# Engineering Economics

## Presentation on theme: "Engineering Economics"— Presentation transcript:

Engineering Economics
4/14/2017 5:03 AM Concepts 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 8 Copyright (c) , K. D. Douglas & D. H. Jensen

Example 1 How 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 = \$2 000 i = 12% FIND P: 1 2 3 n=4 P? \$2 000 DIAGRAM: P = F4(P/F,i,n) = 2 000(P|F,12%,4) = 2 000(0.6355) = \$1 271

Different 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 compounded If you could get 13% on your money, would you rather have the \$1271 today, or \$2000 in 4 years ?

Example 2 Tuition 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?

Engineering Economics 4/14/2017 5:03 AM Present 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 period Note: cash flow starts with A1 at Time 1, increases by constant g% per period n A1 P ? 1 2 3 P = A1(P/A,g,i,n) g = % Copyright (c) , K. D. Douglas & D. H. Jensen

Example 2 Tuition 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?

Example 2 - Concept If 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.

Complex Cash Flows Complex 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!)

Problem 3 A construction firm is considering the purchase of an air compressor. The compressor has the following expected end of year maintenance costs: Year \$800 Year \$800 Year \$900 Year 4 \$1000 Year 5 \$1100 Year 6 \$1200 Year 7 \$1300 Year 8 \$1400 What is the present equivalent maintenance cost if the interest rate is 12% per year compounded annually?

Problem 3 – Alt Soln 1 GIVEN: MAINT COST1-8 PER DIAGRAM
i = 12%/YR, CPD ANNUALLY FIND 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) = \$5014 DIAGRAM: P ? 1 2 3 n=8 \$700 \$100 4 \$200 \$300 PA ? 1 2 3 n=8 \$700 4 \$100 PF ? n=1 PG ? 1 2 3 n=8 \$100 \$700 \$200 \$300 4 NOTE: CAN BREAK INTO 3 CASH FLOWS: ANNUAL, LINEAR GRADIENT, AND FUTURE

Problem 3 – Alt Soln 2 GIVEN: MAINT COST1-8 PER DIAGRAM
i = 12%/YR, CPD ANNUALLY FIND 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) = \$5014 DIAGRAM: P ? 1 2 3 n=8 \$800 \$100 \$600 4 \$200 PA ? 1 2 3 n=8 \$800 4 PG ? PPG ? 1 PG ? 1 2 n=7 3 \$100 \$600 \$200 NOTE: PG MUST BE OFFSET ONE YEAR – SO BRING THE OFFSET YEAR BACK TO TIME ZERO

Problem 4 A 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)? 17 21 yrs \$7 000 A ? DIAGRAM: 4 18 GIVEN: WITHDRAWALS = \$7 000 i = 8%/YR, CPD YEARLY FIND 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 = \$957 STRATEGY: CAN BREAK INTO 2 CASH FLOWS, SO PICK A CONVENIENT POINT IN TIME AND SET DEPOSITS EQUAL TO WITHDRAWALS…