Design of discrete Dutch auctions with an uncertain number of bidders

Abstract

The focus of this paper is on Dutch auctions where the bidding prices are restricted to a finite set of values and the number of bidders follows a Poisson distribution. The goal is to determine what the discrete bid levels should be to maximize the auctioneer’s expected revenue, which is the same as the average selling price of the object under consideration. We take a new approach to the problem by formulating the descending-price competitive bidding process as a nonlinear program. The optimal solution indicates that the interval between two successive bids should be wider as the Dutch auction progresses. Moreover, the auctioneer’s maximum expected revenue increases with the number of bid levels to be set as well as the expected number of bidders. Numerical examples are provided to illustrate the key results from this study and their managerial implications are discussed.

Appendices

Appendix A: Proof of Proposition 1

It suffices to show that the leading principal minor of the objective function’s Hessian of order i has the same sign as (−1)ⁱ (Bazaraa et al. 2006). Let $L_{i} = \frac{l_{i}}{\bar{v}}$, i=1,2,…,m, and the NLP in (2) can be rewritten as

$$ \everymath{\displaystyle }\begin{array}{@{}l@{\quad}l} \mathrm{Maximize} & Z = \bar{v}e^{ - \lambda }\sum_{i = 1}^{m} L_{i} \bigl[ e^{\lambda L_{i + 1}} - e^{\lambda L_{i}} \bigr] \\ \noalign{\vspace{5pt}} \mathrm{subject\ to{:}} & L_{i + 1} \ge L_{i},\quad i = 1, \ldots, m \\ \noalign{\vspace{5pt}} & L_{1} = 0 \\ \noalign{\vspace{5pt}} & L_{m + 1} = 1 \end{array} $$

(A.1)

For each L _i , i=2,3,…,m, we have

Let H(Z) be the Hessian of the objective function in (A.1), it follows that

$$H( Z ) = \bar{v}\lambda e^{ - \lambda } \arraycolsep=1.5pt \left[ \begin{array} {ccccc} e^{\lambda L_{2}} [ - 2 - \lambda( L_{2} - L_{1} ) ] & e^{\lambda L_{3}} & 0 & \cdots & 0 \\ \noalign{\vspace{2pt}} e^{\lambda L_{3}} & e^{\lambda L_{3}}[ - 2 - \lambda( L_{3} - L_{2} ) ] & e^{\lambda L_{4}} & \cdots & 0 \\ \noalign{\vspace{2pt}} 0 & e^{\lambda L_{4}} & e^{\lambda L_{4}}[ - 2 - \lambda( L_{4} - L_{3} ) ] & \cdots & 0 \\ \noalign{\vspace{2pt}} \cdots & \cdots & \cdots & \cdots & \cdots \\ \noalign{\vspace{2pt}} 0 & 0 & 0 & \cdots & e^{\lambda L_{m}}[ - 2 - \lambda( L_{m} - L_{m - 1} ) ] \end{array} \right] $$

Further let $e^{\lambda L_{i + 1}} = e^{\lambda L_{i}} + \Delta( L_{i},\lambda)$, i=2,…,m−1. The Hessian can be rewritten as

$$H( Z ) = \bar{v}\lambda e^{ - \lambda } \arraycolsep=1.5pt \left[ \begin{array} {ccccc} e^{\lambda L_{2}} [ - 2 - \lambda( L_{2} - L_{1} ) ] & e^{\lambda L_{2}} + \Delta( L_{2},\lambda) & 0 & \cdots & 0 \\ \noalign{\vspace{2pt}} e^{\lambda L_{2}} + \Delta( L_{2},\lambda) & e^{\lambda L_{3}}[ - 2 - \lambda( L_{3} - L_{2} ) ] & e^{\lambda L_{3}} + \Delta( L_{3},\lambda) & \cdots & 0 \\ \noalign{\vspace{2pt}} 0 & e^{\lambda L_{3}} + \Delta( L_{3},\lambda) & e^{\lambda L_{4}}[ - 2 - \lambda( L_{4} - L_{3} ) ] & \cdots & 0 \\ \noalign{\vspace{2pt}} \cdots & \cdots & \cdots & \cdots & \cdots \\ \noalign{\vspace{2pt}} 0 & 0 & 0 & \cdots & e^{\lambda L_{m}}[ - 2 - \lambda( L_{m} - L_{m - 1} ) ] \end{array} \right] $$

Now, H(Z) is decomposed into H ₁(Z) and H ₂(Z) below so that

We will prove that the leading principal minors of each of H ₁(Z) and H ₂(Z) of order i has the same sign as (−1)ⁱ. To begin, let M _i be the leading principal minor of H ₁(Z) of order i. Then

and

$$M_{i} = ( - 1 )^{i}2^{i}\bigl( \bar{v}\lambda e^{ - \lambda } \bigr)^{i}\prod_{k = 1}^{i} e^{\lambda L_{k}},\quad i \ge 5 $$

It is clear that M _i has the same sign as (−1)ⁱ, i=1,2,…,m.

