3360 unit 06.1 2014-i-01

ENGR 3360U Winter 2014Unit 6.1

Annual Cash Flow (worth)

Dr. J. Michael Bennett, P. Eng., PMP, UOIT,

Version 2014-I-01

Unit 6 Annual Cash Flow

Change Record

• 2014-I-01 Initial Creation• Reference Text Chapter 6

2014-I-016-2 Dr. J.M. Bennett, P.Eng., PMP ENGR 3360U Eng Eco


Course Outline1. Engineering Economics2. General Economics

1. Microeconomics2. Macroeconomics3. Money and the Bank of

Canada3. Engineering Estimation4. Interest and Equivalence5. Present Worth Analysis6. Annual Cash Flow7. Rate of Return Analysis8. Picking the Best Choice9. Other Choosing Techniques

10. Uncertainty and Risk11. Income and Depreciation12. After-tax Cash Flows13. Replacement Analysis14. Inflation15. MARR Selection16. Public Sector Issues17. What Engineering should know

about Accounting18. Personal Economics for the

Engineer

2014-I-01 Dr. J.M. Bennett, P.Eng., PMP ENGR 3360U Eng Eco6-3


Unit 6 Road Map

6.1 One Life Cycle6.2 Capital Recovery and Annual Worth6.3 Alternative Selections6.4 Annual Worth of Permanent Investments6.5 Salvage6.6 Mortgages




6.1 Advantages of Annual Cash Flow

Popular analysis technique Easily understood -- results are reported in $

per time period, usually $ per year Eliminates the LCM problem associated

with the present worth method Only have to evaluate one life cycle


• EUAC equivalent uniform annual cost• EUAB = “ “ “ benefits• EUAW =EUAB - EUAC


Also Called EUA*



AW Calculation from PW or FW Computation from PW or FW

AW = PW(A/P, i%, n) or AW = FW(A/F, i%, n)

AW converts all cash flows to their end of period equivalent amounts in $ per year



AW Value from Cash Flows

AW values can be calculated directly from cash flows for only one life cycle

Not necessary to consider the LCM of lives as is in PW or FW analysis

For alternative comparison, select the alternative with the best AW value


Example 6.1

Harry buys some furniture for $1,000. It should last 10 years. At 7%, what is the AW?

AW = 1000(A/P, 7%, 10) = 142.40




AW and Repeatability Assumption

If two or more alternatives have unequal life estimates, only evaluate the 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 (from repeatability assumption)



Repeatability AssumptionGiven alternatives with unequal lives, the

assumptions are:

1. The services provided are needed forever

2. The first cycle of cash flows is repeated for all successive cycles in the same manner

3. All cash flows will have exactly the same estimated values in every life cycle.

Note: The third assumption may be unrealistic in many problems encountered in industry



One or More Cycles

Cycle 1

Cycle 2

Cycle k

Annualize any one of the cycles

AW assumes repeatability of cash flows

Find the annual worth of any given

cycle ($/period)

…



6 Year & 12 Year Alternatives For PW or FW analysis, need an 12 year study period

2 life cycles of the 6 year project 1 life cycles of the 12 year project

Means more calculation effort!

6 year Project 6 year Project

9 year Project



Example 6.2 -- Using AW Analysis If the cash flow patterns are assumed to remain the

same for the 6 and 12 year projects for future cycles, then for AW method

Project A: 6 years

Project B: 12 years

Find the AW of any 6-year cycle

Find the AW of any 12-year cycle

Compare AWA value for 6 years with AWB

for 12 years to select the better alternative


Example 6.2

Two pumps are being considered with the following specs.

PumpA PumpB Init cost 7,000 5,000 Salvage 1,500 1,000 Useful Life 12 6



Compare

AW(pumpA)= -7000(A/P, 7%, 12)+1500(A/F, 7%, 12)

= -881.3+83.9 = -797 AW (pumpB)

= -5000(A/P, 7%, 6) + 1000(A/F, 7%, 6) = -909




Advantages/Applications of AWApplicable to a variety of engineering economy

studies such as: Asset replacement Breakeven analysis Make-or-Buy decisions Studies dealing with manufacturing costs Economic value added (EVA) analysis



6.2 Calculating Capital Recovery and AWAn economic alternative should have the following

cash flow estimates made Initial investment -- P Estimated future salvage value -- S Estimated life -- n Interest rate -- i% (this is usually the MARR) Estimated annual operating costs – AOC

Capital Recovery (CR) is the annualized equivalent of the initial investment P and the future salvage value S for n years at i%



Capital Recovery Cost• It is important to know the equivalent annual cost of owning an asset• This cost is called “Capital Recovery” or CR• CR is determined using {P, S, i, and n}

. . .

P is initial purchase price at time 0

Estimated salvage value at time n is S

0 1 2 …………………………. n-1 n



Capital Recovery Cost

Given:

Convert to:

0 1 2 3 n-1 n

An amount of A per year (CR)

P

0 1 2 3 n-1 n

……….

S

……….



