Calculators

Illustrative purchase decision: assume a 70-billion-parameter model at one byte per parameter, 330 KB of KV cache per token, 32,000-token context, 64 concurrent requests and 15% framework overhead. Memory is (70 + 330 × 32,000 × 64 / 1,000,000) × 1.15 = about 858 GB per replica. Assume 288 GB per GPU and an eight-GPU replica you have tested: the three-GPU memory floor does not replace that qualification.

Assume 50,000 output tokens/s peak demand, 11,200 tokens/s per replica and 70% serving capacity. Round 50,000 / (11,200 × 0.70) up to seven replicas: 56 active GPUs. A 72-GPU purchase unit gives 72 installed GPUs, 16 stranded slots and one 72-GPU rack. At an assumed 135 kW nominal rack input and PUE 1.15, the nominal facility screen is about 155 kW. Fractional rental at an assumed $18/GPU-hour costs $1,008/hour for the active fleet.

Assume a $1,100/hour active-fleet rental allowance. The seven-replica case fits. Above 7 × 11,200 × 0.70 = 54,880 output tokens/s, demand needs an eighth replica: at 60,000 tokens/s, 64 active GPUs cost $1,152/hour and exceed the allowance, while the same 72-GPU purchase unit still fits. Acquire the exact serving qualification before committing the purchase.

Decide whether a GB200 NVL72-class rack fits this hall. Assume 132 kW per rack with 87% of the heat going to liquid (HPE's split), a room that can take 20 kW of air per rack, 30 °C facility water on the hottest day, a CDU with a 5 K approach and a rack that accepts coolant up to 45 °C. Liquid heat is 132 × 0.87 = 114.84 kW, and the room takes the other 17.16 kW, inside its 20 kW. The coolant reaches the rack at 30 + 5 = 35 °C, under the 45 °C limit, so direct liquid with room air works. At OCP's 1.25–2.0 L/min per kW, the CDU must deliver about 144–230 L/min to this rack.

Raise the hottest-day facility water to 41 °C: the coolant arrives at 46 °C and the liquid path fails. The crossover is 40 °C facility water at a 5 K approach; a CDU with a 3 K approach moves it to 42 °C. Get the rack maker's flow curve and pressure limits before you order.

Illustrative capital scope: assume 10 MW of usable IT capacity, facility/land/utility works at $10 million per MW, network infrastructure at $4 million per MW, a quoted 20% facility premium and a 1.10 regional factor. Facility cost is 10 × 10 × 1.20 × 1.10 = $132 million. Network cost is 10 × 4 × 1.10 = $44 million. The total is $176 million, or $17.60 per IT watt, excluding servers.

An illustrative $180 million allowance for those same two scopes passes with $4 million remaining. Raising only the facility premium to 25% raises the total to $181.5 million and fails. The topology drawing and supplier scope establish the premium; the reliability requirement buttons do not price it. Add the server purchase once in the shared program capital model.

The facility-premium crossover is 23.6%: (180 − 44) / (10 × 10 × 1.10) − 1. Above it, reduce scope or raise the same-scope capital allowance.

Decide whether candidate A merits the next diligence spend. Assume judgment scores of 7, 6, 6, 7, 6 and 5 for power, interconnect, price, land, water and incentives. The displayed rubric weights those axes 30%, 20%, 15%, 15%, 10% and 10%. These are an illustrative investment preference, not executed agreements or measurements. The weighted trace is (7×30 + 6×20 + 6×15 + 7×15 + 6×10 + 5×10) / 10 = 63.5/100: “Workable with mitigation.”

Advance A to evidence collection under this rubric; hold a site commitment until the service date, land rights, thermal/water design and workload latency are settled (Chapter 3.13). Change the power score to 2: the score becomes 48.5 and the power kill gate disqualifies A. The crossover is power ≤2, regardless of incentives. Reallocate diligence to a candidate with a viable power path.

Decide how much plant to buy for an 800 kW load from 200 kW modules when the owner wants both concurrent maintainability and ride-through of one module failure. N is 800 ÷ 200 = 4 modules. With one shared distribution path, N+1 rides through a module failure but cannot keep the load up while that path is serviced, so it fails the maintainability requirement. Of the options that meet both, 3N/2 needs the fewest modules: three blocks of two, 6 modules or 1.5× N, each module 67% loaded. 2N needs 8 and 2(N+1) needs 10. At Chapter 12.5's illustrative power-path rates (one failure per 2,000 hours, 8 h repair), 3N/2 models about 1.7 hours of downtime a year before shared causes.

Add a standby distribution path for maintenance: N+1 now meets both requirements with 5 modules, 1.25× N. The trade is a second path and its transfer equipment against one fewer module, so price both before you choose, and prove the transfer works in commissioning.

