Ch5: Present Worth Analysis-Part I
Ch5: Present, Future, &
Capitalized Worth Methods
1
Initial Project Screening
Method - Payback Period
Bank Loan vs. Investment Project
3
Presented by: Dr. Magdy Akladios
1
Ch5: Present Worth Analysis-Part I
Payback Period
 Principle:
How fast can I recover my initial investment?
 Method:
Based on the cumulative cash flow (or accounting
profit)
 Screening Guideline:
If the payback period is less than or equal to some
specified payback period, the project would be
considered for further analysis.
 Weakness:
Does not consider the time value of money
4
Example:
Describing Project Cash Flows
Year
(n)
Cash Inflows
(Benefits)
0
0
1
Cash Outflows
(Costs)
Net
Cash Flows
$650,000
-$650,000
215,500
53,000
162,500
2
215,500
53,000
162,500
…
…
…
…
8
215,500
53,000
162,500
5
Example

How long does it take to recover the initial investment
in this cash flow?
6
Presented by: Dr. Magdy Akladios
2
Ch5: Present Worth Analysis-Part I
Solution
Initial Cost
Uniform annual benefit
$650,000

$162,500
 4 years
Payback Period =
7
Example:
Payback Period
N
Cash Flow
0
1
2
3
4
5
6
Cum. Flow
-$85,000
-$50,000
-$5,000
$45,000
$95,000
$140,000
$175,000
-$105,000+$20,000
$35,000
$45,000
$50,000
$50,000
$45,000
$35,000
Payback period should occurs somewhere
between N = 2 and N = 3.
8
$45,000
$45,000
$35,000
Annual cash flow
$35,000
$25,000
$15,000
0
1
2
Years
3
4
5
4
5
6
Cumulative cash flow ($)
$85,000
150,000
3.2 years
Payback period
100,000
50,000
0
-50,000
-100,000
0
1
2
3
6
Years (n)
9
Presented by: Dr. Magdy Akladios
3
Ch5: Present Worth Analysis-Part I
Example





A company is investing $25,000 into a 5-year project
The project will produce a uniform annual revenue of
$8,000
The salvage value of the project is $5,000
The company has decided that they will accept all
projects returning a MARR of 20% or more/year.
Using the simple payback method, show if your
company should adopt this project
Solution
EOY
Cash flow
0
1
2
3
4
5
-$25,000
8,000
8,000
8,000
8,000
13,000
Cumulative PW
at i=0%
-$25,000
-17,000
-9,000
-1,000
+7,000
+$20,000
Year when –ve turned to +ve
Discounted Payback Period
 Principle:
How fast can I recover my initial investment plus
interest?
 Method:
Based on the cumulative discounted cash flow
 Screening Guideline:
If the discounted payback period (DPP) is less than
or equal to some specified payback period, the
project would be considered for further analysis.
 Weakness:
Cash flows occurring after DPP are ignored
12
Presented by: Dr. Magdy Akladios
4
Ch5: Present Worth Analysis-Part I
Discounted Payback Period Calculation
Period
Cash Flow
Cost of Funds
(15%)*
Cumulative
Cash Flow
0
-$85,000
0
-$85,000
1
15,000
2
25,000
-$82,750(0.15) = -12,413
-70,163
3
35,000
-$70,163(0.15) = -10,524
-45,687
4
45,000
-$45,687(0.15) =-6,853
-7,540
5
45,000
-$7,540(0.15) = -1,131
36,329
6
35,000
$36,329(0.15) = 5,449
76,778
-$85,000(0.15) = -$12,750 -$85,000 + (-)$12,750 +
15,000 = -82,750
* Cost of funds = (Unrecovered beginning balance) X (interest rate)
13
Illustration of Discounted Payback Period
14
Example





