Methods-of-comparison-of-Alternatvies.pptx

This a short explanation about Economical Values

1. METHODS OF COMPARISON OF ALTERNATVIES ECONOMIC METHODS

2. CONCEPT AND FORMULAS

3. PRESENT WORTH METHOD Present Worth (PW) Analysis is a method of comparison of alternatives wherein we convert all cashflows to present worth (time = 0) using interest rate equal to the minimum attractive rate of return (i = MARR). Then, depending on the type of decision, the best alternative will be selected by comparing the present worth amounts of the alternatives

4. Note that in PW analysis, you must always compare alternatives with equal life service. Profit, revenue, salvage value (all inflows to an organization) will be assigned with positive sign. The costs (outflows) will be assigned with negative sign. Cash Inflows → Positive (+) Cash Outflows → Negative (-) EVALUATION ▪ For one project, if PW ≥ 0, it is acceptable ▪ For mutually exclusive (ME) alternatives, select one with the numerically highest PW, that is, less negative or more positive ▪ For independent alternatives, accept all alternatives with PW ≥ 0

5. EXAMPLE PW ANALYSIS- EQUAL LIFE There are two mutually exclusive alternatives : A & B as follows: Alternative A: Has a first cost of ₱ 30,000, an operating cost of ₱ 8,000 per year, and a ₱ 6,000 salvage value after 5 years. Alternative B: will cost ₱ 35,000 with an operating cost of ₱ 4,000 per year and a salvage value of ₱ 7,000 after 5 years. At a MARR of 12% per year, which would be accepted?

6. For a PW analysis, we are going to convert this cash-flow to present worth and select the one with the highest PW (ME alternatives). i = MARR = 12% and n = 5 yea rs SOLUTION Cost will be negative and salvage value will be positive. The formula will be: 𝑃𝑟𝑒𝑠𝑒𝑛𝑡 𝑊𝑜𝑟𝑡ℎ = −𝐹𝑖𝑟𝑠𝑡 𝐶𝑜𝑠𝑡 − 𝑂𝑝𝑒𝑟𝑎𝑡𝑖𝑛𝑔 𝐶𝑜𝑠𝑡 + 𝑆𝑎𝑙𝑣𝑎𝑔𝑒 𝑉𝑎𝑙𝑢𝑒 PW = −FC − OC + SV

7. The operating cost is per year with the same amount therefore we are going to use the formula for annuity. (ordinary annuity) PW = A (P/A, i%, n) The salvage value (scrap value or the value of the asset when it is no longer usable) has a future worth , thus, we need to convert it into present worth using the formula for single cash flow. PW = F (P/F, i%, n) *The underlined value is the function symbol notation. Refer to the table of summary of formulas in ppt page 3.

8. Since Alternative B (𝑃𝑊 = −₱ 45,447) is numerically higher than Alternative A (𝑃𝑊 = −₱ 55, 434), select Alternative B The PW analysis must always compare alternatives for equal life service (i.e. alternatives must end at the same time). However, in real life study, not all the alternatives that is analyzed has equal life.

9. PW ANALYSIS- DIFFERENT LIFE (SPECIFIC STUDY PERIOD) Techniques to analyze alternatives with different life using PW method are as follows: Least Common Multiple (LCM) ▪ This technique is always used in PW analysis to compare alternatives with different life unless the problem specified otherwise. In this technique, you use the LCM of the service life of the alternatives that is being compared. Specific Study Period ▪ In this technique, you decide how many years you want to be studied.

10. Compare the machines below using PW analysis at MARR = 10% per year. Select which machine is better using LCM technique. Machine A Machine B First Cost (FC) in ₱  20,000 30,000 Annual Cost (AC) in ₱/yr 9,000 7,000 Salvage Value (SV) in ₱ 4,000 6,000 Life in years 3 6 EXAMPLE PW ANALYSIS- DIFFERENT LIFE (LCM)

11. Notice that the alternatives being compared have different life. So, we’ll use the LCM method. Since the service life of the machines are 3 and 6 its LCM = 6, thus, we will repurchase Machine A after 3 years. As a result, within the 6 years there will be two Machine A. For a PW analysis, we are going to convert the cash-flow to present worth and select the one with the highest PW (ME alternatives). i = MARR = 10% and n = 6 years SOLUTION

12. for PW analysis we need to move all the values in the cash flow into time zero for it the be in present worth. Cost will be negative and salvage value will be positive. The formula will be: 𝑃𝑊𝐴 = −𝐹𝐶 − 𝐴𝐶 − (𝑟𝑒𝑝𝑢𝑟𝑐ℎ𝑎𝑠𝑒 𝑐𝑜𝑠𝑡) + 𝑆𝑉 𝑓𝑜𝑟 1𝑠𝑡 𝑚𝑎𝑐ℎ𝑖𝑛𝑒 + 𝑆𝑉 𝑓𝑜𝑟 2𝑛𝑑 𝑚𝑎𝑐ℎ𝑖𝑛𝑒

