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?
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 $n=2$: