Chapter 40: General-Purpose Technologies, Diffusion, Learning, and Complementary Assets. The Encyclopedia of Economic Principals, interactive companion

Every example here is a constructed teaching example: it uses the hypothetical numbers of the chapter's worked examples, plus a few values added for comparison and labelled as such in each panel. Nothing here measures a real market, firm or household.

Demonstration 1 of 4

The Bass diffusion curve

How do outside influence p and imitation q shape the path of adoption?

The external term p seeds adoption; the imitation term q N/m grows with the installed base. Early on the growing hazard outruns depletion and sales accelerate; later the shrinking pool of prospects dominates.

$Equation, written in LaTeX: \frac{dN(t)}{dt} =(p+q\frac{N(t)}{m})[m-N(t)].$

$Equation, written in LaTeX: N^*=\frac{m(q-p)}{2q}.$

$Equation, written in LaTeX: (0.02+0.30\times0.02)(980)=25.48,$

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m = 1,000 is market potential, N cumulative adopters, p the external coefficient and q the imitation coefficient. Each year's flow is (p + q N/m)(m - N), an annual discrete version of the continuous model. N* is the continuous-time cumulative level at peak sales.

Predict first. With q = 0, do sales ever accelerate?

Your prediction Yes No

Innovation coefficient p

Imitation coefficient q

Figure: The Bass diffusion curve. Annual new adopters and cumulative adopters over ten years with p = 0.02 and q = 0.3: flows start at 20.00 and cumulative adoption reaches 536.2 of 1,000. — Innovation coefficient p: 0.02, Imitation coefficient q: 0.3
Constructed example: the chapter's hypothetical household device (m 1,000, p 0.02, q 0.30 and 0); p of 0.01 and 0.04 and q of 0.50 are added for comparison.

Year 1 flow: 20.00
Year 2 flow: 25.48
Year 3 flow: 32.11
Cumulative N after 10 years: 536.2
Continuous peak N*: 466.7

Year 1: (0.02 + 0.3 x 0)(1,000) = 20.00, so N_1 = 20.00. Year 2: (0.02 + 0.3 x 0.02)(980.00) = 25.48. Sales accelerate because the installed-share term grows faster than the market depletes. In continuous time sales peak at N* = 1,000 x (0.3 - 0.02) / (2 x 0.3) = 466.7.

Worked steps

Flow 1 = (0.02 + 0.3 x 0) x 1,000 = 20.00
N_1 = 20.00, share 0.02
Flow 2 = (0.02 + 0.3 x 0.02) x 980.00 = 25.48
Flow 3 = (0.02 + 0.3 x 0.04548) x 954.52 = 32.11

Use the idea

Fit p and q to early sales before forecasting a takeoff, and remember price cuts and quality gains can masquerade as imitation.

Where the conclusion applies

A fixed market potential, first purchases only, constant coefficients and an annual discrete approximation that differs slightly from the continuous model.

Check your understanding: With p = 0.04 and q = 0.30, what is the year-two flow?

N_1 = 40; (0.04 + 0.30 x 0.04)(960) = 0.052 x 960 = 49.92.

Chapter 40 source: section "Bass diffusion model".

Demonstration 2 of 4

Relieve the bottleneck, not the slack stage

Which equipment upgrade raises completed repairs enough to pay for itself?

Output is set by the smallest capacity. Expanding a stage pays only while it is the binding one; once another stage or demand binds, extra capacity there adds nothing.

$Equation, written in LaTeX: \min(18,10,15,20)=10,$

$Equation, written in LaTeX: X\leq\min_i c_i.$

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Each device passes intake (18 a day), diagnosis, repair (15 a day) and is limited by daily demand. X is throughput. Diagnosis 14 needs a tool costing $300 a day; diagnosis 17 costs $150 more, $450 in all. Each completed repair contributes $120.

Predict first. Does the upgrade from diagnosis 14 to 17 pay, with demand 20?

Your prediction Yes No

Diagnosis capacity

Daily demand

Figure: Relieve the bottleneck, not the slack stage. Bars for intake 18, diagnosis 10, repair 15 and demand 20, with the binding limit highlighted and throughput 10. — Diagnosis capacity: 10, Daily demand: 20
Constructed example: the chapter's hypothetical repair center (intake 18, diagnosis 10, 14 or 17, repair 15, demand 20 or 12, $120 per repair); the combinations of demand 12 with diagnosis 10 and 17 are added for comparison.

Throughput: 10
Binding: Diagnosis
Contribution: $1,200
Equipment cost: $0
Net contribution: $1,200
Change from the $1,200 base: $0

Throughput is min(18, 10, 15, 20) = 10, so contribution is 10 x 120 = 1,200. After 0 of equipment the net is 1,200 - 0 = 1,200, equal to the 1,200 base by 0. The binding limit is diagnosis.

Worked steps

X = min(18, 10, 15, 20) = 10
Contribution = 10 x 120 = 1,200
Net = 1,200 - 0 = 1,200
Change = 1,200 - 1,200 = 0

Use the idea

Before buying capacity, find the stage that limits completed output and check where the limit moves after the change.

