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 EGR 312 - 6

Combining Factors – Shifted Uniform Series P = ?   0 1 2 3 4 n+4 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. EGR 312 - 6

Combining Factors – Shifted Uniform Series P = ?   0 1 2 3 4 14 $200 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 = 0.5% per month. A = $200 P4 = ________________________________ P = _________________________________ A = $200 P4 = $200(P/A,0.5%,10) = $200 x 9.7304 = $1946.08 P = P4(P/F,0.5%,4) = $1946.08 x 0.9802 = $1907.55 P = $200(P/A,0.5%,10) (P/F,0.5%,4) EGR 312 - 6

Combining Factors – Shifted Uniform Series P = ?   i = 6% 0 1 2 3 4 10 $2000 Example: What is the present worth of 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 = ____________________________ P = $2000 + $2000(P/A,6,10) P = $2000 + $2000(7.3601) = $16720.2 EGR 312 - 6

Combining Factors – Shifted Uniform Series   0 1 2 3 4 14 $200 Example: What is F for the cash flow shown above, in which installments of $200 are paid at the end of periods 5 through 14? Assume i = 5%. A = _________________ F = _________________ What is the account worth in period 20 (no installments made after period 14)? A = $200 F = $200(F/A,5%,10) F = $200(12.5779) = $2515.58 F20 = 2515.58(F/P, 5%, 6) = 2515.58 (1.3401) = 3371.13 EGR 312 - 6

Combining Factors – Single Amounts and Uniform Series P = ? i = 10%   0 1 2 3 4 10 $200 $400 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. P1 = ? P2 = ? 0 1 2 3 4 10 5 $200 $200 EGR 312 - 6

Combining Factors – Single Amounts and Uniform Series P = ?   0 1 2 3 4 10 $200 $400 P = $200(P/A,10%,10) + $200(P/F,10%,5) = _________ P = 200(6.1446) + 200(0.6209) = 1353.10 P1 = ? P2 = ? 0 1 2 3 4 10 5 $200 $200 EGR 312 - 6

Combining Factors – Multiple Uniform Series   0 1 2 3 4 10 $200 $400 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. P1 = ? P2 = ? 0 1 2 3 4 10 5 $200 $200 EGR 312 - 6

Combining Factors – Multiple Uniform Series   0 1 2 3 4 10 $200 $400 P = $200(P/A,i%,10) + $200(P/A,i%,3)(P/F,i%,4) P1 = ? P2 = ? 0 1 2 3 4 10 5 $200 $200 EGR 312 - 6

Combining Factors – Shifted Gradients $175 $150 $125 $100   0 1 2 3 4 5 6 7 8   P EGR 312 - 6

Your turn – draw the cash flow diagram(s) …. EGR 312 - 6

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 = ________ P = ___________________________________ Note, P2 term is for ____ periods. P = P1+ P3 P = $100(P/A,i%,8) + 25(P/G,i%,4)(P/F,i%,4) Note, P2 term is for 4 periods. EGR 312 - 6

Combining Factors – Shifted Decreasing Gradients $1000 $850 $700 $550 0 1 2 3 4 5 6 7 8 P EGR 312 - 6

Draw the cash flow diagrams …   EGR 312 - 6

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 P1 -   $450   P $300 $150 P3 P2 EGR 312 - 6

Combining Factors – Shifted Decreasing Gradients $1000 - $450 $300 $150   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. P = _______ P = _______________________________ Note, P2 term is for ___ periods. EGR 312 - 6