1 process analysis and design cost accounting and profitability analysis
TRANSCRIPT
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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.