<p>A first price standard and reverse auction are formally equivalent to each other, and the same method can be used to solve both:</p>

<hr>

<h2>First Price Auction</h2>

<p>In a first price auction, $n$ bidders choose their bid, $b_i$, as a function of their value $v_i$ (distributed according to $F$. They seek to maximise their expected payoff:
$$[v_i-b_i(v_i)]\Pr(b_i\geq\max_j b_j).$$
If we look for a symmetric equilibrium in which bidders with higher values bid more then the probability that I have the highest bid is just the probability I have the highest value:
$$\Pr(b_i\geq\max_j b_j)=F(v_i)^{n-1}.$$</p>

<p>A useful trick is to think of a bidder as choosing what kind of value to "pretend to have". If we call their choice $\widetilde{v}$ then a bidder's probablem is to
$$\max_{\widetilde{v}}[v_i-b(\widetilde{v})]F(\widetilde{v})^{n-1}.$$</p>

<p>Differentiation yields the first-order condition:
$$(n-1)F'(\widetilde{v})F(\widetilde{v})^{n-2}[v_i-b(\widetilde{v})]-F(\widetilde{v})^{n-1}b'(\widetilde{v})=0.$$</p>

<p>In equilibrium, we know that a bidder should not want to pretend to be anyone other than his true type, so this first-order condition must hold at $\widetilde{v}=v_i$:
$$(n-1)F'(v_i)F(v_i)^{n-2}[v_i-b(v_i)]-F(v_i)^{n-1}b'(v_i)=0.$$
This is a differential equation that can be solved for the equilibrium bid function, $b(v_i)$. In, particular, if values are uniformly distributed on $[0,1]$, we have $F(v_i)=v_i$ and the equation becomes
$$(n-1)v_i^{n-2}[v_i-b(v_i)]-v_i^{n-1}b'(v_i)=0,$$
which can be simplified to
$$(n-1)\frac{1}{v_i}[v_i-b(v_i)]=b'(v_i).$$</p>

<p>The solution to this differential equation is
$$b(v_i)=v_i\frac{n-1}{n},$$
which is what you have in your question.</p>

<hr>

<h2>Reverse Auction</h2>

<p>In a reverse auction, $n$ bidders choose their bid, $b_i$, as a function of their cost $c_i$ (distributed according to $F$. They seek to maximise their expected payoff:
$$[b_i(c_i)-c_i]\Pr(b_i\leq\min_j b_j).$$</p>

<p>If we look for a symmetric equilibrium in which bidders with higher costs bid more then the probability that I have the lowest bid is just the probability I have the lowest cost:
$$\Pr(b_i\leq\min_j b_j)=[1-F(c_i)]^{n-1}.$$</p>

<p>A useful trick is to think of a bidder as choosing what kind of cost to "pretend to have". If we call their choice $\widetilde{c}$ then a bidder's probblem is to
$$\max_{\widetilde{c}}[b(\widetilde{c})-c_i][1-F(\widetilde{c})]^{n-1}.$$</p>

<p>Differentiation yields the first-order condition:
$$-(n-1)F'(\widetilde{c})[1-F(\widetilde{c})]^{n-2}[b(\widetilde{c})-c_i]+[1-F(\widetilde{c})]^{n-1}b'(\widetilde{c})=0.$$</p>

<p>In equilibrium, we know that a bidder should not want to pretend to be anyone other than his true type, so this first-order condition must hold at $\widetilde{c}=c_i$:
$$-(n-1)F'(c_i)[1-F(c_i)]^{n-2}[b(c_i)-c_i]+[1-F(c_i)]^{n-1}b'(c_i)=0.$$
This is a differential equation that can be solved for the equilibrium bid function, $b(c_i)$. In, particular, if costs are uniformly distributed on $[0,1]$, we have $F(c_i)=c_i$ and the differential equation can be simplified to
$$-(n-1)[1-c_i]^{n-2}[b(c_i)-c_i]+[1-c_i]^{n-1}b'(c_i)=0.$$
$$(n-1)\frac{1}{1-c_i}[b(c_i)-c_i]=b'(c_i).$$
The solution to this differential equation is
$$b(c_i)=\frac{1+c_i(n-1)}{n}.$$</p>

