Learning path 1 · Topic 5 of 30

Scale a plant without scaling every dollar

A plant twice as large need not cost twice as much. Shared infrastructure can produce economies of scale, but the model is valid only within a comparable range.

Change a scenario. Watch the consequence.

● Current scenarioScaled costProportional benchmark · dashed

Known plant cost400000 $New capacity / old capacity2 ratioCapacity exponent0.6 dimensionless

estimated new plant cost · $

C_2=C_1\left(\frac{K_2}{K_1}\right)^b=400000(2)^{0.6}=606286.626604

Cost rises less than proportionally with capacity: economies of scale in this range.

Follow the reasoning

Use comparable technologies and price dates before scaling.

Cash-flow dates and signs are part of the answer. Model values retain precision; displays round money to cents. Illustrative assumptions exclude taxes and inflation unless stated.

What the model means

The power model C=aKᵇ becomes log C=log a+b log K. Two comparable points give b=log(C₂/C₁)/log(K₂/K₁); then a=C₁/K₁ᵇ. If 0<b<1, total cost rises while average cost per unit capacity falls.

Build the calculation

Power sizing changes capacity; a cost index changes price date: C_new=C_old(index_new/index_old). If both date and size change, apply both adjustments consistently. Do not automatically use a memorized exponent for a different process or capacity range.

C_2=C_1(K_2/K_1)^b,\quad b=\frac{\ln(C_2/C_1)}{\ln(K_2/K_1)}

Practice with feedback

Set up. Calculate. Explain.

Money: enter dollars

Rebuild the starting scenario

Known plant cost: 400000 $; New capacity / old capacity: 2 ratio; Capacity exponent: 0.6 dimensionless. Calculate the estimated new plant cost.

Setup hint

Use comparable technologies and price dates before scaling.

Open the visual solution

C_2=C_1\left(\frac{K_2}{K_1}\right)^b=400000(2)^{0.6}=606286.626604

Use comparable technologies and price dates before scaling.
Capacity ratio = 2; exponent = 0.6.
Multiply the known cost by ratioᵇ. A cost index adjusts date; the exponent adjusts size.

● Current scenarioScaled costProportional benchmark · dashed

Known plant cost400000 $New capacity / old capacity2 ratioCapacity exponent0.6 dimensionless

estimated new plant cost · $

C_2=C_1\left(\frac{K_2}{K_1}\right)^b=400000(2)^{0.6}=606286.626604

Cost rises less than proportionally with capacity: economies of scale in this range.

Visual solution

Use comparable technologies and price dates before scaling.

Cash-flow dates and signs are part of the answer. Model values retain precision; displays round money to cents. Illustrative assumptions exclude taxes and inflation unless stated.

Checks allow 0.1% numerical rounding (minimum $0.05 or 0.001 in the stated nonmonetary unit). Interpolated factors and rates are approximations; keep table precision consistent.

Money: enter dollars

Transfer the idea to a new scenario

A known plant costs $600,000. Estimate a plant with 2.5 times capacity using b=0.65.

Setup hint

Use comparable technologies and price dates before scaling.

Open the visual solution

C_2=C_1\left(\frac{K_2}{K_1}\right)^b=600000(2.5)^{0.65}=1088459.454413

Use comparable technologies and price dates before scaling.
Capacity ratio = 2.5; exponent = 0.65.
Multiply the known cost by ratioᵇ. A cost index adjusts date; the exponent adjusts size.

● Current scenarioScaled costProportional benchmark · dashed

Known plant cost600000 $New capacity / old capacity2.5 ratioCapacity exponent0.65 dimensionless

estimated new plant cost · $

C_2=C_1\left(\frac{K_2}{K_1}\right)^b=600000(2.5)^{0.65}=1088459.454413

Cost rises less than proportionally with capacity: economies of scale in this range.

Visual solution

Use comparable technologies and price dates before scaling.

Cash-flow dates and signs are part of the answer. Model values retain precision; displays round money to cents. Illustrative assumptions exclude taxes and inflation unless stated.

Checks allow 0.1% numerical rounding (minimum $0.05 or 0.001 in the stated nonmonetary unit). Interpolated factors and rates are approximations; keep table precision consistent.

Explain the distinction

For 0<b<1 and increasing capacity, what happens?

Can you defend your setup?

Can you place each amount on the timeline, match the interest period, calculate the equivalent quantity, and state the decision or limitation?

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