One more bidder: a fourth supplier does more than the best price ceiling

Three suppliers with costs from 80 to 120 face an alternative at 130. A ceiling of 105 saves 30.40; a fourth supplier without it saves 34.00. At cost 90, another rival trims the bid from 100.00 to 97.50 and the expected margin from 5.63 to 3.16.

Live result

With three bidders the ceiling of 105 lifts the expected saving from 30.00 to 30.40; a fourth bidder lifts it to 34.00.

Expected price with no ceiling100.00
Saving with no ceiling30.00
Best ceiling105.00
Saving with the ceiling30.40
Saving with one more bidder34.00

One more genuine bidder beats the best ceiling at every count while costs are private; with a shared unknown cost it can reverse.

Givens

Your seat: Buyer

Three tenders

Maintenance suppliers estimate the same scope with different labour and subcontract costs.

Low cost: Low cost80
Both edges run from 0.01 to 1,000,000,000, in one unit of price.
High cost: High cost120
Suppliers bidding: Suppliers bidding3
A whole count from 1 to 8, of suppliers who actually bid.
The alternative price: The alternative price130
Your fallback price if no bid is accepted, above the low cost.
Costs: Private to each supplier, independent and spread evenly across the range.
Award: The lowest bid wins.
Bidding: Every supplier bids as the maths says.
Without a ceiling: The buyer accepts any bid up to its alternative price.

One more bidder and the ceiling

No ceilingBest ceiling

Each step holds the cost range and alternative fixed; each flag reads no ceiling first and the best ceiling second.

This page shows what a mechanism does under stated rules and says nothing about any public procurement regulation. The costs and the alternative are your own estimates. No count, bid or ceiling is recommended.

Every count from 1 to 8

Count	Expected price	Saving with no ceiling	Saving with the best ceiling	Gain from one more bidder
1	130.00	0.00	15.63	23.33
2	106.67	23.33	24.74	6.67
3typed	100.00	30.00	30.40	4.00
4	96.00	34.00	34.12	2.67
5	93.33	36.67	36.70	1.90
6	91.43	38.57	38.58	1.43
7	90.00	40.00	40.00	1.11
8	88.89	41.11	41.11	0.89

Here is the maths

The mechanism

A ceiling can leave the buyer using its fallback. Another genuine bidder changes the cost competition itself. Expected saving counts both purchases and fallback outcomes.

The common cost distribution is:

C\sim U[L,H]

where

C

is a supplier's private cost and

L

the low edge and

H

the high edge.

With at least two bidders and an alternative at or above the range, the expected price is:

E[p]=L+\frac{2(H-L)}{n+1}\quad(n\ge2,\ v\ge H)

\begin{gathered}E[p]=L+\frac{2(H-L)}{n+1} \\ (n\ge2,\ v\ge H)\end{gathered}

where

p

is the price paid and

n

the bidder count and

v

the alternative price.

Saving is measured against the fallback:

S_0(n)=v-E[p]\qquad S_0(1)=0

\begin{gathered}S_0(n)=v-E[p] \\ S_0(1)=0\end{gathered}

where

S_0

is expected saving without a separate ceiling.

The uniform virtual cost gives this ceiling:

r^*=\frac{v+L}{2}

where

r^*

is the optimal threshold within the cost range.

With at least two bidders, the ceiling saving is:

S(n,r)=vP(c_{(1)}\le r)-E[\min(c_{(2)},r)\mathbf1_{\{c_{(1)}\le r\}}]

where

S

is expected saving with a ceiling and

r

the ceiling and

c_{(1)}

the lowest cost and

c_{(2)}

the second lowest cost.

With one bidder the ceiling is the payment if its cost qualifies:

S(1,r)=(v-r)F(r)

where

F

is the cumulative cost distribution.

If the alternative is inside the range, its acceptance limit still applies without a separate ceiling:

S_0(n)=S(n,v)\quad(n\ge2)\qquad E[p]=v-S_0(n)

\begin{gathered}S_0(n)=S(n,v) \\ (n\ge2) \\ E[p]=v-S_0(n)\end{gathered}

The expected price then includes the fallback whenever no cost qualifies.

A worked example

The ceiling changes the chance of buying as well as the payment. The extra supplier increases saving by more than that ceiling does.

At costs 80 to 120 with three bidders, the expected price is:
$E[p]=80+\frac{2(120-80)}{3+1}=100$
Against the alternative 130, saving is:
$S_0(3)=130-100=30.00$
The best ceiling gives saving that rounds to 30.40:
$r^*=\frac{130+80}{2}=105\qquad S(3,105)=30.3955078125$ $\begin{gathered}r^*=\frac{130+80}{2}=105 \\ S(3,105)=30.3955078125\end{gathered}$
A fourth bidder without a ceiling gives:
$S_0(4)=130-\left(80+\frac{80}{5}\right)=34.00$

Show the full maths

The costs in order

Inside the range, the distribution and density are:

F(x)=\frac{x-L}{H-L}\qquad f(x)=\frac1{H-L}\quad(L\le x\le H)

\begin{gathered}F(x)=\frac{x-L}{H-L} \\ f(x)=\frac1{H-L} \\ (L\le x\le H)\end{gathered}

Here

x

is a cost being integrated and

f

its density. The cumulative distribution is zero below the range and one above it.

The second lowest cost has density:

g_2(x)=n(n-1)F(x)(1-F(x))^{n-2}f(x)

where

g_2

is that order statistic's density.

