Money multiplier with drainage

Why idoes the money multiplier become (1+c)/(r+c) when part of the additional deposits remain as cash with people? Say a bank receives $1m in new deposits. If the reseve ratio is 10% then the maximum increase in money supply is $1m/0.10 = $10m. But if 45% of deposits do not get loaned out, the money multiplier becomes (1+c)/(r+c), where c=drainage percentage! In this example, the money multiplier is: (1+0.45)/(0.10+0.45)= 2.64 times (i.e., 2.64 * $1m = $2.64m) Doing it the other way: Out of the $1m deposit, only 0.55$1m=$550k get multiplied, so it should lead to $550/0.10= $5.5m (not $2.64m). What am i missing?

a bank recieves $1 mil in new deposits, loans out (1-0.1-0.45) mil -> those get reinvested again and a 45% gets loaned out again, etc money that flows back into the system 1 + (1 - r)(1-c) + ((1-r)*(1-c))^2 + … = 1/(1-(1-r)(1-c)) money that stays in cash 0 + (1-r)*c+(1-r)^2(1-c)c+ … = (1-r)*c(1+(1-r)(1-c) + … ) = (1-r)*c/(1-(1-r)(1-c)) sum those two = (1+ c - rc)/(1-(1-r)*(1-c)) = (1+c - rc)/(r + c - rc)= approx = (1+c)/(r+c)

I wasn’t looking for the math behind this…only to make sense of it. But it seems that the answer lies in the fact that the drainage occurs at every step in the process, not only at the beginning as I thought! So, 1) First bank will have $550k net deposit out of which $495k ($550k*0.90) will be loaned out. 2) The second bank receives $495k, drains 45%, and keeps 10% in reserves, thus it only lends $495k * (0.55) * (0.90) = $245k, and so on. Cool!

Thanks maratikus, i just saw your answer.

my understanding is this: the first bank gets $1 mil, keeps 10% as reserve, lends out 0.9 million people borrow 0.9 million and don’t do anything with 45% of that amount (draining happens), put in the bank 55%, etc …

That’s correct, but would it make a difference if you multiply by 0.55 first, then by 0.90, or multiply by 0.9 first, then by 0.55? I think not.

no difference if you end up counting all the money generated :slight_smile: