Total Capital Investment, TCI Or I (Total Capital Investment)=(Fixed Capital Investment)+ (Working Capital) TCI = FCI + WC

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. BASIC ECONOMIC TERMS • Product Cost, C

C=CI+CQ+CO +CG

• Fixed Charges, CI Do not depend on production level (insurance, property taxes, depreciation, rent etc.)

• Direct Production Cost, CQ Labor, utilities, raw materials, maintenance, supplies, royalties etc.

• Plant Overhead, CO Recreation, employee facilities, packaging etc.

• General Expenses, CG Administration, marketing, R&D, distribution. BASIC ECONOMIC TERMS

• 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, e Taxation purposes, d Straight line depreciation, e=1/n i Depreciation with capital reinvestment e = ()+ n − (sinking fund method) 1 i 1 BASIC ECONOMIC TERMS

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. BASIC ECONOMIC TERMS Depreciation

Reduction in value due to any causes.

Example: Pump

Cost : CV= $12,000

Scrap 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 reasonable allowance for obsolescence” BASIC ECONOMIC TERMS Service Life The 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.) BASIC ECONOMIC TERMS S, Income from sales

ID Direct C Costs Product Cost II R Indirect Costs (R-d IF) t I F CF Taxes IW Working Capital D=e IF Depreciation Total Capital Investment, I P, Net Earnings BASIC ECONOMIC TERMS

Cumulative Cash Position $

0 n time 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  index value now  Present Cost=(original cost at time t)*    index value 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”. 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. =[ ()+ + ()+ + + ] C ∑∑E EL fxMx fyML ∑∑feHe fddn fF

See p. 181 in P&T. (10-20% accuracy, definitive or preliminary estimate) Methods For Estimating Capital Investment 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% ) =[ + ()+ + +l] C ∑∑E f1E f2E f3E fI

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) Methods For Estimating Capital Investment

4. Estimation based on “Lang” factors .

=[ (+ + + )+ + ] C E1 fF fp fm Ei A fI fF: cost factor for field labor fp: cost factor for piping materials fm: cost factor for miscellaneous (insulation, foundations etc.) fI: indirect cost factor Ei: cost of already installed equipment A: incremental cost for corrosion resistant materials Methods For Estimating Capital Investment 5. Power factor applied to plant-capacity .

Order of magnitude estimates based on the fixed capital investment for a similar plant. = ()x C Cold R 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:

C=[D()R x +I]f Methods For Estimating Capital Investment 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 cost What is the cost for a similar plant in Los Angeles? SW labor rate=0.88 PC labor rate=1.22 SW productivity=1.04 PC productivity=0.89 (Tab. 20) Relative Labor=(PC/SW)=1.22/0.88/=1.3864 Relative Productivity=(PC/SW)=0.89/1.04=0.8558  PC SW =    =1.62*100000=162,000 f SWCost SWCost SWlabor PCproduc Methods For Estimating Capital Investment 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 Cost Table 4 Table 17 Table 26 Fluid Direct Costs Processing Onsite Plant Purchased Equipment E E

Installation 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 Offsite

Buildings 3-18% IF 18 % E 10-70 % E Yard Improvement 2-5% IF 10 % E Included in Service Facilities

Service 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 Processing Plant

Engineering 4-21% IF 33 % E 5-30% D Construction 4-16 % IF 41% E Included in Contractor’s fee

Contractor’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 TERMS Must 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 TERMS Must Be Done Projects: CONCEPT: You borrow the Fixed Capital Investment. The annual payments need to recover the FCI. • Straight line depreciation: AFC=FCI/n where 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 i (1+ i)n i AFC = S = FCI (1+ i)n −1 (1+ i)n −1 BASIC ECONOMIC TERMS Must Be Done Projects • Sinking fund method:

(1+ i)n i Thus TAC = FCI + OC (1+ i)n −1

YOU WILL NEED THIS FOR DESIGN LAB. BASIC ECONOMIC TERMS • Interest - Return on Investment, ROI • Simple Interest: I=P * i * n I : Total interest paid P : Principal or Capital Borrowed i : 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 1P P i P(1+i) 2 P(1+i) P(1+i)i P(1+i)2 3P(1+i)2 P(1+i)2iP(1+i)3 …….. nP(1+i)n-1 P(1+i)n-1iP(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 = 30

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

Iafter 1 year= P (1 + 0.06/2) 0.06 / 2 = 30.90 Safter 1 year= P(1 + 0.06/2)(1 + 0.06/2) = 1060.9 Nominal vs. Effective Interest Rate • Nominal Interest: General Formula: m Safter 1 year= P*(1 + r/m) r : Nominal annual Interest m: Number of periods of compounding per year. m*n Safter n years= P*(1 + r/m) • Effective Annual Interest Rate: Simple interest that will produce the same total interest at the end of one year. m Safter 1 year = P*(1 + r/m) Nominal = P*(1 + ieff) Effective

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

Example : P = $1000 Interest = 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)

12 •Effective Rate: ieff=(1+0.02) -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)

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

Effective Annual Interest Rate:

i ieff = e -1 INTEREST

Repayment (Sn= P + I)

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

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

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

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

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

Present Worth: Solve for P (Note: P is S0) BASIC ECONOMIC TERMS • 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 )

n Discrete compounding Sn = P*(1 + i ) S S = P = n o (1+ i)n 1 : Discrete single-payment present-worth factor. (1+ i)n

Discount (used in bonds): Sn -So 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: $6,000 Assume i=15% and n = 6 years Present worth CS Reactor SS Reactor

So = Sn *exp( -0.15*n ) n=0 10,000 30,000 n=2 7,408 0 n=4 5,488 0 Salvage value (income) 0 - 2,440 Maintenance 6 = ()()−0.15n So ∫ 0.05 10,000 e dn 1,978 0 0 Total 24,874 27,560 BASIC ECONOMIC TERMS • 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 : Payment S : Amount of Annuity i : Interest rate Total Interest accumulated 0 1 2 3 (n-1) n R R R R R (1+i)n-1 R (1+i)n-2 R (1+i) Annuities - Payment Calculation

n − Total amount repaid: S= R ∑(1+ i)n k k =1 Multiply both sides by (1+i) to get

n n−k +1 S(1+i)=R ∑(1+ i) and subtract: k =1 Si = R(1+ i)n − R i Then R = S (1+ i)n −1

If S is a loan, then S = P*(1+i)n i(1+ i)n and the payment is R = P (1+ i)n −1 Annuities - Payment Calculation This is the same formula obtained assuming a revolving loan with zero principal at the end of n years. 0 1 2 3 (n-1) n

Principal left 1 P(1+i)-R 2 (P(1+i)-R)(1+i)-R : nP(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. (1+ i)n −1 P = R i(1+ i)n 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 Cost Example : 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 C Then P = R (1+ i)n −1

The total cost now is : CV+ P= $24,650. C Capitalized Cost, K K= C +P= C + R V V (1+ i)n −1 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 reactors Which one should be installed? Comparison based on Capitalized cost.

Reactor Capitalized Cost 5,000 Mild Steel K=5,000 + =$ 31,180 (1.06)3 −1 15,000 SS K=15,000 + =$ 31,180 (1.06)n −1 SS reactor is preferred. If SS reactor < 11.3 years, Mild Steel reactor is preferred.