Likewise, let N _i be the leading principal minor of H ₂(Z) of order i. A similar approach may be followed to demonstrate that

$$N_{i} = ( - 1 )^{i} \bigl( \bar{v}\lambda^{2}e^{ - \lambda } \bigr)^{i}\prod_{k = 1}^{i} e^{\lambda L_{k}} ( L_{k} - L_{k - 1} ) $$

It is seen that N _i has the same sign as (−1)ⁱ, i=1,2,…,m.

Since the respective leading principal minors of H ₁(Z) and H ₂(Z) of order i have the same sign as (−1)ⁱ, the leading principal minor of H(Z)=H ₁(Z)+H ₂(Z) of order i has the same sign as (−1)ⁱ. This implies that the objective function in (A.1) is concave in L ₁,L ₂,…,L _m and hence in l ₁,l ₂,…,l _m since $L_{i} = \frac{l_{i}}{\bar{v}}$, i=1,2,…,m.

Appendix B: Proof of Proposition 2

A necessary condition for $( L_{1}^{*},L_{2}^{*}, \ldots,L_{m}^{*} )$ to be an optimal solution to the NLP in (A.1) is $\frac{\partial Z}{\partial L_{i}^{*}} = 0$, i=2,3,…,m. Thus, we have

$$ e^{\lambda L_{i + 1}^{*}} - e^{\lambda L_{i}^{*}} - \lambda e^{\lambda L_{i}^{*}} \bigl( L_{i}^{*} - L_{i - 1}^{*} \bigr) = 0 $$

(B.1)

or

$$\frac{e^{\lambda L_{i + 1}^{*}}}{e^{\lambda L_{i}^{*}}} - 1 - \lambda \bigl( L_{i}^{*} - L_{i - 1}^{*} \bigr) = 0 $$

or

$$ e^{\lambda ( L_{i + 1}^{*} - L_{i}^{*} )} = 1 + \lambda \bigl( L_{i}^{*} - L_{i - 1}^{*} \bigr) $$

(B.2)

As a Taylor series representation, we have

(B.3)

It follows from (B.2) and (B.3) that

$$e^{\lambda ( L_{i + 1}^{*} - L_{i}^{*} )} < e^{\lambda ( L_{i}^{*} - L_{i - 1}^{*} )} $$

or

$$L_{i + 1}^{*} - L_{i}^{*} < L_{i}^{*} - L_{i - 1}^{*} $$

Thus, $l_{i + 1}^{*} - l_{i}^{*} < l_{i}^{*} - l_{i - 1}^{*}$, i=2,3,…,m.

Appendix C: Proof of Proposition 3

Let k∈{0}∪N, x>0, and $LN( k,x ) = \underbrace{\ln ( 1 + \ln ( 1 + \cdots + \ln ( 1 + x ) ) )}_{k \ln ( \cdot )\mbox{\scriptsize 's involved}}$. Note that LN(0,x)=x. A necessary condition for $( L_{1}^{*},L_{2}^{*}, \ldots,L_{m}^{*} )$ to be an optimal solution to the NLP in (A.1) is $\frac{\partial Z}{\partial L_{i}^{*}} = 0$, i=2,3,…,m, or, $e^{\lambda L_{i + 1}^{*}} - e^{\lambda L_{i}^{*}} - \lambda e^{\lambda L_{i}^{*}}( L_{i}^{*} - L_{i - 1}^{*} ) = 0$.

For i=2, the following holds since $L_{1}^{*} = 0$:

$$e^{\lambda L_{3}^{*}} - e^{\lambda L_{2}^{*}} - \lambda e^{\lambda L_{2}^{*}}L_{2}^{*} = 0 $$

Thus,

(C.1)

For i=3, one has

$$e^{\lambda L_{4}^{*}} - e^{\lambda L_{3}^{*}} - \lambda e^{\lambda L_{3}^{*}} \bigl( L_{3}^{*} - L_{2}^{*} \bigr) = 0 $$

or

$$e^{\lambda L_{4}^{*}} = e^{\lambda L_{3}^{*}} \bigl( \lambda L_{3}^{*} - \lambda L_{2}^{*} + 1 \bigr) $$

So we have

or

$$L_{4}^{*} = \frac{1}{\lambda} \sum _{k = 0}^{2} LN \bigl( k,\lambda L_{2}^{*} \bigr) $$

Continuing in the same fashion, one can obtain the general expression below and see that $L_{i}^{*}$ is a function of $L_{2}^{*}$ as well as λ, i=3,4,…,m.

