mme 4272 sem 2 2014-2015

INVESTMENT ANALYSIS 1

MME 4272 Engineering Management Part II

INTERNATIONAL ISLAMIC UNIVERSITY MALAYSIA

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Category of Projects
To help formulate alternatives, a project is categorized as one of the following:Mutually exclusive: Only one of the viable projects can be selected by the economic analysisIndependent: More than one viable project may be selected by the economic analysis
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Tools for Evaluating Alternatives
Various tools/methods to evaluate alternatives economically (economic equivalence) using the factors learned Purpose:Compare mutually exclusive alternativesBasis: present worth, future worth & annual worth analysis
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Tools for Evaluating alternatives
Present Worth AnalysisFormulating Mutually Exclusive AlternativesPresent Worth Analysis of Equal-life AlternativesPresent worth Analysis of DifferentLife AlternativesFuture Worth AnalysisPayback Period AnalysisAnnual Worth AnalysisRate of Return AnalysisBenefit/Cost Ratio Analysis
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PW Analysis of Equal-Life Alternatives
One alternative: Calculate PW at the MARRIf PW 0, the requested MARR is met or exceeded (alternative is financially viable)Two or more alternatives: Calculate the PW of each alternativeSelect alternative with largest PW value (alternative with less negative or more positive)
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If the projects are independent, the selection guideline is as follows:

For one or more independent projects, select all projects with PW 0 at the MARR

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PW1PW2Selected alternative$ -1500$ -5002 -500+10002+2500-5001+2500+15001

Perform a present worth analysis of equal-service machine with costs shown below, if the MARR is 10% per year. Revenue for all the alternatives are expected to be the same
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Electricpowered Gas poweredSolar poweredFirst cost, $Annual operating cost (AOC), $Salvage value S, $Life, years- 2500 - 900 200 5- 1500 - 700 350 5- 6000 - 50 100 5

These are service alternatives: The PW of each machine is calculated at i = 10% for n = 5 years
PWE = -2500 - 900(P/A,10%,5) + 200(P/F,10%,5)= $-5788

PWG = -3500 - 700(P/A,10%,5) + 350(P/F,10%,5)= $-5936

PWS = -6000 - 50(P/A,10%,5) + 100(P/F,10%,5)= $-6127

[See the calculations in excel file]

The electric-powered machine is selected since the PW of its costs is the lowest (has numerically the largest PW value)

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PW Analysis of Different-Life Alternatives
A project engineer with EnvironCare is assigned to start up a new office in a city where a 6-year contract has been finalized to take and to analyze ozone-level readings. Two lease options are available, each with a first cost, annual lease cost, and deposit-return estimates as shown below:
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Location ALocation BFirst cost, $Annual lease cost, $ per yearDeposit return, $Lease term, years- 15,000 -3,500 1,000 6- 18,000 -3,100 2,000 9

Determine which lease option should be selected on the basis of a PW comparison, if MARR is 15% per year

EnvironCare has a standard practice of evaluating all projects over a 5-year period. If a study period of 5 years is used & the deposit returns are not expected to change, which location should be used?

Determine location over a 6-year study period if deposit return at location B is estimated to be $6000 after 6 years

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PW Analysis of Different-Life Alternatives

Break-Even Point
Breakeven AnalysisSingle-product CaseMulti-product CaseAssumptionsGraphical and Algebraic ApproachDetermining BEP for single and multi-product cases
[Reference: Operations Management, Heizer & Render, 8th ed (p-287)]

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Break-Even Analysis
A critical tool for determining capacity a facility must have to achieve profitabilityObjective is to find the point in dollars or ringgits & units at which, cost equals revenueRequires estimation of fixed costs, variable costs, & revenueFixed costs are costs that continue even if no units are produced (depreciation, taxes, debt, mortgage payments)Variable costs are costs that vary with the volume of units produced (labor, materials, portion of utilities)
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Break-Even Analysis
Variable costs are costs that vary with the volume of units produced (labor, materials, portion of utilities)Contribution is the difference between selling price and variable costAssumptions:Costs and revenue are linear functions (In reality, the case is not so)There is no time value of money
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Break-Even Analysis

