HP 35s User Manual

							 
hp calculators 
 
HP 35s Sinking Funds 
 
hp calculators - 3 - HP 35s Sinking Funds - Version 1.0 
5) After you press : for the last time, the value of the unknown variable will be calculated and displayed. 
6) To do another calculation with the same or changed values, go back to step 2 above. 
 
The SOLVE feature will work effectively without any initial guesses being supplied for the unknown variable with the 
exception noted above about the variable I in this equation. This equation follows the standard convention that money in 
is considered positive and money out is negative. 
 
The practice problems below illustrate using this equation to solve a variety of sinking fund problems. 
 
Practice solving for payment required to achieve a goal 
 
Example 1: How much would you need to save at the end of every month to accumulate $10,000 in 6 years?  
 Assume the funds would earn 6%, compounded monthly, and that you begin with nothing in the account. 
 
Solution: First, enter the time value of money equation into the HP 35s solver as described earlier in this document.  
 
 Then press !-and press 8or 9 to scroll through the equation list until the time value  
 of money equation is displayed. Then press:-
 
-56# 
 
 The HP 35s SOLVER displays the first variable encountered in the equation as it begins its solution.  
 The value of 0.0000 is displayed below if this is the first time the time value of money equation has been  
 solved on the HP 35s calculator. If any previous equations have used a variable used in the time value of  
 money equation, they may already have been assigned a value that would be displayed on your HP 35s  
 display. Follow the keystrokes shown below and the solution should be found as described. 
 
 Figure 1 
 
-In RPN mode, press: ;2%
						

							 
hp calculators 
 
HP 35s Sinking Funds 
 
hp calculators - 4 - HP 35s Sinking Funds - Version 1.0 
 Figure 4 
 
-In either RPN or algebraic mode, press: &: 
 
 Figure 5 
 
Answer: The required monthly deposit is $115.73. 
 
Example 2: John wants to retire as a millionaire. He is 25 years old. How much would he need to deposit each month 
 beginning one month from now and continuing until his 65th birthday in order to achieve his goal? Assume  
 the funds would earn 5%, compounded monthly, and that John begins with nothing in the account. 
 
Solution: First, enter the time value of money equation into the HP 35s solver as described earlier in this document.  
 
 Then press !-and press 8or 9 to scroll through the equation list until the time value  
 of money equation is displayed. Then press:-
 
-56# 
 
 The HP 35s SOLVER displays the first variable encountered in the equation as it begins its solution.  
 The value of 0.5000 is leftover from the immediately preceding example. If your HP 35s has been used  
 to make changes to the value stored in any of the time value of money equation variables, the initial values  
 displayed may vary from what is shown below. If this example is worked immediately after the preceding  
 example, the displays below will be shown on an HP 35s. Follow the keystrokes shown below and the  
 solution should be found as described. 
 
 Figure 6 
-
-In RPN mode, press: =2%
						

							 
hp calculators 
 
HP 35s Sinking Funds 
 
hp calculators - 5 - HP 35s Sinking Funds - Version 1.0 
 Figure 8 
 
-In either RPN or algebraic mode, press: %&&&&&&: 
 
 Figure 9 
 
-In either RPN or algebraic mode, press: &: 
 
 Figure 10 
 
Answer: $655.30  
 
Example 3: How much money should you deposit each year into an account, beginning one year from today, to have 
 $30,000 in the account after 15 years? Assume the funds would earn 6%, compounded annually, and  
 that the account begins with a balance of $1,000. 
 
Solution: First, enter the time value of money equation into the HP 35s solver as described earlier in this document.  
 
 Then press !-and press 8or 9 to scroll through the equation list until the time value  
 of money equation is displayed. Then press:-
 
-56# 
 
 The HP 35s SOLVER displays the first variable encountered in the equation as it begins its solution.  
 The initial values shown in the figures below assume this example is worked immediately after the  
 preceding example. Follow the keystrokes shown below and the solution should be found as described. 
 
 Figure 11 
-
-In either RPN or algebraic mode, press: ;: 
 
 Figure 12 
 
-In either RPN or algebraic mode, press: %=: 
   
						

							 
hp calculators 
 
HP 35s Sinking Funds 
 
hp calculators - 6 - HP 35s Sinking Funds - Version 1.0 
 Figure 13 
 
-In either RPN or algebraic mode, press: >&&&&: 
 
 Figure 14 
 
-In either RPN or algebraic mode, press: %&&&.: 
 
 Figure 15 
 
Answer: $1,185.92     
						

							 
 
hp calculators 
 
 
 
 
HP 35s Present value 
 
 
 
 
Present value 
 
The Time Value of Money on the HP 35s 
 
Practice solving for the present value of future 
cash flows 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
   
						

							 
hp calculators 
 
HP 35s Present value 
 
hp calculators - 2 - HP 35s Present value - Version 1.0 
Present value 
 
Many problems involving the time value of money require the conversion of monies to be received in the future into the 
equivalent monies today. This conversion is the computing of the present value of the monies being received in the 
future. There are many benefits from this conversion to a present value, including the ability to better visualize the real 
magnitude of future expenditures or receipts as well as the direct comparison using values today of alternative future 
receipts or expenditures. 
 
The Time Value of Money on the HP 35s 
 
To solve time value of money problems on the HP 35s, the formula below is entered into the flexible equation solver built 
into the calculator. This equation expresses the standard relationship between the variables in the time value of money 
formula. The formula uses these variables: N is the number of compounding periods; I is the periodic interest rate as a 
percentage (for example, if the annual interest rate is 15% and there are 12 payments per year, the periodic interest 
rate, i, is 15÷12=1.25%); B is the initial balance of loan or savings account; P is the periodic payment; F is the future 
value of a savings account or balance of a loan. 
 
