how to solve the disequilibrium in balance of payment?
115258
Ask the exact question, you'd get more chances of getting an answer.
861435
You can try paying off the best one you have and just pay the minimum on the others until you do.\nOne I did, was just overcharge the limit, and then keep it paid. They just up my limit to my new self-induced limit, nothing said. ;-)
409087
balance of payment is the document which has a complete record of the transactions that take place between the home country and rest of the world(ROW)\nit consists of the balance of trade which can be adverse , favourable or balanced , and thus by studying it we can correct deficits if any.\ni hope this helps!
151954
teekshi is wrong.  When you put something in layaway, you don't add the layaway payment to the total payment.  In this case, you put down the 10%, which has been correctly calculated to be $14.  You still owe the difference 140-14=126.  So, you can't take the item with you until you pay the difference ($126)
922602
define i = apr/12\nP = quantity (principle)\nn = time (months in this case)\n\nThe calculation formula for payments is\n\np = iP/(1-1/((1+i)^n))\n\nplugging in the numbers,\n\np = (0.01/12)(17,500)/(1 - 1/((1+0.01/12)^72))\n\nOh, if you just want future amount at compound interest, \n\nA(i,n) = A(1+i)^n\n\nA = 17,500(1+0.01/12)^72
817896
to  move balances from one card to another
1049686
He is asking for mechanism....Mechanism is how it works. This is a phase equilibria problem. Please refer to A Level option booklet of phase equilibria.
778679
I believe you will find the solution to your problem here:\n\nhttp://www.engineeringtoolbox.com/bernouilli-equation-d_183.html
1047731
Assume dc/dz is constant, n(t=T)=-DA(dc/dz) T\nBut in reality dc/dz often changes.  Normally dc/dz should decrease with time.  If it is a close system, it will reach the equilibrium state.
310560
That's something you have to ask your credit card company.  When you make payment, try to pay at least $5 more than the minimum due.  It will help you a bit.
941356
The answer here is simple. Go back to paying with cash.
139495
Let's define some variables for the equation.\n\nD = monthly deposit ($200)\nr = annual interest rate (10% is 0.1)\nn = compound frequency (12 months in a year)\nA = account balance\ni = compounded interest\n\nLet's start the account this month, or month 0.\n\nA(0) = D\n\nOne month later, you make another deposit and earn some interest. But you only earn 1/12 of the 10% with you existing balance, which is r/12.  \n\nA(1) = D + A(0) + i(1)\n\nwhere, i(1) = A(0) * r/n. Which makes\n\nA(1) = D + A(0) + A(0) * r/n , or\nA(1) = D + A(0) * (1 + r/n ), or\nA(1) = D + A(0)*R, if R = 1 + r/n, but A(0) is D, so\nA(1) = D + D*R\n\nTwo months later, you make another deposit and earn some interest on the balance A(1). Similarly calculating,\n\nA(2) = D + A(1) * R, or\nA(2) = D + (D + D*R) * R, or\nA(2) = D + D*R + D*R^2\n\nKeep on going to three months later, \n\nA(3) = D + A(2) * R, or\nA(3) = D + D*R + D*R^2 + D*R^3, or\nA(3) = D * (1 + R + R^2 + R^3)\n\nFollowing this pattern, k months later, we get\n\nA(k) = D * (1 + R + R^2 + ... + R^k)\n\nNow, let's calculate the general expression (1 + R + R^2 + ... + R^k) in a closed form.\n\nLet S = (1 + R + R^2 + ... + R^k). Multiply R on both sides, we get\nS*R = (1 + R + R^2 + ... + R^k) * R, or\nS*R = R + R^2 + ... + R^k + R^(k + 1)\n\nSubtracting the two equations S*R and S on both sides, we get\n\nS*R - S = R + R^2 + ... + R^k + R^(k + 1)  - (1 + R + R^2 + ... + R^k)\n\nNotice the terms R all the way through R^k on the right hand side are cancelled and only left with R^(k + 1) - 1. So we have,\n\nS*R - S = S(R - 1) = R^(k + 1) - 1\n\nDividing (R - 1) on both side, we get\n\nS = [R^(k + 1) - 1] / (R - 1)\n\nRewriting the account balance at month k,\n\nA(k) = D * (1 + R + R^2 + ... + R^k), or\nA(k) = D*S = D * [R^(k + 1) - 1] / (R - 1)\n\nNow, 26 years make 26 * 12 = 312 months. But since you start with 0, so your k is actually 311. R = 1 + r/n = 1 + 0.1/12, and D = 200. So\n\nA(311) = 200 * [(1 + 0.1/12)^(311 + 1) - 1] / (1 + 0.1/12 - 1), or\nA(311) = 295667.15330119169 = 295667.15 (round to nearest penny)
381442
Say the grace period is 25 days.  If you buy something and then pay everything off within 25 days, there is no interest.  That only works if you pay the entire balance.  Otherwise if you pay in 26 days or only pay part of the balance, you get charged interest for every day.
477413
Let us suppose interest rate is ‘r’ and the principal amount was ‘P’\nThen for two years;\n\nSimple Interest = 2rP\nCompound interest = P(1+r)^2 – P\nSince we know both the amount of Simple interest and the compound interest we get a set of two simultaneous equations\n\n2rP = 400(1)\nP(1+r)^2 –P = 420(2)\n\nFrom first equation we get \nP=400/2r or P = 200/r\n\nSubstituting in Eq (2)\n\n(200/r)(1+2r+r^2)-200/r = 420\n200/r(1+2r+r^2-1) = 420\n1/r(2r+r^2) = 420/200\n2+r = 420/200\nr = 420/200-2\nr = 2.10-2\nr = 0.10\nSolving for P in eq(1)\nP = 400/2r\nP = 200/r\nP = 200/0.10\nP = 2000\n\nSo the principal amount is 2000.
327033
you can pay double the minimum payments. you can transfer the balance to a lower interest rate card and pay double the minimum payments.
393533
I don't feel like working it out, but you just discount back the cash flows assuming they're constant, and then do some simple algebra.\n\nIt looks kind of like\n\n(785000 - 137) = (x / 1.07) + (x / 1.07^2) + ... (x / 1.07 ^10)\n\nSolve for x.
496828
google search "amortize payments algorithm". I found both the monthly and semi-annual compound algorithms on this huge text file. Good luck on finding it.