A company is investing $25,000 into a 5-year project
The project will produce a uniform annual revenue of
$8,000
The salvage value of the project is $5,000
The company has decided that they will accept all
projects returning a MARR of 20% or more/year.
Using the discounted payback method, show if your
company should adopt this project
Presented by: Dr. Magdy Akladios
5
Ch5: Present Worth Analysis-Part I
Solution (using an i of 20%
EOY Cash
0
-$25,000
1
8,000
2
8,000
3
8,000
4
8,000
5
8,000
Cumul. CF
-$25,000
-$25,000(1+i) + 8,000 = -$22,000
-$18,400
-$14,080
-$8,896
+$2,325
Year when –ve turned to +ve
NFW of this project
Another way is to find the F/P for one year for each transaction and follow the
transactions to the end
EXAMPLE




Capital Investment (I) = $10,000
Uniform annual revenue = $5,310
Annual expenses = $3,000
Therefore, net annual revenue =
$5,310 - $3,000 = $2,310


Investment
Balance, $

Salvage value = $2,000
MARR = 5% per year
N = 5 years
MARR = 5%
5,000
$2,001  FW
’
Years
0
1
- 5,000
- 10,000
2
3
Area of Negative
Investment
Balance
- $4,294
4
- $2,199
+$2,310
5
+$2,310 +$2,000
- $2,309
- $6,290
+$2,310 - $4,509
- $8,190
+$2,310
- $6,604
+$2,310
- $8,600
-$10,500
Presented by: Dr. Magdy Akladios
6
Ch5: Present Worth Analysis-Part I
Discounted Cash Flow
Analysis
Net Present Worth Measure
 Principle: Compute the equivalent net surplus at n = 0 for a given
interest rate of i.
 Decision Rule for Single Project Evaluation: Accept the project if the net
surplus is positive.
Decision Rule for Comparing Multiple Alternatives: Select the alternative
with the largest net present worth.
Inflow
0
1
2
3
4
5
Outflow
Net surplus
PW(i)inflow
0
PW(i) > 0
PW(i)outflow
20
Example:
Tiger Machine Tool Company
inflow
$24,400
0
outflow
1
$55,760
$27,340
2
3
$75,000
P W (15% ) inflow  $24 , 400 ( P / F ,15% ,1)  $ 27 ,34 0 ( P / F ,1 5% ,2 )
 $ 55,760 ( P / F ,15 % ,3 )
 $78,5 53
P W (15% ) outflow  $75,0 00
P W (15% )  $78,5 53  $7 5, 000
 $3,553  0 , A ccept
21
Presented by: Dr. Magdy Akladios
7
Ch5: Present Worth Analysis-Part I
Excel Solution:
A
B
1
Period
Cash Flow
C
2
0
($75,000)
3
1
$24,400
4
2
$27,340
5
3
$55,760
6
7
PW(15%)
$3,553.46
=NPV(15%,B3:B5)+B2
22
Present Worth Amounts at Varying Interest Rates
i (%)
PW(i)
PW(i)
i(%)
0
$32,500
20
-$3,412
2
27,743
22
-5,924
4
23,309
24
-8,296
6
19,169
26
-10,539
-12,662
8
15,296
28
10
11,670
30
-14,673
12
8,270
32
-16,580
-18,360
14
5,077
34
16
2,076
36
-20,110
0
38
-21,745
-751
40
-23,302
17.45*
18
*Break even interest rate – also called: MARR
23
MARR





Minimum attractive Rate of Return
All companies should consider the return that a
given project should produce
That return is used to find out whether a project
is profitable or not
That return is called the “MARR”
It is an interest rate that is used to convert cash
flows into equivalent worth at some point(s) in
time
Presented by: Dr. Magdy Akladios
8
Ch5: Present Worth Analysis-Part I
Present Worth Profile
25
What does $3,553 really mean?
1.
2.
Project Balance Concept
Investment Pool Concept
Project Balance Concept