Equation: P x 100 x ( 1 - ( 1 + I ! 100 )^ -N)  ! I + F x ( 1 + I ! 100 ) ^ -N + B 
 
To enter this equation into the calculator, press the following keys on the HP 35s: 
 
!#$%&&$4%4%()*%&&+,-
./+*)(0$4%()*%&&+,.-
/(12 
 
To verify proper entry of the equation, press 
 
34 
 
and hold down the 4-key. This will display the equation’s checksum and length. The values displayed should be a 
checksum of CEFA and a length of 41. 
 
To solve for the different variables within this equation, the 56 button is used. This key is the right shift of the 
! key. 
 
Notes for using the SOLVE function with this equation: 
 
1) If your first calculation using this formula is to solve for the interest rate I, press %57)-before beginning. 
2) Press !. If the time value of money equation is not at the top of the list, press 8or 9 to scroll through the 
list until the equation is displayed. 
3) Determine the variable for which you wish to solve and press: 
a) 56/ to calculate the number of compounding periods. 
b) 56) to calculate the periodic interest rate. Note: this will need to be multiplied by the number of 
compounding periods per year to get the annual rate. If the compounding is monthly, multiply by 12. 
c) 561 to calculate the initial balance (or Present Value) of a loan or savings account. 
d) 56# to calculate the periodic payment. 
e) 560 to calculate the future value of a loan or savings account. 
4) When prompted, enter a value for each of the variables in the equation as you are prompted and press :. The 
solver will display the variables’ existing value. If this is to be kept, do not enter any value but press : to   
						

							 
hp calculators 
 
HP 35s Present value 
 
hp calculators - 3 - HP 35s Present value - Version 1.0 
continue. If the value is to be changed, enter the changed value and press :. If a variable had a value in a 
previous calculation but is not involved in this calculation (as might happen to the variable P (payment) when solving 
a compound interest problem right after solving an annuity problem), enter a zero for the value and press :. 
5) After you press : for the last time, the value of the unknown variable will be calculated and displayed. 
6) To do another calculation with the same or changed values, go back to step 2 above. 
 
The SOLVE feature will work effectively without any initial guesses being supplied for the unknown variable with the 
exception noted above about the variable I in this equation. This equation follows the standard convention that money in 
is considered positive and money out is negative. 
 
The practice problems below illustrate using this equation to solve a variety of problems involving present values. 
 
Practice solving for the present value of future cash flows 
 
Example 1: If you are to pay $50,000 in 6 years, what is this worth in today’s dollars, assuming interest is applied 
 at 8%, compounded quarterly? 
 
Solution: First, enter the time value of money equation into the HP 35s solver as described earlier in this document.  
 
 Then press !-and press 8or 9 to scroll through the equation list until the time value  
 of money equation is displayed. Then press:-
 
-561 
 
 The HP 35s SOLVER displays the first variable encountered in the equation as it begins its solution.  
 The value of 0.0000 is displayed below if this is the first time the time value of money equation has been  
 solved on the HP 35s calculator. If any previous equations have used a variable used in the time value of  
 money equation, they may already have been assigned a value that would be displayed on your HP 35s  
 display. Follow the keystrokes shown below and the solution should be found as described. 
 
 Figure 1 
 
-Since this is a compound interest example and does not have a series of equal-sized, equal-spaced  
 payments, the value or P is zero.  
 
 In either RPN or algebraic mode, press: &: 
 
 Figure 2 
 
-In RPN mode, press: ;2
						

							 
hp calculators 
 
HP 35s Present value 
 
hp calculators - 4 - HP 35s Present value - Version 1.0 
 Figure 3 
 
-In RPN mode, press: =2&&.: 
 
 Figure 7 
   
						

							 
hp calculators 
 
HP 35s Present value 
 
hp calculators - 5 - HP 35s Present value - Version 1.0 
-In RPN mode, press: =2%?*: 
 In algebraic mode, press: =*%?2:-
 
 Figure 8 
 
-In RPN mode, press: ?&2%?$: 
 In algebraic mode, press: ?&$%?2:-
 
 Figure 9 
 
-In either RPN or algebraic mode, press: &: 
 
 Figure 10 
 
Answer: The equivalent amount in today’s dollars is $69,790.39. 
 
Example 3: Dan will receive $40 per month for the next five years and a single payment 60 months from today of  
 $2,000. If interest is 5.5%, compounded monthly, what is the present value of these cash flows? 
 
Solution: First, enter the time value of money equation into the HP 35s solver as described earlier in this document.  
 
 Then press !-and press 8or 9 to scroll through the equation list until the time value  
 of money equation is displayed. Then press:-
 
-561 
 
 The HP 35s SOLVER displays the first variable encountered in the equation as it begins its solution.  
 The displays for these prompts are not shown in this example. Follow the keystrokes shown below and  
 the solution should be found as described.   
						

							 
hp calculators 
 
HP 35s Present value 
 
hp calculators - 6 - HP 35s Present value - Version 1.0 
 In RPN mode, press: @>2%?*: (Enters I) 
-->2%?$:  (Enters N) 
--?&&&:   (Enters F) 
 
 In algebraic mode, press: @>*%?2: (Enters I) 
-->$%?2:  (Enters N) 
--?&&&:   (Enters F) 
 
 Figure 11 
 
Answer: The equivalent amount in today’s dollars is $3,614.21.