# Reverse auction formula

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?

• Exactly the same method works. Where did you run into trouble? Feb 8, 2015 at 20:25
• I would like to see the resulting formula or some references. I have some doubt in the convergence of the distribution. Feb 8, 2015 at 20:31
• What is reverse auction? What are the rules? Are the rules different from those of a first price auction? Feb 9, 2015 at 0:51
• Reverse auction are auctions where the bidder is the final seller of the goods. Tendering is a reverse auction. Feb 9, 2015 at 3:37

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 $n=2$: • 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. Feb 9, 2015 at 13:53
• 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. Feb 9, 2015 at 13:56
• @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. Feb 9, 2015 at 14:11
• @Ubiquitus sorry but i don't understand. $b'(c_i)<0$ if and only if $c_i<1$ Feb 9, 2015 at 14:36
• ok. i get it, but why cost are limited in $[0,1]$ range? Feb 9, 2015 at 14:42