Engineering Economy PDF

Engineering Economy Engineering Economy  ✔ Topic 7 – Annuities ✔ Topic 8 – Gradients ✔ Topic 9 – Application of Money Time Relationship Engineering Economy How was your Prelim Grades?  Cash Flow Cash flow is the sum of money recorded as...

Engineering Economy Engineering Economy  ✔ Topic 7 – Annuities ✔ Topic 8 – Gradients ✔ Topic 9 – Application of Money Time Relationship Engineering Economy How was your Prelim Grades?  Cash Flow Cash flow is the sum of money recorded as receipts or disbursements in a project’s financial records. A cash flow diagram presents the flow of cash as arrows on a timeline scaled to the magnitude of the cash flow, where expenses are down arrows and receipts are up arrows. Engineering Economy  Cash Flow (Example) A mechanical device will cost $20,000 when purchased. Maintenance will cost $1000 per year. The device will generate revenues of $5000 per year for 5 years. The salvage value is $7000. Engineering Economy  Introduction to Annuity What is an Annuity? An annuity is a series of payments made at equal intervals. Example: You get $200 a week for 10 years. Engineering Economy  Types of Annuities 1. Ordinary Annuity – is one where the equal payments are made at the end of each payment period starting from the first period. 2. Deferred Annuity – is one where the payment of the first amount is deferred a certain number of periods after the first. Engineering Economy  Types of Annuities 3. Annuity due – is one where the payments are made at the start of each period, beginning from the first period. 4. Perpetuity – is an annuity where the payment periods extend forever or in which the periodic payments continue indefinitely. Engineering Economy  Ordinary Annuity P n periods 0 1 2 3 4 5 6 7 Payments F Engineering Economy  Deferred Annuity P n periods 0 1 2 3 4 5 6 7 8 9 Deferment period Payments F Engineering Economy  Annuity Due P n periods 0 1 2 3 4 5 6 7 Payments F Engineering Economy  Perpetuity P n periods n 0 1 2 3 4 5 6 …… Payments Engineering Economy  Annuity Formula UNIFORM SERIES PRESENT WORTH FACTOR (P/A) P = A (P/A, i%, n) UNIFORM SERIES CAPITAL RECOVERY FACTOR (A/P) A = P ( A/ P, i%, n) Engineering Economy  Annuity Formula UNIFORM SERIES COMPOUND AMOUNT FACTOR (F/A) F = A ( F/ A, i%, n) UNIFORM SERIES SINKING FACTOR (P/A) A = F (A/F, i%, n) Engineering Economy Engineering Economy Engineering Economy Engineering Economy  Sample Problem -Perpetuity Company “DITO” pays P8.00 in dividends annually and estimates that they will pay the dividends indefinitely. How much are investors willing to pay for the dividend with a required rate of return of 5%? P = 8/5% = P160 An investor will consider investing in the company if the stock price is P160 or less. Engineering Economy  Sample Problem – Breaking into 2 Cash Flows 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)? Engineering Economy  Sample Problem – Breaking into 2 Cash Flows DIAGRAM: Solution: $7 000 4 17 0 18 21 yrs A? Given: WITHDRAWALS18-21 = $7 000 ; i = 8%/yr. com. 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(24.2149) = 7 000(3.3121) ⇒ A = $957 Engineering Economy  Practice Problems 1. Determine the value of each of the following annuity factors: (P/A, 4%,8) (A/P,14.5%,10) (F/A,9.8%,21) (A/F,6.3%,15) 2. What are the present worth and the accumulated amount of a 10- year annuity paying P1,500 at the end of each year, with interest at 12% compounded annually? Engineering Economy  Cont 3. How much money would you have to deposit for five consecutive years starting one year from now if you want to be able to withdraw P50,000 ten years from now? Assume the interest is 14% compounded annually. 4. A one bagger concrete mixer can be purchased with a down payment of P8,000 and equal installment of P600 each paid at end of every month for the next 12 months. If money is worth 12% compounded monthly, determine the equivalent cash price of the mixer. Engineering Economy  Cont. 5. A person buys a piece of property for P100,000 down payment and ten deferred semi-annual payments of P8,000 each starting three years from now. What is the present value of the investment if the rate of interest is 12% compounded semi-annually? Engineering Economy  Cont. 6. A farmer bought a tractor costing P25,000 payable in 10 semi annual payments, each installment payable at the beginning of each period. If the rate of interest is 26% compounded semi-annually, determine the amount of each installment. Engineering Economy  Cont. 7.What lump sum of money must be deposited in a bank account at present time so that Php 500 per monthly can be withdrawn for five years with the first withdrawal scheduled six years from today? Interest rate is 9% compounded quarterly. Engineering Economy Drag your dot to how you are feeling: I understand I’m a little confused I need help!