7. annual cash flow analysis

7. Annual Cash Flow Analysis

Dr. Mohsin Siddique

Assistant Professor

[email protected]

Ext: 29431

Date: 18/11/2014

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

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� 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 EUAC OR 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-years life and a scrap value equal to 10% of its original cost. If interest is 8%, which alternative should be adopted.


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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


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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 # 7

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� 6.7, 6.16, 6.21, 6.29, 6.42

� Date of Submission: _____________

� Assignment should be hand written

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Thank You

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[email protected]

Tel. +971 6 5050943 (Ext. 2943)