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Combining Factors – Shifted Uniform Series. Question: 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+4.
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Combining Factors – Shifted Uniform Series Question: 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+4
Combining Factors – Shifted Uniform Series $P Approaches 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. 0 1 2 3 4 n+4
Combining Factors – Shifted Uniform Series $P Example: 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 = $200 P4 = $200(P/A,5%,10) = $200 x 7.7217 = $1544.34 P = P4(P/F,5%,4) = $1544.34 x 0.8227 = $1270.53 P = $200(P/A,5%,10) (P/F,5%,4) 0 1 2 3 4 14 $200
Combining Factors – Shifted Uniform Series $P Example: How much will be in an account in which 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 0 1 2 3 4 10 $2000
Combining Factors – Shifted Uniform Series $F Example: What is F for the following cash flow. Installments of $200 are paid for periods 5 through 14. Assume i = 5%. A = $200 F = $200(F/A,5%,10) F = $200(12.5779) = $2515.58 What is the account worth in period 20 (no installments made after period 14)? 0 1 2 3 4 14 $200
Combining Factors – Single Amounts and Uniform Series $P How might you approach the above cost flow? $200 paid in periods 1,2,3,4,8,9,10; and $400 paid in periods 5. 0 1 2 3 4 10 $200 $P1 $P2 0 1 2 3 4 10 0 1 2 3 4 5 10 $200 $200
Combining Factors – Single Amounts and Uniform Series $P 0 1 2 3 4 10 $200 P = $200(P/A,i%,10) + $200(P/F,i%,5) $P1 $P2 0 1 2 3 4 10 0 1 2 3 4 5 10 $200 $200
Combining Factors – Multiple Uniform Series $P How 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. 0 1 2 3 4 10 $200 $P1 $P2 0 1 2 3 4 10 $200 $200
Combining Factors – Multiple Uniform Series $P 0 1 2 3 4 10 $200 P = $200(P/A,i%,10) + $200(P/A,i%,3)(P/F,i%,4) $P1 $P2 0 1 2 3 4 10 $200 $200
Combining Factors – Shifted Gradients $175 $150 $125 $100 0 1 2 3 4 5 6 7 8 $P
Combining Factors – Shifted Gradients $175 $150 $125 $100 0 1 2 3 4 5 6 7 8 $P P = P1+ P3 $75 $100 $50 $25 0 1 2 3 4 5 6 7 8 $P3 $P2 $P1
Combining Factors – Shifted Gradients $75 $100 $50 $25 0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 8 $P3 $P2 $P1 P = P1+ P3 P = $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.
Combining Factors – Shifted Decreasing Gradients $1000 $850 $700 $550 0 1 2 3 4 5 6 7 8 $P
Combining Factors – Shifted Decreasing Gradients $1000 $1000 $850 $700 $550 0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 8 => - $450 $P $P1 $300 $150 0 1 2 3 4 5 6 7 8 $P3 $P2
Combining Factors – Shifted Decreasing Gradients $1000 $450 $300 - $150 0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 8 $P3 $P2 $P1 P = P1- P3 P = $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 periods.