化學(xué)工藝最優(yōu)化方法

1、Hao LingDepartment of Petroleum ProcessingEast China University of Science and TechnologyFormulation of the Objective Function2021-10-163.1 Economic Objectives FunctionsTransfer a verbal statement or concept of desired objective into mathematic terms.This section treats two major components of econo

2、mic objective functions: capital costs and operating costs.Three simple examplesI.Operating costs and income ;II.Capital costs;III.both2021-10-16Example3.1Operating profits as the objective functionREACTOR1REACTOR2REACTOR3X11SPLIT1X12SPLIT2SPLIT3X1X2X3X4X5X6X8X9X10X7ABCEFGGCBAocessFBAocessEBAocess23

3、:3Pr:2Pr: 1Pr2021-10-16Example3.1ProcessProductReactant requirements(Kg/Kg product)Process cost(cents/Kg)Selling price(cents/Kg)1E2/3A;1/3B1.542F2/3A;1/3B0.53.33G1/2A;1/6B;1/3C13.8Raw materialMax. availableCost(cents/Kg)A40,0001.5B30,0002.0C25,0002.52021-10-16Example3.1Raw material costs: 0.015A+0.0

4、2B+0.025CProcessing costs: 0.015E+0.005F+0.01GTotal costs: 0.015A+0.02B+0.025C+ 0.015E+0.005F+0.01GTotal income (Objective function): f(x)=(0.04E+0.033F+0.038G)-(0.015A+0.02B+0.025C+ 0.015E+0.005F+0.01G) =0.025x8+0.028x9+0.028x10-0.015x11-0.02x12-0.025x72021-10-16Example3.1Restraints:000,250000,3000

5、00,400333. 0167. 0333. 0667. 05 . 0667. 0667. 071211107109812109811xxxAlsoxxxxxxxxxx2021-10-16Example3.2Capital costs as the objective functionMinimizes the capital costs of cylindrical pressure vessels. Volume of the tank V is fixed, find the optimal (L/D) that minimizes the cost function.Assumptio

6、ns:Both ends are closed and flat;Constant thickness t with density , and the thickness is not a function of pressure;The cost of material is the same for both the sides and ends, and is S (dollar per unit weight);No waste of materialsLD2021-10-16Example3.2The surface area of the tank is equal to:DVD

7、DVDDLDVCosttDLDSfWeighttDLDfAreaDLDfDLDDLD42424)()2()()2()(22422222232221222021-10-16Example3.23/12240)4(42VDDVDDVDfoptDifferentiation of f , from 1)/(4443/122optDLVLDVLLDV2021-10-162021-10-16Ex.3 Optimum Thickness of InsulationOptimum thickness of insulationConsidering both the costs of the insulat

8、ion and the value of the energy saved by adding the insulation. A surface area of pipe x thickness of pipe k thermal conductivity of insulation hc outside convective heat transfer coeffientHot fluidQInsulation of thickness xT=T(fluid)-T(air)chkxTAQ/1/2021-10-16Ex.3 Optimum Thickness of InsulationThe

9、 cost of installed insulation per unit can be presented byLife time of insulation is 5 years;The funds to purchase the insulation could be borrowed from a bank and paid back in five annual installments;Let r be the fraction of the installed cost to be paid each year to the bank.Let the value of the

10、heat lost from the pipe be Ht. Y be the number of hours per year of operation .xCCfc102021-10-16Ex.3 Optimum Thickness of InsulationFormulate an objective function to maximize the saving in operating cost, savings expressed as the difference between the value of the heat conserved less the annualize

11、d cost of the insulation;Obtain an analytical solution for x;Eq. bcchkxTATAhQQ/1/02021-10-16Ex.3 Optimum Thickness of InsulationThe objective function to be maximized is the value of heat conserved per year less the annualized capital costSubstitute Eq. (b) into above eq., differential f with respec

12、t to x, and solve for optimumyeardollarsArxFFBtudollarsHyearhYhBtuQQft)(*)(100ctopthrkFTYHkx1)10(2/1162021-10-163.2 Consideration of the Time Value of MoneyA payment of $5000 now is not equivalent to a similar payment of $5000 fifteen years later.100*(1+0.05)15=207.9Future worthT=2 yearsT=3 years If

13、 interest is compounded q times per yearT=3 yearsT=n yearsnniPFiPFiPFiPF)1 ()1 ()1 ()1 (443322nqqiPFn)1 ( 2021-10-163.2 Consideration of the Time Value of MoneynqqiPFn)1 ( inPeFn nnnniFPiPF)1 ()1 (If a monthly interest rate is given(=i/q), then set q=12 in the exponent ekspunnt in the Eq. For contin

