wc - oucoecs.ou.edu/dimitrios.v.papavassiliou/che4253/econ-lec.pdf · • marshall and swift. 1....

ECONOMIC DESIGN CRITERIA

BASIC ECONOMIC TERMS

• Total Capital Investment, TCI or I(Total Capital Investment)=(Fixed Capital Investment)+

(Working Capital)TCI = FCI + WC

or: I= IF+IW

reduced risk →• Total Fixed Capital Investment, FCI or IF

FCI = (Direct Costs) + (Indirect Costs)

• Working Capital, WC or IW

Cash, raw materials, stock, etc. About 10-20% of TCI.


• Product Cost, C C=CI+CQ+CO +CG

• Fixed Charges, CIDo not depend on production level (insurance, propertytaxes, depreciation, rent etc.)

• Direct Production Cost, CQLabor, utilities, raw materials, maintenance, supplies,

royalties etc. • Plant Overhead, CO

Recreation, employee facilities, packaging etc.

• General Expenses, CGAdministration, marketing, R&D, distribution.


• Income from Sales, S in ($/yr) • Gross Earnings, R in ($/yr)

R = S - C• Net Earnings, P in ($/yr)

P = R - eIF - (R - d IF) t

(Net Profit) = (Gross) - (Depreciation) - (Taxes)• Depreciation rate

Recovery of Investment, eTaxation purposes, d

Straight line depreciation, e=1/nDepreciation with capital reinvestment (sinking fund method) ( ) 11 −+

= niie


Salvage Value

Net cash obtainable from the sale of used property (above charges for removal and sale)

Scrap value: Salvage value after dismantling a unit.

Present Value

Book Value : (Total Capital Investment) - (All Depreciation)

Market Value : Cash obtainable from selling the unit.

Replacement Value : Cost of obtaining the same property.


Depreciation

Reduction in value due to any causes.

Example: Pump

Cost : CV= $12,000Scrap value : VS= $2,000

Depreciation : CV - VS = $10,000

For engineers, depreciation is considered as a cost for using the equipment.

DEPRECIATION

Types Of Depreciation

Physical: Wear and Tear, corrosion, accidents, age deterioration.

Functional: All other causes.

Obsolescence: Due to technological advances.

Depletion: Loss due to materials consumed. Applicable to Natural Resources (timber, mineral, oil deposits)

IRS: “A reasonable allowance for the exhaustion, wear and tear of property used in the trade or business including a reasonableallowance for obsolescence”


Service LifeThe IRS has determined various values(See P&T for complete list).

Group 1: General Business Assets. (Office furniture, Land, Buildings, etc)

Group 2: Non-manufacturing activities: (Agriculture, Fishing, Mining, etc.)

Group 3: Manufacturing, e.g. Petroleum Refining: 16 years. Chemicals 11 years.

Group 4: Transportation, Communication and Public Utilities: (Electrical, Gas, Motor transport, Radio and TV broadcasting, railroad, etc.)


Total Capital Investment, I

IDDirect Costs

IIIndirectCosts

IWWorking Capital

S, Income from sales

R

CProduct Cost

(R-d IF) tTaxes

D=e IF

Depreciation

P, Net Earnings

IF CF


Cumulative Cash Position

time

$

n0

COST ESTIMATION

Fixed Capital Investment : Cost of equipment and facilities

FCI = (Direct Costs) + (Indirect Costs)Direct Costs: 1. Purchased equipment: Columns, Heat Exchangers, pumps, tanks, etc. 2. Equipment Installation 3. Piping (includes insulation) 4. Instruments and Control 5. Electrical Equipment. 6. Buildings: Process, Administration, Maintenance shops, etc. 7. Site Preparation 8. Service Facilities: steam, water, air, fuel, etc. Waste treatment, fire control, offices, etc. 9. Land

COST ESTIMATION

Indirect Costs:

1. Engineering and Supervision: Administrative and Design. Supervision and Inspection. 2. Construction Expenses 3. Contractor's fee 4. Contingency. 5. Start up expenses

