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LECTURE 7

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LECTURE 7

    LECTURE 7

    Equivalent Uniform Annual Cost

The following lecture provides a review on the previous lectures and gives examples on the

    economic factors from real life and how these factors could be useful. Also an explanation for

    the meaning of the Equivalent Uniform Annual Cost (EUAC) is included with examples

    aided by spreadsheet examples.

7.1 Review

In the previous lectures we dealt with the various engineering economy factors which may

    represent a lot of cases in the real life, we’ll try to revise these factors relating them to

    corresponding real life situation.

    P/F,i%,nSingle-payment present worth factor (SPPWF) ??This factor enables engineers to know the present amount of a payment in the future for

    example, if one wants to know the present amount of an overhaul that might be done to

    installed equipment in his company after a certain period, and the overhaul cost is known to

    be F in the future an equivalent amount P in the present could be determined with a known

    interest rate i%.

    F/P,i% ,nSingle-payment compound amount factor (SPCAF) ??This factor is the reciprocal of the Single-payment present worth factor (SSPWF) which means if an amount P in the present is paid an equivalent amount F in the future could be

    determined for example, if a person borrows a loan P from a bank now and he wants to know

    how much will be paid in the future for this loan at an interest rate i%.

    P/A,i%,nUniform series present worth (USPWF) ??

    It is used to calculate the present worth P equivalent to a number of payments A over a

    certain period, for example if someone is going to buy a car but he’s going to pay its price

    over n periods with equal payments A and interest rate i% per month this factor allows him to know the equivalent price of the car in the present amount P.

    A/P,i%,nCapital recovery (CRF) ??

    This factor is the reciprocal of the above one for example, if someone borrows a loan P from

    a bank and he wants to know how much he is going to pay in equal monthly payments A over

    n months with interest rate i%.

    A/F,i%,nSinking fund (SFF) ??

    This factor is used to calculate the equal payments A over n interest periods for an interest rate i% equivalent to a payment F in the future. For example if someone wants to have an

    amount F in the future and he wants to know how much he should pay(equal payments A)

    every interests period over n number of interest periods for an interest rate i%.

    F/A,i%,n Uniform series compound amount (USCAF)??This factor is used to calculate the future amount F equivalent to an equal payments A over

    interest periods n for an interest rate i%, for example if a person deposits annually an equal payments A for a number of years n for an interest rate i% in a saving account the USCAF

    enables him to calculate the future amount F equivalent to such deposits.

Concerning the Arithmetic Progression Series

Uniform-Gradient compound amount

    nn??1111????ii????1 FAGn???'??iii????

Uniform Gradient Present Worth

    nn??1111????ii????1 PAGn???'??nniiiii??11????????

Uniform Series Equivalent to a Uniform Gradient

    ??1n A?A'?G???ni??1?i?1??

The factors have the same meaning as the Uniform series amount except that arithmetic

    gradient is used instead of the uniform series.

Similarly for the Geometric Progression Series

Compound amount of a geometric progression

    nn??11???Ei????FAEi??'for ?? Ei?????

    n?1FAnEEi???'1for ??

Present worth of a geometric progression

    nn??11???Ei1????PAEi??'for ??nEi?1?i??????

    1PAnEi??'for 1?E

Uniform series equivalent to a geometric progression

    nn??11???Eii????AAEi??'for ?? nEi?11??i??????

    n?11?EE??AAnEi??' for n11??E??

7.2 ANNUAL CASH FLOW

    The annual cash flow is the sum of all incoming and outgoing cash flows. Incoming flows

    might be revenue or savings, and outgoing flows are typically material costs, labor wages and

    taxes. Summing both flows gives the net cash flow for each year. Table 7.1 below gives

    examples to revenues and outgoing flows for typical engineering situations.

Table 7.1: Examples of Cost/Revenue Items

     Project

    Parameter New Product Power Plant Machine Upgrade Revenue Unit sales Electricity sales Savings per unit

    Capacity sales Byproduct sales operation

     Thermal energy sale Incremental - more

     units/hr

    Reduced maintenance

    and

    reduced labor Variable Costs Fuel, Water, Reduced labor (negative, Material

    (or Costs of Reagent Chemicals or shown as a positive Labor

    Goods Sold) Electricity (internal revenue) Manufacturing

     overhead use)

    Product engineering Waste disposal

    Waste disposal

    Fixed Costs Maintenance material Reduced maintenance Factory lease (if any)

     Sales personnel Maintenance labor (negative shown as

    Maintenance Operators Supervisory positive revenue item)

    Stocking Personnel

    Product line Chemical cleaning

    management Accounting, HR,

    Purchasing & other

    functions

    Fixed Cost Performance tests to None Ongoing

    (R&D) optimize system development &

     product support

    Capital New machine Capital equipment Upgraded components

     Installation Installation and installation not

    Development cost Property included in maintenance

     Engineering cost budget include

     engineering and project

    management costs

7.3 Equivalent Uniform Annual Cost

The annual cash flow analysis criteria is based on converting all the expenses of a project or

    an equipment over its entire life to an equivalent uniform annual expenses using the

    compound factors derived in Lecture 5 and 6, which means if a projects X has a capital cost P

    in the present as initial cost for the land, buildings and equipments, beside an annual expenses

    A for operating and maintenance, in addition to an amount of money F to be paid every

certain period for overhauling, it is possible to convert all these expenses (P, A, and F) to

    annual amount A’ to be paid over the life of the project or enterprise.

    The annual cash flow analysis is sometimes called the Equivalent Uniform Annual Cost

    EUAC, which holds the same meaning.

    The idea of annual cash flow analysis may be clearly identified through the following

    example.

Example 7.1:

If a person purchased a new car for 6000 m.u. and sold it 3 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.

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))

     = 2801.92 m.u.per year.

    This means that the above cash flow scheme is equivalent to the payment of 2801.92 m.u.

    per year. Figure 1 below illustrates this equivalence.

    2000 m.u.

    12301230

    =

     750 m.u.

     6000 m.u.2801.92 m.u.

    Figure7.1 Annual cash flow representation for example 7.1

7.4 Alternative selection using EUAC:

In this section the annual cash flow analysis is introduced as a method for selection between

    alternatives as it provides information about a project or equipment from the annual expenses

    point of view. It means that all disbursements (irregular and uniform) must be converted to an

    equivalent uniform annual cost, that is, a year-end amount which is the same each year. When

    the EUAC method is used, the equivalent uniform annual cost of the alternative must be

    calculated for one life cycle only, because, as its name implies, the EUAC is an equivalent

    annual cost over the life of the project. If the project is continued for more than one cycle, the

    equivalent annual cost for the next cycle and all succeeding cycles would be exactly the same

    as for the first, assuming all cash flows were the same for each cycle.

    Therefore EUAC for one cycle of an alternative represents the equivalent uniform annual cost

    of that alternative forever.

Example 7.2:

Compare the following machines, using cash flows shown in table 7.2, on the basis of their

    equivalent uniform annual cost. Use an interest rate of 18% per year.

Table 7.2: Cash outflows of the two machines.

    Comparison point New Machine Used Machine Capital cost 44000 m.u. 23000 m.u. O & M cost ** 7210 m.u./year 9350 m.u./year Overhauling 2500 m.u. every 5 years * 1900 m.u. every 2 years * Salvage value 4000 m.u. 3000 m.u.

* Hint: Concerning the overhauling it is canceled if it is required at the end of the equipment

    life.

    ** O & M is the Operating and maintenance cost.

Solution:

    New machine4000m.u.

    i= 18%

    0123456789101112151314

    7210

    2500m.u/year 2500m.u 2500m.u

    44000m.u

    Figure 7.2 Cash flow diagram for example 7.2

EUAC = 7,210 + (44000-2500) (A/P, 18%, 15) + 2500(A/P, 18%, 5) - 4000(A/F, 18%, 15) new151555 = 7210 + 41500 (0.18(1.18)/ (1.18-1)) + 2500 (0.18(1.18) / (1.18-1)) 15- 4000 (0.18/ (1.18-1))

     = 16094.55 m.u.per year.

    Used machine3000m.u.i= 18 %

    0

    9350m.u./year81234567

    1900m.u.1900m.u.1900m.u.

    23000m.u.

    Figure 7.3 Cash flow diagram for example 7.2

EUAC = (23000-1900) (A/P, 18%, 8) + 9350 + 1900(A/P, 18%, 2) - 3000(A/F, 18%, 8) used8822 = 21100 (0.18 (1.18) / (1.18-1)) + 9350 +1900 (0.18 (1.18)/(1.18-1)) 8-3000 (0.18 / (1.18-1))

     = 15542.4 m.u. per year.

Now comparing between the EUAC for both machines we find that EUAC usednew

    Then it would be more economical to purchase the used machine instead of the new one.

Hint! In this analysis, the reliability of the machines was not taken into considerations, thus

    technical selection may refuse the used machines although economic privilege.

This problem can be easily solved using spreadsheet applications as can be seen from figure

    7.4 below. Intrinsic functions of Excel are used to facilitate solution.

    (PMT(18%,C10,C3-C7,0,0))+(PMT(18%,5,C7,0,0))-(PMT(18%,C10,0,C8,0))+(-7210)

    Figure 7.4 Spreadsheet solutions for example 7.2

    Refer to the following link to use the spreadsheet solutions: EX 7.2.xls

Example 7.3:

A moving and storage company is considering two possibilities for warehouse operations.

    Proposal 1 requires the purchase of a fork lift for 5000 m.u. and 500 pallets that cost 5 m.u.

    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 9000 m.u. annually and spend 600 m.u. per year in

    maintenance and operation, the life of the fork lift is expected to be 12 years, with 700 m.u.

    salvage value. Alternatively, proposal 2 requires that the company hire two people to operate

    power-driven hand trucks at a cost of 7500 m.u. per person. One hand truck will be required

    at a cost of 900 m.u. and the hand truck will have a life of 6 years with no salvage value. If

    the interest rate is l2% per year, which alternative should be selected?

