annual cash flow analysis
TRANSCRIPT
Week 8: Annual Cash Flow Analysis
Dr. Mohsin Siddique
Assistant Professor
Ext: 29431
Date: 14/04/2014
Engineering Economics
University of SharjahDept. of Civil and Env. Engg.
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Part I
Outcome of Today’s Lecture
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� After completing this lecture…
� The students should be able to:
� Define equivalent uniform annual cost (EUAC) and Equivalent uniform annual benefits (EUAB)
� Resolve an engineering economic analysis problem into its annual cash flow equivalent
� Conduct an equivalent uniform annual worth (EUAW) analysis for a single investment
� Use EUAW, EUAC, and EUAB to compare alternatives with equal, common multiple, or continuous lives, or over some fixed study period.
Techniques for Cash Flow Analysis
4
� Present Worth Analysis
� Annual Cash Flow Analysis
� Rate of Return Analysis
� Incremental Analysis
� Other Techniques:
� Future Worth Analysis
� Benefit-Cost Ratio Analysis
� Payback Period Analysis
Annual Cash Flow Analysis
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� Concepts of Annual Cash Flow Analysis
� Comparing Alternatives using Annual Cash Flow Analysis:
� Same-Length Analysis Period
� Different-Length Analysis Periods
� Infinite-Length Analysis Period
� Other Analysis Periods
Techniques for Cash Flow Analysis
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Problem 6-1
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� Compute the value of C for the following diagram, based on “10% interest rate.
C = $15 + $15 (A/G, 10%, 4)= $15 + $15 (1.381) = $35.72
0
$15
4 0
G=$15
4
0= +
Problem 6-8
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� As shown in the cash flow diagram, there is an annual disbursement of money that varies from year to year from $100 to $300 in a fixed pattern that repeats forever. If interest is 10%, compute the value of A, also continuing forever, that is equivalent to the fluctuating disbursements.
Problem 6-8
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Pattern repeats infinitely
There is a repeating series:; 100 – 200 – 300 – 200. Solving this series for A gives us the A for the infinite series.
Problem 6-8
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A= $100 + [$100 (P/F, 10%, 2) + $200 (P/F, 10%, 3) + $100 (P/F, 10%, 4)] (A/P, 10%, 4)= $100 + [$100 (0.8254) + $200 (0.7513) + $100 (0.6830)] (0.3155)
= $100 + [$301.20] (0.3155)= $195.03
0
$100
4 0
200
4+
100100
Annual Cash Flow Analysis
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� The basic idea is to convert all cash flows to a series of EUAW (equivalent uniform annual worth):
Net EUAW = EUAB -EUAC� EUAC: Equivalent Uniform Annual Cost
� EUAB: Equivalent Uniform Annual Benefit
� An expenditure increases EUAC and a receipt of money decreases EUAC.
� To convert a PW of a cost to EUAC, use:
EUAC = (PW of cost) (A/P, i%, n)
� Where there is salvage value?
A = F(A/F, i%, n)� A salvage value will reduce EUAC and increase EUAB
� When there is an arithmetic gradient, use the (A/G, i%, n) factor.
� If there are irregular cash flows, try to first find PW of these flows; then, EUAC may be calculated from this PW.
� Criteria for selection of an alternative: � Maximize Net EUAW (EUAB –EUAC)
� Minimize EUACOR Maximize EUAB
Analysis Period
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� Five different analysis-period situations occur:
� 1. Analysis Period Equal to Alternative Lives
� 2. Analysis Period a Common Multiple
� 3. Analysis Period for a Continuing Requirement
� 4. Infinite Analysis Period
� 5. Some Other Analysis Period:
� Analysis period may be equal to life of the shorter-life alternative, the longer-life alternative, or something different. In this case, terminal values at the end of a specific year become very important.
Analysis Period Equal to Alternative Lives
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� We have an ideal situation (rarely the case in ‘real-life’ ):
� Study period = life-cycle of any of the alternatives
� Example 6-6: In addition to the do-nothing alternative, three alternatives are being considered for improving the operation of an assembly line. Each of the alternatives has a 10-year life and a scrap value equal to 10% of its original cost. If interest is 8%, which alternative should be adopted.