$$ L_{i}^{*} = \frac{1}{\lambda} \sum _{k = 0}^{i - 2} LN \bigl( k,\lambda L_{2}^{*} \bigr) $$

(C.2)

For the Dutch auction with m=2, we have $L_{m + 1}^{*} = L_{3}^{*} = 1$. It follows from (C.1) that

$$\lambda L_{2}^{*} + \ln \bigl( 1 + \lambda L_{2}^{*} \bigr) = \lambda $$

or

$$\lambda L_{2}^{*} = e^{\lambda ( 1 - L_{2}^{*} )} - 1 $$

When m=3, $L_{m + 1}^{*} = L_{4}^{*} = 1$ and we see from (C.2) that

$$\lambda L_{2}^{*} + \ln \bigl( 1 + \lambda L_{2}^{*} \bigr) + \ln \bigl( 1 + \ln \bigl( 1 + \lambda L_{2}^{*} \bigr) \bigr) = \lambda $$

or

$$\lambda L_{2}^{*} = e^{\frac{e^{\lambda ( 1 - L_{2}^{*} )}}{1 + \lambda L_{2}^{*}} - 1} - 1 $$

When m=4, $L_{m + 1}^{*} = L_{5}^{*} = 1$ and it is seen from (C.2) again that

$$\lambda L_{2}^{*} + \ln \bigl( 1 + \lambda L_{2}^{*} \bigr) + \ln \bigl( 1 + \ln \bigl( 1 + \lambda L_{2}^{*} \bigr) \bigr) + \ln \bigl( 1 + \ln \bigl( 1 + \ln \bigl( 1 + \lambda L_{2}^{*} \bigr) \bigr) \bigr) = \lambda $$

or

$$\lambda L_{2}^{*} = e^{e^{\frac{e^{\lambda ( 1 - L_{2}^{*} )}}{ ( 1 + \lambda L_{2}^{*} ) ( 1 + \ln ( 1 + \lambda L_{2}^{*} ) )} - 1} - 1} - 1 $$

In general, for a Dutch auction with m bid levels, one has

$$\lambda L_{2}^{*} = e^{\scriptstyle\cdot^{\scriptstyle\cdot^{\scriptstyle\cdot^{\overbrace{\scriptstyle e^{\frac{e^{\lambda ( 1 - L_{2}^{*} )}}{ ( 1 + \lambda L_{2}^{*} ) ( 1 + \ln ( 1 + \lambda L_{2}^{*} ) ) \cdots (1+\ln (1+\cdots + \ln(1+\lambda L_2^*)))} - 1} } ^{m-1\ \mathrm{EXP}(\cdot)\mbox{\scriptsize's involved}}- 1}}}}- 1 $$

It follows that, for i=2,3,…,m,

$$LN\bigl(k,\lambda L_{2}^{*}\bigr) = {\underbrace{\ln\bigl( \ln \cdots \ln}_{k-1\ \ln(\cdot)\mbox{\scriptsize 's involved}}} \bigl( e^{\scriptstyle\cdot^{\scriptstyle\cdot^{\scriptstyle\cdot^{\overbrace{\scriptstyle e^{\frac{e^{\lambda ( 1 - L_{2}^{*} )}}{ ( 1 + \lambda L_{2}^{*} ) ( 1 + \ln ( 1 + \lambda L_{2}^{*} ) ) \cdots (1+\ln (1+\cdots + \ln(1+\lambda L_2^*)))} - 1}} ^{m-2\ \mathrm{EXP}(\cdot)\mbox{\scriptsize's involved}}}- 1}}} +1\bigr) +1\bigr) $$

Since the increasing rate of the exponential function is greater than the decreasing rate of $\frac{1}{\lambda}$, $\frac{1}{\lambda} LN( k, \lambda L_{2}^{*} )$ is an increasing function of λ. As a result, $L_{i}^{*} = \frac{1}{\lambda} \sum_{k = 0}^{i - 2} LN( k,\lambda L_{2}^{*} )$ is an increasing function of λ, so is $l_{i}^{*}$, i=2,3,…,m.

