1. 1 Process Analysis and Design Cost Accounting and Profitability Analysis

2. 2 Introduction Fixed Costs –Direct investment as well as overhead and management associated with this investment –Capital investment costs Variable Costs –Raw material, labor, utilities, and other costs that are dependent on operations –Manufacturing costs

3. 3 –Definitions Gross profit = Gross sales – manufacturing cost Net profit before taxes = Gross profit – SARE (Sales, Administration, Research and Engineering) expenses (10% sales) Net annual earnings = Net profit before taxes – taxes on net profit –Economic Measures Return on investment (ROI) Payout time Simple Measures to Estimate Earnings

4. 4 Return on Investment Definition ROI = (net annual earnings) / (fixed and working capital) Characteristics –Typical minimum desired ROI  15% (or 30% before taxes) –ROI does not take time value of money (i.e., the timing of expenses and incomes) into account. –It is only useful for a mature plant project when startup cost is not significant.

5. 5 Payout Time Definition Payout time = (total capital investment) / (net annual profit before taxes + annual depreciation) Characteristics –The depreciation that was part of the manufacturing cost is added back and cancelled. –This measure represents the total time to recover investment based on the net income without depreciation. –Like ROI, the payout time does not take time value of money (i.e., the timing of expenses and incomes) into account.

6. 6 Time Value of Money The value of money changes due to: –Interest, which reflects rent paid on the use of money –Returns received from competing investments. Consequently, the investment must compensate the loss of opportunity to invest elsewhere. –Inflation, which can be compensated in the interest rate What is the correct interest rate for a company to choose? –The rate that the company receives for its money when the money is sitting in reverse –A guaranteed rate with enough fluidity

7. 7 Compounded Interest Future Worth F = P (1 + i ) n where i = nominal interest rate, n = number of periods (years), P = present value, and F = future worth Present Value of a Future Value P = F / (1 + i ) n where 1 / (1 + i ) n = discount factor Example Future worth F = $10 6 in 100 years Present value P = $10 6 / (1 + i ) 100 If i = 0.05, P = $7,604 If i = 0.2, P = $0.012

8. 8 Nominal and Effective Interest Rates Future Worth F = P (1 + i /m) mn where i = nominal interest rate n = number of periods (years) for nominal rate m = number of compounding intervals per nominal period Example –For i = 6%, m = 4, and n = 1 year, F = P (1 + 0.06 / 4) 4 = P (1.0614) Effective rate = 6.14%

9. 9 Continuous Interest Future Worth –As m  , F = P (1 + i /m) mn  P e in Effective rate = e i – 1 Example Nominal interest rate i = 6% Effective rate = e 0.06 – 1 = 6.18%

10. 10 Annuities Present value (P) of distributing an equal payment on a regular basis (R) –When payment is at the end of period, P = R  k=1 n 1 / (1 + i ) k = R [1 – (1 + i ) –n ] / i R = P i / [1 – (1 + i ) –n ] = P  (capital recovery factor) = [F / (1 + i ) n ] i / [1 – (1 + i ) –n ] = F i / [(1 + i ) n – 1] RRRR n P F n P Figure 1. Timelines for payments.

11. 11 Annuities Present value (P) of distributing an equal payment on a regular basis (R) –When payment is at the beginning of period, P = R  k=1 n 1 / (1 + i ) k – 1 = R [(1 + i ) – (1 + i ) 1 – n ] / i R = P i / [(1 + i ) – (1 + i ) 1 – n ] = [F / (1 + i ) n ] i / [(1 + i ) – (1 + i ) 1 – n ] = F i / [(1 + i ) n + 1 – (1 + i )] Figure 2. Present and future value of annuities. RRRR n P RF n RRRR

12. 12 A $10,000 loan at a nominal (annual) rate of 12% is to be repaid in 60 monthly installments at the end of each month. What is the monthly payment? –For i = 0.12/12 = 0.01, n = 60 and P = $10,000 R = P i / [1 – (1 + i ) –n ] = $222.4/month Example 1. Annuity Payments

13. 13 For a life insurance policy with a lump sum payment starting at 65, monthly payments start at 21 by a premium of $10 at the beginning of each month. If the nominal rate is 3%, what is the value of the lump sum? –For i = 0.03/12 = 0.0025, n = 44  12 = 528 and R = $10 F = R [(1 + i ) n + 1 – (1 + i )] / i = $10,976 Example 2. Future Value of Regular Payments

14. 14 Continuous Payment over a Fixed Period Present Value P = R* [1 – (1 + i* ) –n* ] / i* = (  R / m)[1 – (1 + i / m) –mn ] / (i / m) where  R = average yearly payment =  R dt –As m  , P  R [1 – e –in ] / i Future Worth F = P e in =  R [e in – 1] / i

15. 15 The energy bill for a boiler is prorated at $1,000 per month. For a nominal annual rate of 10%, what is the present value of energy cost for a two year operation? –For i = 0.10 / 12 = 0.00833, n = 24, and  R = 1,000 P =  R [1 – e –in ] / i = $21,752 Example 3. Continuous Payments

