《化学工程设计技术》课程教学资源（参考资料，英文版）10 Engineering Economics 1

ENGINEERING ECONOMICS I TO INVEST OR NOT TO INVEST THATIS THE OUESTION (With apologies to william Shakespeare Given one or more potential projects, how do we decide in which to invest, if any? o What criteria do we use to evaluate potential projects? 10/1599 Engineering economics 1

10/15/99 Engineering Economics 1 1 ENGINEERING ECONOMICS I TO INVEST, OR NOT TO INVEST, THAT IS THE QUESTION . (With apologies to William Shakespeare) t Given one or more potential projects, – how do we decide in which to invest, if any? t What criteria do we use to evaluate potential projects?

THE TWO BASIC INVESTMENT CONCERNS ◆ PROFITABⅠTY (HOW MUCH MONEY WILL I MAKE, ◆RISK (WHAT ARE MY CHANCES OF MAKING IT?) 10/1599 Engineering economics 1

10/15/99 Engineering Economics 1 2 THE TWO BASIC INVESTMENT CONCERNS uPROFITABILITY (HOW MUCH MONEY WILL I MAKE?) uRISK (WHAT ARE MY CHANCES OF MAKING IT?)

PROFITABILITY EVALUATION okeY to PROFITABLITY ANAL YSIS E RETURN ONINVESTMENT keY tO RETURN ON INVESTMENT E TIME VALUE OF MONEY 10/1599 Engineering economics 1 3

10/15/99 Engineering Economics 1 3 PROFITABILITY EVALUATION t KEY to PROFITABLITY ANALYSIS ð RETURN ON INVESTMENT uKEY to RETURN ON INVESTMENT ð TIME VALUE OF MONEY

TIME VALUE OF MONEY I SMPLE INTEREST- SINGLE PAYMENT 9 Example: $100 invested at 5% per annum At beginning of Year l, invest $100 Atend of Year l, receive(0.05)(100)=$5.00 At beginning of Year 2, leave the $100 on deposit Atend of Year 2, receive another $5.00 And so on, as long as we leave the s100 on deposit. 10/1599 Engineering economics 1

10/15/99 Engineering Economics 1 4 TIME VALUE OF MONEY I SIMPLE INTEREST - SINGLE PAYMENT t Example: $100 invested at 5% per annum At beginning of Year 1, invest $100 At end of Year 1, receive (0.05)(100) = $5.00 At beginning of Year 2, leave the $100 on deposit At end of Year 2, receive another $5.00 And so on, as long as we leave the $100 on deposit

TIME VALUE OF MONEY II COMPOUND INTEREST- SINGLE PAYMENT A Example: $100 invested at 5% per annum At beginning of Year 1, invest $100 At end of rear l, add(0.05)(100)=$5.00 At beginning of Year 2, have $105.00 on deposit At end of year2,dld(0.05)(105.0)=S5.25 At beginning of Year 3, have $110.25 on deposit At end of Year 3, add(0.05)(110.25)=$5.51 And so on 10/1599 Engineering economics 1

10/15/99 Engineering Economics 1 5 TIME VALUE OF MONEY II COMPOUND INTEREST - SINGLE PAYMENT t Example: $100 invested at 5% per annum At beginning of Year 1, invest $100 At end of Year 1, add (0.05)(100) = $5.00 At beginning of Year 2, have $105.00 on deposit At end of Year 2, add (0.05)(105.00) = $5.25 At beginning of Year 3, have $110.25 on deposit At end of Year 3, add (0.05)(110.25) = $5.51 And so on

TIME VALUE OF MONEY III COMPOUND INTEREST- MULTIPLE PAYMENTS A Example: $100 invested initially at 5% per annum At end of Year l, have $105.00 on deposit Pay in another $200. Begin Year 2 with $305.00 At end of Year2,dd(0.05)(305.00)=S15.2 Pay in another S300. Begin Year 3 with S620.25. At end of year3,ad(0.05)620.25)=S31.01. Pay in so. Begin year 4 with s651 26 And so on 10/15/99 Engineering economics 1

10/15/99 Engineering Economics 1 6 TIME VALUE OF MONEY III COMPOUND INTEREST - MULTIPLE PAYMENTS t Example: $100 invested initially at 5% per annum At end of Year 1, have $105.00 on deposit Pay in another $200. Begin Year 2 with $305.00 At end of Year 2, add (0.05)(305.00) = $15.25 Pay in another $300. Begin Year 3 with $620.25. At end of Year 3, add (0.05)(620.25) = $31.01. Pay in $0. Begin Year 4 with $651.26. And so on

THE EQUATION OF MONEY Let e(n =the amount of money available at the beginning of the nth year F(n=the amount of money paid in at the end of the nth Jy ear i= the interest rate Then, the equation of Money is Q(n+1)=(1+i)Qm)+F(m),n=0,…,N Note: g(0=the initial payment 10/1599 Engineering economics 1

10/15/99 Engineering Economics 1 7 THE EQUATION OF MONEY Let Q(n )= the amount of money available at the beginning of the nth year F(n) = the amount of money paid in at the end of the nth year i = the interest rate Then, the Equation of Money is Q(n+1) = (1+i) Q(n ) + F(n ), n = 0, …, N Note: Q(0) = the initial payment

SOLVING THE EQUATION OF MONEY 1)Use a spreadsheet 2) For constant cash flows, there is an analytical solution, i.e., for F(n+1)=F(m)=F then Q(N=Q)++F∥1+i-1B 10/1599 Engineering economics 1

10/15/99 Engineering Economics 1 8 SOLVING THE EQUATION OF MONEY 1) Use a spreadsheet 2) For constant cash flows, there is an analytical solution, i.e., for F(n+1) = F(n) = F then Q(N) = Q(0) [(1+i)N] + F {[(1+i)N] - 1}/i

ANNUITIES AND MORTGAGES ◆ MORTGAGE Repay a loan obl in n equal payments, i.e determine fso that Q(N=0 F=-{iQ)+N(1+iy-1 ◆ ANNUITY Accumulate a specified amount /e( in N equal Payments Fstarting with 200=0. F=-{iQ/(1+i-1 10/1599 Engineering economics 1

10/15/99 Engineering Economics 1 9 ANNUITIES AND MORTGAGES t MORTGAGE Repay a Loan [Q(0)] in N Equal Payments, i.e., determine F so that Q(N) = 0 F = -{ i Q(0) [(1+i)N] } / {[(1+i)N] - 1} t ANNUITY Accumulate a Specified Amount [Q(N)] in N Equal Payments F starting with Q(0) = 0. F = - { i Q(N)} / {[(1+i)N] - 1}

PRESENTⅤ ALUE AND FUTUREⅤALUE ◆ PRESENT VALUE(PV Let Fy be a amount of money to be received n years from now? What is Fv worth today? Py=FV/(+i ◆ FUTURE VALUE(F What will an investment Py made today be worth n time periods in the future? FV=PY(+ in 10/1599 Engineering economics 1

10/15/99 Engineering Economics 1 10 PRESENT VALUE AND FUTURE VALUE t PRESENT VALUE (PV) Let FV be a amount of money to be received N years from now? What is FV worth today? PV = FV/(1 + i)N t FUTURE VALUE (FV) What will an investment PV made today be worth N Time periods in the future? FV = PV (1 + i)N