Suppose that the firm has no internal funds to
finance the project, so will borrow the entire
investment from a bank at an interest rate of 15%.
Then, any proceeds from the project will be used
to pay off the bank loan.
Then, our interest is to see if how much money
would be left over at the end of the project period.
30
Presented by: Dr. Magdy Akladios
9
Ch5: Present Worth Analysis-Part I
Project Balance Concept
1
2
Beginning
Balance
N
0
3
-$75,000
-$61,850
-$43,788
Interest
-$11,250
-$9,278
-$6,568
Payment
-$75,000
+$24,400
+$27,340
+$55,760
Project
Balance
-$75,000
-$61,850
-$43,788
+$5,404
Net surplus
PW(15%) = $5,404 (P/F, 15%, 3) = $3,553
31
Project Balance Diagram
60,000
Terminal project balance
(net Future Worth, or
project surplus)
40,000
Project balance ($)
20,000
$5,404
0
Discounted
payback period
-$43,788
-20,000
-40,000
-60,000
-$75,000
-$61,850
-80,000
-100,000
-120,000
0
1
2
3
Year(n)
32
Investment Pool Concept


Suppose the company only has $75,000 to invest.
It has two options:
1.
2.
Take the money out and invest it in the project, or
Leave the money in the pool and continue to earn a 15%
interest.
33
Presented by: Dr. Magdy Akladios
10
Ch5: Present Worth Analysis-Part I
Meaning of Net Present Worth
N=3
How much would you have if the
Investment is made?
Investment pool
$24,400(F/P,15%,2) = $32,269
$27,340(F/P,15%,1) = $31,441
$55,760(F/P,15%,0) = $55,760
$119,470
$75,000
$55,760
How much would you have if the
investment was not made?
$75,000(F/P,15%,3) = $114,066
$27,340
$24,400
Project
What is the net gain from the
investment?
0
1
2
3
$119,470 - $114,066 = $5,404
PW(15%) = $5,404(P/F,15%,3) = $3,553
34
What Factors Should the Company Consider in Selecting
a MARR in Project Evaluation?
Cost of capital

The required return necessary to make
an investment project worthwhile.
Viewed as the rate of return that a firm
would receive if it invested its money
someplace else with a similar risk
Risk premium

The additional risk associated with the
project if you are dealing with a project
with higher risk
Risk
premium

Cost of capital

MARR

35
Example






A company is investing $10,000 into a 5-year project
The project will produce a uniform annual revenue of
$5,310
The salvage value of the project is $2,000
Annual expenses will be $3,000/year
The company has decided that they will accept all
projects returning a MARR of 10% or more/year.
Using the PW method, show if your company should
adopt this project
Presented by: Dr. Magdy Akladios
11
Ch5: Present Worth Analysis-Part I
Solution





PW(10%) = -$10,000 + ($5,310-$3,000)(P/A,
10%, 5) + $2,000 (P/F, 10%, 5)
PW(10%) = -$10,000 + $2,310(3.7908) +
$2,000(0.6209)
PW(10%) = -$10,000 + $8,756.75 + $1,241.80
PW(10%) = -$1.45 = $0
Project is boarder line acceptable
Example





A company is investing $25,000 into a 5-year
project
The project will produce a uniform annual
revenue of $8,000
The salvage value of the project is $5,000
The company has decided that they will accept all
projects returning a MARR of 20% or more/year.
Using the PW method, show if your company
should adopt this project
Solution




PW(20%) = -$25,000 + $8,000(P/A, 20%, 5)
+ $5,000 (P/F, 20%, 5)
PW(20%) = $934.29
Since PW (20%) >0
Therefore, project is economically justifiable.
Presented by: Dr. Magdy Akladios
12
Ch5: Present Worth Analysis-Part I
Example:






