Cash Management Models: Baumol and Miller-Orr

A company that keeps too little cash must keep selling investments, paying brokers’ charges each time, or borrowing in a hurry. A company that keeps too much loses the interest it could have earned. Somewhere between the two is a best balance, and for the finance manager the question is how to find it with arithmetic and not guesswork. Two models do this.

The Baumol model treats cash like inventory: you sell securities to raise cash, and the best amount to sell each time balances the cost of selling against the interest lost on idle cash. The Miller-Orr model is for businesses whose cash flows jump about unpredictably: it sets an upper and a lower limit and a return point, and lets the balance wander between the limits until it hits one.

This topic explains both, their assumptions and formulae, and worked calculations, and compares them. The next topic covers working capital investment policy and determining the working capital needed.

The formulae for both models are printed on the formula sheet of the paper, so the skill examined is applying them, interpreting the result and knowing the assumptions. Read short-term investment and treasury management first.

The Baumol model

The Baumol model is based on the same principles as the economic order quantity model. It assumes that the company regularly sells short-term investments to raise cash for its operating payments, and it finds the amount to raise each time, Q, that minimises the total cost. The assumptions are:

• cash is used at a constant, predictable rate through the year;
• cash inflows are known and regular;
• cash can be replenished immediately when it runs out;
• holding cash has an opportunity cost, the interest lost by not investing it;
• cash is replenished by selling short-term investments; and
• each sale has a transaction cost (brokerage, commission).

The average cash held is Q ÷ 2, so the annual holding cost = Q ÷ 2 × CH, and the annual transaction cost = D ÷ Q × Co. Setting them equal gives:

Q = √(2 × Co × D ÷ CH)

where D is the cash needed in the period (annual if CH is annual), Co the transaction cost of each sale and CH the holding cost as a rate. The result shows that when interest rates are high the cash balance should be low. The weakness is that the assumptions are unrealistic: payments are rarely constant.

Example: a company pays suppliers ₦36m a year. Each sale of securities costs ₦200 and the interest rate is 8%. Q = √(2 × 200 × 36,000,000 ÷ 0.08) = ₦424,264. The number of sales a year is 36,000,000 ÷ 424,264 = 84.9. Annual transaction cost = 84.85 × ₦200 = ₦16,971 and annual holding cost = 424,264 ÷ 2 × 8% = ₦16,971, equal, as they must be.

When the demand is monthly it must be annualised: a company needs ₦2.5m a month, so D = ₦30m; the transaction cost is ₦150 and the interest rate 10%. Q = √(2 × 150 × 30,000,000 ÷ 0.10) = ₦300,000, giving 100 sales a year, a transaction cost of 100 × ₦150 = ₦15,000 and a holding cost of 150,000 × 10% = ₦15,000.

The Miller-Orr model

The Miller-Orr model finds the best cash balance when cash flows are uncertain. Its assumptions are that cash flows are uncertain but can be measured statistically, that upper and lower control limits can be set, that transactions in securities have a cost and that holding cash has an opportunity cost. Management decides the lower limit, the minimum cash it will hold (a safety level).

The model then sets the spread, the distance between the limits, and the return point.

Spread = 3 × [ (¾ × transaction cost × variance of daily cash flows) ÷ daily interest rate ]^1/3

• Upper limit = lower limit + spread.
• Return point = lower limit + one third of the spread (so the return point is one third of the way up from the lower limit and two thirds below the upper limit).

The variance is the square of the standard deviation of daily net cash flows, and the interest rate must be on the same daily basis. The balance is allowed to move freely between the limits.

If it reaches the upper limit, securities are bought with enough cash to bring the balance back to the return point; if it falls to the lower limit, securities are sold to bring it up to the return point.

Example: the minimum balance is ₦40,000, the standard deviation of daily cash flows is ₦3,000, each transaction costs ₦90 and the annual interest rate is 9%. Variance = 3,000² = 9,000,000; daily interest rate = 0.09 ÷ 365 = 0.000247. Spread = 3 × (0.75 × 90 × 9,000,000 ÷ 0.000247)^1/3 = ₦40,519. Upper limit = 40,000 + 40,519 = ₦80,519; return point = 40,000 + 40,519 ÷ 3 = ₦53,506. If the balance rises to ₦80,519 the company buys ₦27,013 of securities (to bring it to ₦53,506); if it falls to ₦40,000 it sells ₦13,506 of securities.

Comparing the models

The Baumol model is simple and gives the amount to transfer, but it needs a steady drain on cash.

The Miller-Orr model fits real businesses better, because it allows for variability:

• the greater the variance of cash flows, the wider the spread and the higher the average balance
• the higher the interest rate, the narrower the spread and the lower the balance
• the higher the transaction cost, the wider the spread

Both ignore other sources of cash such as an overdraft, and both depend on estimates of cost, interest and variability. The lower limit in Miller-Orr is a management judgement and may be set by the bank’s requirements or the company’s attitude to risk.

The ideas in more depth

Why Miller-Orr uses a cube root

The spread depends on the cost of transactions, the variance of cash flows and the interest rate, and the best trade-off between them turns out to grow with the cube root of the combination. A fourfold increase in variance raises the spread by only about 59%, so a larger uncertainty does not require a proportionately larger buffer.

Why the return point is nearer the bottom

Interest is lost on every naira held above the minimum, so a return point close to the floor keeps the average balance, and the interest forgone, low. The one-third and two-thirds split is the result of minimising the expected cost of transactions together with the cost of holding cash.

Sources and references

• No accounting standard governs the models. They derive from Baumol (1952) and Miller and Orr (1966).

These points are stated as at 5 November 2026.

Worked examples

1. Q.
   √(2 × 150 × 30,000,000 ÷ 0.10) = ₦300,000.
2. Sales and cost. 100 sales; transaction cost ₦15,000 plus holding cost ₦15,000 = ₦30,000.

1. Upper.
   40,000 + 40,519 = ₦80,519.
2. Return point.
   40,000 + 40,519 ÷ 3 = ₦53,506.

1. Action.
   Buy securities of 80,519 − 53,506 = ₦27,013.

How to answer in the exam

• Use the formula from the formula sheet and keep the units consistent: annual with annual, daily with daily.
• For Baumol, annualise the demand and present the number of sales and the equal costs as a check.
• For Miller-Orr, square the standard deviation and convert the interest rate to a daily rate before using the formula.
• Explain the action at each limit and state the assumptions and limits of the model.

Common mistakes in the exam

• Using a monthly demand with an annual interest rate.
• Forgetting to square the standard deviation.
• Using an annual rate in the Miller-Orr formula with a daily variance.
• Placing the return point at the middle of the spread.
• Applying Baumol to a business with very uneven cash flows.
• Not saying what to do at each limit.

Quick revision

• Baumol: Q = √(2CoD ÷ CH); transaction cost = holding cost at Q; constant use.
• Miller-Orr.
  spread = 3 × [¾ × cost × variance ÷ daily rate]^1/3.
• Upper = lower + spread; return = lower + spread ÷ 3.
• At the upper limit buy securities; at the lower limit sell them.

Practice questions

These are original questions written in the style of the Institute’s exams. Try each one before you open the answer. Section A is a recall check, because the paper itself is written.

Recall check: multiple choice

Section B style: written questions

A further written question

Model answers

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