Table 1, p. 159 in P&T

COST ESTIMATION

Types Of Cost Estimates

1. Order of Magnitude estimate. Extrapolate similar plant cost Accuracy: over 30%

2. Study Estimate. Knowledge of major pieces of equipment Accuracy: ± 30%

3. Preliminary Estimate. Enough for budget authorization. Accuracy: ± 20%

4. Definitive Estimate. Based on basic Engineering and quotes from suppliers and contractors. Accuracy: ± 10%

5. Detailed Estimate. Based on Detailed Engineering.Accuracy: ± 5%

COST ESTIMATION

Cost Indexes

Present Cost=(original cost at time t)*

• Marshall and Swift.1. All industry-equipment index. Arithmetic average of 47

equipment types. 2. Process-industry equipment index. Weighted average of 8

of these: cement 2% paint 5%chemicals 48% paper 10%clay products 2% petroleum 22%glass 3% rubber 8%

M&S was 100 in 1926. Published in “Chemical Engineering”.

ttimeatvalueindexnowvalueindex

COST INDEXES• Engineering News-Record Construction Cost index.

Steel, lumber, labor, concrete.Published in “Engineering News-record”.ENR value reported based on 100 in 1913, 1949 or 1967.

• Nelson-Farrar Refinery Construction Cost index.Skilled and common labor, iron and steel, building materials, miscellaneous equipment.Published in “Oil and Gas Journal”.N-R value of 100 in 1946.

• Chemical Engineering Plant Cost Index. Chemical Plants.Equipment, machinery, Engineering and supports 61% supervision 10%Installation labor 22% Buildings, material, labor 7% Published in “Chemical Engineering”.PCI value of 100 in 1957-59.

Methods For Estimating Capital Investment

1. Detailed-Item Estimate.

All items in the direct and indirect cost are evaluated with as much detail as possible. All specs are known.(± 5% accuracy, contractor’s estimate)

2. Unit-Cost Estimate.

Prices from quotations or index-corrected records.

See p. 181 in P&T.(10-20% accuracy, definitive or preliminary estimate)

( ) ( )[ ] FndeeLyxxL fdfHfMfMfEEC ∑ ∑ ∑ ∑+++++=


3. Percentage of Delivered-Equipment Cost.

All items in the direct and indirect cost are evaluated as a percentage of the delivered-equipment cost.(definitive estimate in certain cases, ± 10% )

See Table 17, p. 183 in P&T.

4. Estimation based on “Lang” factors .

Named after Lang (1947). The Fixed Capital Investment is found by multiplying equipment cost by a factor (see Table 18, p. 184 in P&T)(± 30% accuracy, order of magnitude estimate)

( )[ ] IfEfEfEfEC ∑ ∑ ++++= l321


4. Estimation based on “Lang” factors .

fF: cost factor for field laborfp: cost factor for piping materialsfm: cost factor for miscellaneous (insulation, foundations etc.)fI: indirect cost factorEi: cost of already installed equipmentA: incremental cost for corrosion resistant materials

( )[ ] IimpF fAEfffEC +++++= 1


5. Power factor applied to plant-capacity .

Order of magnitude estimates based on the fixed capital investment for a similar plant.

x: between 0.6 and 0.7 R: Capacity ratio, (new facility)/(old facility)

If the direct, D, and indirect, I, costs are known, then:

( )xold RCC=

( )[ ] fIRDC x +=


5. Power factor applied to plant-capacity .

The factor f is a composite of the geographical labor cost index, the area productivity index and a material and equipment index.

Example:Plant in Dallas: $100,000 costWhat is the cost for a similar plant in Los Angeles?SW labor rate=0.88 PC labor rate=1.22SW productivity=1.04 PC productivity=0.89 (Tab. 20)

Relative Labor=(PC/SW)=1.22/0.88/=1.3864Relative Productivity=(PC/SW)=0.89/1.04=0.8558

=1.62*100000=162,000Cost

produclaborCost SW

PCSW

SWPCSWf

=


6. Turnover ratios.

Very rapid and very crude estimation. Can be off by 100% or more.

Turnover Ratio=(gross annual sales)/FCI

It can be anywhere between 0.2 and 5. Assumption for CPI (Chemical Process Industry): TR=1

Fixed Capital Investment CostTable 4 Table 17 Table 26

FluidDirect Costs ProcessingOnsite PlantPurchased Equipment E EInstallation 6-14% IF 47 % E 22-55 % E Instrumentation 2-8% IF 18 % E 6-30 % E Piping 3-20% IF 66 % E 10-80 % E Electrical 2-10% IF 11 % E 10-40 % E

OffsiteBuildings 3-18% IF 18 % E 10-70 % E Yard Improvement 2-5% IF 10 % E Included in

