1 Process Analysis and Design Cost Accounting and Profitability Analysis

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Process Analysis and DesignProcess Analysis and Design

Cost Accounting and Cost Accounting and Profitability AnalysisProfitability Analysis

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IntroductionIntroduction• 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

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– 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 EarningsSimple Measures to Estimate Earnings

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Return on InvestmentReturn 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.

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Payout TimePayout 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.

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Time Value of MoneyTime 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

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Compounded InterestCompounded 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 = $106 in 100 years Present value P = $106 / (1 + i ) 100 If i = 0.05, P = $7,604 If i = 0.2, P = $0.012

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Nominal and Effective Interest RatesNominal 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%

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Continuous InterestContinuous 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%

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AnnuitiesAnnuities• Present value (P) of distributing an equal payment

on a regular basis (R)– When payment is at the end of period,

P = R k=1n 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]R R R R

n

P

F

n

P

Figure 1. Timelines for payments.

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AnnuitiesAnnuities• Present value (P) of distributing an equal payment

on a regular basis (R)– When payment is at the beginning of period,

P = R k=1n 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.

R R R R

n

P

RF

n

R R RR

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• 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 PaymentsExample 1. Annuity Payments

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• 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 = 4412 = 528 and R = $10

F = R [(1 + i ) n + 1 – (1 + i )] / i = $10,976

Example 2. Future Value of Regular Example 2. Future Value of Regular PaymentsPayments

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Continuous Payment over a Fixed PeriodContinuous Payment over a Fixed Period• Present Value

P = R* [1 – (1 + i* ) –n*] / i*

= (R / m)[1 – (1 + i / m) –mn] / (i / m)

whereR = average yearly payment = R dt

– As m , P R [1 – e –in ] / i

• Future Worth F = P e in =R [e in – 1] / i

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• 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, andR = 1,000

P =R [1 – e –in ] / i = $21,752

Example 3. Continuous PaymentsExample 3. Continuous Payments

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PerpetuitiesPerpetuities• 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 = C0 + C / [(1 + i ) z – 1] where C0 = original price

Figure 3. Replacement cost into perpetuity.

C C C C

P

C

Z Z

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Example 4. Comparison of Two ReactorsExample 4. Comparison of Two Reactors• A stainless steel and a carbon steel reactor Reactor A Reactor B (SS) (CS)

Original cost (C0) $10,000 $5,000

Life (years) 8 3

Replacement (C = C0 – salvage) $8,000 $5,000

K (@ i = 10%) $16,995 $20,105

– Based on the capitalized cost into perpetuity, reactor A is actually cheaper.

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Using Time Value of Money for Cost and Using Time Value of Money for Cost and Project ComparisonsProject 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.

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Example 5. Project ComparisonExample 5. Project Comparison• Two 5 year projects

A B Capital, fixed & working ($) 3106 300,000 Income before taxes ($/yr) 106 200,000

• Net present values NPV (A) = –3106 + 106 [1 – (1 + i )–5] / i NPV (B) = –3105 + 2105 [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.

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• Income on annualized basis R (A) = 106 – 3106 i / [1 – (1 + i )–5] R (B) = 2105 – 3105 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.

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Example 6. Cost Comparison for Equal Example 6. Cost Comparison for Equal LifetimesLifetimes• 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.

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Cost Comparison for Different Project Cost Comparison for Different Project LivesLives• 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.

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Example 7. Cost Comparison with Example 7. Cost Comparison with Different LivesDifferent Lives• A carbon steel and a stainless pump

CS SS Purchase price (C0) $5,000 $8,000 Salvage value (C0 – 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

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• Compare projects into perpetuity. NPV = C0 + R / i + C / [(1 + i ) z – 1]

CS SS C0 $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.

R R RR R RC0 C C C C C

z = 4 or 8

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• 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

C0 = 5000 C = C0

200 200 200 200 200 200 200

C0 = 8000

150 150 150 150 150 150 150 150

2000

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• Annualized costs for each project NPV = C0 + R [1 – (1 + i ) –z] / i – (C0 – C) / (1 + i ) z

X = NPV i / [1 – (1 + i ) –z]

CS SS C0 $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.