# Money-Time Relationships Part II – Tricks and Techniques.

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Money-Time Relationships Part II – Tricks and Techniques

Overview Important assumptions and how to break them Deferred annuities –Slow (but simple) use only P/F –Faster - Combining P/F and P/A Linear Gradients –P/G, A/G, etc. –Sometimes P/F is easier Exponential growth (or decay) Compounding intervals more often than cash flows Variable interest rates require F/P or P/F

Important Assumptions FactorAssumptionViolations All (P/F,F/P,P/A,etc.)Constant interest rate for N periods Variable interest rates (5% for 3 years, then 6% for 2 years, then 4% for 6 years) (P/A,i%,N) (F/A,i%,N) Constant cash flowsCash Flows that grow or shrink (P/A,i%,N) A/P (F/A,i%,N) A/F Cash flow interval matches interest compounding interval Interest compounded more often (monthly) than cash flows (quarterly/yearly) (P/A,i%,N) A/P (F/A,i%,N) A/F Cash flows start at end of year 1 and finish at end of year N Cash flow starts at a future year (year 10)

Dealing with Violations I ProblemSolution Techniques Changes in i%Find breakdown of problem where i% is piecewise-constant. Use P/F or F/P only. Avoid using P/A, etc. Variable Cash Flows (\$A grows or shrinks) If \$A is piecewise-constant, with changes in the level, use deferred annuity technique. If \$A changes every period, shortcuts exist only for special cases (linear, exponential). For the general case, you can always use P/F or F/P and sum up the results.

Dealing with Violations II ProblemSolution Techniques Interest compounded more often (monthly) than cash flows (quarterly/yearly) Use time periods that match the cash flows. Adjust the monthly interest rate I up to the new period. I* = (1+I) 4 -1 I*=(1+I) 12 -1 First Payment of Annual Cash flow starts at a future year, not year 1 (example: need PV of cash flow that starts at year 10 with last payment in year 20) Use deferred annuity formula. Apply appropriate combinations of P/F and P/A. example: (P/F,i%,9)*(P/A,i%,10)

Deferred annuities Cash flow starts at beginning of year J+1 Last payment at end of year N 0 J J+1 N \$A/year Present Value \$P = (P/F,i%,J)*(P/A,i%,N-J)*A \$P ?

Why not just P/A? 0 J J+1 N \$P ? \$A/year (P/A,i%,N-J)*\$A gives the value of the cash flow in units of year J dollars. This is probably not what you wanted as your final result. Remember: P/A gives a dollar value that is timed one year before the start of the cash flows. You need to use P/F or F/P to move this value to other years! Thats why \$P = (P/F,i%,J)(P/A,i%,N-J)*\$A

Variable cash flows For general cases, use P/F or F/P For linear cases (e.g. –3000,-1000,1000,3000,5000), there is a gradient method involving gradient factors P/G, F/G, etc. For exponential cases, (e.g. 1000,1100,1210,…) there is a convenience interest rate method Use of excel together with P/F or F/P is often the best solution technique. If you are confused, then use P/F or F/P together with a table. This keeps the analysis simple and easy to follow.

Exponential Decay Example A watch manufacturer expects a revenue of \$100,000 for the first month. The revenue declines by 10% each month and ends after the 12 th month. Calculate the Present Value given i=1%/month. 0123456789101112 \$P? \$100000 \$59049 \$31,381

Exponential Growth/Decay Slow but simple (Excel + P/F)

Exponential Growth/Decay Convenience Interest Rate Method Another way to calculate annuities with a growth (or decay) factor is to adjust the interest rate factor and use a special formula. Common ratio =f = (A k - A k-1 )/ A k-1 = -0.10 Convenience rate i cr =[(1+i)/(1+f)]-1 =[1.01/0.90]-1=0.1222 Special formula PV = A 1 (P/A, i cr %,N)/(1+f) = = (\$100000)(6.132)/(0.90)=\$681,333

Exponential Growth/Decay Convenience Interest Rate Method Common ratio =f = (A k - A k-1 )/ A k-1 = -0.10 Convenience rate i cr =[(1+i)/(1+f)]-1 =[1.01/0.90]-1=0.1222 PV = A 1 (P/A, i cr %,N)/(1+f) = = (\$100000)(6.132)/(0.90)=\$681,333 Compare with more careful excel+P/F method: \$681,235 (difference is due to rounding 6.132)

Linear Gradients Cash flow increases or decreases _linearly_ (by the same _amount_ each period) 12 3 45 -3000 -1000 1000 3000 50007000 6 Investment loses 3000 in year 1, 1000 in year 2, but earns 1000,3000,5000,7000 in years 3-6. What is the PV at i=8%?

