 (gradient omitted)
$244,420 (gradient treated as uniform)

Reference: FE Reference Handbook — Engineering Economics → G��Uniform gradient amount per interest period

Example 9

Future worth of a single payment — G��Uniform gradient amount per interest period (4)

$60,000 is invested at 7.0% compounded annually for 19 years. What is the future worth?

Given

P = $60,000
i = 7.0%
n = 19 years

Find

Start with the thinking

Single sum moved forward — use (F/P, i, n).
i and n must be in the same period.

Step-by-step solution

Factor
$F = P (1 + i)^{n}$
Substituting — F = $60,000(1 + 0.070)^19
Factor value
$(1 + i)^{n} = 3.6165$
Result — F = $216,992

Answer: F ≈ $216,992

Why the other options are there

$139,800 (simple interest)
$16,590 (factor inverted)

Reference: FE Reference Handbook — Engineering Economics → G��Uniform gradient amount per interest period

Example 10

Present worth of an arithmetic gradient — G��Uniform gradient amount per interest period (4)

Maintenance is $28,000 in year 1 and grows by $5,000 each year for 6 years. At 8.5%, what is the present worth?

Given

A₁ = $28,000
G = $5,000/yr
n = 6
i = 8.5%

Find

Total present worth

Start with the thinking

Split the cash flow into a uniform base plus a gradient.
The gradient factor starts in year 2 by definition.

Step-by-step solution

Uniform part
$(P / A, i, n) = [(1 + i)^{n} - 1] / [i (1 + i)^{n}]$
Evaluate
$(P / A) = 4.5536$
Gradient part
$(P / G, i, n) = (1/ i) [(P / A) - n / (1 + i)^{n}]$
Evaluate
$(P / G) = 10.3049$
Combine — PW = $28,000(4.5536) + $5,000(10.3049) = $179,025

Answer: PW ≈ $179,025

Why the other options are there

$127,500 (gradient omitted)
$264,108 (gradient treated as uniform)

Reference: FE Reference Handbook — Engineering Economics → G��Uniform gradient amount per interest period

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given an alternative being compared over a study period, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

G��Uniform gradient amount per interest period contains 0 relations; you must be able to find this page in under 15 seconds.
Exam style: one cash-flow diagram converted with one factor.
Unit rule: i per period must match n periods.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

i per period must match n periods
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.