module 4 present worth analysis
TRANSCRIPT
Module 4: Present Worth Analysis
SI-4251 Ekonomi Teknik
Muhamad Abduh, Ph.D.
SI-4251 Ekonomi Teknik
Outline Module 4 Proposals for Investment Alternatives Selection of Alternatives Present-worth comparison Future Worth Analysis Capitalized-cost calculation
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Proposals for Investment Alternatives An investment is usually established from
(engineering) proposal. Every proposal can be considered as an investment
alternatives, but An investment alternative can consist of a group or
set of proposals, which in turn may include option to “do nothing” Independent Proposal
The condition at which the acceptance of a proposal from a set of proposals has not effect on the acceptance of any of the other proposals in the set.
Dependent Proposal The condition at when the acceptance of a proposal from the set will influence the
acceptance of the others Mutually exclusive the acceptance of a proposal from the set precludes the acceptance
of any others
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Selection of Alternatives Decision criteria in an economic analysis
can be done comparing mutually exclusive alternatives: The differences between alternatives
highest inflow or lowest outflow The minimum attractive rate of return
highest rate of return Payback period
shortest payback period Do nothing
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Selection of Alternatives Selection of two or more (investment)
alternatives by comparing their economic values. Method for comparison of alternatives:
present values, future worth, capitalized cost, annual values, rate of returns, or payback period
Conditions of comparison: equal lives different lives
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MARR Minimum Acceptable Rate of Return
Minimum amount the investor is willing to accept for the use of money
Different from lending rates More akin to Opportunity Cost of the money Determined by Company Policy
Rate of Return
Expected Rate of Return on New Proposal
Range for ROROn accepted proposals
MARR
ROR on Safe investment
All proposals must offer at Least MARR
to be considered
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Present-worth comparison
The comparison of alternatives is made by transforming all future receipts and expenditures into equivalent today’s rupiah
Example: Two types of production systems are being considered based on
MARR of 12% per year and the following characteristics:
Muhamad Abduh, Ph.D.4-7
system A system B
Initial cost Rp. 625.000.000,- Rp. 570.000.000,-
Monthly expenses
Rp. 45.750.000,- Rp. 55.750.000,-
Receipts Rp. 152.000.000,- / quarter
Rp. 140.000.000,- / 4 months
Salvage value Rp. 225.000.000,- Rp. 190.000.000,-
Life 3 tahun 3 tahun
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Present-worth comparison
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System A:
I
0 1 2 3
R R R R R R R R R R R R
E
SV
?PA = -I - E(P/A, iA, 36) + R(P/F, iA, 3) + R(P/F, iA, 6) + … + R(P/F, iA, 36) +
SV(P/F, iA, 36)
Effective monthly interest rate, iA = i / 12 = 1%
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SI-4251 Ekonomi Teknik
Present-worth comparison
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System B:
I
0 1 2 3
E
SV
R R R R R RR R R
Muhamad Abduh, Ph.D.
PB = -I - E(P/A, iB, 36) + R(P/F, iB, 4) + R(P/F, iB, 8) + … + R(P/F, iB, 36) + SV(P/F, iB, 36)
Select System A if PA > PB
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Present-worth comparisonTo have a fair comparison of alternatives with different lives, the time span over must made equal:
a)The time period of comparison is made equal to the least common multiple (LCM) for their lives.Cash flows of the shorter period will be extended up to the remaining time period of comparison
b)At any time span to be considered (the study period approach or planning horizon approach), when LCM is impossible to perform. Only cash flows up to the time span is to be considered;
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Present-worth comparison
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Example:Two types of production systems are being considered based on MARR of 12% per year and the following characteristics:
system A system B
Initial cost Rp. 625.000.000,- Rp. 770.000.000,-
Monthly expenses
Rp. 45.750.000,- Rp. 55.750.000,-
Monthly receipts
Rp. 32.000.000,- Rp. 40.000.000,-
Salvage value Rp. 225.000.000,- Rp. 110.000.000,-
Life 2 tahun 4 tahun
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SI-4251 Ekonomi Teknik
Present-worth comparison
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System A:
1 2
I
0
E
SVR
1 2
I
0
E
R
3 4
SV
System B:
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Present-worth comparison (LCM)
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System A:
1 2
I
0
E1
SVR1
1 2
I
0
E
R
3 4
SV
System B:
3 4
I2E2
SV2R2
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SI-4251 Ekonomi Teknik
Present-worth comparison (Study Periods: 2 years)
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System A:
1 2
I
0
E
SVR
1 2
I
0
E
R
3 4
SV
System B:
Muhamad Abduh, Ph.D.
