Systematic Economic Analysis Technique for analysis

1
John D. Lee
Industrial Engineering
Engineering Economy Review

2
Main concepts
 Models are approximations of reality (THINK)
 Time value of money, cash flow diagrams, and
equivalence
 Comparison of alternatives
 Depreciation, inflation, and interest rates

3
Suggestions for solving problems
 Lookup unfamiliar terms in the index
 Draw cash flow diagrams
 Identify P, A, F, i
 Be flexible in using equations and tables
 Check with alternate methods

4
Cash flows
 Cash flows describe income and outflow of money
over time
 Disbursements =outflows “-”
 Receipts =inflows “+”
 Beginning of first year is traditionally defined as
“Time 0”

5
Equivalence
 Translating cashflows over time into common units
 Present values of future payments
 Future value of present payments
 Present value of continuous uniform payments
 Continuous payments equivalent to present
payment

6
Single Payment Compound Interest
 P= (P)resent sum of money
 i= (i)nterest per time period (usually years)
 MARR=Minimal Acceptable Rate of Return
 n= (n)umber of time periods (usually years)
 F= (F)uture sum of money that is equivalent to P
given an interest rate i for n periods
 F=P(1+i)n P=F(1+i)-n
 F=P(F/P,i,n) P=F(P/F,i,n)

7
Bank example
 You 1000 deposit
 12% per year
 5 years
 How much do you have at end if compounded
yearly?
 How much do you have at end if compounded
monthly?

8
5.47 Income from savings
 $25,000 deposited
 Account pays 5% compounded semiannually
 Withdrawals in one year and continuing forever
 Maximum equal annual withdrawal equals?

9
5.47 Capitalized cost problem
 P=25,000
 A=?
 r=5%
 i=?
 A=iP
0 1 2
25,000
3

1
0
Key points to remember
 Time value of money
• $1000 today is not the same as $1000 one hundred years from
now
• Money in the future is worth less than present sums
 Cash flow diagrams
• Starts at year zero
• Superposition to convert to standard forms
 Equivalence
• Functional notation, F=P(F/P,i,n)
• i and n must match units
• Capitalized cost, A=Pi, P=A/i

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1
Comparison of alternatives
 Present/Future worth
 Cash flow
 Rate of return
 Cost benefit
 Payback period
 Breakeven analysis

1
2
Present/Future worth
 Determine time period for analysis, least common
multiple
 Calculate present value for each alternative
• Draw cashflow diagram
• Identify/calculate A, i, P, F, n
• Use present value equations to determine P
 Compare costs

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3
Tomato peeling machines
Machine A Machine B
Purchase cost=$52,000 $63,000
Annual cost=$15,000/year $9,000/year
Annual benefit= $38,000/year $31,000 /year
Salvage value= $13,000 $19,000
Useful life= 4 years 6 years

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4
Present cost of A
0 1 2 3
52,000
15,000
13,000
P4= -52000+(38,000-15,000)(P/A,12%,4)+13,000(P/F,12%,4)
P12= P4+ P4(P/F,12%,4) + P4(P/F,12%,8)
P12 =$53,255
52,000 52,000
15,000 15,000
5 6 7
15,000 15,000 15,000
38,000 38,000 38,000 38,000 38,000 38,00013,000
9 10 11
15,000 15,000 15,000
13,000
12
38,000 38,000 38,000
P4 P4 P4 A=38,000-15,000
i= MARR=12%
n=4
F=13,000

1
5
Present cost of B
0 1 2 3
63,000
9,000
P6= -63000+(31,000-9,000)(P/A,12%,6)+19,000(P/F,12%,6)
P12= P6+ P6(P/F,12%,6)
P12 =$55,846
63,000
9,000 9,000
5 6 7
9,000 9,000 9,000
31,000 31,000 31,000 31,000
9 10 11
9,000 9,000 9,000
13,000
12
9,000
31,000
19,000
31,000 31,000 31,000 31,000
31,000
4
9,000
P6 P6
A=31,000-9,000
i= MARR=12%
n=6
F=19,000

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6
Cash flow analysis
 Determine time period for analysis: common
multiple OR continuing operation then doesn’t
require least common multiple
 Calculate annual cost/benefit/profit for each
alternative
• Draw cashflow diagram
• Identify/calculate A, S, i, P, F, n
• Use uniform payment equations to determine A
 Compare annual costs