Its mean follows by integration:

E[c_{(2)}]=\int_L^H xg_2(x)\,dx=L+\frac{2(H-L)}{n+1}

Separate outcomes with at least two qualifying costs from those with only one:

S(n,r)=\int_L^{\min(r,H)}(v-x)g_2(x)\,dx+(v-r)nF(r)(1-F(r))^{n-1}

The last term pays the ceiling when exactly one supplier qualifies.

Normalize the interval:

D=H-L\qquad u=F(r)\qquad q=1-u

\begin{gathered}D=H-L \\ u=F(r) \\ q=1-u\end{gathered}

where

D

is the cost spread and

u

the fraction below the ceiling and

q

the fraction above it.

For a ceiling within the range, integrating gives:

S(n,r)=(v-L)(1-q^n)-\frac{2D}{n+1}\left(1-(n+1)q^n+nq^{n+1}\right)\quad(L\le r\le H)

\begin{gathered}S(n,r)=(v-L)(1-q^n)-\frac{2D}{n+1}\left(1-(n+1)q^n+nq^{n+1}\right) \\ (L\le r\le H)\end{gathered}

The computation evaluates the same integral as a finite sum of positive terms to retain precision near the low edge.

Why that ceiling

The buyer's virtual cost adds the distribution to cost through the density:

\psi(x)=x+\frac{F(x)}{f(x)}=2x-L

where

\psi

is virtual cost.

Buying remains worthwhile until virtual cost reaches the alternative:

\psi(r^*)=v\qquad r^*=\frac{v+L}{2}

\begin{gathered}\psi(r^*)=v \\ r^*=\frac{v+L}{2}\end{gathered}

The bidder count does not enter that threshold.

With one bidder, a threshold above the range is capped at the highest cost:

r^*_{1}=\min\left(H,\frac{v+L}{2}\right)

where

r^*_{1}

is the optimal ceiling for a lone bidder. With two or more bidders, every ceiling at or above the high cost gives the same saving, so the displayed threshold can stay above the range.

Where the shade comes from

A supplier can bid as though it had another cost:

\Pi(z;c)=(b(z)-c)(1-F(z))^{n-1}

where

\Pi

is its expected margin from the deviation and

z

the cost it imitates and

c

its true cost and

b

the equilibrium bid.

At its own cost the first derivative with respect to that imitation is zero:

b'(c)(1-F(c))^{n-1}=(b(c)-c)(n-1)(1-F(c))^{n-2}f(c)

Rearranging gives the derivative of equilibrium margin:

\frac{d}{dc}\left[(b(c)-c)(1-F(c))^{n-1}\right]=-(1-F(c))^{n-1}

Integrate up to the highest cost, where margin is zero:

b(c)=c+\frac{\int_c^H(1-F(t))^{n-1}\,dt}{(1-F(c))^{n-1}}=c+\frac{H-c}{n}

Here

t

is the integration variable. At the highest cost the limiting bid equals cost, with zero shade and zero chance of being lowest against a rival.

Averaging the winning bid gives the same expected payment as the second lowest cost:

E[b(c_{(1)})]=L+\frac{2(H-L)}{n+1}=E[c_{(2)}]

This revenue equivalence instance explains why the buyer's price and the bidder's bid are two readings of the same private cost tender when the acceptance limit does not bind.

The comparison and its boundary

The private values result compares a seller's revenue with an extra bidder and no reserve against the optimal reserve with the original field, and mirrors it as buyer saving:

R_{n+1}(0)\ge R_n(\rho^*)\qquad S_0(n+1)\ge S(n,r^*)

\begin{gathered}R_{n+1}(0)\ge R_n(\rho^*) \\ S_0(n+1)\ge S(n,r^*)\end{gathered}

where

R

is expected seller revenue and

\rho^*

the seller's optimal reserve.

The comparison needs independent private values and the same distribution for genuine bidders. In this buyer's mirror, the alternative must cover the whole cost range. Inside the range the ceiling can do better, as the live reading shows. The comparison does not price the work of attracting a bidder or predict entry.

A shared unknown value can reverse it

Klemperer's common value example makes the value the largest of three signals. With only an uninformed bidder, expected seller revenue is the first of these amounts; with that bidder against one, two and three informed bidders, it is the next three in order:

\frac9{12},\quad\frac8{12},\quad\frac7{12},\quad\frac6{12}

\begin{gathered}\frac9{12}, \\ \frac8{12}, \\ \frac7{12}, \\ \frac6{12}\end{gathered}

Extra informed competition makes the uninformed bidder more cautious about winning, and revenue falls. That sale example marks the boundary of the private cost tender shown here.

Sources

Paul Milgrom, Putting Auction Theory to Work
Paul Klemperer, Auctions: Theory and Practice
Tim Roughgarden, Twenty Lectures on Algorithmic Game Theory

Scale: Whole counts, amounts to 2 decimals with scientific notation for very large amounts, table chances to whole percentages
Basis: Independent uniform private costs, genuine bidders and equilibrium bids
Model version: 1.0.0

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This model reads stated assumptions and your estimates. It does not predict a tender outcome or recommend a bidder count, bid or price ceiling.

Givens

One more bidder and the ceiling

Every count from 1 to 8

Here is the maths

The mechanism

A worked example

The costs in order

Why that ceiling

Where the shade comes from

The comparison and its boundary

A shared unknown value can reverse it

Sources

Related