An electrical motor rated at 15HP needs to be
purchased for $1,000.
The service life of the motor is known to be 10
years with negligible salvage value.
Its full load efficiency is 85%.
The cost of energy is $0.08 per kwh.
The intended use of the motor is 4,000
hours/year.
Find the total cost of owning and operating the
motor at 10% interest.
40
Solution
1 H P = 0 .7 4 5 7 k W
1 5 H P = 1 5  0 .7 4 5 7 = 1 1 .1 8 5 5 k W
R e q u ire d in p u t p o w e r a t 8 5 % e f fic ie n c y ra tin g :
1 1 .1 8 5 5 k W
 1 3 .1 5 9 4 k W
0 .8 5
R e q u ire d to ta l k W h p e r y e a r
1 3 .1 5 9 4 k W  4 ,0 0 0 h o u rs /y e a r = 5 2 ,6 3 8 k W h /y r
T o ta l a n n u a l e n e r g y c o s t to o p e ra te th e m o to r
5 2 ,6 3 8 k W h  $ 0 .0 8 /k W h = $ 4 ,2 1 1 /y r
T h e to ta l c o s t o f o w n in g a n d o p e ra tin g th e m o to r
P W (1 0 % )  $ 1, 0 0 0  $ 4 , 2 1 1( P / A , 1 0 % ,1 0 )
= $ 2 6 ,8 7 5
41
Cash Flow Series Associated with Owning and
Operating the Motor
PW (10%)  $1,000  $4, 211( P / A,10%,10)
 $26,875
0
1
2
3
4
5
6
7
8
9
10
$1,000
$4,211
42
Presented by: Dr. Magdy Akladios
13
Ch5: Present Worth Analysis-Part I
Variations of Present Worth
Analysis: The Future Worth
Method
Chapter 5 - Part II
Future Worth Criterion



Given: Cash flows
and MARR (i)
Find: The net
equivalent worth at a
specified period other
than “present”,
commonly the end of
project life
0
Decision Rule: Accept
the project if the
equivalent worth is
positive.
$55,760
$24,400
1
$27,340
2
3
$75,000
Project life
44
Example:
Net FW at the End of the Project
45
Presented by: Dr. Magdy Akladios
14
Ch5: Present Worth Analysis-Part I
Alternate Way of Computing the NFW
FW(15%)inflow  $24,400(F / P,15%,2)  $27,340(F / P,15%,1)
$55,760(F / P,15%,0)
 $119,470
FW(15%)outflow  $75,000(F / P,15%,3)
 $114,066
FW(15%)  $119,470  $114,066
 $5,404  0, Accept
46
Excel Solution:
A
B
C
1
Period
Cash Flow
2
0
($75,000)
3
1
$24,400
4
2
$27,340
5
3
$55,760
6
PW(15%)
$3553.46
7
FW(15%)
$5,404.38
=FV(15%,3,0,-B6)
47
Example:
Future Equivalent at an Intermediate Time
49
Presented by: Dr. Magdy Akladios
15
Ch5: Present Worth Analysis-Part I
Example:
Project’s Service Life is Extremely Long
• Mr. Bracewell Built a hydroelectric plant using his
personal savings ($800,000)
• Power generating capacity of 6 million kwhs
• Estimated annual power sales after taxes - $120,000
• Expected service life of 50 years
Was Bracewell's $800,000 investment a wise one?
How long does he have to wait to recover his initial
investment, and will he ever make a profit?
50
Mr. Bracewell’s Hydroelectric Project
V1  V2  $1,101K  $1, 468 K
 $367 K  0
V2  120 K ( P / A,8%,50)
 $1, 468K
V1  $50 K ( F / P,8%,9)  $50 K ( F / P,8%,8)
  $100 K ( F / P,8%,1)  60 K
 $1,101K
51
FW Example





A company is investing $25,000 into a 5-year
project
The project will produce a uniform annual
revenue of $8,000
The salvage value of the project is $5,000
The company has decided that they will accept
all projects returning a MARR of 20% or
more/year.
Using the FW method, show if your company
should adopt this project
52
Presented by: Dr. Magdy Akladios
16
Ch5: Present Worth Analysis-Part I
Solution





FW(20%) = -$25,000(F/P, 20%, 5) +
$8,000(F/A, 20%, 5) + $5,000
FW(20%) = -$25,000(2.4883) + $8,000(7,4416)
+ $5,000 = $2,325.30
Alternative solution:
FW = PW(20%) X (F/P, 20%, 5) = $934.29
(2.4883) = $2,324.80
Since FW (20%) >0, therefore, project is
economically justifiable.
53
The Capitalized Worth
Method
Chapter 5 - Part II
COMPARING ALTERNATIVES USING
THE CAPITALIZED WORTH METHOD

