Presentation on theme: "Engineering Economics"— Presentation transcript:
1Engineering Economics 4/14/2017 5:03 AMConcepts from the example problems( F / P, i, N) Problem 1( P / F, i, N) Problem 2( F / A, i, N) Problem 3( A / F, i, N) Problem 4( P / A, i, N) Problem 5( A / P, i, N) Problem 6( P / G, i, N) Problem 7( A / G, i, N) Problem 8Copyright (c) , K. D. Douglas & D. H. Jensen
2Example 1How much would you have to deposit today to have $2000 in 4 years if you can get a 12% interest rate compounded annually?GIVEN:F4 = $2 000i = 12%FIND P:123n=4P?$2 000DIAGRAM:P = F4(P/F,i,n)= 2 000(P|F,12%,4)= 2 000(0.6355) = $1 271
3Different Ways of Looking at P/F From previous example, if you can earn 12% compounded annually, you need to deposit $1271 to have $2000 in 4 years.You are indifferent between $1271 today and $2000 in 4 years, assuming you can earn a return on your money of 12%.The present worth of $2000 in 4 years is $1271 (i = 12% cpd ‡ annually).‡ my abbreviation for compoundedIf you could get 13% on your money, would you rather have the $1271 today, or $2000 in 4 years ?
4Example 2Tuition costs are expected to inflate at the rate of 8% per year. The first year’s tuition is due one year from now and will be $10,000. To cover tuition cost for 4 years, a fund is to be set up today in an account that will earn interest at the rate of 5% per year, compounded annually. How much must be deposited into the fund today in order to pay the 4 years of tuition expenses?
5Present Given Gradient (Geometric) Engineering Economics4/14/2017 5:03 AMPresent Given Gradient (Geometric)g is the geometric gradient over the time period(time period: Time 0 to Time n, 1st flow at Time 1)P is the present value of the flow at Time 0(n periods in the past)i is the effective interest rate for each periodNote: cash flow starts with A1 at Time 1, increases by constant g% per periodnA1P ?123P = A1(P/A,g,i,n)g = %Copyright (c) , K. D. Douglas & D. H. Jensen
6Example 2Tuition costs are expected to inflate at the rate of 8% per year. The first year’s tuition is due one year from now and will be $10,000. To cover tuition cost for 4 years, a fund is to be set up today in an account that will earn interest at the rate of 5% per year, compounded annually. How much must be deposited into the fund today in order to pay the 4 years of tuition expenses?
7Example 2 - ConceptIf your rich Aunt Edna wanted to put a sum of money in the bank today to pay for your next four years of tuition, that sum would be $39,759 assuming 5% return on investment and tuition that begins at $10,000 increasing by 8% per year. This problem assumes tuition is due at the end of the year.
8Complex Cash FlowsComplex Cash Flows – Break apart (or separate) complex cash flows into component cash flows in order to use the standard formulas.Remember: You can only combine cash flows if they occur at the same point in time.(This is like building with LEGOs!)
9Problem 3A construction firm is considering the purchase of an air compressor.The compressor has the following expected end of year maintenance costs:Year $800Year $800Year $900Year 4 $1000Year 5 $1100Year 6 $1200Year 7 $1300Year 8 $1400What is the present equivalent maintenance cost if the interest rate is 12% per year compounded annually?
10Problem 3 – Alt Soln 1 GIVEN: MAINT COST1-8 PER DIAGRAM i = 12%/YR, CPD ANNUALLYFIND P:P = PA + PG + PF = A(P/A,i,n) + G(P/G,i,n) + F(P/F,i,n)= $700(P/A,12%,8) + $100(P/G,12%,8) + $100(P/F,12%,1)= $700(4.9676) + $100( ) + $100(0.8929) = $5014DIAGRAM:P ?123n=8$700$1004$200$300PA ?123n=8$7004$100PF ?n=1PG ?123n=8$100$700$200$3004NOTE: CAN BREAK INTO 3 CASH FLOWS: ANNUAL, LINEAR GRADIENT, AND FUTURE
11Problem 3 – Alt Soln 2 GIVEN: MAINT COST1-8 PER DIAGRAM i = 12%/YR, CPD ANNUALLYFIND P:P = PA + PG(PPG) = A(P/A,i,n) + G(P/G,i,n-1)(P/F,i,1)= $800(P/A,12%,8) + $100(P/G,12%,7)(P/F,12%,1)= $800(4.9676) + $100( )(0.8929) = $5014DIAGRAM:P ?123n=8$800$100$6004$200PA ?123n=8$8004PG ?PPG ?1PG ?12n=73$100$600$200NOTE: PG MUST BE OFFSET ONE YEAR – SO BRING THE OFFSET YEAR BACK TO TIME ZERO
12Problem 4A young couple has decided to make advance plans for financing their 3 year old daughter’s college education. Money can be deposited at 8% per year, compounded annually.What annual deposit on each birthday, from the 4th to the 17th (inclusive), must be made to provide $7,000 on each birthday from the 18th to the 21st (inclusive)?1721 yrs$7 000A ?DIAGRAM:418GIVEN:WITHDRAWALS = $7 000i = 8%/YR, CPD YEARLYFIND A4-17:P17 = A(F/A,i,n) = A(P/A,i,n)= A(F/A,8%,14) = 7 000(P/A,8%,4)= A( ) = 7 000(3.3121) A = $957STRATEGY: CAN BREAK INTO 2 CASH FLOWS, SO PICK A CONVENIENT POINT IN TIME AND SET DEPOSITS EQUAL TO WITHDRAWALS…