<hr>

<h2>Comparing the two</h2>

<p>The two solutions,
$$b(v_i)=v_i\frac{n-1}{n},\qquad b(c_i)=\frac{1+c_i(n-1)}{n}$$
are equivalent in the sense that a bidder in the first price auction shades (bids below his true valuation) by exactly the same amount as a bidder in a procurement auction bids above his true cost. You can confirm this by, for example, plotting these two bid functions for some value of $n$. Here's an example with $n=2$:</p>

<p><img src="https://i.sstatic.net/TI7JF.png" alt="Example for $n=2$"></p>

CC BY-SA 3.0
2015-02-09T13:53:26.070Z
The second solution cannot be true. The value of the offer must be smaller by increasing the number of the bidders, otherwise public tendering would be meaningles.
emanuele
CC BY-SA 3.0
2015-02-09T13:56:33.353Z
If I am the only seller,i can set the price i like. If am in a crowded market i must lowering my price. So your solution has an error somewhere.
emanuele
CC BY-SA 3.0
2015-02-09T14:11:16.460Z
@emanuele There's no error: differentiating $b(c_i)$ with respect to $n$, one obtains $b'(c_i)=(c_i-1)/n^2<0$ so the solution does predict that sellers will offer to perform the task for less if the number of bidders is large—just as your intuition suggests it should.
Ubiquitous
CC BY-SA 3.0
2015-02-09T14:36:34.183Z
@Ubiquitus sorry but i don't understand. $b'(c_i)<0$ if and only if $c_i<1$
emanuele
CC BY-SA 3.0
2015-02-09T14:42:08.790Z
ok. i get it, but why cost are limited in $[0,1]$ range?
emanuele
CC BY-SA 3.0
2015-02-09T14:50:47.780Z
...even more. If I am the only seller  I should be able to set the price I like. Your solution don't state this.
emanuele
CC BY-SA 3.0
2015-02-09T15:00:52.020Z
@emanuele Costs are limited to $[0,1]$ range because this is the solution to the differential equation assuming that costs are uniformly distributed on this interval (I chose that distribution because it is the one behind the formula in your original question). If, instead, costs follow a different distribution then the solution will be different but you will still find that $b$ is decreasing in $n$ for any possible cost.
Ubiquitous
CC BY-SA 3.0
2015-02-09T15:04:50.047Z
On the second point: if $n=1$ then the solution predicts that you can charge $1$, and not more. You are right that if you are the only seller then one might think that you would be able to charge more than this. Technically, this limit comes from the fact that we need a boundary condition to solve the differential equation and we obtain one by assuming that the auctioneer will not accept bids above 1. Although you might not like this much, note that the same kind of condition is at work in the standard auction example: bidders are not allowed to submit negative bids.
Ubiquitous
CC BY-SA 3.0
2015-02-09T15:05:41.137Z
One intuition is this: by the same token that a seller should be able to dictate any price if he is alone, the buyer (auctioneer, who is also a monopolist) should be able to refuse to pay more than $1$ since he knows that the bidder's cost is definitely not above this level.
Ubiquitous
CC BY-SA 3.0
2015-02-09T15:34:48.493Z
The last comment is very interesting.
emanuele
CC BY-SA 3.0
2015-02-10T01:18:32.343Z
Why $\Pr(b_i\leq\max_j b_j)=[1-F(c_i)]^{n-1}$ does not depends on $b_i(v_i)$?
emanuele
CC BY-SA 3.0
2015-02-10T09:04:15.650Z
@emanuele There was a typo. There shouldn't be any $b_i(v_i)$ in the second (reverse auction) case. In that case, your bid depends only on your cost.
Ubiquitous
CC BY-SA 3.0
2015-02-09T12:30:06.303Z
2015-02-10T07:55:00.943Z
2015-02-10T07:55:00.943Z
Ubiquitous
<p>I am studing a little bit of auction theory. I found the optimal bid value in the Milgrom paper for the first price auction that is
$$
P=v \frac{n-1}{n}
$$
where $P$ is the optimal bid, $v$ is the true value and $n$ is the number of bidders.
Now I am wondering if there exist an equivalent formula for reverse auctions.
What is the formula for the optimal bidding in reverse auctions?</p>

