What is Baumol Cash Model: EOQ Applied to Corporate Treasury?
Mathematical Foundation
Laws & Principles
- The EOQ Analogy: In inventory theory, EOQ minimizes ordering costs vs. holding costs. In the Baumol model: 'ordering cost' = transaction fee F for each securities liquidation; 'holding cost' = opportunity cost i of idle cash. The mathematics are identical; only the context changes. At C*, marginal transaction cost equals marginal opportunity cost — the equilibrium that proves the optimum.
- Number of Replenishments Per Year: N = T / C*. With C* determined, the treasurer knows exactly how many times per year to transfer from securities to cash. If T = $12M/yr and C* = $400k, then N = 30 replenishments/year — approximately every 12 days. Each replenishment is triggered when the cash balance approaches zero.
- Miller-Orr Extension: The Baumol model assumes perfectly predictable, smooth cash outflows — appropriate for large corporations with stable operational spending. The Miller-Orr model (1966) extends Baumol to include random daily cash flow variance. Miller-Orr introduces a lower bound (L), an upper bound (H), and a return-to-target point (Z), making it more suitable for firms with volatile daily cash needs like retailers or manufacturers with lumpy payments.
Step-by-Step Example Walkthrough
" A mid-size manufacturer spends $6M/year on operating expenses evenly. Each securities liquidation costs $250 in broker/wire fees. The current T-bill rate is 5.2% annually. "
- Inputs: T = $6,000,000, F = $250, i = 0.052.
- Optimal C*: √(2 × $6,000,000 × $250 / 0.052) = √($3,000,000,000 / 0.052) = √($57,692,307,692) = $240,192.
- Replenishments per year: N = $6,000,000 / $240,192 = 24.98 ≈ 25 per year (every ~14.6 days).
- Average cash balance: C*/2 = $120,096.
- Annual opportunity cost: $120,096 × 0.052 = $6,245.
- Annual transaction cost: 25 × $250 = $6,250.
- Total annual cost: $6,245 + $6,250 = $12,495 (note: costs balance exactly at the optimum — proof the formula is correct).