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