13. The annual cost (AC) is per year with the same amount therefore we are going to use the formula for annuity. PW = A (P/A, i%, n) PW = 9,000 (P/A, 10%, 6) The repurchase cost has a future worth since we will repurchase after 3 years, thus, we need to convert it into present worth using the formula for single cash flow. PW = F (P/F, i%, n) PW = 20,000 (P/F, 10%, 3)

14. The salvage value (SV) (scrap value or the value of the asset when it is no longer usable) has a future worth, thus, we need to convert it into present worth using the formula for single cash flow. PW = F (P/F, i%, n) For the 1st machine: PW = 4,000 (P/F, 10%, 3) For the 2nd machine: PW = 4,000 (P/F, 10%, 6)

15. PW for the Machine A is – 68,960

16. For Machine BPW analysis we need to move all the values in the cash flow into time zero for it the be in present worth Cost will be negative and salvage value will be positive. The formula will be: 𝑃𝑊𝐵 = −𝐹𝐶 − 𝐴𝐶 + 𝑆𝑉

17. The annual cost (AC) is per year with the same amount therefore we are going to use the formula for annuity. PW = A (P/A, i%, n) PW = 7,000 (P/A, 10%, 6) The salvage value (SV) (scrap value or the value of the asset when it is no longer usable) has a future worth, thus, we need to convert it into present worth using the formula for single cash flow. PW = F (P/F, i%, n) PW = 6,000 (P/F, 10%, 3)

18. Since Alternative B (𝑃𝑊 = −₱ 57,100) is numerically higher than Alternative A (𝑃𝑊 = −₱ 68,960), select Alternative B .

19. Compare the machines below using PW analysis at MARR = 10% per year. Select which machine is better for a study period of 3 years. EXAMPLE PW ANALYSIS- DIFFERENT LIFE ( SPECIFIC STUDY PERIOD) Machine A Machine B  First Cost (FC) in ₱  20,000 30,000 Annual Cost (AC) in ₱/yr 9,000 7,000 Salvage Value (SV) in ₱ 4,000 6,000 (after 6 years) 10,000 (after 3 years) Life in years 3 6

20. For a PW analysis, we are going to convert the cash-flow to present worth and select the one with the highest PW (ME alternatives). i = MARR = 10% and n = 3 years Since the problem has a specific study period of 3 years, n = 3. Disregard all estimates after 3 years. The PW analysis we need to move all the values in the cash flow into time zero for it the be in present worth SOLUTION

21.  Machine A Cost will be negative and salvage value will be positive. The formula will be:

22. THE FUTURE WORTH METHOD Future worth (FW) analysis over a specified study period is often utilized if the asset (equipment, a building, etc.) might be sold or traded at some time before the expected life is reached. Analysis of alternatives using FW values is especially applicable to large capital investment decisions when a prime goal is to maximize the future wealth of a corporation’s stockholders. ”FW is exactly like PW analysis except that we should calculate the future worth” Computation of this type of analysis is somewhat the same with the present worth analysis except that instead of comparing the alternatives in its present worth, we will compare alternatives by its future worth

23.  Machine B Cost will be negative and salvage value will be positive. The formula will be: Marginally (for a period of 6 years), Alternative A (𝑃𝑊 = −₱ 39,376) is numerically higher than Alternative B (𝑃𝑊 = −₱ 39,895), select Alternative A.

24. General guidelines for FW analysis are the same with PW analysis: Profit, revenue, salvage value (all inflows to an organization) will be assigned with positive sign. The costs (outflows) will be assigned with negative sign. Cash Inflows → Positive (+) Cash Outflows → Negative (-) EVALUATION ▪ For one project, if FW ≥ 0, it is acceptable ▪ For mutually exclusive (ME) alternatives, select one with the numerically highest FW, that is, less negative or more positive. ▪ For independent alternatives, accept all alternatives with FW ≥ 0

25. EXAMPLE FW ANALYSIS- DIFFERENT LIFE Two processes can be used for producing a polymer that reduces friction loss in engines. Process T will have a first cost of ₱ 750,000, an operating cost of ₱ 60,000 per year, and a salvage value of ₱ 80,000 after its 2-year life. Process W will have a first cost of ₱ 1,350,000, an operating cost of ₱ 25,000 per year, and a ₱ 120,000 salvage value after its 4-year life. Process W will also require updating at the end of year 2 at a cost of ₱ 90,000. Which process should be selected on the basis of a future worth analysis at an interest rate of 12% per year? Process T Process W  First Cost (FC) in ₱  750,000 1,350,000 Operating Cost (OC) in ₱/yr 60,000 25 ,000 Salvage Value (SV) in ₱ 80,000 120,000 Updating Cost n/a 90,000 ( in year 2) Life in years 2 4

26. SOLUTION Notice that the alternatives being compared have different life. So, we’ll use the LCM method. Since the service life of the machines are 2 and 4 its LCM = 4, thus, we will repurchase Process T after 2 years and have 2 life cycle of this process. For a FW analysis, we are going to convert the cash-flow to future worth, that is year 4, and select the one with the highest FW (ME alternatives). i = MARR = 12% and n = 4 years.

