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Transcript
Strategy 1-Bid $264.9M
-Bid your signal. What will happen? Give reasoning for your analysis
Had all companies bid their signals, the
losses would have been huge, ranging
from 14.1 to 37.3 million dollars!
Evidently, bidding one’s signal is a
disastrous strategy. Strategy 1 not optimal
because the extra profit values for the
historic auctions are all negative, indicating
that each winner paid more for the lease than
it was worth to the company.
Strategy 1-Bid $264.9M
Strategy 1-Bid $264.9M
“Winner’s Extra Profit” is almost always negative.
That is, the winning company will not get its
needed fair return on the lease.
Strategy 1 is not optimal because,
the highest signal will almost always be well
above the value of the lease and the winning
company will have paid too much for the drilling
rights. This is called the winner’s curse(we will
learn how to calculate after simulation of
Normal Errors).
Strategy 2-First plan-Company 1 will
2 is not optimal
Bid=$264.9M-$25M=$239.9M Strategy
because the difference
between the highest bid &
second highest bid
wasted(E(B))
Subtract Winner’s curse from your signal to obtain bid. Assume all other companies do
the same process. What will happen? Give reasoning for your analysis
1)Equal probability of winning
2)Company 1 expected value
is very small & other
companies expected value is
negative
Strategy 2-First plan-Company 1 will
Bid=$264.9M-$25M=$239.9M
• If all companies bid 25 million dollars less than
their signals, the expected value of the
difference between the top two bids will still be
6.38 million dollars. Hence, the winning
company will, on average, pay an unnecessary
premium of 6.38 million.
Strategy 3 is not optimal
because companies tend to
improve its expected value &
we need to incorporate other
companies actions
Strategy 3-Second plan-Company 1 will
Bid=$264.9M-$31.38M=$233.52M
Subtract Winner’s curse & Winner’s blessing from your signal to obtain bid. Assume all
other companies do the same process. What will happen? Give reasoning for your
analysis
1)Equal probability of winning
2) expected values very close
• Strategy 4
-Find a optimal adjustment for company 1.
Assume all other companies Subtract
Winner’s curse and Winner’s blessing
from their signals to obtain their bids.
Strategy 4
Strategy 4
-Find a optimal adjustment for company 1. Assume all
other companies Subtract Winner’s curse and Winner’s
blessing from their signals to obtain their bids
.
• steps
• 1. Enter the sum of the WC & WB as the signal adjustment for all
other companies cell
• 2. Change company 1 signal adjustment cell to get a set of
expected values for company 1.
• 3. record results of expected value for company 1
• good adjustment points(10 points ) Use
17,19,21,23,25,27,29,31,33,35
• 4. enter each good adj. ->hit F9(to recalculate)->manually record
the expected values & create a table for company 1
Strategy 4-Constructing f(a)
Bidding on
function
an Oil Lease
Probability, Mathematics,
on the project
for Company 1 must
find the maximum
expected value of
adjustment(assuming
that all other
companies subtract
both the curse and
blessing)
this best adjustment,
acb
Tests, Homework, Computers
Strategy 4-
f(a) function
COMPANY 1: CURSE & BLESSING FOR ALL OTHERS
0.6
Expected Value
0.5
0.4
0.3
0.2
y = -0.00000878x4 + 0.00111529x3 - 0.05242985x2 + 1.05475476x
- 7.09416335
0.1
0.0
0
4
8
12
16
20
24
28
32
36
Signal Adjustment
copy from the sheet Strategy in Auction Focus.xls to a new book
 Let f(a) be the expected value for Company 1 for subtracting a million
dollars from its signal, assuming that all other companies adjust their signals
by both the curse and blessing.
Fit a 4th degree polynomial trend line, which we will use as an approximate
formula for the unknown function f.
Use solver to find the best adjustment
Strategy 4- Company 1 will
Bid=$264.9M-$21.28M=$243.62M
Is Strategy 4 optimal ?
• No
• This is the real world of business, where we expect our competitors
to be well-managed companies
• other 18 companies are sitting in their offices and boardrooms
making the same calculations that we have just performed. Given
the results, other companies will also elect to subtract less than
31.38million dollars from their signals.