Estimated
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Example 1: Three Alternatives
Assume i = 10% per year
A1Electric Power
First Cost: -2500Ann. Op. Cost: -900Sal. Value: +200Life: 5 years
A2Gas Power
First Cost: -3500Ann. Op. Cost: -700Sal. Value: +350Life: 5 years
A3Solar Power
First Cost: -6000Ann. Op. Cost: -50Sal. Value: +100Life: 5 years
Which Alternative – if any, should be selected based upon a present worth analysis?
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Example 1: Cash Flow Diagrams
0 1 2 3 4 5
0 1 2 3 4 5
0 1 2 3 4 5
-2500
-3500
-6000
A = -900/Yr.
A = -700/Yr.
A = -50/Yr.
FSV = 200
FSV = 350
FSV = 100
A1:
Electric
A2: Gas
A3:Solar
i = 10%/yr and n = 5Muhamad Abduh, Ph.D.4-16
SI-4251 Ekonomi Teknik
Calculate the Present Worth's
Present Worth's are:
1. PWElec. = -2500 - 900(P/A,10%,5) +
200(P/F,10%,5) = $-5788
2. PWGas = -3500 - 700(P/A,10%,5) +
350(P/F,10%,5) = $-5936
3. PWSolar = -6000 - 50(P/A,10%,5) +
100(P/F,10%,5) = $-6127Select “Electric” which has the min. PW Cost!
Muhamad Abduh, Ph.D.4-17
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Example 2:
Two Location Alternatives, A and B where one can lease one of two locations.
Which option is preferred if the interest rate is 15%/year?
Location ALocation BFirst cost, $ -15,000 -18,000Annual lease cost, $ per year -3,500 -3,100Deposit return,$ 1,000 2,000Lease term, years 6 9
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Use LCM, where “n” = 18 yrs.
The Cash Flow Diagrams are:
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Unequal Lives: 2 Alternatives
i = 15% per year
A
LCM(6,9) = 18 year study period will apply for present worth
Cycle 1 for A Cycle 2 for A Cycle 3 for A
Cycle 1 for B Cycle 2 for B
18 years
6 years
6 years
6 years
9 years 9 yearsB
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LCM Present Worth's
Since the leases have different terms (lives), compare them over the LCM of 18 years.
For life cycles after the first, the first cost is repeated in year 0 of the new cycle, which is the last year of the previous cycle.
These are years 6 and 12 for location A and year 9 for B.
Calculate PW at 15% over 18 years.
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SI-4251 Ekonomi Teknik22
PW Calculation for A and B -18 yrs
PWA = -15,000 - 15,000(P/F,15%,6) +
1000(P/F,15%,6)- 15,000(P/F,15%,12) + 1000(P/F,15%,12) + 1000(P/F,15%,18) - 3500(P/A,15%,18)= $-45,036
PWB = -18,000 - 18,000(P/F,15%,9) +
2000(P/F,15%,9) + 2000(P/F,15%,18) - 3100(P/A,15 %,18)
= $-41,384 Select “B”: Lowest PW Cost @ 15%
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Use The Study Period Approach
An alternative method; Impress a study period (SP) on all of the alternatives;
A time horizon is selected in advance; Only the cash flows occurring within
that time span are considered relevant; May require assumptions concerning
some of the cash flows. Common approach and simplifies the
analysis somewhat.