27. Process T (𝐹𝑊 = −₱ 2,227,347) is numerically higher than Process W (𝐹𝑊 = −₱ 2,236,630), select Process T.

28. THE ANNUAL WORTH METHOD Annual worth (AW), is also known as equivalent annual worth (EAW), equivalent annual cost (EAC), annual equivalent (AE), and equivalent uniform annual cost (EUAC). The AW value, which has the same interpretation as A used thus far, is the economic equivalent of the PW and FW values at the MARR for n years.

29. It is a method of comparison that is often desired in engineering economic study compared to PW analysis, FW analysis, and rate of return, since annual worth method offers an advantage because the AW value needs to be calculated for only one life cycle. The AW value determined over one life cycle is the AW for all future life cycles. 𝑨𝑾𝟏 𝒄𝒚𝒄𝒍𝒆 = 𝑨𝑾𝟐 𝒄𝒚𝒄𝒍𝒆𝒔 = 𝑨𝑾 𝟑 𝒄𝒚𝒄𝒍𝒆𝒔 … Therefore, it is not necessary to use the LCM technique to satisfy the equal-service requirement.

30.  EVALUATION ▪ One alternative (Single Project): If AW ≥ 0, the requested MARR is met or exceeded and the alternative is economically justified. ▪ Two or more alternatives (ME alternatives): Select the alternative with the AW that is numerically largest, that is, less negative or more positive. This indicates a lower AW of cost for cost alternatives or a larger AW of net cash flows for revenue alternatives. ▪ If the projects are independent, the AW at the MARR is calculated. All projects with AW ≥ 0 are acceptable.

31. EXAMPLE ANNUAL WORTH ANALYSIS RR Racing and Performance Motor Corporation wishes to evaluate two alternative CNC machines for NHRA engine building. Use the AW method at 10% per year to select the better alternative. Machine R Machine S  First Cost (FC) in ₱  250,000 370,000 Operating Cost (OC) in ₱/yr 40,000 50,000 Salvage Value (SV) in ₱ 20,000 30,000 Life in years 3 5

32. SOLUTION For AW analysis, we are going to convert the cash-flow to its annual equivalent worth, and select the one with the numerically highest AW (ME alternatives). i = MARR = 10% AW = −Annual equivalent of First Cost − Annual Operating Cost + Annual equivalent of Salvage Value

33. Machine R (AW =−₱ 134,486) is numerically higher than Machine S (AW =−₱ 142,691), select Machine R.

34. EXAMPLE ANNUAL WORTH (ONE OR TWO LIFE CYCLES) A machine has a first cost of ₱ 20,000, an annual operating and maintenance cost of ₱ 5,000, and a salvage value of ₱ 3,000 after 2 years. Calculate the AW at i = 10% for (a.) one life cycle (b.) two life cycle.

35. SOLUTION To find the AW, we have to convert all cash flows into annual amount “A”. Since the annual operating and maintenance cost is already in annual value, we will leave it as it is.

36. Note that we have to move the cash flows in year 2 to present worth (year 0) before we convert it to annual worth for 4 years. Also, the cost of buying the machine which is ₱ 20,000 will be subtracted by the salvage value of ₱ 3,000 of the first machine, that will be ₱ 17,000.

37. Notice that the annual worth of one cycle is also equal to the annual worth of 2 cycles. This proves our previous statement, “AW value determined over one life cycle is the AW for all future life cycles” holds true.

38. ACTIVITY 1. National Homebuilders, Inc., plans to purchase new cut-and-finish equipment. Two manufacturers offered the estimates below:  Vendor A Vendor B First Cost (FC) in ₱ 15,000 18,000 Annual Cost (AC) in ₱ 3,500 3,100 Salvage Value (SV) in ₱ 1,000 2,000 Life in years 6 9 Determine which vendor should be selected on the basis of a present worth comparison, if the MARR is 15% per year.

39. 2. An engineer is considering two robots for purchase by a fiber-optic manufacturing company. Robot X will have a first cost of ₱80,000, an annual maintenance and operation (M&O) cost of ₱30,000, and a ₱40,000 salvage value. Robot Y will have a first cost of ₱97,000, an annual M&O cost of ₱27,000, and a ₱50,000 salvage value. Which should be selected on the basis of a future worth comparison at an interest rate of 15% per year? Use a 3-year study period.