• It is worth noting that the 21.28million dollar adjustment that is our
appropriate response to a larger reduction by the other companies,
is not itself a stable strategy. Since this is less than the winner’s
curse of 25 million dollars, there would be a negative expected
value for extra profit if all companies adjusted downward by
21.28million dollars.
Strategy 5
-Find a optimal adjustment for company 1.
Assume all other companies Subtract
Winner’s curse from their signals to
obtain their bids.
Strategy 5- Company 1 will
Bid=$264.9M-$30.81M=$234.09M
change
Strategy 5
-Find a optimal adjustment for company 1. Assume all
other companies Subtract Winner’s curse from their
signals to obtain their bids
.
• steps
• 1. Enter the WC as the signal adjustment for all other companies
cell
• 2. Change company 1 signal adjustment cell to get a set of
expected values for company 1.
• 3. record results of expected value for company 1
• good adjustment points(11 points ) Use
19,21,23,25,27,29,31,33,35,37,39
• 4. enter each good adj. ->hit F9(to recalculate)->manually record
the expected values & create a table for company 1
Strategy 5- Constructing
Bidding on
an Oil Lease
g(a)
function
on
the project
Probability, Mathematics,
Tests, Homework, Computers
for Company 1 must
find the maximum
expected value of
adjustment(assuming
that all other
companies subtract
both the curse and
blessing)
this best adjustment,
ac
Simulating,
Focus
Auction
Focus.xls
Class Project
(material continues)
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Strategy 5 -g(a) function
USE the sheet Strategy in Auction Focus.xls.
 Let g(a) be the expected value for Company 1 for subtracting a million dollars from its signal,
assuming that all other companies adjust their signals by curse.
Fit a 4th degree polynomial trend line, which we will use as an approximate formula for the
unknown function g.
Use solver to find the best adjustment
the use of Solver in Strategy shows that g(30.8) = 0. Hence, ac = 30.8 million dollars.
Company 1’s best response to an adjustment of 25 million dollars by all other
companies is to lower its signal by the considerably larger amount of $30.8M
Strategy 5- Company 1 will
Bid=$264.9M-$30.81M=$234.09M
Is Strategy 5 optimal?
if we know what all other companies plan to
do. Moreover, this same information is available
to all of the bidders.
Need a stable strategy???
If all companies made such a stable
adjustment to their signals, then there would
be no incentive for anyone to alter the
strategy. A stable bidding strategy is also called
a Nash equilibrium
A signal adjustment of as would be a stable strategy if Company
1’s best response to an adjustment of as by all other companies, would be to
also adjust by as. If all companies made such a stable
adjustment
Bidding
onto their
Probability, Mathematics,
Tests, Homework,
Computers
signals, then there would be no incentive
for anyone to alter
the
LeaseThe
strategy. A stable bidding strategy is also called a an
NashOil
equilibrium.
development of this game theoretic concept earned a share in the 1994 Nobel
Prize in Economics for the mathematician and economist John F. Nash (1928
- ).
on the project
It seems to be quite possible that there is a Nash equilibrium, or
stable strategy, for our auctions. We have seen that Company 1 should
respond with a smaller adjustment to a large adjustment by all other
companies. Conversely, Company 1 should respond with a larger
adjustment to a small adjustment by all other companies. We are looking for
a universal intermediate strategy, as.
Strategy-4
Strategy-5
Company 1: Best Adjustment
21.28
Company 1:
as
22.0
30.2
30.8
All Other Companies:
as
25
31.2
24.9
31.38
We know We want We know
Simulating, Focus
Auction
Focus.xls
Class Project
(material continues)

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Strategy 6
Finally, each team should experiment with
Auction Equilibrium.xls to determine a stable
Nash equilibrium bidding strategy for its auction
scenario. This will lead to a modification of
your team’s signal and a specific bid in the
upcoming auction.
Optimal
Adjustment, a max
$27.031M
24.4423
Adjustment
Subtracted From
Signal
24.8
$27.031M
Need to enter different
values for the blue cell and
experiment until the two
cells are equal(this will take
a LONG time. We will use
10 iterations and get the
average
Strategy 6
Strategy 6
Strategy 6
(2wc+wb)/2
Excel
method
for
Strategy
6
How?
(a) Use Auction Equilibrium.xls(.