14、uous compounding ,let q,and the above Eq. reduces toPresent worth: Suppose the value of the saving account will be $150 five years from now. whats its corresponding present worth?Reverse2021-10-163.2 Consideration of the Time Value of MoneySuppose that an investment yields a constant cash flow equal

15、 to F every year for n years. Then, the total present worth of all of the cash flows is found by summing the present worths for each year, namely 等款還貸：nnnniFiFiFiFP)1 ()1 ()1 (111221Let Fi=F, multiply by (1+i)12121)1 ()1 ()1 ()1 (nnnniFiFiFFiPMinus the two Eq.sPiiiFiFFiPnnn1)1 ()1 ()1 (2021-10-163.2

16、 Consideration of the Time Value of Money1)1 ()1 (nniiirr: Series present worth factor (現(xiàn)價(jià)系數(shù)).r*P-The annual charge.2021-10-163.2 Consideration of the Time Value of Money2021-10-16Ex.3.4 Payment scheduleSuppose you obtain a $100,000 mortgage on a house, agreeing to pay 10 percent interest and to rep

17、ay the load over 25 years. What is the annual payment you must make if the interest is compounded just once per year?From Table 3.1, the repayment multiplier is 0.110, hence, the yearly payment is yearF/000,11$)000,100)($110. 0(2021-10-16Ex.3.4 Payment scheduleExperience a variable annual return fro

18、m an investment.Suppose that you were asked to loan $100 to an acquaintance. He wants to repay the loan over a period of 5 years and is offering you the following payback schedule:1st $50;2nd $25;3rd $25;4th $15;5th $10The present worth for a series of unequal annual returns can be calculated by:202

19、1-10-16Ex.3.4 Payment schedule554433221)1 ()1 ()1 ()1 (1iFiFiFiFiFPAssuming an 8% interest rate each year:100$41.105$43.21$)08. 1 (25)1 (29.46$08. 1501222211PiFPiFP2021-10-16Ex.3.4 Payment scheduleWould the same outcome be true for an interest rate of 10%? An alternative method to evaluate the propo

20、sed load versus investment is the find the value of i such that the above equation is satisfied when P=$100.The value of i is called the internal rate of return (IRR, 內(nèi)部收益率內(nèi)部收益率 ).PiiiFnn1)1 ()1 (2021-10-163.3 Measure of ProfitabilityCash flowCash flow is the movement of money into or out of a busin

21、ess, project, or financial product. It is usually measured during a specified, finite period of time. A cash flow in a given year is defined as the sum of the net profit after taxes are deducted plus depreciation(貶值).2021-10-163.3 Measure of ProfitabilityPBP (Payback years): Take an integrated view

22、of profits over a period of several years. The PBP is one criterion and defined as:flowcashannualaverageinvestmentfixededepreciablPBP Note that no weighting of earlier cash flows and later cash flows;No consideration of project earning after the initial investment has been recovered.No assumptions r

23、egarding equipment life.2021-10-163.3 Measure of ProfitabilityROI（return on investment，投資回報(bào)率）: relative profit criterion divided the investmentcapitalworkinginvestmentcapitaltaxafterincomenetROInjjjIiFNPV10)1 (njnjjjjjiIiFNPV110)1 ()1 (NPV (net present value，凈現(xiàn)值) Ignore taxes and computer a before-t

24、ax cash flow2021-10-163.3 Measure of ProfitabilitynjjjIiFNPV10)1 (Fj: each year cash flow;i: the stated interest rate for capital;I0: the initial investment.Let NPV=0, solve the above equation to find i, which is IRR內(nèi)部收益率內(nèi)部收益率.2021-10-163.3 Measure of Profitability2021-10-16Example 3.5Payout period

25、of A=100,000/20,000=5yearPayout period of B=100,000/15,000=6.67year2021-10-16Example 3.5Applying Eq.(3.12) for NPV=0 and n=10, we find that the IRR for A is 10%, B is 8%. For i=0.1,NPV of A is -$640, NPA of B is -$7840. 2021-10-163.4 Opt. Profitability（盈利能力最優(yōu)化）（盈利能力最優(yōu)化）In order to do the job, you fi

26、rst need to determine the independent variables which influence:（1）Capital investment;（2）Income;（3）Operating costs;（4）Cash flowAssuming there are p independent variables, denote (1) as I(x1,x2,.xp) and the cash flow F(x1,x2,.xp) is obtain from (1),(2),(3)2021-10-163.4 Opt. ProfitabilityDefine the sp

27、ecific profitability measure, such as payback time PBP, return on investment ROI (投資回報(bào)率）, net present value NPV, or internal rate of return IRR (內(nèi)部收益率）.The first three objective functions are explicit functions（顯函數(shù)） of F and I; while IRR is an implicit functions （隱函數(shù)） of F and I; once F and I are sp

28、ecified, an iterative solution must be used to find i. In order to maximize the IRR, you must vary the independent unknowns and computer I and F for each trial set of (x1,x2,.xp). Next solve equation for i and continue the procedure until you find the value of x which maximizes the IRR. 2021-10-163.

