The Encyclopedia of Economic Principals

Chapter 35

Entry, Scale, Contestability, and Strategic Deterrence

Market room, the cost of size, the pull of entry and the pull of attention.

Four of the chapter's worked examples, made interactive: how many plants fit at minimum efficient scale, when to split an overloaded hub, how a license cap keeps profit alive, and how rank-sensitive attention makes superstars. Change one value at a time and watch the figure, the numbers and the hand calculation respond.

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

Minimum efficient scale and market room

How many equal firms can each reach the bottom of the cost curve?

The ratio, not either number alone, sets the room: technology that lowers MES and growth that raises demand both shrink it, for different reasons.

Equation, written in LaTeX: \frac{40{,}000}{150{,}000}=0.2667.

Equation, written in LaTeX: \frac{25{,}000}{150{,}000}=0.1667,

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MES is the smallest annual output at which unit cost reaches its $50 floor. The MES ratio divides MES by annual market demand; its whole-number inverse is the most equal firms that can each reach MES.

Predict first. Halving MES from 50,000 to 25,000 or doubling demand from 150,000 to 300,000: which leaves room for more efficient firms?

Your prediction

Choose an example

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Figure: Minimum efficient scale and market room. Left: unit cost falls to a $50 floor that starts at 40,000 units. Right: demand of 150,000 units split into 3 blocks of 40,000.
MES (units): 40,000, Market demand (units): 150,000
Constructed example: the chapter's hypothetical component plant (MES 40,000 and 25,000, demand 150,000, cost floor $50); MES 50,000 and demand 75,000 and 300,000 are added for comparison. The cost curve away from the floor is drawn for illustration through the book's $72 and $56 points.

Calculated values

MES ratio
0.2667
Most equal firms at MES
3
Output each with 4 equal firms
37,500

MES ratio = 40,000 / 150,000 = 0.2667. 3 x 40,000 = 120,000 fits within demand, but 4 equal firms would each sell 150,000 / 4 = 37,500, below MES. So at most 3 equal firms can reach the cost floor of $50.

Worked steps

  1. MES ratio = 40,000 / 150,000 = 0.2667
  2. 3 x 40,000 = 120,000, within 150,000
  3. 150,000 / 4 = 37,500, below 40,000
  4. Most equal efficient firms: 3

Use the idea

Before predicting how many efficient suppliers a market can hold, divide the plant size that reaches minimum unit cost by the demand it will serve.

Where the conclusion applies

Equal firms, a known flat cost floor and fixed demand. Unequal shares, ramp-up losses and product differences can change the count.

Check your understanding: At MES 50,000 and demand 300,000, how many equal efficient firms fit?
300,000 / 50,000 = 6, and the MES ratio is 0.1667.

Chapter 35 source: section "Minimum efficient scale".

Demonstration 2 of 4

Splitting an overloaded hub

When does duplicating a fixed setup cost pay for itself?

Average cost falls while the setup is spread and rises once congestion dominates. Splitting pays only when the hub is far enough past Q* that the congestion saved exceeds the extra setup.

Equation, written in LaTeX: C(Q)=80{,}000+4Q+0.0008Q^2,

Equation, written in LaTeX: Q^*=\sqrt{\frac{80{,}000}{0.0008}}=10{,}000,

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Q is parcels per month and C(Q) is a hub's monthly cost in dollars: setup 80,000, 4 per parcel, and a quadratic congestion term b Q^2. Q* is the hub size with the lowest average cost.

Predict first. With the quadratic coefficient at 0.0003, is splitting 16,000 parcels into two hubs still worth it?

Your prediction

Choose an example

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Figure: Splitting an overloaded hub. Average cost curve of one hub with quadratic coefficient 0.0008, lowest at 10,000 parcels. With 1 hub each hub handles 16,000.00 parcels at $21.80 each.
Quadratic coefficient b: 0.0008, Number of hubs: 1
Constructed example: the chapter's hypothetical sorting hub (80,000 + 4Q + 0.0008Q^2, 16,000 parcels, one or two hubs, coefficient 0.0003); a coefficient of 0.0012 and three hubs are added.

Calculated values

Parcels per hub
16,000.00
Cost per hub
$348,800.00
Total cost
$348,800.00
Cost per parcel
$21.80
One-hub total
$348,800.00
Saving against one hub
$0.00
Q* (minimum average cost)
10,000

Each of 1 hub handles 16,000 / 1 = 16,000.00 parcels. 80,000 + 4 x 16,000.00 + 0.0008 x 16,000.00^2 = 80,000 + 64,000.00 + 204,800.00 = $348,800.00. Total cost is 1 x $348,800.00 = $348,800.00, or $21.80 per parcel. Average cost is lowest at Q* = sqrt(80,000 / 0.0008) = 10,000. This is the single-hub benchmark.

Worked steps

  1. q = 16,000 / 1 = 16,000.00
  2. Cost per hub = 80,000 + 4 x 16,000.00 + 0.0008 x 16,000.00^2 = 80,000 + 64,000.00 + 204,800.00 = $348,800.00
  3. Total = 1 x $348,800.00 = $348,800.00
  4. Per parcel = $348,800.00 / 16,000 = $21.80
  5. One hub: $348,800.00; difference 0.00

Use the idea

Compare the total cost of one large unit with the cost of several smaller ones at the volume you actually serve, not the average cost of the unit you already have.

Where the conclusion applies

Identical hubs, an equal split and no coordination cost between hubs. A change in management technology moves the quadratic coefficient and with it the best size.