CC BY-SA 3.0
2015-02-08T20:25:41.413Z
Exactly the same method works.  Where did you run into trouble?
Steven Landsburg
CC BY-SA 3.0
2015-02-08T20:31:52.820Z
I would like to see the resulting formula or some references. I have some doubt in the convergence of the distribution.
emanuele
CC BY-SA 3.0
2015-02-09T00:51:57.967Z
What is reverse auction? What are the rules? Are the rules different from those of a first price auction?
Herr K.
CC BY-SA 3.0
2015-02-09T03:37:43.603Z
Reverse auction are auctions where the bidder is the final seller of the goods. Tendering is a reverse auction.
emanuele
<expected-utility><optimization><academic-graduate><auctions>
Reverse auction formula
emanuele
# Reverse auction formula

(Asked by: username_0 on 2015-02-08)

I am studing a little bit of auction theory. I found the optimal bid value in the Milgrom paper for the first price auction that is
$$
P=v \frac{n-1}{n}
$$
where $P$ is the optimal bid, $v$ is the true value and $n$ is the number of bidders.
Now I am wondering if there exist an equivalent formula for reverse auctions.
What is the formula for the optimal bidding in reverse auctions?

**Tags:** expected-utility, optimization, academic-graduate, auctions

> **Comments:**
> **username_1**: Exactly the same method works.  Where did you run into trouble?  
> **username_0**: I would like to see the resulting formula or some references. I have some doubt in the convergence of the distribution.  
> **username_2**: What is reverse auction? What are the rules? Are the rules different from those of a first price auction?  
> **username_0**: Reverse auction are auctions where the bidder is the final seller of the goods. Tendering is a reverse auction.  

---

## 1 answers

### Answer by username_3 on 2015-02-09. Score: 5

A first price standard and reverse auction are formally equivalent to each other, and the same method can be used to solve both:

---

## First Price Auction

In a first price auction, $n$ bidders choose their bid, $b\_i$, as a function of their value $v\_i$ (distributed according to $F$. They seek to maximise their expected payoff:
$$[v\_i-b\_i(v\_i)]\Pr(b\_i\geq\max\_j b\_j).$$
If we look for a symmetric equilibrium in which bidders with higher values bid more then the probability that I have the highest bid is just the probability I have the highest value:
$$\Pr(b\_i\geq\max\_j b\_j)=F(v\_i)^{n-1}.$$

A useful trick is to think of a bidder as choosing what kind of value to "pretend to have". If we call their choice $\widetilde{v}$ then a bidder's probablem is to
$$\max\_{\widetilde{v}}[v\_i-b(\widetilde{v})]F(\widetilde{v})^{n-1}.$$

Differentiation yields the first-order condition:
$$(n-1)F'(\widetilde{v})F(\widetilde{v})^{n-2}[v\_i-b(\widetilde{v})]-F(\widetilde{v})^{n-1}b'(\widetilde{v})=0.$$

In equilibrium, we know that a bidder should not want to pretend to be anyone other than his true type, so this first-order condition must hold at $\widetilde{v}=v\_i$:
$$(n-1)F'(v\_i)F(v\_i)^{n-2}[v\_i-b(v\_i)]-F(v\_i)^{n-1}b'(v\_i)=0.$$
This is a differential equation that can be solved for the equilibrium bid function, $b(v\_i)$. In, particular, if values are uniformly distributed on $[0,1]$, we have $F(v\_i)=v\_i$ and the equation becomes
$$(n-1)v\_i^{n-2}[v\_i-b(v\_i)]-v\_i^{n-1}b'(v\_i)=0,$$
which can be simplified to
$$(n-1)\frac{1}{v\_i}[v\_i-b(v\_i)]=b'(v\_i).$$

The solution to this differential equation is
$$b(v\_i)=v\_i\frac{n-1}{n},$$
which is what you have in your question.