Solution:

Proposal 1:

    i= 12%

    0123456789101112

    m.u.9600/year

    m.u.2500m.u.2500m.u.2500m.u.2500m.u.2500

    m.u.5000+ m.u.2500

    Figure 7.5 Cash flow diagram for example 7.3

    EUAC= 5000(A/P, 12%, 12) + 2500(A/P,12%,2) + 9600 - 700(A/F,12%,12) 1 121222 =5000 (0.12 (1.12) / (1.12-1)) + 2500 (0.12 (1.12)/(1.12-1)) +9600 12- 700 (0.12 / (1.12-1)) EUAC = 11857.423 m.u. per year. 1

Proposal 2:

    i= 12%

    0123456Years

    m.u.900

    m.u.15000/year

    Figure 7.6 Cash flow diagram of proposal 2 for example 7.3

    EUAC= 900(A/P, 12%, 6) + 7500 × 2 2 66 = 900 (0.12 (1.12) / (1.12- 1)) + 7500× 2

     = 15218.903 m.u. per year.

    Select proposal 1 because EUAC1

    Spreadsheet solution is also available at this link, EX 7.3.xls

Example 7.4:

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 urethane foam and fiber-glass insulation. The initial cost of the foam insulation will

    be 35000 m.u. with no salvage value. The foam will have to be painted every 3 years at a cost

    of 2500 m.u. the energy saving is expected to be 6000 m.u. per year. Alternatively, fiber-

    glass batts can be installed for 12000 m.u. the fiber-glass batts would not be salvageable

    either, but there would be no maintenance costs. If the fiber-glass batts would save 2500 m.u.

    per year in energy costs, which method of insulation should the company use if the interest

    rate is 15% per year? Use a 24-year study period and an equivalent uniform annual-cost

    analysis.

Solution:

    Foam:

    = (35000-2500) (A/P, 15%,24)+2500(A/P,15%,3)-6000 EUACfoam 242433 = 32500 (0.15(1.15)/ (1.15-1)) +2500(0.15(1.15)/(1.15-1))-6000

     = 146.412 m.u.per year (costs)

Fiber-glass insulation:

    EUAC=12000(A/P,15%,24)-2500 batts2424 =12000(0.15(1.15)/(1.15-1))-2500

     =- 634.84 m.u.per year (saving)

    Then the company should use fiber glass batts for insulation as this insulation method would

    result in savings for the company.

    Spreadsheet solution also is available at this link, EX 7.4.xls

Example 7.5:

The following costs are proposed for two equal-service tomato-peeling machines in a food

    canning plant:

Table 7.3: Cash out flows for the two machines in example 7.5

    Item Machine A Machine B

    First cost, m.u. 26,000 36,000

    Annual maintenance cost, m.u./year 800 300

    Annual labour cost, m.u./year 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 interest rate is 15%, which machine should be selected?

Solution:

    AA, 15%,611,800, 15%, 6EUACPSV???????APF n????ii1???i???????26,00011,8002,000nn????????1111????ii???? 6????0.151.15??0.15?????26000+11800-200066??????1.15-11.15-1?????? EUAC=18442m.u./year A

    Machine A

    i= 15 %

    S.V.=m.u.20000123456

    years

    A = m.u.11800/year

    m.u.26000

    Figure 7.7 Cash flow diagram for machine A in example 7.5

    ??0.15A??EUACP???, 15%, 109,9003000??B10P??1.151????? EUAC=16925m.u./year B

    Machine B

    i= 15 %

    S.V.=m.u.3000

    012345698107years

    A = m.u.9900/year

    m.u.36000

    Figure 7.8 Cash flow diagram for machine B in example 7.5

???EUAC machine BEUACSelectBA

Example 7.6:

Shown in Table 7.3 the information of two plans for cash flows, using the annual cash flow

    analysis Compare between the two plans at i=15%.

Table 7.3: the Cash flows for the two plans in example 7.6

     Plan A Plan B

    Machine 1 Machine 2

    First cost, m.u. 90,000 28,000 175,000

    Annual operating cost, m.u./year 6,000 300 2,500

    Salvage value, m.u. 10,000 2,000 10,000

    Life, years 8 12 24

Solution:

Plan A:

EUAC=EUAC+EUACA12

    AAEUAC=P, 15%, 8+6000-10000, 15%, 8????11PF 8??0.15(1.15)0.15?? ???90000600010000????88(1.15)1(1.15)1??????

    ?25328m.u./year

    AAEUAC=P, 15%, 12+300-2000, 15%, 12????22PF 12????0.151.15??0.15????=28000+300-2001212????????1.15-11.15-1????

    ?5397m.u./year = m.u.5397 /year

    ?EUAC=25,328+5,397=m.u.30725/yearA

Plan B:

EUAC=PA/P, 15%, 24+2500-10000A/F, 15%, 24????B 24????0.151.15??0.15????=175000+2500-100002424????????1.15-11.15-1????

    = m.u.29646/year?29646m.u./year

    ?Select plan B, since EUACBA

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