Capitalized worth or capitalized cost method
is a convenient basis for comparing MEA
when:


A period of needed services is indefinitely long;
and
The repeatability assumption is applicable
55
Presented by: Dr. Magdy Akladios
17
Ch5: Present Worth Analysis-Part I
THE CAPITALIZED WORTH METHOD


Capitalized Worth (CW) method -Determining the PW of all revenues and/or
expenses over an infinite length of time.
If only expenses are considered over an
infinite length of time, it is sometimes called
“Capitalized Cost” -- Determining the PW of
costs only.
56
How Would You Find P for a Perpetual
Cash Flow Series, A?
57
Capitalized Equivalent Worth
 Principle: PW for a project with an annual
receipt of A over infinite service life
Equation:
CE(i) = A(P/A, i,
) = A/i
A
0
P = CE(i)
58
Presented by: Dr. Magdy Akladios
18
Ch5: Present Worth Analysis-Part I
Example

Given the following 2 MEA, use the CW
method to determine which project you
should invest in (MARR = 15%):
I
Salvage
Annual Expenses
Useful Life
A
B
-$12,000 -$40,000
0
10,000
-2,200
-1,000
10 yrs
25 yrs
59
Solution (Part I)

First, find AW of each alternative:


AW(15%)A = -$12,000(A/P, 15%, 10) - $2,200 = $4,592
AW(15%)B = -$40,000(A/P, 15%, 25) - $1,000 +
$10,000(A/F, 15%, 25) = -$7,141
60
Solution (Part II)

Second, find CW of each alternative:



CW(15%)A = AWA/i = -$4,592/0.15 = -$30,613
CW(15%)B = AWB/i = -$7,141/0.15 = -$47,607
Since CW of alternative A yields less cost, it
should be selected.
61
Presented by: Dr. Magdy Akladios
19
Ch5: Present Worth Analysis-Part I
Final Solution



-$30,613 < -$47,607
CW(15%)A < CW(15%)B
Since CW of alternative A yields less cost, it
should be selected.
62
Example:
Given: i = 10%, N = ∞
Find: P or CE (10%)
$2,000
$1,000
0
10
∞
P = CE (10%) = ?
63
Solution
$2,000
$1,000
0
10
∞
P = CE (10%) = ?
$1, 000 $1, 000

( P / F ,10%,10 )
0.10
0.10
 $10, 000 (1  0.3855 )
CE (10%) 
 $13,855
Presented by: Dr. Magdy Akladios
64
20
Ch5: Present Worth Analysis-Part I
A Bridge Construction Project
 Construction cost = $2,000,000
 Annual Maintenance cost = $50,000
 Renovation cost = $500,000 every 15 years
 Planning horizon = infinite period
 Interest rate = 5%
65
Cash Flow Diagram for the Bridge Construction
Project
Years
15
30
45
60
$500,000
$500,000
$500,000
$500,000
0
$50,000
$2,000,000
66
Solution:
 Construction Cost
P1 = $2,000,000
 Maintenance Costs
P2 = $50,000/0.05 = $1,000,000
 Renovation Costs
P3 = $500,000(P/F, 5%, 15)
+ $500,000(P/F, 5%, 30)
+ $500,000(P/F, 5%, 45)
+ $500,000(P/F, 5%, 60)
+--= {$500,000(A/F, 5%, 15)}/0.05
= $463,423
 Total Present Worth
P = P1 + P2 + P3 = $3,463,423
67
Presented by: Dr. Magdy Akladios
21
Ch5: Present Worth Analysis-Part I
Alternate way to calculate P3
 Concept: Find the effective interest rate per payment period
0
15
$500,000
30
$500,000
45
60
$500,000 $500,000
 Effective interest rate for a 15-year cycle
i = (1 + 0.05)15 - 1 = 107.893%
 Capitalized equivalent worth
P3 = $500,000/1.07893
= $463,423
68
Presented by: Dr. Magdy Akladios
22