Gradient Factors 012345 ………….N \$0 \$1 \$2 \$3 \$4 \$(N-1) (P/G,i%,N) is the present value (units: year 0\$) of this cash flow (F/G,i%,N) is the future value (units: year N\$) of this cash flow (A/G,i%,N) is the annuity value (units: \$/year) of this cash flow Year

Linear Gradients 12 3 45 -3000 -1000 1000 3000 50007000 6 = + -3000 20004000 6000800010000 annuity Simple gradient

Linear Gradients 12 3 45 -3000 -1000 1000 3000 50007000 6 Cash flow = annuity (-3000) + gradient (2000/year) Gradient always starts at year 2. PV = -3000 (P/A,8%,6) + 2000(P/G,8%,6) (p.632) = (-3000*4.6229) + 2000 (10.523) =-\$13869 + \$21046 = \$7177

60 Seconds Investment Challenge Let i=2%/month. There are two investments, A and B. An investment costs \$300 in terms of todays dollars. 01234….N…..24months \$0 \$1 \$2 \$3 ….\$N… \$36 (end) INVESTMENT A INVESTMENT B 0123….N…..36months \$21 \$0 Do you want to swap \$300 for the PV of A, or the PV of B? \$21 (end)

Investment Challenge: Analysis of A Let i=2%/month. The investment costs \$300 in terms of todays dollars. \$0 \$1 \$2 \$3 ….\$N… \$36 (end) INVESTMENT A 0123….N…..36months PV = \$1 * (P/G,2%,36) + \$1 * (P/A,2%,36) = \$392.04 + \$25.49 = \$417.53 Did you forget that cash flow for P/G begins in year 2? In this example, this mistake cost you money!

Investment Challenge: Analysis of B Let i=2%/month. The investment costs \$300 in terms of todays dollars. 01234….N…..24months INVESTMENT B \$21 \$0 Evaluating B is simple, because it has a constant cash flow of \$21. It starts in year 1, so we can use the P/A formula. \$PV = \$21 * (P/A,24,2%) = \$21 * 18.9139 = \$397.20

Compounding intervals more often than cash flows Solution: Change interest rate period to match cash flow period Example: Interest compounded every month at 1%/month, but cash flows are every six months. Use i*=[(1+i)^N]-1=1.01^6-1 =1.06152-1 =6.152%/6-month period

Savings account example: Problem Every month your bank pays 0.25% interest on your savings account balance. Every 3 months you deposit \$5000. How much do you have after 3 years? Note: this is a F/A problem, except that the compounding interval of 1 month does not match the deposit (cash flow) interval of 3 months.

Savings account example: Analysis Step 1: Change interest rate to 3-month rate: i*=[(1+.0025) 3 -1]=.0075187 Step 2: Determine N. If a period is 3-months, then we have N=12 periods in 3 years. Step 3: Notice that the cash flow every period is constant, so we can use the FV formula, \$FV = \$5000 * (F/A,i*,12) Final Step: Calculate the F/A formula. \$FV =\$5000 * [(1.0075187) 12 -1]/(0.0075187) =\$5000*12.50888=\$62544.42

Variable interest rates and F/P (or P/F) Rule: Use a separate F/P (or P/F) for each group of years or periods where the interest rate is constant. Example: You have \$5000 today. For 3 years you invest it at 4% per year, then for 5 more years at 5% per year, then 2 more for 3% per year. The future value is \$FV = \$5000 * (F/P,4%,3) * (F/P,5%,5) * (F/P,3%,2) = \$5000 * 1.1249 * 1.2763 * 1.0609

Summary We learned some tricks for finding present and future values in special situations. The trick that always works is to make a table of all cash flows in excel, apply appropriate P/F or F/P factors, and add up the result. To apply shortcuts for linear or geometric cases, one must pay careful attention to detail. Next week – Chapter 4 –more applications –Return on investment (finding the i%) –comparing machines, investment plans, etc.

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