Profit corridor

Loss corridor

Total revenue line

Total cost line

Variable cost

Fixed cost

Break-even point

Total cost = Total revenue

900

800

700

600

500

400

300

200

100

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010020030040050060070080090010001100

Cost in dollars

Volume (units per period)

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Break-Even Analysis

TR = TC

or

Px = F + Vx

Break-even point occurs when

Profit= TR - TC

= Px - (F + Vx)

= Px - F - Vx

= (P - V)x - F

BEPx=Break-even point in units

BEP$=Break-even point in dollars

P=Price per unit (after all discounts)

x=Number of units produced

TR=Total revenue = Px

F=Fixed costs

V=Variable costs per unit

TC=Total costs = F + Vx

BEPx =

F

P - V

F

(P - V)/P

F

P - V

F

1 - V/P

BEP$= BEPx P

= . P

=

=

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Fixed costs = $10,000 Material = $0.75/unit

Direct labor = $1.50/unit Selling price = $4.00 per unit

Break-Even Analysis

F

1 - (V/P)

$10,000

1 - [(1.50 + 0.75)/(4.00)]

BEP$ = =

$10,000

1 - [(1.50 + 0.75)/(4.00)]

F

1 - (V/P)

$10,000

.4375

= = $22,857.14

F

P - V

$10,000

4.00 - (1.50 + 0.75)

BEPx = =

= 5,714

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Break-Even Example

50,000

40,000

30,000

20,000

10,000

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02,0004,0006,0008,00010,000

Dollars

Units

Fixed costs

Total costs

Revenue

Break-even point

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Break-Even Example

Multi-product Case

whereV= variable cost per unit

P= price per unit

F= fixed costs

W= percent each product is of total dollar sales

i= each product

Fixed costs = $3,500 per month

Annual Forecasted

ItemPriceCostSales Units

Sandwich$2.95$1.257,000

Soft drink.80.307,000

Baked potato1.55.475,000

Tea.75.255,000

Salad bar2.851.003,000

Vi

Pi

1 - x (Wi)

F

BEP$ =

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Multiproduct BEP Example

Sandwich$2.95$1.25.42.58$20,650.446.259

Soft drink.80.30.38.625,600.121.075

Baked 1.55.47.30.707,750.167.117

potato

Tea.75.25.33.673,750.081.054

Salad bar2.851.00.35.658,550.185.120

$46,3001.000.625

AnnualWeighted

SellingVariableForecasted% ofContribution

Item (i)Price (P)Cost (V)(V/P)1 - (V/P)Sales $Sales(col 5 x col 7)

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Multiproduct Example

1 - x (Wi)

Vi

Pi

F

BEP$ =

$3,500 x 12

.625

= = $67,200

$67,200

312 days

= = $215.38

Daily sales

.446 x $215.38

$2.95

= 32.6 33

sandwiches

per day

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Problems for practice (to be solved in the class)
Given the following data, calculate BEP(x), BEP ($), and the profit at 100,000 units:
P= $8/unit, V = $4/unit and F =$50,000.

(2) A prolific author is considering starting her own publishing company. She will call it DSI Publishing, Inc. DSIs estimated costs are

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Fixed$250,000.00

Variable cost per book $20.00

Selling price per book $30.00

How many books must DSI sell to break even? What is its break-even point in dollars?

Depreciation - Terminology
Depreciation - reduction in value of an asset (usually annual depreciation is tax deductible - it is subtracted from income when calculating amount of taxes due each yearFirst cost (or unadjusted basis) - the delivered & installed cost of the asset including purchase priceBook value (BV)represents the remaining (undepreciated) capital investment in the books after total amount of depreciation charges to date have been subtracted from the basis
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Depreciation - Terminology
Book value (BV) - usually determined at the end of each year; consistent with the end-of-year conventionRecovery period (n)- the depreciable life, n of the asset in yearsSalvage value (S)- the estimated trade-in or market value at the end of the assets useful lifeDepreciation rate or recovery rate (dt)- the fraction of first cost removed by depreciation each year (may be same for straight line method or different for each year of the recovery period)
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Depreciation Methods
Straight Line (SL) DepreciationBook value decreases linearly with timeDepreciation rate (dt = 1/n = d)is the same each year of the recovery period, nConsidered the standard against which any depreciation model is comparedThe annual SL depreciation is determined by multiplying the first cost minus salvage value by d
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Straight Line (SL) Depreciation
In equation form,
where,

t = year (t=1,2,n)