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Example Problem with a 5-yr SP
Assume a 5- year Study Period for both options:
For a 5-year study period no cycle repeats are necessary. PWA = -15,000 - 3500(P/A,15%,5) +
1000(P/F,15%,5)= $-26,236
PWB = -18,000- 3100(P/A,15%,5) +
2000(P/F,15%,5) = $-27,397Location A is now the better choice.Note: The assumptions made for the A and B
alternatives! Do not expect the same result with a study period approach vs. the LCM approach!
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Future Worth Analysis
In some applications, management may prefer a future worth analysis;
Analysis is straight forward: Find P0 of each alternative:
Then compute Fn at the same interest rate
used to find P0 of each alternative.
For a study period approach, use the appropriate value of “n” to take forward.
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Future Worth Approach (FW)
Applications for the FW approach: Wealth maximization approaches; Projects that do not come on line
until the end of the investment (construction) period: Power Generation Facilities Toll Roads Large building projects Etc.
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Capitalized Cost Calculations
CAPITALIZED COST- the present worth of a project which lasts forever. Government Projects; Roads, Dams, Bridges, project that possess
perpetual life; Infinite analysis period; “n” in the problem is either very long,
indefinite, or infinity.
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Derivation of Capitalized Cost
We start with the relationship: P = A[P/A,i%,n] Next, what happens to the P/A factor
when we let n approach infinity. Some “math” follows.
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P/A where “n” goes to infinity
The P/A factor is:
(1 ) 1
(1 )
n
n
iP A
i i
On the right hand side, divide both numerator and denominator by (1+i)n 1
1(1 )ni
P Ai
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CC Derivation…
Repeating: 11(1 )ni
P Ai
If “n” approaches the above reduces to:
AP
i
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CC Explained
For this class of problems, we can use the term “CC” in place of P.
Restate:
Or,
AW: annual worth
ACC
i
AWCC
i
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CC Problem: Public Works Example
Problem Parameters
The suspension bridge will cost $50 million with annual inspection and maintenance - costs of $35,000. In addition, the concrete deck would have to be resurfaced every 10 years at a cost of $100,000. The truss bridge and 'approach roads' are expected to cost $25 million and have annual maintenance costs of $20,000. The bridge would have to be painted every 3 years at a cost of $40,000. In addition, the bridge would have to be sandblasted every 10 years at a cost of $190,000. The cost of purchasing right-of-way is expected to be $2 million for the suspension bridge and $15 million for the truss bridge. Compare the alternatives on the basis of their capitalized cost if the interest rate is 6% per, year.
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Two, Mutually Exclusive Alternatives: Select the best alternative based upon a CC analysis
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Bridge Alternatives: Suspension
Cash Flow Diagrams
0 1 2 3 4 . . . . . 9 10 11 ……..
$50 Million
$35,000/yr
$100,000$2 Million
Suspension Bridge Alternative
i = 6%/year
Muhamad Abduh, Ph.D.4-33
Suspension Bridge Analysis
CC1= -52 million at t = 0.
1
2
1 22
A $35,000
A 100,000( / ,6%,10) $7,587
35,000 ( 7,587)$709,783.