---

## Reverse Auction

In a reverse auction, $n$ bidders choose their bid, $b\_i$, as a function of their cost $c\_i$ (distributed according to $F$. They seek to maximise their expected payoff:
$$[b\_i(c\_i)-c\_i]\Pr(b\_i\leq\min\_j b\_j).$$

If we look for a symmetric equilibrium in which bidders with higher costs bid more then the probability that I have the lowest bid is just the probability I have the lowest cost:
$$\Pr(b\_i\leq\min\_j b\_j)=[1-F(c\_i)]^{n-1}.$$

A useful trick is to think of a bidder as choosing what kind of cost to "pretend to have". If we call their choice $\widetilde{c}$ then a bidder's probblem is to
$$\max\_{\widetilde{c}}[b(\widetilde{c})-c\_i][1-F(\widetilde{c})]^{n-1}.$$

Differentiation yields the first-order condition:
$$-(n-1)F'(\widetilde{c})[1-F(\widetilde{c})]^{n-2}[b(\widetilde{c})-c\_i]+[1-F(\widetilde{c})]^{n-1}b'(\widetilde{c})=0.$$

In equilibrium, we know that a bidder should not want to pretend to be anyone other than his true type, so this first-order condition must hold at $\widetilde{c}=c\_i$:
$$-(n-1)F'(c\_i)[1-F(c\_i)]^{n-2}[b(c\_i)-c\_i]+[1-F(c\_i)]^{n-1}b'(c\_i)=0.$$
This is a differential equation that can be solved for the equilibrium bid function, $b(c\_i)$. In, particular, if costs are uniformly distributed on $[0,1]$, we have $F(c\_i)=c\_i$ and the differential equation can be simplified to
$$-(n-1)[1-c\_i]^{n-2}[b(c\_i)-c\_i]+[1-c\_i]^{n-1}b'(c\_i)=0.$$
$$(n-1)\frac{1}{1-c\_i}[b(c\_i)-c\_i]=b'(c\_i).$$
The solution to this differential equation is
$$b(c\_i)=\frac{1+c\_i(n-1)}{n}.$$

---

## Comparing the two

The two solutions,
$$b(v\_i)=v\_i\frac{n-1}{n},\qquad b(c\_i)=\frac{1+c\_i(n-1)}{n}$$
are equivalent in the sense that a bidder in the first price auction shades (bids below his true valuation) by exactly the same amount as a bidder in a procurement auction bids above his true cost. You can confirm this by, for example, plotting these two bid functions for some value of $n$. Here's an example with $n=2$:

![Example for $n=2$](https://i.sstatic.net/TI7JF.png)

> **Comments:**
> **username_0**: The second solution cannot be true. The value of the offer must be smaller by increasing the number of the bidders, otherwise public tendering would be meaningles.  
> **username_0**: If I am the only seller,i can set the price i like. If am in a crowded market i must lowering my price. So your solution has an error somewhere.  
> **username_3**: @username_0 There's no error: differentiating $b(c_i)$ with respect to $n$, one obtains $b'(c_i)=(c_i-1)/n^2<0$ so the solution does predict that sellers will offer to perform the task for less if the number of bidders is large—just as your intuition suggests it should.  
> **username_0**: @Ubiquitus sorry but i don't understand. $b'(c_i)<0$ if and only if $c_i<1$  
> **username_0**: ok. i get it, but why cost are limited in $[0,1]$ range?  
> **username_0**: ...even more. If I am the only seller  I should be able to set the price I like. Your solution don't state this.  
> **username_3**: @username_0 Costs are limited to $[0,1]$ range because this is the solution to the differential equation assuming that costs are uniformly distributed on this interval (I chose that distribution because it is the one behind the formula in your original question). If, instead, costs follow a different distribution then the solution will be different but you will still find that $b$ is decreasing in $n$ for any possible cost.  
> **username_3**: On the second point: if $n=1$ then the solution predicts that you can charge $1$, and not more. You are right that if you are the only seller then one might think that you would be able to charge more than this. Technically, this limit comes from the fact that we need a boundary condition to solve the differential equation and we obtain one by assuming that the auctioneer will not accept bids above 1. Although you might not like this much, note that the same kind of condition is at work in the standard auction example: bidders are not allowed to submit negative bids.  
> **username_3**: One intuition is this: by the same token that a seller should be able to dictate any price if he is alone, the buyer (auctioneer, who is also a monopolist) should be able to refuse to pay more than $1$ since he knows that the bidder's cost is definitely not above this level.  
> **username_0**: The last comment is very interesting.  
> **username_0**: Why $\Pr(b_i\leq\max_j b_j)=[1-F(c_i)]^{n-1}$ does not depends on $b_i(v_i)$?  
> **username_3**: @username_0 There was a typo. There shouldn't be any $b_i(v_i)$ in the second (reverse auction) case. In that case, your bid depends only on your cost.  

---

