Combining Factors – Shifted Uniform SeriesSlide 2Slide 3Slide 4Slide 5Combining Factors – Single Amounts and Uniform SeriesSlide 7Combining Factors – Multiple Uniform SeriesSlide 9PowerPoint PresentationSlide 11Slide 12Slide 13Slide 14Slide 15Combining Factors – Shifted Uniform SeriesQuestion: What is P for the following cash flow, a shifted uniform series of n equal installments? The first installment occurs at the end of period 5.$P 0 1 2 3 4 n+4Combining Factors – Shifted Uniform SeriesApproaches for finding P:•Use (P/F) for each of the n payments.•Use (F/P) for each of the payments to find FT, then use FT(P/F,i%,n+4) to find P.•Use (P/A) to find the P4, then use P4(P/F,i%,4) to find P.$P 0 1 2 3 4 n+4Combining Factors – Shifted Uniform SeriesExample: What is P for a computer you purchase in which installments of $200 are paid for 10 months, with the first payment deferred until the 5th month after purchase. Assume i = 5%.A = $200P4 = $200(P/A,5%,10) = $200 x 7.7217 = $1544.34P = P4(P/F,5%,4) = $1544.34 x 0.8227 = $1270.53P = $200(P/A,5%,10) (P/F,5%,4) $P 0 1 2 3 4 14$200Combining Factors – Shifted Uniform SeriesExample: What is P if you invest $2000 beginning now and at the end of each year for 10 years? The account pays interest at 6%.P = $2000 + $2000(P/A,6,10)P = $2000 + $2000(7.3601) = $16720.2$P 0 1 2 3 4 10$2000Combining Factors – Shifted Uniform SeriesExample: What is F for the following cash flow. Installments of $200 are paid for periods 5 through 14. Assume i = 5%.A = $200F = $200(F/A,5%,10) F = $200(12.5779) = $2515.58What is the account worth in period 20 (no installments made after period 14)?$F 0 1 2 3 4 14$200Combining Factors – Single Amounts and Uniform SeriesHow might you approach the above cost flow? $200 paid in periods 1,2,3,4,6,7,8,9,10; and $400 paid in periods 5.$P 0 1 2 3 4 10$200$P1$200$P2$200 0 1 2 3 4 10 0 1 2 3 4 5 10Combining Factors – Single Amounts and Uniform Series$P 0 1 2 3 4 10$200$P1$200$P2$200 0 1 2 3 4 10 0 1 2 3 4 5 10P = $200(P/A,i%,10) + $200(P/F,i%,5)Combining Factors – Multiple Uniform SeriesHow might you approach the above cost flow? $200 paid in periods 1,2,3,4,8,9,10; and $400 paid in periods 5,6,7.$P 0 1 2 3 4 10$200$P1$200$P2$200 0 1 2 3 4 10Combining Factors – Multiple Uniform Series$P 0 1 2 3 4 10$200$P1$200$P2$200 0 1 2 3 4 10P = $200(P/A,i%,10) + $200(P/A,i%,3)(P/F,i%,4)$125$150$1750 1 2 3 4 5 6 7 8 $100Combining Factors – Shifted Gradients$P$125$150$1750 1 2 3 4 5 6 7 8 $100Combining Factors – Shifted Gradients$P$25$50$750 1 2 3 4 5 6 7 8 $100$P1$P2$P3P = P1+ P3Combining Factors – Shifted Gradients$25$50$750 1 2 3 4 5 6 7 8 $100$P1$P2$P3P = P1+ P3P = $100(P/A,i%,8) + 25(P/G,i%,4)(P/F,i%,4) Note, P2 = $25(P/G,i%,4) term is for 4 periods. 0 1 2 3 4 5 6 7 80 1 2 3 4 5 6 7 8 $1000Combining Factors – Shifted Decreasing Gradients$P$850$700$5500 1 2 3 4 5 6 7 8 $1000Combining Factors – Shifted Decreasing Gradients$P$850$700$5500 1 2 3 4 5 6 7 8 $1000$P1=> -$150$300$450$P2$P30 1 2 3 4 5 6 7 8Combining Factors – Shifted Decreasing Gradients0 1 2 3 4 5 6 7 8 $1000$P1$150$300$450$P2$P30 1 2 3 4 5 6 7 8 -P = P1- P3P = $1000(P/A,i%,7) - 150(P/G,i%,4)(P/F,i%,3) Note, P2 = $150(P/G,i%,4) term is for 4