Check your understanding: Two hubs at b = 0.0003: what is total cost?
Each costs 80,000 + 32,000 + 0.0003 x 8,000^2 = 131,200; total 262,400, against 220,800 for one hub, so splitting loses 41,600.

Chapter 35 source: section "Diseconomies of scale".

Demonstration 3 of 4

Free entry versus license caps

What happens to profit when the number of bakeries is capped below the free-entry count?

Positive profit draws entry, which adds supply and lowers price until it reaches minimum average cost. A cap stops that process and leaves a rent with each license holder.

Equation, written in LaTeX: Q_D=1200-30p,

Equation, written in LaTeX: 30(\frac{p}{2})=1200-30p.

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Market demand is Q = 1200 - 30p loaves. Each bakery has cost C(q) = 144 + q^2, so it supplies q = p/2 and its average cost is lowest, 24, at q = 12. The control is the number of bakery licenses.

Predict first. Does a cap of 35 bakeries leave profit positive?

Your prediction

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Figure: Free entry versus license caps. Demand Q = 1200 - 30p and the supply of 40 bakeries cross at p = 24.00; minimum average cost is 24.
Bakery licenses: 40
Constructed example: the chapter's hypothetical bakeries (demand 1200 - 30p, cost 144 + q^2, 30 and 35 firms and the free-entry 40); a non-binding cap of 45 is added.

Calculated values

Bakeries operating
40
Price p
24.00
Output per bakery q
12.00
Profit per bakery
$0.00

Forty bakeries is the free-entry count. Market clearing 40(p/2) = 1200 - 30p gives p = 2400 / (60 + 40) = 24.00 and q = p/2 = 12.00. Profit = p^2/4 - 144 = 144.00 - 144 = $0.00.

Worked steps

  1. 40(p/2) = 1200 - 30p
  2. p = 2400 / (60 + 40) = 24.00
  3. q = p / 2 = 12.00
  4. Revenue p q = p^2/2 = 288.0000; cost 144 + q^2 = 144 + 144.0000 = 288.0000
  5. Profit = 288.0000 - 288.0000 = $0.00

Use the idea

When a licensed trade earns returns above its costs, check whether a limit on entry, rather than skill, explains them; the rent is often built into the license price.

Where the conclusion applies

Identical firms, price taking, and long-run costs; a cap above 40 does not bind. The firm count is treated as continuous in the free-entry calculation.

Check your understanding: With 30 bakeries, what is profit per firm?
p = 2400 / 90 = 26.67, q = 13.33, profit = p^2/4 - 144 = 177.78 - 144 = 33.78.

Chapter 35 source: section "Free-entry zero-profit condition".

Demonstration 4 of 4

Rank-sensitive attention and superstars

How can a 3 percent quality gap turn into an 11 to 1 audience gap?

The share rule raises each score to the power kappa. A small ratio of scores becomes a large ratio of audiences when kappa is high, because scalable delivery lets the leader serve everyone.

Equation, written in LaTeX: s_A=\frac{100^{80}}{100^{80}+97^{80}}=\frac{1}{1+0.97^{80}}\approx0.9196.

Equation, written in LaTeX: s_A=\frac{1}{1+0.97^4}\approx0.5304.

Scroll sideways for the whole equation

Two instructors charge the same price; A's quality score is 100 and B's is the rival score. kappa measures how strongly 500,000 viewers favor the higher score. s_A is A's audience share; revenue is $3 per viewer.

Predict first. With kappa = 4 and B's score 97, roughly what share does A get?

Your prediction

Choose an example

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Figure: Rank-sensitive attention and superstars. Left: A's share rises with attention sensitivity; at kappa 80 and rival score 97 it is 0.9196. Right: audiences of 459,793 for A and 40,207 for B.
Attention sensitivity kappa: 80, Rival quality score: 97
Constructed example: the chapter's hypothetical course market (scores 100 and 97, kappa 80 and 4, 500,000 viewers, $3 each); kappa 1 and 20 and rival scores 90 and 99 are added. The book rounds audiences to hundreds (459,800); the exact value is shown here.

Calculated values

Share of A, s_A
0.9196
Viewers of A
459,793
Viewers of B
40,207
Revenue of A
$1,379,379
Revenue of B
$120,621
Audience ratio A to B
11.4

(97/100)^80 = 0.087446, so s_A = 1 / (1 + 0.087446) = 0.919586. A draws 500,000 x 0.91958610 = 459,793.0 viewers and B 40,207.0, a ratio of 1 / 0.087446 = 11.4 to 1. At $3 a viewer the revenues are $1,379,379 and $120,621. A score gap of 3 points out of 100 becomes this audience gap because attention is rank sensitive.

Worked steps

  1. (97/100)^80 = 0.087446
  2. s_A = 1 / (1 + 0.087446) = 0.919586
  3. A: 500,000 x 0.91958610 = 459,793.0; B: 500,000 - 459,793.0 = 40,207.0
  4. Revenue: 3 x 459,793.0 = $1,379,379; 3 x 40,207.0 = $120,621
  5. Ratio = s_A / (1 - s_A) = 1 / 0.087446 = 11.4

Use the idea

Before reading a winner-take-most outcome as a large quality gap, ask how rank sensitive the attention channel is and whether the leader faces any capacity limit.

Where the conclusion applies

Equal prices, a fixed audience, an illustrative share rule and no capacity limit. The book notes that a live-teaching cap breaks the concentration.

Check your understanding: At kappa = 20 and B's score 97, what is s_A?
0.97^20 = 0.5438, so s_A = 1 / 1.5438 = 0.648.

Chapter 35 source: section "Superstar Markets, Power Laws, and Attention Economics".