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New Vision of Engineering Economy Teaching. تنمية المهارات الاقتصادية للمهندسين. Engineering Economy Module 1: Lectures 1-8. Dr. Mohamed F. El-Refaie (Sunday and Tuesday, L3,L4,L5 and L6) Dr. Sayed Kaseb (Saturday and Wednesday, L1,L2,L7 and L8). Module 1 Teaching Team. - PowerPoint PPT Presentation
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المهارات تنميةاالقتصادية للمهندسين
New Vision of
Engineering Economy
Teaching
2
Engineering Engineering Economy Economy Module 1: Module 1:
Lectures 1-8Lectures 1-8 Dr. Mohamed F. El-RefaieDr. Mohamed F. El-Refaie
(Sunday and Tuesday, L3,L4,L5 and (Sunday and Tuesday, L3,L4,L5 and L6)L6)
Dr. Sayed KasebDr. Sayed Kaseb (Saturday and Wednesday, L1,L2,L7 (Saturday and Wednesday, L1,L2,L7
and L8)and L8)
Module 1 T
eaching T
eam
3
Module 1 - Lecture 7Module 1 - Lecture 7
Equivalent Uniform Annual
Cost (EUAC)Dr. Sayed KasebDr. Sayed Kaseb
Associate Professor, MPED, FECUAssociate Professor, MPED, FECU
Manger of Pathways to Higher Manger of Pathways to Higher EducationEducation
Grant Coordinator of VISION Grant Coordinator of VISION ProjectProject
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Module 1 - Lecture 7 Overview
7.1 Review7.2 Annual cash flow7.3 Alternative selection using EUAC7.4 EUAC for infinite period
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Standard factor notationsP, A, F, i, n
Factor name Standard notation
Single-payment present worth (SPPWF)
Single-payment compound amount (SPCAF)
Uniform series present worth (USPWF)
Capital recovery (CRF)
Sinking fund (SFF)
Uniform series compound amount (USCAF)
/ , %,P F i n
/ , %,F P i n
/ , %,P A i n
/ , %,A P i n
/ , %,A F i n
/ , %,F A i n
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SINGLE-PAYMENT COMPOUND AMOUNT FACTOR (SPCAF)
1 2 3 4 n-1n-20
P= given
F= ??
(SPCAF)
n
1n
F P i
/ / , , 1n
F P F P i n i
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1 2 3 4 n-1n-20
P= ??
F= given
(SPPWF)
n
SINGLE-PAYMENT PRESENT WORTH FACTOR (SPPWF)
/ 1n
P F i
1/ / , ,
1nP F P F i n
i
8
1 2 3 4 n-1n-20
F= ??(USCAF)
n
A =given
UNIFORM-SERIES COMPOUND-AMOUNT FACTOR (USCAF)
/ , ,F A i n 1 1
*( )n
iF A
i
9
11
ni
iFA
, ,1 1
n
A iA i nFF i
1 2 3 4 n-1n-20
A = ??
