# 9.14 The Optimal Cash Order Quantity Solution

Until now, “C” was an undefined quantity.  Using some algebra (see below), we will derive C*, the optimal cash withdrawal amount, or the optimal “order quantity.”

Let us see how this works in our prior example.

Given:

T = Total required yearly outlay = \$25mm (yearly)

F = \$100

i = .05

Due to the fact that, at the bottom of the “U” curve, Opportunity and Transaction Costs will be the same (take a look back at the table at around 300,000 units) , we may solve for C* using the formula: Transaction Costs = Opportunity Costs.

 Formula and (Alternative) Solutions C* Proof (T/C) (F) = (C/2) (i) (\$25,000,000 / C) (100) = (C/2) (0.05) C* = \$316,227 (T/C) (F) = (C/2) (i) 2 T F = c2 i C2 = (2 T F) ÷ i C* = [(2 × T × F) ÷ i] 0.5 C* = {(2 × \$25,000,000 × \$100} ÷ 0.05] 0.5 = \$316,227

Summary: At an order quantity of \$316,227 (or C*), (T/C × F) = (C/2 × i).

Now that we know C*, the ordering interval, i.e., the number of times per year that “cash orders” must be made, also becomes known.  (It is the annual cash requirement divided by C*.)

Question: Every how many days will the firm have to re-order cash?

Answer: (\$25 million) ÷ (316,277) ~ 79 times a year – or about every 4 days

Notes:

1. C* will vary with F – if F is high, reduce number of transactions
2. The average balance will increase inversely with i: if i is high, decrease size of transactions
• Therefore, if F increases, C* will increase; if i increases, C* will decrease
1. (This analysis assumes that market risk – in the liquidation of ST securities for meeting cash requirements, is nil.)
2. This model has also assumed a linear (i.e., smooth) reduction in cash over the period and, hence, equal intervals between re-order dates.