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7
Cash flow analysis
 Provides a shortcut for long/infinite analysis
periods or when least common multiple might be a
long time period with lots of calculations
 Compare on the basis of annual cost if EITHER
• Common multiple (e.g., 2 years and 8 years)
OR
• Continuing operation (e.g., business will keep
operating indefinitely with ability to replace equipment)

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Rate of return analysis
 Draw cash flow diagram of each alternative
 Draw combined cash flow diagram
(higher initial cost- lower initial cost)
 Convert to Present worth and Present costs
OR
Convert to EUAB and EUAC
 Write equation
 Solve for i
 If RORMARR, choose higher-cost alternative

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9
7-52: Purchase vs. Lease
 Purchase machine:
 $12,000 initial cost
 $1,200 salvage value
 Lease machine
 $3,000 annual payment
 15% MARR, 8 year useful life

2
0
n=8
Purchase-Lease
3000
4200
PW of Benefits-PW of Costs=0
3000(P/A,i,7)+4200(P/F,i,8)-12,000= 0
i=17% 3000(3.922)+4200(0.2848) -12,000= 962
i=18% 3000(3.812)+4200(0.2660) -12,000= 553
i=20% 3000(3.605)+4200(0.2326) -12,000= -208

2
1
Interest rate (%)
i *
NPW
+
-
553
18% 20%
-208
i=18%+2%(553/761)
i=19.45%
13,200
-12,000

2
2
 Internal rate of return =17.6%
 17.6%>15% therefore choose purchase option

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3
Evaluation of multiple alternatives
 Identify all alternatives
 Compute rate of return of all alternatives
• Delete alternatives with a return< MARR
 Arrange remaining alternatives in order of increasing
investment (find alternative where investing component
dominates)
 Evaluate first two alternatives
 Repeat evaluation until all alternatives have been
evaluated

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4
Repeated evaluation of alternatives
A
B
C
D
Multiple comparisons of return on incremental investment

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5
General suggestions
 Think about alternatives
• i<0
• i=0
• A=Pi when salvage value equals initial cost
• P=Ai = Capitalized cost
• Infinite analysis period EUAB-EUAC=NPWi
 Consider using Present Worth AND EUAB to
frame rate of return calculation

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6
Payback period analysis
 Approximate rather than exact calculation
 All costs and profits are included without
considering their timing
 Economic consequence beyond payback period
are ignored (salvage value, gradient cash flow)
 May select a different alternative than other
methods
 Focus is speed versus efficiency

2
7
Benefit cost ratio
 Benefit cost ratio analysis
• (PW of benefit/PW of cost  1)
• Compare incremental investment, similar to rate of
return analysis

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9.9 Three alternatives
A B C
 Initial cost 50 150 110
 AB first 28.8 39.6 39.6
 Useful life 2 6 4
 Rate of Return 10% 15% 16.4%
 Compare using MARR=12%
• Future worth
• Benefit cost
• Payback period

2
9
Future worth: Option C
39.6
110
F
F=-110(F/P,12,4)
-110(F/P,12,8)
-110(F/P,12,12)
+39.6(F/A,12,12)
F=81.61

3
0
Future worth analysis
A B C
 AB first 28.8 39.6 39.6
 Future worth -18.94 75.17 81.61
• Benefit costs
• Payback period

3
1
Benefit-cost ratio analysis
Year C A C-A
0 -110 -50 -60
1 39.6 28.8 10.8
2 39.6 28.8-50 60.8
3 39.6 28.8 10.8
4 39.6 28.8 10.8
Present worth of Cost=60
Present work of benefit=10.8(P/A,12,4)+50(P/F,12,2)
B/C=72.66/60>1
Reject A

3
2
Year B C B-C
0 -150 -110 -60
1-4 39.6 39.6 0
4 0 -110 110
5-6 39.6 39.6 0
6 -150 0 -150
7-8 39.6 39.6 0
8 0 -110 110
9-12 39.6 39.6 0

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3
 PW of cost=40+150(P/F,12%,6)
 PW of cost=115.99
 PW of benefits= 110(P/F,12%,4)+110(P/F,12%,8)
 PW of benefits=114.33
 B/C=114.33/115.99<1
 Reject B

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4
Payback period
 A 50/28.8 = 1.74 years
 B 150/39.6= 3.79 years
 C 110/39.6= 2.78 years
 Select A

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5
Summary
A B C
 AB first 28.8 39.6 39.6
 Future worth -18.94 75.17 81.61
 Benefit cost C-A=1.21 B-C=0.98
 Payback period 1.74 3.79 2.78

3
6
Motor comparison
Graybar Blueball
 Initial cost $7,000 $6,000
 Efficiency 89% 85%
 Maintenance 300/year 300/year
 Electricity cost $0.072/kW-hour
 200 hp
 20 year useful life, No salvage value
 Interest rate =10%
 Hours used to justify expense

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7
Motor comparison
 Graybar-Blueball>0
 NPC of Graybar-Blueball=
1000+(300-300)+
(P/A,10%,20)200*0.746kW/hp*0.072$/kWhr*HRS(1/0
.89)-
(P/A,10%,20)200*0.746kW/hp*0.072$/kWhr*HRS(1/0.85)
 1000= 8.514*0.568*HRS
 206.7 hrs

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8
Key points to remember
 Present/Future worth
• Use least common multiple
 Cash flow
• Useful for infinite analysis periods
 Rate of return
• Do not use rate of return, but incremental rate of return as criterion
• Set up cash flow as investment
 Cost benefits
• Use incremental comparison similar to rate of return analysis
 Payback period
• Approximate method that makes huge assumptions
 Breakeven analysis

3
9
Interest rates, depreciation, and inflation
 Concepts that allow more precise modeling of
economic decisions
 Nominal vs effective
 Depreciation
• Straight line
• MACRS (Modified Accelerated Cost Recovery System)
• Book value
 Inflation moderates value of rate of returns

4
0
Nominal and effective interest rates
Effective interest rate, iP, (period of
compounding=period of interest) is used in formulas:
i=iP=(1+ is)m-1
i=iP=(1+rP/m)m-1
is=interest per subperiod
m=number of subperiods in period P
rP=nominal interest per period P
Nominal interest rate, rP=m X is
Continuous compounding: ia=er -1
F = P(1+ ia ) n = P* ern

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1
Depreciation
 Depreciation basis=
Initial cost(C)- Salvage value (S)
 Book value = C-Accumulated depreciation
 Straight line depreciation
• Di=(C-S)/n
• n= service life
 MACRS
• Di =C X Factor from table

4
2
Methods for depreciation
 Book value=cost-depreciation charges
 Straight line (SL)
• Same amount each year
• Proportional to years of useful life and (initial cost-
salvage)
 Sum-of-years (SOYD)
• Initial rate is faster than SL
• Proportional to sum of digits in useful life and (initial
cost-salvage)

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3
Methods for depreciation
 Declining balance, double declining balance (DDB)
• Double declining = 200% of straight line
• Proportional to years of useful life and book value
• Salvage value not considered
 Declining balance/conversion to straight line
(DDB/SL)
• Optimal switch year CANNOT be determined from a
simple comparison of depreciation schedules
 Unit of production (UOP)
 Modified Accelerated Cost Recovery System
(MARCS)

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4
Depreciation calculations
Method Annual Book value (year J)
Straight line (P-S)/N P- (P-S) J/N
SOYD (P-S)[(N-J+1)/(N(N+1)/2)] P-sum of dep.
DDB 2(Book value)/N
2P/N(1-2/N)j-1 P-P(1-(1-2/N) j )
UOP (P-S)Prod. in year/Total prod. P-sum of dep.
MARCS Table lookup (Property class, year) P-sum of dep.

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5
Depreciation of machine
 Initial cost of $50,000
 Salvage value of $10,000
 Service life of 10 years
 Straight line depreciation=
 dn=(P-S)/N
 dn =(50,000-10,000)/10
 dn =4,000/year

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6
10.3 Capsulating machine
 Initial cost= $76,000
 Five year useful life
 No salvage value
 Find depreciation schedule
• Straight line
• Sum of years digits
• Double declining balance
• DDB with conversion

4
7
10.3 Straight line
Year Dep/year Cumulative Dep
P-S/N
0 0 0
1 76,000/5=15,200 15,200
2 15,200 30,400
3 15,200 45,600
4 15,200 60,800
5 15,200 76,000

4
8
10.3 Sum of year digits
(P-S)[(N-J+1)/(N(N+1)/2)]
0 0 0
1 76,000(5)/15=25,333 25,333
2 20,267 45,600
3 15,200 60,800
4 10,133 70,933
5 5,067 76,000

4
9
10.3 Double declining balance
2P/N(1-2/N)j-1 OR
2/N(Cost-cumulative dep)
0 0 0
1 76,000(2/5)=30,400 30,400
2 (76,000-30,400)(2/5)=18,240 48,640
3 10,944 59,584
4 6,566 66,150
5 3,940 70,090

5
0
10.3 Summary of depreciation schedules
Year SL SOYD DDB
1 15,200 25,333 30,400
2 15,200 20,267 18,240
3 15,200 15,200 10,944
4 15,200 10,133 6,566
5 15,200 5,067 3,940
What is best year to switch from DDB to SL depreciation?

5
1
Straight line depreciation if DDB has been
used in previous years
 Book value in year three for DDB =
 76,000 - 30,400 - 18,240 = 27,360
 SL depreciation = Book value/ remaining useful life
Switch year BV SL dep
3 27,360 27,360/3= 9,120<10,940 from DDB
4 16,416 8,208> 6,566
5 9,850 9,850

5
2
Inflation
 Interest rate adjusted for computing present worth
and other values
 Increases the value of the MARR to account for
loss in value of future dollars due to inflation
 Inflation adjusted interest rate = i + f + if
 f= rate of inflation

5
3
13.33 Value of a 10,000 investment
 Interest rate 10%
 General price inflation is projected to be:
• 3% for next 5 years
• 5% for five years after that
• 8% for following five years
 Calculate future worth of investment:
• in terms of actual dollars
• in terms of real dollars at that time
• real growth in purchasing power

5
4
 A) Future value of actual $
=10,000 (F/P,10%,15)= $41,770
 B) Future value in real $, constant value
=41,770 (P/F,8%,5)(P/F,5%,5)(P/F,3%,5)
0.6806 0.7835 0.8626
=19,213
 C) Real growth rate of investment
=19213=10,000(1+i)15 =4.45%
13.33 Value of a 10,000 investment

5
5
Alternate solution solving for real dollars
 Use real rather than market interest rate
 Real interest rates; i’=(i - f)/(1+f)
• First five years: 6.796%
• Second five years: 4.762%
• Third five years: 1.9608%
 Real dollar value in 15 years
• 10,000*(1.06796)5 *(1.04762)5 *(1.019608)5
• 19,318

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6
13.30 Comparison of alternatives with
inflation
 3 year lives with no salvage value
 Inflation = 5%
 Income tax rate of 25%
 Straight line depreciation
 MARR=7%
 Using rate of return analysis which is preferable?

5
7
13.30 Cash flow
Year A B
0 -420 -300
1 200 150
2 200 150
3 200 150

5
8
Cash flow for option A
Year A Actual Dep Tax Inc Tax ATCF ATCF Y0$
0 -420 -420 -420 -420
1 200 210 140 70 -17.5 192.5 183.3
2 200 220.5 140 80.5 -20.1 200.4 181.8
3 200 231.5 140 91.5 -22.9 208.6 180.2

5
9
Cash flow for option B
Year A Actual Dep Tax Inc Tax ATCF ATCF Y0$
0 -300 -300 -300 -300
1 150 157.5 100 57.5 -14.4 143.1 136.3
2 150 165.4 100 65.4 -16.4 149.0 135.1
3 150 173.6 100 73.6 -18.4 155.2 134.1

6
0
Incremental ROR analysis A-B
Year A B Y0$ A-B
0 -420 -300 -120
1 183.3 136.3 47
2 181.8 135.1 46.7
3 180.2 134.1 46.1
Guessing 7%
NPW = -120 + 47(P/F,7%,1) +46.7(P/F,7%,2) + 46.1(P/F,7%,3)
= 2.3, therefor ROR >7% choose more expensive alternative

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