Dt= annual depreciation charge

B = first cost or unadjusted basis

S = estimated salvage value

n = recovery period

d = depreciation rate = 1/n

Book value

time

B

S

t

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Since the asset is depreciated by the same amount each year, the book value after t years of service (BVt) will be equal to the first cost B minus the annual depreciation times t
BVt = B tDt
Earlier we defined dt as a depreciation rate for a specific year t. However, the SL model has the same rate for all years, that is
d = dt = 1/n
Excel function is SLN(B,S,n)

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If an asset has a first cost of $50,000 with a $10,000 estimated salvage value after 5 years, (a) calculate the annual depreciation and (b) compute and plot the book value of the asset after each year, using straight line (SL) depreciation

Solution

(a) The depreciation each year for 5 years can be found by

= (50,000 10,000)/5 = $8000

Enter SLN(50000,10000,5) in any cell to get Dt of $8000

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The BVt values are plotted for years 1 and 5 as
Year t

BV,x $1000


(b) The book value after each year t is computed by using equation

BVt = B tDt.

BV1 = 50,000 1(8,000) = $42,000

BV5 = 50,000 5(8,000) = $10,000 = S

50

40

30

20

10

1

2

3

4

5

0

Dt=$8,000

S=$10,000

Declining Balance (DB) Depreciation
Declining balance method is commonly applied (also known as the fixed percentage or uniform percentage method)Accelerates write-off of asset value because annual depreciation is determined by multiplying the book value at the beginning of a year by a fixed (uniform) percentage, dIf d = 0.1 then 10% of the book value is removed each year so, the depreciation amount decreases each year
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Double Declining Balance (DDB) Depreciation
Maximum annual depreciation rate for DB method can be twice the straight line rate, that is dmax = 2/nA special title called double declining balance (DDB) methodIf n = 10 years, the DDB rate is 2/10 = 0.2 of the book value is removed each yearThe depreciation for year, t is the fixed rate d times the book value at the end of the previous year, Dt = (d)BVt-1
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DB and DDB Depreciation
The actual depreciation rate for each year t, relative to the first cost B, is dt = d(1-d)t-1If BVt-1 is not known, the depreciation in year t can be calculated using B and d, i.e Dt = dB(1-d)t-1Book value in year t is determined in one of two ways:
by using the rate d and the first cost B or

by subtracting current depreciation charge from the previous book value

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The equations are
BVt = B(1-d)t and BVt = BVt-1 Dt
Note that the book value for DB method never goes to zero. Why?The implied salvage value after n years is BVn amount. Thus, implied salvage value = Implied S = BVn = B(1-d)nIf a salvage value is estimated for the asset, this estimated S value is not used in the DB or DDB method to calculate annual depreciation
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However, if the implied S < estimated S, it is correct to stop charging further depreciationSo stop depreciating when the book value is at or below the estimated salvage valueIn most cases, the estimated S is in the range of zero to the implied S valueIf the fixed percentage d is not stated, it is possible to determine the implied fixed rate using the estimated S value, if S > 0
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The range for d is 0 < d < 2/nImplied d = 1- (S/B)1/n [as we know, Implied S = BVn = B(1-d)n]
A fiber optics testing device is to be DDB depreciated. It has a first cost of $25,000 and an estimated salvage value of $2500 after 12 years

Calculate the depreciation and book value for years 1 and 4

Calculate the implied salvage value after 12 years

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SolutionThe DDB fixed depreciation rate is d = 2/n = 2/12 = 0.1667 per year.
So, for year 1,