Service FacilitiesService Facilities 8-20% IF 70 % E 40-100 % E Land 1-2% IF 6 % E 1-2 % IF

(or 4-8 %E)

Fixed Capital Investment Cost

Table 4 Table 17 Table 26 Fluid

Indirect Costs ProcessingPlant

Engineering 4-21% IF 33 % E 5-30% DConstruction 4-16 % IF 41% E Included in

Contractor’s feeContractor’s Fee 2-6 % IF 5 % 6-30 % D

(direct+eng+const)

Contingency 5-15% IF 10 % 5-15% IF(direct+eng

+const)Working Capital

10-20% IF 15 % TCI 10-20% TCI

BASIC ECONOMIC TERMSMust Be Done Projects: A MUST BE DONE Project is a project that has no specific revenue or savings identified. Examples •Replacement of obsolete equipment •Waste treatment/ Pollution Prevention units mandated by law. •Safety improvement. How do we calculate a Total Annualized Cost for these type of Projects? Easy!!

TAC= AFC + OC TAC : Total Annualized Cost AFC :Annualized Fixed Cost OC : Operating Costs

BASIC ECONOMIC TERMSMust Be Done Projects: CONCEPT: You borrow the Fixed Capital Investment. The annual payments need to recover the FCI. • Straight line depreciation:

AFC=FCI/nwhere n is the number of service years expected.

• Sinking fund method:Need to recover S=FCI*(1+i)n

AFC is similar to an annuity payment

1)1()1(

1)1( −++=

−+= n

n

n iiiFCI

iiSAFC

BASIC ECONOMIC TERMSMust Be Done Projects

• Sinking fund method:

Thus

YOU WILL NEED THIS FOR DESIGN LAB.

OCi

iiFCITAC n

n

+−+

+=1)1(

)1(


• Interest - Return on Investment, ROI• Simple Interest:

I=P * i * n

I : Total interest paid P : Principal or Capital Borrowedi : Interest rate for one period of time n : Number of periods.

Repayment is

S = P + I = P * (1 + i * n)

Usually : 1 period = 1 year . For less than 1 year we have:

Ordinary Simple Interest = P* i * d/360

Exact Simple Interest = P* i * d/365

BASIC ECONOMIC TERMS - ROI

• Compound Interest:At the end of each interest period the interest is added to the principal.

Period Principal Interest earned Compound amount at start of S at the end of period period

1 P P i P(1+i)2 P(1+i) P(1+i)i P(1+i)2

3 P(1+i)2 P(1+i)2i P(1+i)3

……..n P(1+i)n-1 P(1+i)n-1i P(1+i)n

Repayment is S=P*(1 + i )n

(1 + i )n : Discrete single payment compound-amount factor

Nominal vs. Effective Interest

• Nominal Interest:

Interest rate for 1-year period but compounded for periods different than one year.

Example : P=1000, at 6% compounded every 6 months. At the end of six months the interest is:

Iafter 1/2 year= P 0.06 / 2 = 30Safter 1/2 year= P (1 + 0.06/2) = 1030

ThenIafter 1 year= P (1 + 0.06/2) 0.06 / 2 = 30.90Safter 1 year= P(1 + 0.06/2)(1 + 0.06/2) = 1060.9

Nominal vs. Effective Interest Rate

• Nominal Interest:General Formula:

Safter 1 year= P*(1 + r/m)m

r : Nominal annual Interestm: Number of periods of compounding per year.

Safter n years= P*(1 + r/m)m*n

• Effective Annual Interest Rate:Simple interest that will produce the same total interest at the end of one year.

Safter 1 year = P*(1 + r/m)m Nominal= P*(1 + ieff) Effective

Then ieff=(1+r/m)m-1

INTEREST - ROI

Example :

P = $1000Interest = 2% monthly Total Time = 2 years.

•Simple : S=1000 (1+0.02*24) = $1480

•Compounded : S= 1000(1+0.02)24 =$1608

•Nominal Interest Rate: 2 x 12 = 24 % (annual)

•Effective Rate: ieff=(1+0.02)12-1=0.268 (26.8%)

Continuous Compound Interest

At time n S = P + i*P*n

At time n+dn S+dS = S+i*(P + i*P*n)dn

Then dS=i*S*dn

Integrate from time zero (S(0)=P) to time n to get

ln(Sn/P) = i*n and Sn = P*exp(i*n)

Compare to Sn = P*(1 + ieff)n

Effective Annual Interest Rate:

ieff = ei-1

INTEREST

Repayment (Sn= P + I)

Simple Interest: Sn= P * (1 + i * n)

Compound Interest Sn= P*(1 + i )n

Nominal Interest Sn= P*(1 + r/m)m*n

Continuous Compound Interest Sn = P*exp(i*n)

Effective Interest Rate Sn = P*(1 + ieff)n

Present Worth: Solve for P (Note: P is S0)


• Present Worth

Present principal that will yield a desired amount in the future.

