You know you should reduce token costs but not how much you would actually save, or whether the setup is worth it. Take three real numbers - monthly calls, average input, average output - apply a unit-price table, and you get a monthly figure, a savings range and a payback period. Here is the arithmetic, plus a logging script you can adapt.

💡 What You Will Learn

You know you should reduce token costs but not how much you would actually save, or whether the setup is worth it. Take three real numbers - monthly calls, average input, average output - apply a unit

📜 Table of Contents

• 1. Get three real numbers
• 2. The cost formula
• 3. A script to measure instead of guessing
• 4. Express savings as a range
• 5. Payback period
• 6. Three ways the estimate goes wrong
• FAQ

Reduce AI Token Cost Calculator: Estimating Your Savings in 2026

"How much can I save" has no universal answer. It depends on three numbers you already own: calls per month, average input tokens, average output tokens. Once you have those, the rest is multiplication.

1. Get three real numbers

Do not guess "around 500 tokens per call". Each number has a source:

If you mix models, group by model. An average across models is not a number.

2. The cost formula

Per call = input tokens x input price + output tokens x output price (converted from a per-million-token quote).

Monthly = sum over groups of (per-call cost x monthly calls), adjusted for cached input billed at the cached rate.

Treat prices as variables. Keep a sheet: model, input price, output price, average tokens, monthly calls - cost falls out.

3. A script to measure instead of guessing

Most SDK responses carry a usage object; log it:

import collections
stats = collections.defaultdict(lambda: {"calls": 0, "in": 0, "out": 0})

def record(tag, model, usage):
    k = (tag, model)
    stats[k]["calls"] += 1
    stats[k]["in"] += usage.prompt_tokens
    stats[k]["out"] += usage.completion_tokens

def report(price_in, price_out):   # dicts: USD per 1M tokens, your own quotes
    rows = []
    for (tag, model), s in stats.items():
        cost = s["in"]/1e6*price_in.get(model, 0) + s["out"]/1e6*price_out.get(model, 0)
        rows.append((cost, tag, model, s["calls"]))
    for row in sorted(rows, reverse=True):
        print(row)

After a week the distribution is visible, and the few tags that dominate the bill tell you where to aim.

4. Express savings as a range

Savings depend on how many calls can be replaced.

Savings = current monthly cost x replaceable share x reduction on that share, plus the cached-input discount. Multiply those layers, and only then state a number.

5. Payback period

Payback (months) = one-off engineering cost / monthly saving. If a week of work saves a few dollars a month, skip it. If it saves days of work every month, start with the biggest line item.

6. Three ways the estimate goes wrong

Ranking by unit price instead of total. A cheap model used heavily can cost more in total. Sort by price times volume.

Ignoring failures and retries. Timeouts and retried calls are billed; count them separately.

Ignoring cache hit rate. A low hit rate means your fixed prefix keeps changing and the discount never lands.

FAQ

Only a total bill, no logs? Back out average cost per call (spend / calls), then group tasks roughly and sample a few real usage objects per group. Less precise, good enough to decide.

How do I price a local model? Electricity plus hardware amortisation. If the hardware is already bought, amortise over remaining life; marginal cost is close to electricity, plus your maintenance time.

Can I promise a savings percentage? Not publicly. State the assumptions (replaceable share, cache hit rate) rather than a flattering number.

Note: this article gives arithmetic and data structures only, no prices - check official pricing pages and use the quotes on your own invoice.

❓ FAQ

No call logs?

Divide the total bill by call count, group tasks roughly, and sample real usage objects per group.

How to price a local model?

Electricity plus hardware amortisation over remaining life, plus maintenance time.

Can savings be promised?

No - state the assumptions, since the same measures save very different amounts in different workloads.