(SFF)
n
F = given
SINKING FUND FACTOR (SFF)
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Uniform-series present-worth factor (USPWF)
-1- 1P P= , i, n =AA
ni
i
11
Capital-recovery factor (CRF)
1, ,
1 -1
n
n
i iA A i nPP i
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Summary of Compound Interest Factors
Name Abbreviation
Notation Formula
Single-payment present worth factor
SPPWF
Single-payment compound-amount factor
SPCAF
Uniform-series compound-amount factor
USCAF
Sinking fund factor SFF
Uniform-series present-worth factor
USPWF
Capital recovery factor CRF
/ , ,P F i n 1n
i
/ , ,F P i n 1n
i
/ , ,F A i n 1 1n
i
i
/ , ,A F i n 1 1
n
i
i
/ , ,P A i n 1 1n
i
i
/ , ,A P i n 1 1n
i
i
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ARITHEMETIC PROGRESSION SERIES
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10 2 3 4 5 n-1 n
A’
10 2 3 4 5 n-1 n
G2G 3G
(n-1)G(n-2)G
15
1 1 1 11'
n ni i
A G ni i i
F
1 1 1 11P '
1 1
n n
n n
i iA G n
ii i i i
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GEOMETRIC PROGRESSION SERIES
10 2 3 4
E=10 %
A’
10 2 3 4
E= -10 %
A’1000
1100
1210
1331 1000
900
810
729
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Compound amount of a geometric progression
10 2 3 4
E%
A’
n
F
1' 1
nF A n E for E i
1 1'
n nE i
F AE i
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Present worth of a geometric progression
10 2 3 4
E%
A’
n
P
The present worth can be found from
ni
FniFPFP
1
1,,/
for
1 11'
1
n n
n
E i
E iP A
E ii
for
1'
1
E i
P A nE
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Uniform series equivalent to a geometric progression
10 2 3 4
E%
A’
n
A A A A A
The equivalent uniform series can be obtained as
A / , ,1 1
n
iF A F i n F
i
for
1 1A = A '
1 1
n n
n
E i
E ii
E ii
1
for
1A = A '
1 1
n
n
E i
E En
E
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The annual cash flow analysis criteria is based on converting all the
cost of a project or an equipment over its entire life to an Equivalent
Uniform Annual Cost (EUAC) using the compound factors
7.2 ANNUAL CASH FLOW
P
A
F
EUAC = A’
≡Project X
Project X
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If a person purchased a new car for 6000 m.u. and sold it 7
years later for 2000 m.u., what is the equivalent uniform
annual cost if he spent 750 m.u., per year for upkeep and
operation? Use an interest rate of 15 % per year.
Example 7.1
Solution:
EUAC = 750 + 6000 (A/P, 15%, 3) – 2000 (A/F, 15%, 3)
= 750 + 6000 (0.015 (1.15)3 / (1.153 – 1)) – 2000 (0.15 / (1.153 – 1))
= m.u. 2801.92 per year.
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7.3 Alternative selection
using EUAC
1. Disbursements (irregular and uniform) must be converted to
an equivalent uniform annual cost
2. EUAC are calculated for each alternative for one life cycle.
NB: EUAC for one cycle of an alternative represents the equivalent uniform annual cost of that alternative forever.
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New plant design Upgrade old plant
Alternative 1
Description
Cash flows over some time period
Analysis using an engineering
economy modelEvaluated
alternative 1
Noneconomic issues-environmental considerations
Alternative2
Description
Cash flows over some time period
Analysis using an engineering
economy modelEvaluated
alternative2
•Income, cost estimations•Financing strategies•Tax laws
•Planning horizon•Interest•Measure of worth
•Calculated value of measure of worth
I select alternative 2
Rate of return (Alt 2) >Rate of return (Alt 1)
Alternatives
Methods of Economic Selection
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Compare the following machines on the basis of their equivalent
uniform annual cost. Use an interest rate of 18% per year.
Comparison pointNew Machine Used Machine
Capital cost 44000 m.u. 23000 m.u.
Annual operating cost
7000 m.u. 9000 m.u.
Annual repair cost210 m.u. 350 m.u.
Overhauling 2500 m.u. every 5 years 1900 m.u. every 2 years
Salvage value 4000 m.u. after 15 years 3000 m.u. after 8 years
Example 7.2
Cash flows of the two machines.
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EUACnew = 7,210 + (44000 – 2500) (A/P, 18%, 15) + 2500 (A/P, 18%, 5) – 4000 (A/F, 18%, 15)
= 7210 + 41500 (0.18 (1.1815) / (1.1815 – 1)) + 2500 (0.18 (1.185) / (1.185 – 1))
– 4000 (0.18/ (1.1815-1))
EUACnew = 16094.55 m.u. per year.
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EUACused = 9350 + (23000 – 1900) (A/P, 18%, 8) + 1900 (A/P, 18%, 2) – 3000 (A/F, 18%, 8)
= 21100 (0.18 (1.18)8 / (1.188 – 1)) + 9350 + 1900 (0.18 (1.18)2 / (1.182 – 1))
– 3000 (0.18 / (1.188 – 1))
= 15542.4 m.u. per year.