Continuous compounding Sn = P*exp(i*n)

So = P = Sn *exp( - i*n )

Discrete compounding Sn = P*(1 + i )n

So = P =

: Discrete single-payment present-worth factor.

Discount (used in bonds): Sn - So

nn

iS

)1( +

ni)1(1+

Present Worth - example:a) CS Reactor at $10,000 and 2 year life.

Maintenance: 5% of equipment cost per year.b) SS Reactor at $30,000 and 6 year life. No maintenance.

Salvage value: $6,000Assume i=15% and n = 6 yearsPresent worth CS Reactor SS ReactorSo = Sn *exp( -0.15*n )

n=0 10,000 30,000n=2 7,408 0n=4 5,488 0

Salvage value (income) 0 - 2,440MaintenanceSo 1,978 0

Total 24,874 27,560

( )( )∫−=

6

0

15.0000,1005.0 dne n


• Annuities

Series of equal payments occurring at equal intervals of time.

Common type : Payment at the end of each interest interval.

Annuity Term : Time from beginning of the first to the end of the last payment periods.

Amount Of Annuity : Sum of all the payments and the interest accumulated at the end of the period.

Annuities - Payment Calculation

Let R : PaymentS : Amount of Annuityi : Interest rate

Total Interestaccumulated

0 1 2 3 (n-1) n

R R R R R (1+i)n-1

R (1+i)n-2

R (1+i)


Total amount repaid: S= R

Multiply both sides by (1+i) to get

S(1+i)=Rand subtract:

Then

If S is a loan, then S = P*(1+i)n

and the payment is

∑=

−+n

k

kni1

)1(

∑=

+−+n

k

kni1

1)1(

RiRiS n −+= )1(

1)1( −+= ni

iSR

1)1()1(−+

+= n

n

iiiPR


This is the same formula obtained assuming a revolving loan withzero principal at the end of n years.

0 1 2 3 (n-1) n

Principal left1 P(1+i)-R2 (P(1+i)-R)(1+i)-R: n P(1+i)n-R(1+i)n-1-....- R(1+i)-R

Present Worth Of An Annuity

Principal needed to be invested NOW at compounded interest i to yield the amount of the annuity, S, at the end of the annuity term.

n

n

iiiRP

)1(1)1(

+−+=

Perpetuities and Capitalized Cost

• Perpetuity

Annuity in which periodic payments continue indefinitely.

Example :Asssume a piece of equipment costs $12,000. It lasts for 10 years, after which it is going to be replaced. The scrap value (VS) is $2,000.

Therefore, there is a one time cost of CV=$12,000 now and an amount P needed to obtain 10,000 every 10 years. Assume an interest rate of 6%. How much is P?

Need S=10,000 +P at the end of 10 years. Then:

S=10,000+P = P (1+i)10

Perpetuities and Capitalized CostExample :

Then, P=$12,650 (present value of the renewable perpetuity)

In general:S=P(1+i)n

P=S-CR

CR :Replacement cost =CV-VS

Then

The total cost now is : CV+ P= $24,650.

Capitalized Cost, K K= CV+P= CV +

1)1( −+= n

R

iCP

1)1( −+ nR

iC

Capitalized Cost - example:a) Mild Steel Reactor at $5,000 and 3 year life. b) SS Reactor at $15,000 and 12 year life.

Scrap value: 0 for both types of reactorsWhich one should be installed?

Comparison based on Capitalized cost.

Reactor Capitalized Cost

Mild Steel K=5,000 + =$ 31,180

SS K=15,000 + =$ 31,180

SS reactor is preferred. If SS reactor < 11.3 years, Mild Steel reactor is preferred.

1)06.1(000,5

3 −

1)06.1(000,15

−n

wc - oucoecs.ou.edu/dimitrios.v.papavassiliou/che4253/econ-lec.pdf · • marshall and swift. 1....

Documents