Math 373 test 1 12 payments are monthly so we need we

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Math 373 Test 1

Fall 2012

September 27, 2012

1. Meng takes out a loan to buy a new motorcycle. The amount of the loan is 12,500. Meng will repay the loan with 9 monthly payments of Q at a nominal interest rate of 12% compounded monthly.

Calculate Q.

Solution

Payments are monthly so we need

i(12) .

We are given i(12).

12

i(12) 12% 1% 12 12

Use our calculator N 9; I / Y 1; PV 12,500

CPT PMT=>1,459.25

March 9, 2014

2. Yu invests 10,000 in an account earning simple interest. During the 10th year, the amount of interest that Yu earns is 1000. Jay invests X into an account earning compound interest of at an annual effective rate of i. During the 7th year, Yu and Jay earn the same annual effective interest rate. At the end of 20 years, Yu has the same amount in her account as Jay has in his account. Determine X .

Solution

Amount of Interest in the 10th year = A(10) A(9) K a(10) K a(9) K[a(10) a(9)]

For Yu with simple interest

Amount of interest in 10th year = 1000 K[a(10) a(9)] 10,000[(1 s 10) (1 s 9)] 10,000 s s 1000 0.10

10, 000

Annual effective interest rate in 7th year for Jay under simple interest =

s

0.10

0.10

in 1 (n 1)s i11 1 (7 1)0.10 1.60 0.0625

Annual effective interest rate for 7th year for Jay under compound interest = in i i7 Since i7 for Yu = i7 for Jay, i 0.0625 For Yu, the amount after 20 years = 10,000(1 st) 10,000(1 (0.10)(20)) 30,000

For Jay, the amount after 20 years = X (1 i)20 X (1 0.0625)20

Since amount after 20 years for Yu = the amount for Jay after 20 years

30,000 X (1.0625)20

30, 000 X (1.0625)20 8923.65

March 9, 2014

3. Ahmad is the winner of a lottery. He has the option to take his prize in any of the following three options:

a. A lump sum payable now of 20 million (20,000,000); b. A perpetuity immediate with annual payments of 1.25 million; or c. An annuity due with level monthly payments of P for 30 years.

These options all have the same present value when calculated at an annual effective interest rate of i .

Determine P .

Solution

PV of a.=20,000,000

PV of b.=1,250,000a 1, 250,000

i

20,000,000 1, 250,000 i 1, 250,000 0.0625

i

20, 000, 000

Since payments are monthly under c., we need i(12) (1.0625)1/12 1 0.005064835 12

PV of c.=Pa

which is equal to PV of a. = 20,000,000

360 0.005064835

So Set Calc to BGN; N=360; I/Y=0.5064835; PV=20,000,000

CPT PMT=>120,303.02

March 9, 2014

4. Wang Corporation is building a new factory. The cash flows from this factory are expected to be:

Time 0 1 2 3

Cash Flow -10,000 3,000 5,500 X

The Internal Rate of Return for this factory is expected to be 10% based on the above cash flows.

Determine X.

Solution

The Net Present Value at the Internal Rate of Return is zero.

NPV

10,000

3000

1 1.1

1

5500

1 1.1

2

X

1 3 1.1

0

X

10,

000

3000

1 1 1.1

1

3

5500

1 1.1

2

3630

1.1

March 9, 2014

5. A loan of 42,000 is being repaid with level annual payments of 6000 plus a smaller drop payment. The interest rate on the loan is an annual effective interest rate of 8%. Calculate the amount of the drop payment.

Solution We will use our calculator to do this calculation. PV=42,000; I/Y=8; PMT=6000; CPT N=>10.67 years. We round down and will have 10 payments of 6000 and a smaller drop payment at time 11 2nd Amort; P1=10; P2=10; ; Bal=3755.47 Drop Payment=3755.47(1.08) 4,055.91

March 9, 2014

6. Brandon takes out a loan of 25,000 to be repaid with a single payment of B at the end of 6 years. The interest rate on Brandon's loan is equivalent to a nominal discount rate of 8% compounded quarterly.

Leslie takes out a loan of 25,000 to be repaid with a single payment L at the end of 6 years. Leslie pays a force of interest of t 0.02t 0.001t2 .

Determine B L .

Solution

Brandon

a(t)

1

d(4) 4

4t

25, 000 1

0.08 4(6) 4

B

B

40, 598.89

Leslie

t

t

s ds a(t) e0

(0.02s0.001s2 )ds 25,000e0

L L 25, 000e0.01s2 0.001s3 /3

|06

25, 000e0.432

38,508.38

B L 40,598.89 38,508.38 2090.51

March 9, 2014

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