1 تنمية المهارات الاقتصادية للمهندسين New Vision of Engineering Economy Teaching

2 Engineering Economy Module 1: Lectures 1-8 Dr. Mohamed F. El-Refaie Dr. Mohamed F. El-Refaie (Sunday and Tuesday, L3,L4,L5 and L6) (Sunday and Tuesday, L3,L4,L5 and L6) Dr. Sayed Kaseb Dr. Sayed Kaseb (Saturday and Wednesday, L1,L2,L7 and L8) (Saturday and Wednesday, L1,L2,L7 and L8) Module 1 Teaching Team

3 Module 1 - Lecture 7 Module 1 - Lecture 7 Equivalent Uniform Annual Cost (EUAC) Dr. Sayed Kaseb Associate Professor, MPED, FECU Manger of Pathways to Higher Education Grant Coordinator of VISION Project

4 Module 1 - Lecture 7 Overview 7.1 Review 7.2 Annual cash flow 7.3 Alternative selection using EUAC 7.4 EUAC for infinite period

5 Standard factor notations P, A, F, i, n Standard notationFactor name 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)

6 SINGLE-PAYMENT COMPOUND AMOUNT FACTOR (SPCAF)

7 SINGLE-PAYMENT PRESENT WORTH FACTOR (SPPWF)

8 UNIFORM-SERIES COMPOUND- AMOUNT FACTOR (USCAF)

9 SINKING FUND FACTOR (SFF)

10 Uniform-series present-worth factor (USPWF)

11 Capital-recovery factor (CRF)

12 Summary of Compound Interest Factors NameAbbreviatio n NotationFormula Single-payment present worth factor SPPWF Single-payment compound- amount factor SPCAF Uniform-series compound- amount factor USCAF Sinking fund factorSFF Uniform-series present-worth factor USPWF Capital recovery factorCRF

13 ARITHEMETIC PROGRESSION SERIES

14

15

16 GEOMETRIC PROGRESSION SERIES

17 Compound amount of a geometric progression

18 Present worth of a geometric progression The present worth can be found from

19 Uniform series equivalent to a geometric progression The equivalent uniform series can be obtained as

20 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

21 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 = (A/P, 15%, 3) – 2000 (A/F, 15%, 3) = (0.015 (1.15) 3 / ( – 1)) – 2000 (0.15 / ( – 1)) = m.u per year.

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.

23 New plant designUpgrade old plant Alternative 1 Description Cash flows over some time period Analysis using an engineering economy model Evaluated alternative 1 Noneconomic issues-environmental considerations Alternative2 Description Cash flows over some time period Analysis using an engineering economy model Evaluated 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

24 Compare the following machines on the basis of their equivalent uniform annual cost. Use an interest rate of 18% per year. Comparison point New MachineUsed Machine Capital cost44000 m.u m.u. Annual operating cost 7000 m.u.9000 m.u. Annual repair cost 210 m.u.350 m.u. Overhauling2500 m.u. every 5 years1900 m.u. every 2 years Salvage value4000 m.u. after 15 years3000 m.u. after 8 years Example 7.2 Cash flows of the two machines.

25 EUAC new = 7,210 + (44000 – 2500) (A/P, 18%, 15) (A/P, 18%, 5) – 4000 (A/F, 18%, 15) = (0.18 ( ) / ( – 1)) (0.18 ( ) / ( – 1)) – 4000 (0.18/ ( )) EUACnew = m.u. per year.

26 EUAC used = (23000 – 1900) (A/P, 18%, 8) (A/P, 18%, 2) – 3000 (A/F, 18%, 8) = (0.18 (1.18) 8 / ( – 1)) (0.18 (1.18) 2 / ( – 1)) – 3000 (0.18 / ( – 1)) = m.u. per year.

27 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.

28 Spreadsheet Solution for example 7.2

29

30 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

31 First proposal EUAC 1 = 5000 (A/P, 12%, 12) (A/P,12%,2) – 700 (A/F,12%,12) = 5000 (0.12 (1.12) 12 / ( – 1)) (0.12 (1.12) 2 / ( – 1)) – 700 (0.12/( )) EUAC 1 = m.u per year.

32 EUAC 2 = 900 (A/P, 12%, 6) × 2 = 900 (0.12 (1.12) 6 / ( – 1)) × 2 = m.u per year. Second Proposal Select proposal 1 because EUAC1<EUAC2

33 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 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 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

34 Foam: EUAC foam = ( ) (A/P, 15%,24)+2500(A/P,15%,3)-6000 =32500(0.15(1.15) 24 / ( )) +2500(0.15(1.15) 3 /( )) = m.u per year (costs) Fiber-glass: EUAC f =12000(A/P,15%,24)-2500 =12000(0.15(1.15) 24 /( ))-2500 = - m.u per year (saving) Then the company should use fiber glass for insulation as this insulation method would result in savings for the company. Solution:

35 The following costs are proposed for two equal-service tomato-peeling machines in a food canning plant: ItemMachine AMachine B First cost, m.u.26,00036,000 Annual maintenance cost, m.u Annual labour cost, m.u.11,0007,000 Extra income taxes, m.u./year2,600 Salvage value, m.u.2,0003,000 Life, years610 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

36 Solution:

37

38 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 APlan B Machine 1Machine 2 First cost, m.u Annual operating cost, m.u./year Salvage value, m.u Life, years81224 Example 7.6 Table 7.3: the Cash flows for the two plans in example 7.6

39 Solution Plan A:

40 Plan B:

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

42 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 mountainPipeline around mountain Initial costm.u.5.5 millionm.u.5 million Maintenance00 Useful lifePermanent50 years Salvage value00

43 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 Pipeline: EUAC = m.u. 5 million (A/P, 6%, 50) = m.u. 5 million (0.0634) = m.u 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.