Matakuliah
Tahun
: D0762 – Ekonomi Teknik
: 2009
Annual Worth Analysis
Course Outline 5
Outline
•
•
•
•
•
•
•
Principle and Benefit next
Equivalent Annual Worthnext
Capital Ownership Cost next
AW by salvage sinking-fund method next
AW by Salvage present-worth method
next
AW by Capital recovery plus interest method
next
next
Spreadsheet
next
Refererences
- Engineering Economy – Leland T. Blank, Anthoy J. Tarquin p.180199
- Engineering Economic Analysis, Donald G. Newman, p. 141-163
-- Engineering
Economy, William G. Sulivan, p.137-194, p. 212-284
http://www.cs.wright.edu/~snarayan/hfeeconomics/ise481_07_files/v3_document.htm
2
Annual Worth Analysis
• Principle: Measure investment worth on annual basis
• Benefit: By knowing annual equivalent worth, we can:
• Seek consistency of report format
• Determine unit cost (or unit profit)
• Facilitate unequal project life comparison
3
Computing Equivalent Annual Worth
$120
$80
0
$70
1
2
3
4
5
6
$50
$189.43
$100
A = $46.07
0
PW(12%) = $189.43
1
2
3
4
5
6
AE(12%) = $189.43(A/P, 12%, 6)
= $46.07
http://www.cs.wright.edu/~snarayan/hfeeconomics/ise481_07_files/v3_document.htm
4
Annual Equivalent Worth
• Repeating Cash Flow Cycles
$700
$500
$1,000
$800
$700
$400 $400
$500
$800
$400
$400
$1,000
Repeating cycle
http://www.cs.wright.edu/~snarayan/hfeeconomics/ise481_07_files/v3_document.htm
5
Annual Equivalent Worth
• First Cycle:
PW(10%) = -$1,000 + $500 (P/F, 10%, 1)
+ . . . + $400 (P/F, 10%, 5)
= $1,155.68
AE(10%) = $1,155.68 (A/P, 10%, 5) = $304.87
• Both Cycles:
PW(10%) = $1,155.68 + $1,155.68 (P/F, 10%, 5)
+ . . . + $400 (P/F, 10%, 5)
= $1,873.27
AE(10%) = $1,873.27 (A/P, 10%,10) = $304.87
http://www.cs.wright.edu/~snarayan/hfeeconomics/ise481_07_files/v3_document.htm
6
• When only costs are
involved, the AE
method is called the
annual equivalent cost.
• Revenues must cover
two kinds of costs:
Operating costs and
capital costs.
Annual Worth Costs
Annual Equivalent Cost
Capital
costs
+
Operating
costs
http://www.cs.wright.edu/~snarayan/hfeeconomics/ise481_07_files/v3_document.htm
7
Capital (Ownership) Costs
• Definition: The cost of owning
an equipment is associated
with two transactions—
(1)its initial cost (I) and
(2) its salvage value (S).
Capital costs: Taking into
these sums, we calculate
the capital costs as:
S
0
N
I
0 1
2
3
N
CR(i)
CR(i)  I ( A / P, i, N )  S( A / F, i, N )
 ( I  S)( A / P, i, N )  iS
http://www.cs.wright.edu/~snarayan/hfeeconomics/ise481_07_files/v3_document.htm
8
Example - Capital Cost
Calculation
$50,000
•
•
Given:
I = $200,000
N = 5 years
S = $50,000
i = 20%
Find: CR(20%)
0
5
$ 200,000
CRi   I  S  A / P, i, N   iS
CR20%   $200.000  $50.000  A / P,20%,5
 0,20 $50.000
 $60.157
http://www.cs.wright.edu/~snarayan/hfeeconomics/ise481_07_files/v3_document.htm
9
Justifying an investment based on AE
Method
•
•
•
•
Given: I = $20,000, S =
$4,000, N = 5 years, i =
10%
Find: see if an annual
revenue of $4,400 is
enough to cover the
capital costs.
Solution:
CR(10%) = $4,620.76
Conclusion: Need an
additional annual revenue
in the amount of $220.76.
http://www.cs.wright.edu/~snarayan/hfeeconomics/ise481_07_files/v3_document.htm
10
Salvage -Sinking Fund Method
AW  P A / P, i, n  S  A / F , i, n
•
General Equation
•
Example 6.1
Calculate the AW of a tractor attachment that has an initial cost of $8000 and a
salvage value of $500 after 8 years. Annual operating cost for the machine are
estimated to be $900 and an interest rate of 20% per year is applicable.
Solution
The problem indicates there are 2 cashflow
AW = A1 + A2
Where A1 = annual cost of initial investment with salvage value considered Equation
above
= -8000(A/P,20%8) + 500(A/F,20%,8) = $2055
A2 = annual operating cist = $-900
The Annual worth for the attachment is :
AW = -2055 – 900 = $-2955
11
Salvage Present-Worth
Method
• General Equation
AW  [P  S P / F , i, n] A / P, i, n
• The Steps to determine the complete asset AW are
1. Calculate the present worth of the salvage value via the P/F
factor
2. Combine the value obtained in step 1 with the investment cost
P
3. Annualize the resulting difference over the life of the asset
using the A/P factor
4. Combine any uniform annual worth with the value from step 3
5. Convert any other cash flows into an equivalent uniform annual
worth and combine with the value obtained in step 4
12
Example 6.2
• Compute the AW of the attachment detailed in Example
6.1 using the salvage present worth method
Solution
Using the steps outline and equation before
AW = [-8000+500(P/F,20%,8)(A/P,20%,8) – 900
= $-2955
13
Capital Recovery Plus Interest Method
• General equation
AW  ( P  S ) A / P, i, n  S i 
• The steps to be followed for this method are :
1.
2.
3.
4.
5.
6.
Reduce the initial cost by the amount of the salvage value
Annualize the value in step 1 using A/P factor
Multiply the salvage value by the interest rate
Combine the values obtained in steps 2 and 3
Combine any uniform annual amounts
Convert all other cash flows into equivalent uniform amount and combine them
with the value from step 5
Step 1 through 4 are accomplished bya pplying equation before
14
Example 6.1
• Use the value of Example 6.1 to compute the AW using
the capital recovery plus interest
Solution
From equation and steps before :
AW =-(8000-500)(A/P,20%,8)-500(0,20) – 900
= $-2955
15
Comparing Alternatives
•
•
Following costs are estimated for two equal service tomato peeling
machines to evaluated by a canning plant manager’
if the minimum required rate of return is 15% per year, help the manager
decide which machine to select !
Machine A
Machine B
First Cost
26,000
36,000
Annual maintenance cost,$
800
300
Annual Labor cost, $
11,000
7000
Extra annual income taxes, $
-
2,600
Salvage Value
2,000
3,000
Life, years
6
10
Solution
AW A = -26,000(A/P,15%,6) + 2000(A/F,15%,6)-11800 = $-18,442
AW B = -36,000(A/P,15%,10)+3000(A/F,15%,10) = $-16,925
16
Spreadsheet
Example 6.10
If Ms.Kaw Deposits $10,000 now at an interest rate of 7% per year, hom
many year s must the money accumulate before she can withdraw $1400
per year forever
17