Comparing CAPITAL RECOVERY with AWCR is a cost, so it carries a negative sign

CR is the annual equivalent (an A value) that represents the implied ‘cost’ of an asset for n years at i% with a first

cost of P and a salvage value of S in year n

CR does NOT include annual operating costs, AOC

To obtain AW once CR is determined, calculate

AW = - CR - AOC

where AOC itself is an annual equivalent amount (same each year)



Capital Recovery CalculationsCOMPUTING CR FOR INVESTMENTS WITH A SALVAGE VALUE

Method 1 Compute equivalent annual cost of the investment P

and subtract the equivalent annual savings of the salvage value S. This is

P(A/P, i%, n) - S(A/F, i%, n)

Determine CR as the negative (cost) of this relation

CR = -[P(A/P, i%,n) - S(A/F, i%, n)]



More Commonly Used CR Calculation

Method 2 Subtract salvage value S from original cost P and

calculate the equivalent annual cost of (P-S) Add to that the interest which the salvage value would

return each year, S(i)

CR = -[(P - S)(A/P,i,n) + S(i)]{-[P(A /P,i,n)-S(A/F,i,n)] and (A/P,I,n)=(A/F,i,n)+i so {-[P(A/P,I,n)-[A/P,I,n)-Si] =-(P-S)(A/P,I,n)+Si]}



CR Amount: What It MeansCR is the annual cost associated with owning a productive asset over n time

periods at interest rate i% per period

Equivalently, CR may be interpreted as the minimum amount of money an investment must earn each of n years to recover the initial cost at a return of i%

Why? Remember, the purchase of assets to conduct business involves a commitment of the owner's funds. As such, an investment is a commitment of the owner’s funds over n time periods. Thus, the owners expect a return on that investment.



6.3 Evaluating Alternatives Using AWFor mutually exclusive alternatives, select one

with lowest AW of costs(service) or highest AW of net incomes (revenue)

This means, select the numerically largest AW alternative

If AW < 0 at MARR, the (revenue) alternative is not economically justifiable, since initial investment P is not recovered over n years at the required rate of MARR = i% per year



Example 6.3Telesat’s Anik-1 was the world’s first domestic communications

satellite when it was launched in 1972, bringing live broadcasts of Hockey Night in Canada to northern communities. Fourteen launches later in 2004, the Anik F2 was the first satellite to commercialize the “Ka” frequency band for cost-efficient, two-way broadband services. Telesat is interested in a piece of Earth-based tracking equipment, which will require an investment of $13 million. Eight million dollars will be committed now and the remaining $5 million expended at the end of year 1 of the project. Annual operating costs for the system are expected to start the first year and continue at $0.9 million per year. The useful life of the tracker is 8 years with a salvage value of $0.5 million. Calculate the AW value for the system, if the corporate MARR is currently 12% per year.



Solution to 6.3

The cash flows for the tracker system must be converted to an equivalent AW cash flow sequence over 8 years. (All amounts are expressed in $1 million units.) The AOC is A = - $0.9 per year. The present worth P in year 0 of the two separate investment amounts of $8 and $5 is determined before multiplying by the A/P factor.



Cash Flow Diagrams a and b

0 1 2 3 4 5 6 7 8

$8.0$5.0

$0.9

0 1 2 3 4 5 6 7 8

AW = $?

(a) (b)



CR = -{[8.0 + 5.0(P/F, 12%, 1)](A/P, 12%, 8) – 0.5(A/F, 12%, 8))} = -{[12.46](0.2013) – 0.040} = $-2.47

The correct interpretation of this result is very important to Telesat. It means that each and every year for 8 years, the equivalent total revenue from the tracker must be at least $2,470,000 just to recover the initial present worth investment plus the required return of 12% per year. This does not include the AOC of $0.9 million each year. Since this amount, CR = $2.47 million, is an equivalent annual cost, as indicated by the minus sign, total AW is found; AW = -2.47 - 0.9 = $-3.37 million per year

This is the AW for all future life cycles of 8 years, provided the costs rise at the same rate as inflation, and the same costs and services are expected to apply for each succeeding life cycle.



Example 6.4 – PizzaButtPizzaButt, a popular Oshawa pizza joint, competes well in the north end,

attracting many Durham and UOIT students. It also has an on-line virtual store and delivers on demand, which the university kids really like. It turns out that most students order the same pizza, pepperoni with anchovies and shrooms. The owner, who coincidentally is taking ENGR 3360, wants to reduce delivery times by stocking the 5 cars with this common pizza, aiming at grabbing market share by routing the web orders through On-Star to the nearest pizza car on the road thus delivering the pizza in super-fast times To effect this, she proposes installing On-Star in all of her vehicles. On-Star will also give directions to any location in the city. (On-Star is satellite wireless).Each system costs $4,600, has a 5-year life cycle and may be salvaged for $300. Total operating costs are $650 for the first year, increasing by $50 per year thereafter. Assume that the MARR is 10%. The increased income is assumed to be $1200/yr for all 5.