Where the conclusion applies

Deterministic capacities, every device uses every stage, and a constant contribution per repair.

Check your understanding: With demand 12 and diagnosis 17, what is net contribution?

min(18, 17, 15, 12) = 12; 12 x 120 = 1,440, minus 450 = 990, below the 1,200 base.

Chapter 40 source: section "Bottleneck Relief and General-Purpose Enabling Methods".

Demonstration 3 of 4

Amdahl's law and communication cost

When does adding processors stop paying for a fixed job?

The serial share caps the speedup, and communication grows with N, so each extra processor saves less time while the processor-hour bill keeps rising.

$Equation, written in LaTeX: 100(0.20+\frac{0.80}{10})=28\text{ hours},$

$Equation, written in LaTeX: 10(19.95)(\$6)=\$1{,}197$

$Equation, written in LaTeX: 20(18.725)(\$6)=\$2{,}247.$

$Equation, written in LaTeX: T(N)=s+\frac{1-s}{N},$

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Architecture R needs 100 one-processor hours with serial share s = 0.20; architecture P needs 110 with s = 0.05. N is the number of processors; communication adds 0.4N hours when switched on. Processor-hours cost $6 and an hour saved is worth $50.

Predict first. With communication overhead, does doubling processors cut P's runtime by as much as without it?

Your prediction About as much Much less

Architecture

Processors

Communication overhead

Runtime at N = 10: 19.950 h
Cost at N = 10: $1,197
Cut from 10 to 20 processors: 6.1%
Extra bill for 20: $1,050
Value of time saved: $61.25

T = 110 x (0.05 + 0.95 / 10) + 0.4 x 10 = 19.950 hours, costing 10 x 19.950 x 6 = $1,197. Going from 10 to 20 processors cuts runtime from 19.950 to 18.725 hours (6.1%), saving 1.225 hours worth 1.225 x 50 = $61.25 against an extra bill of $1,050.

Worked steps

T(10) = 110 x (0.05 + 0.95 / 10) + 0.4 x 10 = 19.950
Cost = 10 x 19.950 x 6 = $1,197
Saved = 19.950 - 18.725 = 1.225 hours, worth $61.25
Extra bill = 2,247 - 1,197 = $1,050

Use the idea

Price the hours saved before scaling out hardware, and look for redesigns that lower the serial share instead.

Where the conclusion applies

A fixed workload, linear communication cost and processor-hours priced at $6. Architecture R with overhead is not in the book and is added for comparison.

Check your understanding: For P with overhead, how does the extra bill for 20 processors compare with the value of time saved?

2,247 - 1,197 = $1,050 against 1.225 hours x $50 = $61.25: uneconomic.

Chapter 40 source: section "Parallelization Economies and Scale Elasticity".

Demonstration 4 of 4

Staged research as an option

Why can a pilot be worth funding for society but not for a private funder?

Staging buys the option to stop after a bad signal, so development is paid only when it is worth doing. The private funder captures too little of the value for the option to cover the pilot cost.

$Equation, written in LaTeX: -\$100{,}000+0.10\max\{\$3{,}000{,}000-\$500{,}000,0\} =\$150{,}000.$

$Equation, written in LaTeX: 1-(1-0.10)^{10}=1-0.9^{10}\approx65.1\%.$

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A pilot costs C0 = $100,000 and gives a favorable signal with probability p. Development then costs C1 = $500,000. The method is worth $3,000,000 to society but only $600,000 in capturable revenue. Committed means both costs are paid before any signal.

Predict first. Can any success probability in this range make the private funder invest?

Your prediction Yes No

Chance of a favorable signal

Valuation

Staged value: $150,000
Committed value: -$300,000
Funds the project: Yes
P(at least one success in 10): 65.1%

Staged: -100,000 + 0.1 x max(3,000,000 - 500,000, 0) = $150,000. Committed: 0.1 x 3,000,000 - 600,000 = -$300,000. Society gains from a staged pilot. Across 10 independent pilots, the chance of at least one success is 1 - 0.9^10 = 65.1%.

Worked steps

Staged = -100,000 + 0.1 x 2,500,000 = 150,000
Committed = 0.1 x 3,000,000 - 600,000 = -300,000
P(at least one) = 1 - 0.9^10 = 65.1%

Use the idea

Value a research program both ways: with and without the ability to stop, and with social and private payoffs, to see whether the gap is flexibility or appropriation.

Where the conclusion applies

Independent pilots, a known success probability, no discounting and costs that can truly be avoided after failure.

Check your understanding: At a success chance of 0.20, what is the social staged value?

-100,000 + 0.20 x 2,500,000 = $400,000.

Chapter 40 source: section "Public Goods, Knowledge Spillovers, and the Option Value of Basic Research".

The Bass diffusion curve

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Calculated values

Worked steps

Use the idea

Where the conclusion applies

Relieve the bottleneck, not the slack stage

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Calculated values

Worked steps

Use the idea

Where the conclusion applies

Amdahl's law and communication cost

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Calculated values

Worked steps

Use the idea

Where the conclusion applies

Staged research as an option

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Calculated values

Worked steps

Use the idea

Where the conclusion applies