engineering economy

Example 6.1 : Tax

REI sells outdoor equipment and sporting goods through retail outlets, the internet
and catalogs. Assume that for 1 year REI has the following financial results in the
state of Kentucky, USA which has a flat tax rate 6% on corporate taxable income.

Total revenue $19.9 million

Operating expenses $8.6 million

Depreciation and other allowed deductions $1.8 million

(a) Determine the state taxes and federal taxes using table rates

(b) Find the everage federal tax rate paid of the year.

Solution
(a) Kentucky state TI = GI – OE – D = 19.9M – 8.6M – 1.8M = $9.5M

(b) Kentucky state taxes = (0.06)(9,500,000) = $570,000
Federal TI = GI – OE – D – state taxes = 9,500,000 – 570,000

= $8,930,000
Federal taxes = 113, 900 + 0.34(8,930,000 – 335,000) = $3,036,200

Total federal and state taxes = 3,036,200+570,000=3,606,200

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DIY

Workman Tools reported a TI of $90,000 last year. If the state income tax rate is
7%, determine the ( a ) average federal tax rate, ( b ) overall effective tax rate, ( c )
total taxes to be paid based on the effective tax rate, and ( d ) total taxes paid to the
state and paid to the federal government.

Ans

Two companies have the following values on their annual tax returns.

Company A Company B

Sales revenue 1,500,000 820,000

Interest revenue 31,000 25,000

Expenses -754,000 -591,000

Depreciation 48,000 18,000

(a) Calculate the federal income tax for the year for each company.

(b) Determine the percent of sales revenue each company will pay in federal income
tax.

(c) Estimate the taxes using an effective rate of34% of the entire TI.

Determine the percentage error made relative to the exact taxes in

part ( a ).

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6.2 After Tax Analysis
CFBT = gross income – operating expenses – initial investment +salvage value

= GI –OE – P +S [6.7]

CFAT = CFBT – taxes [6.8]

CFAT = GI –OE – P +S – (GI – OE –D)(Te) [6.9]

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Year Gross Operating Investment CFBT Depreciation Taxable Taxes CFAT
Income Expenses and D Income
(4)=
GI OE Salvage (1)+(2)+(3) TI

(1) (2) P and S (5) (6)=(1)+ (T)= (8)=
(2) –(5) Te(6) (4)-(7)
(3)

Depreciation – the monetary value of an asset decrease over time due to use, wear and

tear or obsolences, ex, machinery or equipmennt.

• Straight line depreciation
Straight line depreciation derive its name from the fact that book value
decreases linearly with time. The depreciation rate is the same (1/n) each year
of the recovery period n
= ( − )
where = ( = 1,2, … . )
Dt = annual depreciation charge
B = first cost or unadjusted basis
S = estimated salvage value
n = recovery period
dt = depreciation rate = 1/n

• MACRS

Determine annual depreciation using the relation, Dt = dtB, where the rate can
be found from the table.

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Example 6.2 : After Tax Analysis

ABC company with an effective income tax rate and a capital gains tax of 40% and
a MARR at 12% must choose from 2 mutual exclusive project as shown in table 2.

Alternative X Alternative Y
$33,000
Initial cost $11,000 9,000
MACRS
Uniform annual benefit 3,000 3 years
0
Depreciation method Straight line
5 years
Depreciable life 3 years 2,000

IRS approved salvage value for 2,000
depreciation purpose

Useful life 5 years

Actual market value at end value of 2000
life

Solution

Alternative X

EOY BTCF DEP TI T ATCF PW
0 -11,000 - - -
1 3,000 3,000 0 - -11,000 -11,000
2 3,000 3,000 0 -
3 3,000 3,000 0 - 3,000 2679
4 3,000 - 3000 -1200
5 3,000 - 3000 -1200 3,000 2392
5 2,000 - 0 0
3,000 2135

1800 1144

1800 1021

2000 1135

PW -494

• Book value = 2000

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Alternative Y

EOY BTCF DEP TI T ATCF PW
0 -33,000 -- - -33,000 -33,000
1 9,000
10998.9 -1998.9 -799.56 9799.56 8751

2 9,000 14668.5 -5668.5 -2267.4 11267.4 8982

3 9,000 4913.7 4086.3 -1634.52 7366 5243

4 9,000 2445.3 6554.7 -2621.88 6378 4053
5 9,000 0 9000
5 2,000 - 2000 -3600 5400 3064

-800 1200 681

PW -2,227

Book value =0
Select alternative x to low cost

DIY

A corporation uses the following; before tax MARR of 14% per year, after tax
MARR of 7% per year and Te of 50%. Two machines have the following
estimates.

First cost Machine A Machine B
Salvage value -15,000 -22,000
AOC, $ per year 3,000 5,000
-3,000 -1,500

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Life, years 10 10

The machine is refor tained in use for a total of 10 years, then sold for the
estimated salvage value. Select one machine under the following conditions;

a) Before tax PW analysis
b) After- tax PW analysis, using classical SL depreciation over 10 years life.

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Chapter 7
Breakeven Analysis
7.1 Breakeven Analysis for a Single Project
Breakeven analysis finds the value of a parameter that makes two elements equal. The
breakeven point QBE is determined from mathematical relations,e.g product revenue
and costs or material supply and demand or other parameterst hat involve the parameter
Q. Breakeven analysis is fundamental to evaluations such as make-buy decisions.
Figure 7.1 presents different shapes of a revenue relation identified as R. A linear
revenue relation is commonly assumed, but a nonlinear relation is often more realistic.

Figure 7.1 : Graph linear and nonlinear revenue relation
Cost, which may be linear or nonlinear, usually two components – fixed and variable.
Fixed costs (FC). These include costs such as buildings, insurance, fixed overhead,
some minimum level of labor, equipment capital recovery and information systems.

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Variable costs (VC) . These include costs such as direct labor, materials, indirect costs,
contractors, marketing, advertisement and warranty.

If Q > QBE, there is predictable profit, but if Q < QBE , there is a loss.
Profit is calculated as: Profit = revenue – total cost = R – TC = R – (FC + VC)

A relation for the breakeven point may be derived when revenue and total cost are
linear functions of quantity Q by setting the relations for R and TC equal to each other,
indicating profit of zero.

R = TC
rQ = FC + vQ

where r = revenue per unit, v = variable cost per unit

Solve the breakeven for Q= QBE for linear R and TC functions


= −

Linear Nonlinear

Figure 7.2 : Effect of the breakeven point for linear and nonlinear when the variable
cost per unit reduced

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Example 7.1 : Breakeven analysis
Indira Industries is a major producer of diverter dampers used in the gas turbine
power industry to divert gas exhausts from the turbine to a side stack, thus
reducing the noise to acceptable levels for human environments. Normal
production level is 60 diverter systems per month, but due to significantly
improved economic conditions in Asia, production is at 72 per month. The
following information is available.
Fixed costs
Variable cost per unit Revenue per unit
FC = $2.4 million per month v =$35,000
r = $75,000
(a) How does the increased production level of 72 units per month compare with
the current breakeven point?
(b) What is the current profit level per month for the facility?
(c) What is the difference between the revenue and variable cost per damper that
is necessary
to break even at a significantly reduced monthly production level of
45 units, if fixed costs remain constant
Solution
(a) Use Equation to determine the breakeven number of units. All dollar amounts
are in $1000 units.

2400
= − = 75 − 35 = 60 ℎ

The breakeven value is 60 damper units. Theincreased production level of 72 units
is above the breakeven value.

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