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Since we have found that: EUACused <EUACnew
Then it would be more economical to purchase the used
machine instead of the new one.
Providing that both have same productivity and quality.
NB: The overhauling cost is not taken into consideration at the end of the equipment life.
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Spreadsheet Solution for example 7.2
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New machine Used machineFirst cost 44000 23000 m.u
Annual operating cost
7000 9000 m.u./yr
Annual repair cost 210 350 m.u./yr
Overhaul every 2 years
1900 m.u.
Overhaul every 5 years
2500 m.u
Salvage value 4000 3000Interest rate 18% 18%
Life 15 8 yrsRequired
(EUAC)new (EUAC)old16,094.55 15,542.48
Select old machine
Example 7.2
Data
choose new or old?
Answer
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A moving and storage company is considering two possibilities for warehouse operations. Proposal 1 requires the purchase of a fork lift for m.u.5000 and 500 pallets that cost m.u.3 each. The average life of a pallet is assumed to be 2 years, lf the fork lift is purchased, the company must hire an operator for m.u.9000 annually and spend m.u.600 per year in maintenance and operation, the life of the fork lift is expected to be 12 years, with m.u.700 salvage value. Alternatively, proposal 2 requires that the company hire two people to operate power-driven hand trucks at a cost of m.u.7500 per person. One hand truck will be required at a cost of m.u.900 and the hand truck will have a life of years with no salvage value. If the interest rate is l2% per year, which alternative should be selected?
Example
7.3
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First proposal
EUAC1= 5000 (A/P, 12%, 12) + 2500 (A/P,12%,2) + 9600 – 700 (A/F,12%,12)
= 5000 (0.12 (1.12)12 / (1.1212 – 1)) + 2500 (0.12 (1.12)2 / (1.122 – 1))
+ 9600 – 700 (0.12/(1.1212-1))
EUAC1= m.u. 11857.423 per year.
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EUAC2 = 900 (A/P, 12%, 6) + 7500 × 2
= 900 (0.12 (1.12)6 / (1.126 – 1)) + 7500 × 2
= m.u. 15218.903 per year.
Second Proposal
Select proposal 1 because EUAC1<EUAC2
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The warehouse for a large furniture manufacturing company currently requires too much energy for heating and cooling because of poor insulation. The company is trying to decide between foam and fiber-glass insulation. The initial cost of the foam insulation will be m.u.35000 with no salvage value. The foam will have to be painted every 3 years as a cost of m.u.2500 the energy saving is expected to be m.u.6000 per year. Alternatively, fiber-glass can be installed for m.u.12000 the fiber-glass would not be salvageable either, but there would be no maintenance costs. If the fiber-glass would save m.u.2500 per year in energy costs, which method of insulation should be company use at an interest rate of 15% per year? Use a 24-year study period.
Example 7.4
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Foam:
EUACfoam= (35000-2500) (A/P, 15%,24)+2500(A/P,15%,3)-6000
=32500(0.15(1.15)24/ (1.1524-1)) +2500(0.15(1.15)3/(1.153-1))-
6000
= m.u.146.412 per year (costs)
Fiber-glass:
EUACf =12000(A/P,15%,24)-2500
=12000(0.15(1.15)24/(1.1524-1))-2500
= - m.u.634.84 per year (saving)
Then the company should use fiber glass for insulation as this
insulation method would result in savings for the company.
Solution:
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The following costs are proposed for two equal-service tomato-peeling machines in a food canning plant:
Item Machine A Machine B
First cost, m.u. 26,000 36,000
Annual maintenance cost, m.u. 800 300
Annual labour cost, m.u. 11,000 7,000
Extra income taxes, m.u./year 2,600
Salvage value, m.u. 2,000 3,000
Life, years 6 10
If the minimum required rate of return is 15%, which machine should be selected?
