2 3 sinking-funds
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Transcript of 2 3 sinking-funds
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2.3 Sinking Funds2.3 Sinking Funds
Sinking Funds
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2.3 Sinking Funds2.3 Sinking Funds
• Sinking fund – a fund-accumulation scheme wherein the amount is generated by making periodic deposits
• In amortization problems, we usually want to find the following values:– Periodic deposit– Amount in the fund after any kth deposit– Interest earned in any period– Increase in the fund in any period
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2.3 Sinking Funds2.3 Sinking Funds
• Sinking fund schedule – a table which shows how a target amount is completely attained through periodic deposits as well as interests these deposits earn in the process
• Sinking fund method of paying off a debt – happens when the debtor pays interest periodically and pays the principal in one lump-sum payment at the end of the term from a sinking fund
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2.3 Sinking Funds2.3 Sinking Funds
Formulas:• Amount in the fund after k deposits
• Periodic deposit
€
Sk = D(1+ is)
k −1
is
⎡
⎣ ⎢
⎤
⎦ ⎥
€
D =Skis
(1+ is)k −1
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2.3 Sinking Funds2.3 Sinking Funds
Formulas:• Interest earned
• Principal repayment
• Periodic cost (when a sinking fund is used to pay off a debt)
€
IEk = Sk−1is = D (1+ is)k−1 −1[ ]
€
INCk = Sk − Sk−1 = D(1+ is)k−1
€
C = iAA +isA
(1+ is)n −1
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2.3 Sinking Funds2.3 Sinking Funds
1. In order to have Php800,000 in 5 years, Shane deposits an amount each year in a sinking fund earning 6% effective. Find the annual deposit and construct the sinking fund schedule.
€
D =Skis
(1+ is)k −1
€
=(800,000)(.06)
1.065 −1
€
=Php141,917.12
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2.3 Sinking Funds2.3 Sinking Funds
Period Beginning amount
Interest earned
Regular deposit
Ending amount
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2.3 Sinking Funds2.3 Sinking Funds
Period Beginning amount
Interest earned
Regular deposit
Ending amount
1
2
3
4
5
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2.3 Sinking Funds2.3 Sinking Funds
Period Beginning amount
Interest earned
Regular deposit
Ending amount
1 0 0 141,917.12 141,917.12
2
3
4
5
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2.3 Sinking Funds2.3 Sinking Funds
Period Beginning amount
Interest earned
Regular deposit
Ending amount
1 0 0 141,917.12 141,917.12
2 141,917.12 8,515.03 141,917.12 292,349.27
3
4
5
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2.3 Sinking Funds2.3 Sinking Funds
Period Beginning amount
Interest earned
Regular deposit
Ending amount
1 0 0 141,917.12 141,917.12
2 141,917.12 8,515.03 141,917.12 292,349.27
3 292,349.27 17,540.96 141,917.12 451,807.35
4
5
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2.3 Sinking Funds2.3 Sinking Funds
Period Beginning amount
Interest earned
Regular deposit
Ending amount
1 0 0 141,917.12 141,917.12
2 141,917.12 8,515.03 141,917.12 292,349.27
3 292,349.27 17,540.96 141,917.12 451,807.35
4 451,807.35 27,108.44 141,917.12 620,832.91
5
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2.3 Sinking Funds2.3 Sinking Funds
Period Beginning amount
Interest earned
Regular deposit
Ending amount
1 0 0 141,917.12 141,917.12
2 141,917.12 8,515.03 141,917.12 292,349.27
3 292,349.27 17,540.96 141,917.12 451,807.35
4 451,807.35 27,108.44 141,917.12 620,832.91
5 620,832.91 37,249.97 141,917.12 800,000
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2.3 Sinking Funds2.3 Sinking Funds
3. To raise money for office space expansion, a small business operator estimates that Php15M will be needed in 4 ½ years. He projects that a certain amount must be invested every 3 months in a fund which earns interest at 12% converted quarterly.
a) How much is the quarterly investment?b) How much will be in the fund after the 3rd
deposit?c) How much interest is earned on the 3rd year?
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2.3 Sinking Funds2.3 Sinking Funds
3. To raise money for office space expansion, a small business operator estimates that Php15M will be needed in 4 ½ years. He projects that a certain amount must be invested every 3 months in a fund which earns interest at 12% converted quarterly.
a) How much is the quarterly investment?
€
D =Skis
(1+ is)k −1
€
=(15,000,000)(.03)
1.0318 −1
€
=Php640,630.44
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2.3 Sinking Funds2.3 Sinking Funds
3. To raise money for office space expansion, a small business operator estimates that Php15M will be needed in 4 ½ years. He projects that a certain amount must be invested every 3 months in a fund which earns interest at 12% converted quarterly.
b) How much will be in the fund after the 3rd deposit?
€
S3 = D(1+ is)
3 −1
is
⎡
⎣ ⎢
⎤
⎦ ⎥
€
=640,630.441.033 −1
.03
⎡
⎣ ⎢
⎤
⎦ ⎥
€
=Php1,980,124.63
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2.3 Sinking Funds2.3 Sinking Funds
3. To raise money for office space expansion, a small business operator estimates that Php15M will be needed in 4 ½ years. He projects that a certain amount must be invested every 3 months in a fund which earns interest at 12% converted quarterly.
c) How much interest is earned on the 3rd year?
€
IE12 = 640,630.44(1.0311 −1)
€
=Php246,151.91
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2.3 Sinking Funds2.3 Sinking Funds
5. Caloi and Santi want to pool their money in order to have P1M in 5 years. They will place semi-annual deposits in a fund that earns 5% compounded semi-annually. How much each of these deposits should be?
€
D =Skis
(1+ is)k −1
€
=(1,000,000)(.025)
1.02510 −1
€
=Php89,258.76
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2.3 Sinking Funds2.3 Sinking Funds
9. A Php500,000 loan at 13% interest rate payable semi-annually is to be repaid in 10 years. Find the semi-annual expense if
a) the loan is to be amortized every 6 monthsb) the loan is repaid through a sinking fund
earning at 15% compounded semi-annually.How much does the borrower save semi-
annually by choosing the cheaper method?
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2.3 Sinking Funds2.3 Sinking Funds
9. A Php500,000 loan at 13% interest rate payable semi-annually is to be repaid in 10 years. Find the semi-annual expense if
a) the loan is to be amortized every 6 months.
€
R =(500,000)(.065)
1 −1.065−20
€
=Php45,378.20
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2.3 Sinking Funds2.3 Sinking Funds
9. A Php500,000 loan at 13% interest rate payable semi-annually is to be repaid in 10 years. Find the semi-annual expense if
b) the loan is repaid through a sinking fund earning at 15% compounded semi-annually.
€
C = iAA +isA
(1+ is)n −1
€
=(.065)(500,000) + (.075)(500,000)
1.07520 −1
€
=Php44,046.10
€
The borrower saves Php1,332.10.
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2.3 Sinking Funds2.3 Sinking Funds
11. George borrows Php80,000, with a plan to repay the whole amount in 4 years. Between now and that time, he has to make monthly interest payments at 12% compounded monthly. The Php80,000 will be repaid by making deposits each month in a sinking fund that earns 13% compounded monthly. How much is the monthly cost of the loan?
€
C = iAA +isA
(1+ is)n −1
€
=(.01)(80,000) +.1312( )(80,000)
1+ .1312( )
48−1
€
=Php2,079.53