0.06
A F
A ACC
i
Total CC – suspension bridge is:
-52 million + (-709,783) = -$52.71 million
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Truss Bridge Alternative
For the Truss Bridge Alternative: Cash Flow Diagram:
/ / / / / /0 1 2 3 4 5 6 7 8 9 10
11 ….. A. Maint. = $20,000/yr
-25M +(-15M)
Paint: -40,000
Paint: -40,000
Paint: -40,000
Sandblast: -190,000
Truss Design:
i = 6%/year
n =
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Truss Bridge Alternative
1. CC1 Initial Cost:
-$25M + (-15M) = -$40M
/ / / / / /0 1 2 3 4 5 6 7 8 9 10
11 ….. A. Maint. = $20,000/yr
-25M +(-15M)
Paint: -40,000
Paint: -40,000
Paint: -40,000
Sandblast: -190,000
Truss Design:
i = 6%/year
n =
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Truss Bridge Alternative
2. Annual Maintenance is already an “A” amount so: A1 = -$20,000/year
/ / / / / /0 1 2 3 4 5 6 7 8 9 10
11 ….. A. Maint. = $20,000/yr
-25M +(-15M)
Paint: -40,000
Paint: -40,000
Paint: -40,000
Sandblast: -190,000
Truss Design:
i = 6%/year
n =
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Truss Bridge Alternative
3, A2: Annual Cost of Painting
/ / / / / /0 1 2 3 4 5 6 7 8 9 10
11 ….. A. Maint. = $20,000/yr
-25M +(-15M)
Paint: -40,000
Paint: -40,000
Paint: -40,000
Sandblast: -190,000
Truss Design: i = 6%/year
n =
For any given cycle of painting compute:
A2 = -$40,000(A/F,6%,3) = -$12,564/year
Use A/F,6%,3
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Truss Bridge Alternative
3, A3 Annual Cost of Sandblasting
/ / / / / /0 1 2 3 4 5 6 7 8 9 10
11 ….. A. Maint. = $20,000/yr
-25M +(-15M)
Paint: -40,000
Paint: -40,000
Paint: -40,000
Sandblast: -190,000
Truss Design: i = 6%/year
n =
For any given cycle of Sandblasting Compute
A3 = -$190,000(A/F,6%,10) =-$14,421
Use The A/F,6%,10 to convert to an equivalent $/year amount
Muhamad Abduh, Ph.D.4-39
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Bridge Summary for CC(6%)
CC2 = (A1+A2+A3)/i
CC2 = -(20,000+12,564+14,421)/0.06
CC2 – $783,083
CCTotal = CC1 + CC2 =-40.783 million•CCSuspension = -$52.71 million
•CCTruss - -40.783 million
•Select the Truss Design!
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Exercise:
Muhamad Abduh, Ph.D.4-41
The municipal government of Bandung is considering two proposals to build new toll highway system. The first alternative calls for upgrading the existing
system, which would cost Rp. 225,75 B for construction and additional Rp 210 M and Rp 250 M per year for maintenance and operation cost
The other option is to build a new elevated highway that is estimated to cost Rp. 885 B for construction and annual maintenance cost of Rp 325 M
If either alternatives would yield Rp 815 M revenue per year, and the interest rate is set at 8% p.a., which alternative should the government go about realizing the toll highway system?
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Homework #41. A manufacturing company is trying to decide among three different pieces of equipment that have the
following characteristics:
useful life = 6 years and interest rate of 12% per year.
2. Which of these two machines that have the following costs is to be selected for a continuous production process, if the i = 15% p.a:
3. Ganesha consulting firm is considering to build or lease an office space. For interest rate of 6% compounded semiannualy compare and select alternative.
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equipment A equipment B equipment C
First cost Rp. 975.000.000,- Rp. 854.500.000,- Rp. 1.025.000.000,-
Annual M&O cost Rp. 89.700.000,- Rp. 95.000.000,- Rp. 75.000.000,-
Salvage value Rp. 161.000.000,- Rp. 205.000.000,- Rp. 321.000.000,-
Overhaul cost Rp. 175.000.000,- / 2 years Rp. 135.000.000,- / 3 years Rp. 175.000.000,- / 3 years
machine X machine Y
First cost Rp. 3.800.000.000,- Rp. 1.675.000.000,-
Annual operating cost Rp. 289.700.000,- Rp. 315.000.000,-
Salvage value Rp. 461.000.000,- Rp. 205.000.000,-
Life 5 years 3 years
build own lease
Construction cost Rp. 8.750.000.000,- -
Lease cost - Rp. 415.000.000,- / 3 years
Maintenance cost Rp. 110.000.000,- / year Rp. 75.000.000,- / year
Period ∞ 3 years
Muhamad Abduh, Ph.D.