16. 16 Perpetuities An expenditure for an infinite time period –As n  , P =  R [1 – e –in ] / i  R / i Periodic replacement of process equipment –If C is to be paid at intervals of z years, P =  k=1  C / (1 + i ) kz = C / [(1 + i ) z – 1] where C = replacement cost (cost – salvage value)  Capitalized cost K = C 0 + C / [(1 + i ) z – 1] where C 0 = original price Figure 3. Replacement cost into perpetuity. CCCC P C ZZ  

17. 17 Example 4. Comparison of Two Reactors A stainless steel and a carbon steel reactor Reactor A Reactor B (SS) (CS) Original cost (C 0 )$10,000 $5,000 Life (years) 8 3 Replacement (C = C 0 – salvage) $8,000 $5,000 K (@ i = 10%)$16,995 $20,105 –Based on the capitalized cost into perpetuity, reactor A is actually cheaper.

18. 18 Using Time Value of Money for Cost and Project Comparisons Criteria –Net present value (NPV) of project with a given rate of return (i) Basis for comparison of projects with different payment schedules but similar lifetimes –Annualized payments with a given rate of return (i) For comparison of projects with different lifetimes –Calculated rate of return (i*) with NPV = 0 Interest rate for comparison with a competing investment Magnitudes in the investment are not considered.

19. 19 Example 5. Project Comparison Two 5 year projects A B Capital, fixed & working ($)3  10 6 300,000 Income before taxes ($/yr) 10 6 200,000 Net present values NPV (A) = –3  10 6 + 10 6 [1 – (1 + i ) –5 ] / i NPV (B) = –3  10 5 + 2  10 5 [1 – (1 + i ) –5 ] / i A B NPV (i = 10%) $790,800 $458,200 NPV (i = 20%) –$9,387 $298,120 i* (NPV = 0) 19% 60% –For NPV calculations, a high rate of return favors projects with income payments at beginning.

20. 20 Income on annualized basis R (A) = 10 6 – 3  10 6 i / [1 – (1 + i ) –5 ] R (B) = 2  10 5 – 3  10 5 i / [1 – (1 + i ) –5 ] A B R (i = 10%) $208,600 $120,870 R (i = 20%) –$3,139 $99,685 –For projects with same lives, the conclusions are same as the NPV calculation.

21. 21 Example 6. Cost Comparison for Equal Lifetimes An old car with a higher operating cost and a new car with a lower operating cost Old New Price ($) 2,000 13,000 Operating cost ($/yr)1,000 300 Net present values NPV (old) = 2,000 + 1,000 [1 – (1 + i ) –n ] / i NPV (new) =13,000 + 300 [1 – (1 + i ) –n ] / i Old New NPV (i = 6%, n = 5)$6,212 $14,263 Annualized$1,475/yr$3,386/yr –The old car has a lower NPV.

22. 22 Cost Comparison for Different Project Lives Approaches –Project each project life into perpetuity, then do an NPV calculation. –Put both project lives on the same time basis (use least common multiple, LCM) then do NPV calculations. –Convert all income and costs to an annualized basis.

23. 23 Example 7. Cost Comparison with Different Lives A carbon steel and a stainless pump CS SS Purchase price (C 0 )$5,000 $8,000 Salvage value (C 0 – C) $0 $2,000 Operating cost (R)$200/yr $150/yr Operating life 4 yrs 8 yrs Rate of return = 10% Methods 1. Compare projects into perpetuity 2. Common life for both projects 3. Annualized costs for each project

24. 24 Compare projects into perpetuity. NPV = C 0 + R / i + C / [(1 + i ) z – 1] CS SS C 0 $5,000 $8,000 C $5,000 $6,000 R $200/yr $150/yr z 4 yrs 8 yrs NPV$17,773 $14,747 Figure 4. Payments into perpetuity. RRRRRR C0C0 CCCCC z = 4 or 8

25. 25 Common life for both projects LCM(4, 8) = 8 NPV (CS) = 5,000 + 5,000 / (1 + i ) 4 + 200 [1 – (1 + i ) –8 ] / i = $9,482 NPV (SS) = 8,000 – 2,000 / (1 + i ) 8 + 150 [1 – (1 + i ) –8 ] / i = $7,867 Figure 5. Least common multiple payments. 200 C 0 = 5000C = C 0 200 C 0 = 8000 150 2000

26. 26 Annualized costs for each project NPV = C 0 + R [1 – (1 + i ) –z ] / i – (C 0 – C) / (1 + i ) z X = NPV i / [1 – (1 + i ) –z ] CS SS C 0 $5,000 $8,000 C $5,000 $6,000 R $200/yr $150/yr z 4 yrs 8 yrs NPV $5,634 $7,867 X $1,777 $1,475 –The NPV by itself provides a misleading comparison if the project life is different. –Among the three methods, the first and third incorporate essentially the same results.