29、4 Opt. Profitability2021-10-163.4 Opt. Profitability554433221)1 ()1 ()1 ()1 (1iFiFiFiFiFPNote that F in the equation is fixed. Maximization of rate of return on investment is one alternative to maximizing IRR, and it has the advantage of being an explicit performance criterion.ROI is well approximat

30、ed by a relation mentioned by Happel and Jordan (1975):inROI65. 0/12021-10-16Ex.3.72021-10-16Ex.3.7對(duì)數(shù)平均溫差對(duì)數(shù)平均溫差Tm=（T1-T2）/（T1/T2）2021-10-16Ex.3.72021-10-16Ex.3.72021-10-16Profit Financial EvaluationDetailed evaluation of a project requires specifying the following parameters.Initial investment;Futur

31、e cash flows;Salvage value（殘余價(jià)值）or scrap value;Economic life;Depreciation（貶值，折舊）;Depletion（損耗）Tax credit(抵稅，可退稅額度 )；Taxes；Inflation;Debt/equity ratio（債資比例）2021-10-16Profit Financial EvaluationSalvage value:殘余價(jià)值Salvage value is the price that can be actually obtained or is imputed to be obtained from

32、 the sale of used property if, at the end of its usage, the equipment still has some utility.Economic life: have a major impact on depreciation calculations, which is in turn influence profits and taxes.For economic lives in the 10- and 20 years range, profitability is not very sensitive to the spec

33、ific value of the equipment. However, when n7, it counts.2021-10-16Profit Financial EvaluationDepreciationDepreciation is a measure of decreasing in the value of equipment and improvements over a period of time.The commonly used depreciation methods include:1.Straight lines;2.Declining balance;3.Sum

34、 of years digits（年數(shù)加總折舊法）4.Accelerated cost recovery system2021-10-16Profit Financial EvaluationStraight lines depreciation: it is assumed that the equipment value declines linearly with respect to time.Declining-Balance Method Depreciation methods that provide for a higher depreciation charge in th

35、e first year of an assets life and gradually decreasing charges in subsequent years are called accelerated depreciation methods. This may be a more realistic reflection of an assets actual expected benefit from the use of the asset: many assets are most useful when they are new. 2021-10-16Profit Fin

36、ancial EvaluationOne popular accelerated method is the declining-balance method. Under this method the Book Value is multiplied by a fixed rate.Annual Depreciation = Depreciation Rate * Book Value at Beginning of Year2021-10-16Profit Financial EvaluationBook Value -DepreciationDepreciationAccumulate

37、dBook Value -Beginning of YearRateExpenseDepreciationEnd of Year$1,000 (Original Cost)40%$400 $400 $600 $600 40%$240 $640 $360 $360 40%$144 $784 $216 $216 40%$86.40 $870.40 $129.60 $129.60 $129.60 - $100$29.60 $900 $100 (Scrap Value)2021-10-16Profit Financial EvaluationSum-of-Years Digits is a depre

38、ciation method that results in a more accelerated write-off than straight line, but less than declining-balance method. Under this method annual depreciation is determined by multiplying the Depreciable Cost by a schedule of fractions. Depreciable Cost = Original Cost - Salvage ValueBook Value = Ori

39、ginal Cost - Accumulated Depreciation2021-10-16Profit Financial EvaluationSum-of-Years Digits is a depreciation method that results in a more accelerated write-off than straight line, but less than declining-balance method. Under this method annual depreciation is determined by multiplying the Depre

40、ciable Cost by a schedule of fractions. n years, 1+2+3+.+n=n(1+n)/2=AAt No. j year, the depreciation is n-j+1/AExample. 5 years life, 1st year depreciation is5-1+1/(5*6/2)=5/152021-10-16Profit Financial EvaluationBook Value -TotalDepreciationDepreciationAccumulatedBook Value -Beginning of YearDepreciableRateExpenseDepreciationEnd of YearCost$1,000 (Original Cost)$900 5/15$300 ($900 * 5/15)$300 $700 $700 $900 4/15$240 ($900 * 4/15)$540 $460 $460 $900 3/15$180 ($900 * 3/15)$720 $280 $280 $