Analysis Period Equal to Alternative Lives
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$6000
$25,000
$9,000 $2500
010
$8000
$15,000
$14,000 $1500
010
$6000
$33,000
$14,000 $3300
010
Plan A Plan B
Plan C
Analysis Period Equal to Alternative Lives
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Problem 6-32
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Problem 6-32
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Around the Lake Under the LakeFirst Cost $75,000 $125,000Maintenance $3,000/yr $2,000/yrAnnual Power Loss $7,500/yr $2,500/yrProperty Taxes $1,500/yr $2,500/yrSalvage Value $45,000 $25,000Useful Life 15 years 15 years
Problem 6-32
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0 15
$75,000
$3000
$7500
$1500
$45000
0 15
$125,000
$2000
$2500
$2500
$25000
Around the Lake Under the Lake
Around the LakeEUAC = $75,000 (A/P, 7%, 15) + $12,000 - $45,000 (A/F, 7%, 15)
= $75,000 (0.1098) + $12,000 - $45,000 (0.0398)= $18,444
Under the LakeEUAC = $125,000 (A/P, 7%, 15) + $7,000 - $25,000 (A/F, 7%, 15)
= $125,000 (0.1098) + $7,000 - $25,000 (0.0398)= $19,730
Go around the lake.
Analysis Period a Common Multiple of
Alternative Lives
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� Assume a replacement with an identical item with same cost and performance. When the lives of two alternatives vary, one can use a common multiple of the two lives to determine the better project. Nevertheless, compare alternatives based on their own service lives.
� Example 6-7: Two pumps are being considered for purchase. If interest is 7%, which pump should be bought. Assume that Pump B will be replaced after its useful life by the same one.
Analysis Period a Common Multiple of
Alternative Lives
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(EUAC-EUAB)A
(EUAC-EUAB)B
0
$7,000
$1500
120
$5,000
$1000
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Analysis Period with a Repeatability
Assumption
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� If two or more alternatives have unequal lives, only evaluate the annual worth (AW) for one life cycle of each alternative
� The annual worth of one cycle is the same as the annual worth of all future cycles
Under the circumstances of identical replacement (repeatability):
Problem 6-37
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Problem 6-37
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� Machine X
� EUAC= $5,000 (A/P, 8%, 5)= $5,000 (0.2505) = $1,252
� Machine Y
� EUAC= $8,000(A/P, 8%, 12) + $150 - $2,000 (A/F, 8%, 12)
=8000(0.1327) +150 -2000( 0.0527) = $1,106
� Select Machine Y.
$5,000
0 5
$150
$8,000
$2000
012
Machine X Machine Y
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Part II
Analysis Period for a Continuing Requirement
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� Many civil infrastructure provide a continuing requirement/service. There is no distinct analysis period; therefore, assume it is long but undefined.
� Compare different-life alternatives assuming identical replacement. In this case, compare the annual cash flows computed for alternatives based on their own different service lives.
(EUAC-EUAB)A
(EUAC-EUAB)B
Infinite Analysis Period
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� At times we may have an alternative with a finite useful life in an infinite analysis period situation.
� With identical replacement:
� EUAC for infinite analysis period = EUAC for limited life
Infinite Analysis Period
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� EUAC for infinite analysis period = P (A/P ,i, ∞)+ any other annual (costs-benefits)
� (A/P, i,∞) = i
� (EUAC-EUAB)A= $100(A/P,8%, ∞) -$10.00 = $100* 0.08 -$10.00 = $-2.00
� (EUAC-EUAB)B= $150(A/P,8%,20) -$17.62 = $150 * 0.1019-$17.62 =$-2.34
� (EUAC-EUAB)C= $200(A/P,8%,5) -$55.48 = $200 * 2.505 -$55.48 =$-5.38
� Select alternative C
Problem 6-41
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Problem 6-41
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� Because we may assume identical replacement, we may compare 20 years of B with an infinite life for A by EUAB – EUAC.
� Alternative A
� EUAB – EUAC (for an inf. period) = $16 - $100 (A/P, 10%, ∞)
� = $16 - $100 (0.10)= +$6.00
� Alternative B
� EUAB – EUAC (for 20 yr. period) = $24 - $150 (A/P, 10%, 20)
� = $24 - $150 (0.1175)= +$6.38
� Choose Alternative B.
Summary
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� Popular analysis technique:
� Easily understood -results are reported in $ per time period, usually $ per year
� AW method is often preferred to the PW method
� Only have to evaluate one life cycle of an alternative
� Assumption for AW method: Cash flows in one cycle are assumed to replicate themselves in future cycles
� No need to convert lifetimes of all projects to their least common multiple!
� AW offers an advantage for comparing different-life alternatives
� For infinite life alternatives, simply multiply P by i to get AW value
Assignment # 4
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� 6.7, 6.16, 6.21, 6.29, 6.42
� Date of Submission: _____________
� Assignment should be hand written