What is the CR and AW needed to recover this cost?



PizzaButt Example

Cash Flow Diagram is:

1 2 3 4 5

P=-23,000

S = +$1500

-$650-$700

-$750-$800

-$850

A = +$1200/yr



PizzaButt Example

The Capital Recovery component is:

1 2 3 4 5

P=-23,000

S = +$1500

CR(10%) = -23,000(A/P,10%,5) +1500(A/F,10%,5) = -$5822



PizzaButt Example Cost/Revenue component is seen to equal:

=+1200 - 650 –50(A/G,10%,5)= 550 – 90.50= $459.50

(1.8101)



PizzaButt Example (Revenue – Operating Costs) are:

1 2 3 4 5

Annual Revenue = A = +$1200/yr

-$650-$700

-$750-$800

-$850

Annual Operating Costs



PizzaButt AnswerTotal Annual worth (CR + Cost/Rev)

CR(10%) = -$5822Revenue/Cost Annual amount: $459.50AW(10%) = -$5822+$459.50AW(10%) = -$5,362.50

This is a bad deal at 10% and/or an income of + $1200. Question: what would you have to make in additional profit for this to work?



Special Cases for AW Analysis

If cash flow repeatability assumption can’t be made, specify a study period of n years and perform analysis with this n in all computations (Example 6.4(b) did this)

If projects are independent, select all with AW > 0 at i = MARR, provided no budget limit is defined.



6.4 AW of a Permanent InvestmentIf an investment has no finite cycle (or a very long

estimated life) it is called a perpetual or permanent investment

If “P” is the present worth of the cost of the investment, then the AW value is P times i

AW =A = P(i)AW is actually the amount of interest P would earn each year,

forever.

Remember: P = A/ifrom the previous unit


Example 6.5

A road must get past a mountain. There are 2 options: tunnel through the mountain or build a road around it supported by concrete pillars.The cost for the tunnel is $5.5M and the useful life is infinite. The road would cost $5M but only last 50 years. Assume 6%.



Solution

Tunnel: A = Pi. So A = 5.5(0.06) = $330KRoad: A = 5(A/P,6%,50) = $317K

Select the road.



6.5 Salvage Value• When there is a salvage value at the end of

the life of an asset, it is represented as a one-time cash flow benefit at the end of the asset’s life.– Example: Salvage value (S) of asset with a

4 year life

0 1 2 3 4

S



Salvage Value, cont’d.• When there is an initial cost (P) followed by a salvage

value (S) the annual worth (AW) can be computed by:

AW = P(A/P, i, n) – S(A/F, i, n)ORAW = (P – S)(A/F, i, n) + PiORAW = (P – S)(A/P, i, n) + Si



6.6 Mortgages in Canada

• The legal document:– Long term amortized loan that is used for buying property– If payments are not made, the lender can seize the property

and sell it to recover the debt– Legal document outlines the terms and conditions for

repayment of the loan (amortization period, interest rate, penalties, etc.)

• Amortization:– Length of time it takes to pay off loan assuming:

• Payments are made on time with no additional payments• Interest rate doesn’t change



Mortgages in Canada

• Amortization periods are typically between:– 5 years to 35 years– The norm is 20 or 25 years

• Terms– A mortgage is often made up of smaller periods

called ‘terms’ and a term is the period in which the interest rate ‘term’ is fixed.

– At the end of the term, the mortgage can be renewed for a another term at the current interest rate.



Types of Mortgages• ‘Conventional’:

– For 75% or less of the appraised value• ‘High-ratio’ mortgages:

– Higher than 75% and usually require an outside agency such as the CMHC (Central Mortgage and Housing Corporation) to insure the mortgage.

• Some Others:– Open, variable rate, ARM, capped rate,

convertible rate, second, reverse, and CHIP.



Mortgage Equity and Interest Rates• Equity

– The value remaining in a property after all mortgage and loans registered against the title are subtracted from its value.

• Interest Rate– Interest rate is expressed as:

• % compounded semi-annually, not in advance– An old English term when ledgers were kept by hand– Nominal rate mortgage at 6% is actually 3% semi-annually



Summary

• Unlike present worth analysis, annual cash flow analysis does NOT require a common analysis period between the alternatives.

• However two assumptions are included:– That the actual values of future instances of a

service are the same as the current instance– There is a common multiple of useful lives

between the alternatives• As the value of n increases the capital recovery

factor approaches i.



Summary, cont’d.

• Mortgages in Canada are amortized over a number of years.

• There are a number of types of mortgages.• Equity:

– The appraised property value – value of mortgage owing

• Interest rate is traditionally stated as:– % compounded semi-annually, not in advance




Chapter Summary cont. AW method is often preferred to the PW method AW deals with only one life cycle of an alternative AW offers an advantage for comparing different-life

alternatives Assumption for AW method: Cash flows in one

cycle are assumed to replicate themselves in future cycles

For infinite life alternatives, simply multiply P by i to get AW value

3360 unit 06.1 2014-i-01

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