D1 = dB(1-d)1-1 =(0.1667)(25000)(1-0.1667)1-1

= $4167

BV1 = B(1-d)1 = 25,000(1-0.1667)1 = $20,833

For year 4,

D4 = dB(1-d)4-1 =(0.1667)(25000)(1-0.1667)4-1

= $2411

BV4 = B(1-d)4 = 25,000(1- 0.1667)4 = $12,054

Declining Balance (DB) and Double Declining Balance (DDB) Depreciation

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Solution (contd)DDB function for D1 and D4 are respectively, DDB(25000,2500,12,1) and DDB(25000,2500,12,4)
Assignment: Check with Excel

(b)The implied salvage value after 12 years is

Implied S = B(1-d)n = 25,000(1-0.1667)12

= $2803

Declining Balance (DB) and Double Declining Balance (DDB) Depreciation

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Another Example
A mining company has purchased a computer-controlled gold ore grading unit for $80,000, with an anticipated life of 10 years & a salvage value of $10,000. Use DB & DDB methods to compare the schedule of depreciation & book values for each year. An implied DB depreciation rate is
d = 1 (S/B)1/n = 1 (10,000/80,000)1/10

= 0.1877

Note: 0.1877 < 2/n = 0.2. So the DB model does not exceed twice the straight line rate

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Example 16.3:Schedule of Depreciation & Book values
Year tDeclining BalanceDouble Declining Dt BVt Dt BVt0--$80,000--$80,0001$15,016 64,084$16,000 64,0002 12,197 52,787 12,800 51,2003 9,908 42,879 10,240 40,9604 8,048 34,831 8,192 32,7685 6,538 28,293 6,554 26,2146 5,311 22,982 5,243 20,9727 4,314 18,668 4,194 16,7778 3,504 15,164 3,355 13,4229 2,846 12,318 2,684 10,73710 2,318 10,000 737 10,000

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Another problem
Equipment for immersion cooling of electronic components has an installed value of $182,000 with an estimated trade-in value of $50,000 after 18 yrs. Determine annual depreciation charge using DDB & DB for years 2 &18.Ans:We need to determine the implied salvage value to verify whether the implied S is less than the estimated S (50,000)For DDB, d = 2/n = 2/18 = 0.1111Implied S = Book value after useful life = B(1- d)n
= 182,000(1- 0.1111)18 = $21,848

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Using DDB
It is found that implied S (21,848) < estimated S (50,000)It will be correct to stop charging depreciation beyond $50,000For DDB, depreciation charge, Dt = dB(1-d)t-1For this case, d = 2/n = 2/18 = 0.1111Depreciation charge for year 2, D2=(0.1111)(182,000)(1-0.1111)2-1 =$17,973.74For year 18, D18=(0.1111)(182,000)(1-0.1111)18-1 =$2,730.77So since after 18 years the implied salvage value is lower than the estimated salvage value, D18 should not be charged

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Using DB
For DB, d = 1/n = 1/18 = 0.05555 or We can also use d = 1- (S/B)1/n
= 1- (50000/182000)1/18 = 0.0693
Implied S = B(1-d)n = 182,000(1-0.05555)18
= $65,057

Since the implied S (65,057) > estimated S (50,000), it will be correct to continue charging further depreciation

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Using DB
Given, B=182,000S=50,000For DB, we know Dt = dB(1-d)t-1Now d = 1/n = 1/18 = 0.055555Depreciation charge for year 2, D2=(0.05555)(182,000)(1-0.05555)2-1 =$9,549Similarly, D18=(0.05555)(182,000)(1-0.05555)18-1 =$3,827It should also be correct if someone answers by using implied d = 0.0693

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Example
A copy machine is purchased for $4,217.75. The expected life is 5 years. Use double declining balance to calculate the depreciation. The salvage value of the copy machine is $500.00.

Bibliography

Chang, C. M. (2005). Engineering Management Challenges in the New Millennium, Upper Saddle River, NJ: Prentice Hall

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IQ + EQ + SQ = TQ

The End

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mme 4272 sem 2 2014-2015

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