Appendix D: Proof of Proposition 4

Since the optimal bid levels must satisfy (B.1) for i=2,3,…,m, we have

$$\sum_{i = 2}^{m} \bigl[ e^{\lambda L_{i + 1}^{*}} - e^{\lambda L_{i}^{*}} - \lambda e^{\lambda L_{i}^{*}} \bigl( L_{i}^{*} - L_{i - 1}^{*} \bigr) \bigr] = 0 $$

This leads to

or

$$ \sum_{i = 2}^{m} e^{\lambda L_{i}^{*}} \bigl( L_{i}^{*} - L_{i - 1}^{*} \bigr) = \frac{1}{\lambda} \bigl( e^{\lambda } - e^{\lambda L_{2}^{*}} \bigr) $$

(D.1)

In addition, given $L_{1}^{*} = 0$, one has

$$ \sum_{i = 2}^{m} e^{\lambda L_{i}^{*}} \bigl( L_{i}^{*} - L_{i - 1}^{*} \bigr) = \sum_{i = 2}^{m} L_{i}^{*}e^{\lambda L_{i}^{*}} - \sum _{i = 2}^{m - 1} L_{i}^{*}e^{\lambda L_{i + 1}^{*}} $$

(D.2)

Combining (D.1) and (D.2), we have

$$ \sum_{i = 2}^{m} L_{i}^{*}e^{\lambda L_{i}^{*}} - \sum _{i = 2}^{m - 1} L_{i}^{*}e^{\lambda L_{i + 1}^{*}} = \frac{1}{\lambda} \bigl( e^{\lambda } - e^{\lambda L_{2}^{*}} \bigr) $$

(D.3)

Based on (2) and (D.3), the maximum expected revenue may be expressed as

(D.4)

This completes the proof.

Appendix E: Proof of Proposition 5

To maximize the objective function in (A.1), the following must hold for i=3,4,…,m+1 according to (C.2):

$$L_{i}^{*} = \frac{1}{\lambda} \sum _{k = 0}^{i - 2} LN\bigl( k,\lambda L_{2}^{*} \bigr), $$

where $LN( k,\lambda L_{2}^{*} ) = \underbrace{\ln ( 1 + \ln ( 1 + \cdots + \ln ( 1 + \lambda L_{2}^{*} ) ) )}_{k\ \ln ( \cdot )\mbox{\scriptsize 's involved}}$. Clearly,

$$ L_{i}^{*} = L_{i - 1}^{*} + \frac{1}{\lambda} LN\bigl( i - 2,\lambda L_{2}^{*} \bigr) $$

(E.1)

If m bid levels are to be set, then $L_{m + 1}^{*} = 1$ and (C.2) becomes

$$\frac{1}{\lambda} \sum_{k = 0}^{m - 1} LN \bigl( k,\lambda L_{2}^{*} \bigr) = 1 $$

or

$$\lambda = \sum_{k = 0}^{m - 1} LN \bigl( k, \lambda L_{2}^{*} \bigr) $$

or

$$\lambda - \lambda L_{2}^{*} = \sum _{k = 1}^{m - 1} LN \bigl( k,\lambda L_{2}^{*} \bigr) $$

or

(E.2)

Also, according to (E.1), $L_{m + 1}^{*} = L_{m}^{*} + \frac{1}{\lambda} LN( m - 1,\lambda L_{2}^{*} ) = 1$. Thus, $L_{m}^{*}$ can be written as

$$ L_{m}^{*} = 1 - \frac{1}{\lambda} LN\bigl( m - 1,\lambda L_{2}^{*} \bigr) $$

(E.3)

Based on (D.4), (E.2) and (E.3), we have

Since $LN( k,\lambda L_{2}^{*} ) > 0$, k=0,1,…, the right-hand side of $\sum_{k = 0}^{m - 1} LN( k,\lambda L_{2}^{*} ) = \lambda$ includes more positive items as m increases. It follows that $LN( k,\lambda L_{2}^{*} )$ is a decreasing function of m. This leads to the result that $L_{2}^{*}$ decreases with m, so is $l_{2}^{*} = \bar{v}L_{2}^{*}$. The implication is that each of $LN( m - 1,\lambda L_{2}^{*} )$ and $\prod_{k = 0}^{m - 2} [ 1 + LN( k,\lambda L_{2}^{*} ) ]$ is also decreasing function of m. Thus, Z ^∗ is an increasing function of m. This completes the proof.

Appendix F: Proof of Proposition 6

Observe from (D.4) that

(F.1)

Thus, according to (E.2), (F.1) can be rewritten as

$$Z^{*} = \bar{v}L_{m}^{*} - \bar{v} \frac{\prod_{k = 0}^{m - 2} [ 1 + LN ( k,\lambda L_{2}^{*} ) ] - 1}{\lambda \prod_{k = 0}^{m - 2} [ 1 + LN ( k,\lambda L_{2}^{*} ) ]} $$

Note that $\frac{\prod_{k = 0}^{m - 2} [ 1 + LN( k,\lambda L_{2}^{*} ) ] - 1}{\lambda \prod_{k = 0}^{m - 2} [ 1 + LN( k,\lambda L_{2}^{*} ) ]}$ is a decreasing function of λ. On the other hand, as shown in Proposition 3, $l_{m}^{*}$ is an increasing function of λ, so is $L_{m}^{*}$. Consequently, Z ^∗ is an increasing function of λ. The proof is complete.