Illustrative dense-model run: assume 70 billion parameters, 15 trillion training tokens, 10,000 GPUs at 2.5 dense PFLOPS each, 40% MFU and 90% goodput. Define MFU on productive compute time so goodput does not deduct the same lost time twice. Training work is 6 × 70 billion × 15 trillion = 6.3 × 10²⁴ FLOPs. Effective rate is 10,000 × 2.5 × 10¹⁵ × 0.40 × 0.90 = 9 × 10¹⁸ FLOP/s. The run takes 700,000 seconds, or 8.10 days and about 1.94 million GPU-hours.

At an assumed $18/GPU-hour, rental is $35 million. With assumed 1.2 kW GPU power, a 1.50 total-IT multiplier and PUE 1.20, constant-power energy is 4,200 MWh: $336,000 at $0.08/kWh and 1,680 tonnes at 400 gCO₂/kWh. Do not add electricity to rental when the rental already includes it.

The run passes an illustrative nine-day deadline. At 80% goodput it takes 9.11 days and fails. Change the training-work model for sparse or non-dense workloads; the screening formula does not qualify model placement or convergence.

The nine-day deadline requires at least 81.0% goodput: 6.3 × 10²⁴ / (10,000 × 2.5 × 10¹⁵ × 0.40 × 9 × 86,400). Below that, reserve more capacity or move the deadline.

Illustrative server-ownership decision: assume $30,000 per installed GPU recovered over three years at an 8% cost of capital, $1,000 annual energy, annual opex equal to 5% of purchase cost, and 1.25 installed GPUs per active GPU. The capital-recovery factor at 8% over three years is 0.3880, so capital costs 30,000 × 0.3880 = $11,641 a year. Annual cost per active GPU is (11,641 + 1,000 + 1,500) × 1.25 = $17,676, or $2.02 per calendar GPU-hour.

An assumed $4/hour contract rental bills every hour, used or not, so compare it per calendar hour: owning costs $2.02 against $4.00 and is about 50% cheaper. At 50% productive use both double per useful hour ($4.04 owned, $8.00 rented); utilization cannot flip the answer under a reserved contract. With the cost of capital at zero the capital line falls to $10,000 and owning looks cheaper still, which is why the rental side, which pays the lessor’s return, must be met by a return on the owned side. The facility charge is set to zero here to isolate the server; enter the facility’s capital recovery per kW-month for a full comparison.

At a $1.90 contract rate, renting is about 6% cheaper: $2.02 owned against $1.90. The contract rate, server price, life and cost of capital move the answer; productive use does not.

Illustrative serving case: assume one tested node produces 1,000 useful output tokens/s within the declared latency target and runs productively 50% of each paid hour. Useful output is 1,000 × 3,600 × 0.50 = 1.8 million tokens/hour. Assumed rental of $20/hour costs $11.1 per million output tokens; an assumed owned calendar cost of $10/hour costs $5.56 per million.

Against an illustrative $6 per million output-token cost allowance, ownership passes and rented capacity fails. At 40% productive use the owned cost rises to $6.94 and also fails. Keep the model, precision, prompt/output mix, batching, latency target and cost boundary identical; prompt tokens and cached tokens do not silently enter this output denominator.

The owned case crosses the $6 allowance at 46.3% productive utilization: 10 / (1,000 × 3,600 × 6 / 1,000,000).

Illustrative annual operating case: assume 10 MW nominal installed IT power, 60% annual-average load, energy-weighted PUE 1.20, electricity at $0.08/kWh and site-water WUE 0.40 L/kWh of IT energy. Average IT power is 6 MW; annual IT energy is 6 × 8,760 = 52,560 MWh. Facility energy is 63,072 MWh, costing $5,045,760. Site water is 52,560 × 1,000 × 0.40 = 21,024,000 litres.

Against an illustrative $5.2 million annual electricity allowance, this case passes with $154,240 remaining. The allowance crosses at 61.8% average load. Holding all other assumptions constant at 100% nominal operation raises electricity cost to $8,409,600 and fails that allowance. Productive utilization alone does not change either energy case; change the average-load assumption when the load study changes.

Illustrative retained-fleet case, all amounts in $M: assume capex 100, immediate operation, debt 50 at zero interest amortized over four years, revenue 40/year, opex 25% of revenue, cash tax 20%, tax depreciation over 3.5 years, no sustaining spend, no growth, four-year hold, zero exit proceeds and no reserve. Assume the original fleet supports all four earning years. These tax and earning periods are independent assumptions.

Closing equity is 50. Annual EBITDA is 30. Tax depreciation is 28.5714 in years 1–3 and 14.2857 in year 4, totaling 100. Cash tax is 0.2857 in years 1–3 and 3.1429 in year 4. Annual principal is 12.5; minimum DSCR is 26.8571 / 12.5 = about 2.15. Equity cash flow is 17.2143 in years 1–3 and 14.3571 in year 4. At a 10% hurdle, equity NPV is −50 + Σ[17.2143 / 1.1ᵗ, t=1…3] + 14.3571 / 1.1⁴ = about 2.62.

The case passes the assumed 1.50 covenant and positive-NPV hurdle. A covenant of 2.20 fails. Without an earning period the return is not decided; a tax depreciation schedule does not supply that evidence. The bridge shows the complete annual lines, including zero terminal proceeds.