Example 7.5
Table 7.2: Cash out flows for the two machines in example 7.5
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AEUAC , 15%,6 11,800 , 15%, 6A AP SVP F
126,000 11,800 2,000
1 1 1 1
n
n n
i i i
i i
6
6 6
0.15 1.15 0.1526000 +11800-2000
1.15 -1 1.15 -1
AEUAC m.u.18442/year
Solution:
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B 10
0.15EUAC =P A/P 15%, 10 +9900-3000
1.15 -1
BEUAC =16925 m.u./year
1 2 3 40
m.u.36000
5
i= 15 %
6 7years
8
A = m.u.9900/year
9 10
S.V.=m.u.3000
Machine B
BEUAC machine BAEUAC Select
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Shown in Table 7.3 the information of two plans cash flows, using the annual cash flow analysis Compare between the two plans at i=15%.
Plan A Plan B
Machine 1 Machine 2
First cost, m.u. 90000 28000 175000
Annual operating cost, m.u./year
6000 300 2500
Salvage value, m.u. 10000 2000 10000
Life, years 8 12 24
Example 7.6
Table 7.3: the Cash flows for the two plans in example 7.6
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A 1 2EUAC =EUAC +EUAC
1 1
8
1 8 8
1
A AEUAC = P , 15%, 8 + 6000 - 10000 , 15%, 8P F
0.15 1.15 0.15EUAC = 90000 +6000-10000
1.15 -1 1.15 -1
EUAC = m.u. 25328 /year
Solution
Plan A:
2 2
12
2 12 12
2
A AEUAC = P , 15%, 12 +300-2000 , 15%, 12P F
0.15 1.15 0.15EUAC = 28000 300 - 200
1.15 -1 1.15 -1
EUAC = m.u. 5397 /year
AEUAC 25,328 5,397 . . 30725 / yearmu
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B
24
B 24 24
B
EUAC =P A/P, 15%, 24 +2500-10000 A/F, 15%, 24
0.15 1.15 0.15EUAC 175000 2500 -10000
1.15 -1 1.15 -1
EUAC m.u. 29646 / year
B ASelect plan B, since EUAC <EUAC
Plan B:
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7.4 EUAC FOR INFINITE PERIOD
But for an alternative with an infinite life in a problem with
an infinite analysis period:
EUAC for infinite analysis period = P(A/P, i,∞) + any other annual costs
When n = ∞, we have A = P × i and, hence, (A/P , i, ∞) equals i.
EUAC for infinite analysis period = P × i + any other annual costs
EUAC are calculated for any alternative for one life cycle, however, EUAC for one cycle of an alternative represents the equivalent uniform annual cost of that alternative forever.
EUAC for infinite analysis period = EUAC for limited life n
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Example 7.7
In the construction of the conduit to expand the water supply of a city, there are two alternatives for a particular portion of the conduit. Either a tunnel can be constructed through a mountain, or a pipeline can be laid to go around the mountain. If there is a permanent need for the conduit, should the tunnel or the pipeline be selected for this particular portion of the conduit? Assume a 6% interest rate.
Tunnel through mountain Pipeline around mountain
Initial cost m.u.5.5 million m.u.5 million
Maintenance 0 0
Useful life Permanent 50 years
Salvage value 0 0
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Tunnel:For the tunnel, with its permanent life, we want (A/P,6%,oo). For an infinite life, the capital recovery is simply interest on the invested capital. So (A/P ,6%,oo) = i EUAC = P × i = m.u. 5.5 million × (0.06) = m.u. 330000
Pipeline:
EUAC = m.u. 5 million (A/P, 6%, 50)
= m.u. 5 million (0.0634) = m.u. 317000
For fixed output, minimize EUAC. Select the pipeline
Solution
44
The difference in annual cost between a long life and an infinite
life is small unless an unusually low interest rate is used. In
example 7.7 the tunnel is assumed to be permanent. For
comparison, compute the annual cost if an 85-year life is assumed
for the tunnel?
EUAC = m.u. 5.5 million (A/P,6%,85)
= m.u. 5.5 million (0.0604) = m.u. 332,000
The difference in time between 85 years and infinity is great
indeed, yet the difference in annual costs is very small.