Early Withdrawal

Archive for February, 2011

Question: Should I have 10% withheld from an rollover IRA distribution?

I’ve rolled a significant amount of money from my 401K to a Roll-over IRA to be used as a down payment on a house. I am now ready to transfer the money from the IRA to my savings account.

I have the option to have the 10% Early Withdrawal Penalty withheld or not; with the following warning: “Regardless of your withholding election, you are responsible for the full payment of any federal income taxes, any state or local taxes and any penalties that may apply to your distribution. You may be responsible for estimated tax payments and could incur penalties if your estimated tax payments are not sufficient.”

What potential penalties do I face if I elect not to withhold and put the money in an interest earning savings account until next tax season?

Answer: In addition to all you stated, the IRS will also penalize you for not pre paying at least 90 % of your tax for that particular year, that is the reason for withholding, so they get their share before you file.

Self-Direction — How do I Know This is Legal?

Question: Can I withdraw money out of 457(b) plan and put in Roth IRA w/o paying Early Withdrawal Penalty?

Answer: Yes , if you rollover the money directly from your 457(b) to the Roth IRA. You will have to pay regular income tax, however, since you are going from a pre-tax account to an after-tax account.

Personal Finance & Money Management Tips : Money Market Account Tips & Advice

Question: method of undetermined coefficients?

Find a particular solution to the differential equation

y”+5y’+6y=72t^3

y= ?

Answer: y” + 5y’ + 6y = 72t³

Solve for the complimentary solution, y₍(t).

r² + 5r + 6 = 0
(r + 3)(r + 2) = 0
r = -3, r = -2

y₍(t) = C₁e^(-3t) + C₂e^(-2t)

You have to find the particular solution before the initial conditions can be applied to solve for C₁ and C₂.

Make a guess at the particular solution. Since 72t³ is a polynomial, the particular solution, Yᵨ(t), is also a polynomial.

Yᵨ(t) = At³ + Bt² + Ct + D
Yᵨ’(t) = 3At² + 2Bt + C
Yᵨ”(t) = 6At + 2B

Substituting into the DE

6At + 2B + 5(3At² + 2Bt + C) + 6(At³ + Bt² + Ct + D) = 72t³

6At + 2B + 15At² + 10Bt + 5C + 6At³ + 6Bt² + 6Ct + 6D = 72t³

6At³ + (15A + 6B)t² + (6A + 10B + 6C) + (2B + 5C + 6D) = 72t³

Solve for the coefficients:
6A = 72
—-> A = 12

15A + 6B = 0
—> B = -30

6A + 10B + 6C = 0
—> C = 38

2B + 5C + 6D = 0
—> D = -65/3

So Yᵨ(t) = 12t³ – 30t² + 38t – 65/3

The general solution is the sum of the complimentary and particular solutions.

y(t) = y₍(t) + Yᵨ(t)

y(t) = C₁e^(-3t) + C₂e^(-2t) + 12t³ – 30t² + 38t – 65/3