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In the compound interest example on P.344. Calculate S20 and T240. Which is the better investment after 20 years?

In the compound interest example on P.344. Calculate S20 and T240. Which is the better investment after 20 years?

Pfin (with Mindtap, 1 Term Printed Access Card) (mindtap Course List)
Pfin (with Mindtap, 1 Term Printed Access Card) (mindtap Course List)
7th Edition
ISBN:9780357033609
Author:Randall Billingsley, Lawrence J. Gitman, Michael D. Joehnk
Publisher:Randall Billingsley, Lawrence J. Gitman, Michael D. Joehnk
Chapter4: Managing Your Cash And Savings
Section: Chapter Questions
Problem 7FPE: Calculating interest earned and future value of savings account. If you put 6,000 in a savings...
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In the compound interest example on P.344. Calculate S20 and T240. Which is the better investment after 20 years?

Hence, S40 = (1.11)"(10 090.090 909. ..) (9 090.909 090...) = (65.000 867...)(10 090.090 909...) (9 090.909 090. ..) 655 917.842... (9 090.909 090...) e $646 826. // Can that be right? You put in $40,000 and take out > $600,000 in interest. Now consider Plan B and let T, denote the number of dollars in the plan after (exactly) n months of operation. Then To = $100 and Tn+1 – T,+ interest on T, +$100 - = T,+(1/12) of 10% of T„+$100 = T„[1+0.1/12] +$100. %3D 100 In this RE, a = 12.1/12, c = 100, so 1 - -12000 and -0.1/12 a T, = (12.1/12)"[100+12000] – 12000. Hence, after 40 × 12 months, T480 = (12.1/12)4" (12100) = (1.008 333...)+8° (12100) - (12000) = (53.700 663...)(12100) = 649 778.023 4... e $637 778. - (12000) 480 - (12000) - (12000) Therefore, Plan A has a slightly larger value after 40 years.
Transcribed Image Text:Hence, S40 = (1.11)"(10 090.090 909. ..) (9 090.909 090...) = (65.000 867...)(10 090.090 909...) (9 090.909 090. ..) 655 917.842... (9 090.909 090...) e $646 826. // Can that be right? You put in $40,000 and take out > $600,000 in interest. Now consider Plan B and let T, denote the number of dollars in the plan after (exactly) n months of operation. Then To = $100 and Tn+1 – T,+ interest on T, +$100 - = T,+(1/12) of 10% of T„+$100 = T„[1+0.1/12] +$100. %3D 100 In this RE, a = 12.1/12, c = 100, so 1 - -12000 and -0.1/12 a T, = (12.1/12)"[100+12000] – 12000. Hence, after 40 × 12 months, T480 = (12.1/12)4" (12100) = (1.008 333...)+8° (12100) - (12000) = (53.700 663...)(12100) = 649 778.023 4... e $637 778. - (12000) 480 - (12000) - (12000) Therefore, Plan A has a slightly larger value after 40 years.
Example 8.2.3: Compound Interest Suppose you are offered two retirement savings plans. In Plan A, you start with $1,000, and each year (on the anniversary of the plan), you are paid 11% simple interest, and you add $1,000. In Plan B, you start with $100, and each month, you are paid one-twelfth of 10% simple (annual) interest, and you add $100. Which plan will be larger after 40 years? Consider Plan A and let S, denote the number of dollars in the plan after (exactly) n years of operation. Then So = $1,000 and // Can we apply a recurrence equation? Sn+1 = Sn +interest on S,+ $1000 = S„ +11% of S, = S„(1+0.11) +$1000 + $1000. 1000 In this RE, a = 1.11, c = 1000, so 1 and - a -0.11 1000 1000 Sn = (1.11)" | 1000 -0.11 -0.11 1110 1000 (1.11)" [+0.11] +0.11
Transcribed Image Text:Example 8.2.3: Compound Interest Suppose you are offered two retirement savings plans. In Plan A, you start with $1,000, and each year (on the anniversary of the plan), you are paid 11% simple interest, and you add $1,000. In Plan B, you start with $100, and each month, you are paid one-twelfth of 10% simple (annual) interest, and you add $100. Which plan will be larger after 40 years? Consider Plan A and let S, denote the number of dollars in the plan after (exactly) n years of operation. Then So = $1,000 and // Can we apply a recurrence equation? Sn+1 = Sn +interest on S,+ $1000 = S„ +11% of S, = S„(1+0.11) +$1000 + $1000. 1000 In this RE, a = 1.11, c = 1000, so 1 and - a -0.11 1000 1000 Sn = (1.11)" | 1000 -0.11 -0.11 1110 1000 (1.11)" [+0.11] +0.11
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