Question

Hello! I need help solving these problems for engineering economy. Please show the full solution and all work for BOTH PROBLEMS. please write neatly and large (the resolution of photos on the site are low). please be as detailed as possible and answer BOTH questions. Thank you

6. Develop an equation to solve (NO NEED to solve it) for the equivalent future worth for the year 15t, using a nominal interest rate of 10% compounded. (You must use a gradient factor and you must use an annuity factor!) 1500 A 1200 900 600 A200 EOM 7 8 9 10 11 6 200

Answer 1

6. Here in this cash flow series there are two gradient factors i.e., 300 and 400. The cash flow series first decreases by 300 for the first 3 years and then decreases by 400 for next two years. Therefore, present worth of cash flow series

is PW = 1500-300(P/G,10%,3)-400(P/G, 10%,2) = 1500-300* 2.3291 -400* .8264 =470.71

Therefore, future worth of cash flow series at 15th year is

FW= PW(1+.10)¹⁵ = 470.71(1+.10)¹⁵ =1966.272

7. Gradient factor = 200 and cash flow series increases by 200 for 4 years.

The present worth of cash flow series = -200 -200(P/A, 18%,1) +200(P/A,18%,1) + 200(P/G,18%,4)

                                                   = -200 +200* 4.3781 = 675.62

Therefore, future worth of cash flow series at the end of 7th year

FW= PW(1+.18)⁷ = 675.62(1+.18)⁷ =2152.169