2. (70 points) Consider a credit card that starts with a S2000 balance and for w

Question

2. (70 points) Consider a credit card that starts with a S2000 balance and for which, each month, $500 in new charges are accrued, and interest is collected on any unpaid balance at a rate of 1.5% per month (19.56% APR). (a) Write a recurrence relation for the balance after k months, be, when a fixed payment of p dollars is made each month. (b) Solve this recurrence relation for bk as a function of p. (c) For what value of p does the monthly balance stay at $2000? (d) Find an expression for the month, k, in which the balance becomes zero as a function of ii)How many months does it take to pay off the balance if p $750? What about p = i)How much should be paid each month to have a zero balance after exactly 6 months? p. $1000? After exactly 12 months? Use the bisection code from HW2 to compute the payments.

Explanation / Answer

To solve this equation let us first of all try to develop the recurrance equation.

We will start with month zero (M0)

At M0 , total balance is 2000

At M1 ,500 is added as charges and p is taken out as payment

thus at Month 1 (M1) total balance will be (2000+500-p)

interest of 1.5 % (let us assume it to be r for simplicity sake)

thus after second month the total carry forward backlog will be

(2000+500-p) (1+r)

Now again 500 $ new charge is levied and p payment is made .

Thus total out standing after 2 months

(2000+500-p)(1+r)+500-p

rate r will be implied on it

thus initial opening balance at end of 3rd month

((2000+500-p)(1+r)+500-p )(1+r)

and balance after 3rd month will be

((2000+500-p)(1+r)+500-p )(1+r)+500-p

Thus generalizing it for k months we can get

the recurrance balance as

2000*(1+r)k +(500-p){(1+r)k +(1+r)k-1 +...+(1+r)}

a). thus bk =2000*(1+r)k +(500-p){(1+r)k +(1+r)k-1 +...+(1+r)}

Solving it we get

bk =2000*(1+r)k +(500-p)*(1+r) ((1+r)k -1))/r

Keeping r =0.015

b) .bk =2000*(1.015)k +(500-p)*67.67*(1.015k -1)

c). Keeping bk =2000

2000 =2000*(1.015)k +(500-p)*67.67*(1.015k -1)

It can be solved if numerical value of p is provided.

d). when bk =0
then

0=2000*(1.015)k +(500-p)*67.67*(1.015k -1)

Solving it we get

p=500+30/(1-1/1.015k )

clearly after k =240 months the balance tends to be constant at 530

thus we can assume that after 20 years the balance will become 0 .

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