(b) FOLLOW THE INSTRUCTIONS IN THIS FILE!
(c) Enter appropriate values in cells B10 through E10.
(d) Enter a logical value in cell E39. Run the macro Optimize.
(the first logical value to use- (2wc+wb)/2
(e) Enter another logical value in cell E39 and press the key F9..
record numbers in a table.
(f)
See table
(g) Find the stable adj for strategy 6
Excel method for Strategy 6
1
Company 1
Optimal
Adjustment, amax
(use 4 decimals)
All Other
Companies
Adjustment
Subtracted
From Signal
New logical value
25.3933
28.19
(25.3933+28.19)/2=
26.7916
2
26.7916
.
.
10
Avg of amax=final stable
adj for strategy 6
first logical value to
use for class
project(2wc+wb)/2=28.19
Strategy 6- Company 1 will
Bid=$264.9M-$27.031M=$237.869M
Using Auction Equilibrium.xls to determine 10 stable
strategy values, and averaging the results, we find a
signal adjustment of 27.031 million dollars. This
provides our Nash equilibrium and is the final answer to
our bidding strategy problem. If each company
reduced its signal by $27,031,000 then there would
be no expected gain for any one company, if it
deviated from this plan.
Specifically, we will reduce our signal of
$264,900,000 by $27,031,000 and submit a bid of
$237,869,000.
Finding the Nash Equilibrium for
the exams-Example
Signal Adjustment
Company 1
All Other
Optimum
Companies
25.0755
23.6583
24.3138
24.3669
24.3273
24.3404
24.3116
24.3338
24.3202
24.3227
24.3218
24.3221
Nash equilibrium is when the two values
of the columns are approximately equal
to each other=24.322
Recall- Class Project-Goals
 Determine what would be expected to happen if each
company bid the same amount as its signal.
 Determine the Company 1 bid under several
uniform bidding strategies, and explore the expected
values of these plans.
 Find a stable uniform bidding strategy that could be
followed by all companies, without any chance for
improvement.
Recall-Project Assumptions
Assumption 1.
The same 18 companies
will each bid on future similar leases
only bidders for the tracts.
Assumption 2.
The geologists employed by companies
equally expert
on average, they can estimate the correct
values of leases.
each signal for the value of an undeveloped tract is
an observation of a continuous random variable, Sv,
Mean of Sv  v ( actual value of lease )
Recall-Project Assumptions
Assumption 3. Except for their means, the
distributions of the Sv’s are all identical
(The shape /The Spread)
Assumption 4.
All of the companies have the same profit margins
Strategies for bidding on an Oil
Lease
• Strategy 1
-Bid your signal. What will happen? Give reasoning for your
analysis
• Strategy 2(First Plan)
-Subtract Winner’s curse from your signal to obtain bid.
Assume all other companies do the same process. What
will happen? Give reasoning for your analysis
• Strategy 3(Second Plan)
-Subtract Winner’s curse and Winner’s blessing from your
signal to obtain bid. Assume all other companies do the
same process. What will happen? Give reasoning for
your analysis
Strategies for bidding on an Oil
Lease
• Strategy 4
-Find a optimal adjustment for company 1. Assume all
other companies Subtract Winner’s curse and Winner’s
blessing from their signals to obtain their bids.
• Strategy 5
-Find a optimal adjustment for company 1. Assume all
other companies Subtract Winner’s curse from their
signals to obtain their bids.
• Strategy 6
-Determine a stable Nash equilibrium bid. This stable
strategy is such that any company will not have any
incentive to deviate from
Error random variable, R
We can assume that all of the individual
error random variables represent a
common error random variable, R. This
gives
error = signal  proven value.
Errors –R random variable is a
Normal random varible
Simulation
• Need more than 20 auctions. Since we have to bid now, we
cannot wait for more actual data to accumulate.
• The only practical solution is to use the small amount of
historical data(380 errors) to train a computer to simulate
thousands of similar auctions.
• Monte Carlo method of simulation to amplify the
information that is contained in our original sample of
data from 20 auctions
Important - Generate Errors
•10000 leases
•class project has 19 companies->19 error columns
•Want Normal Errors with Mean 0 & Standard Deviation 13.53
•using =NORMINV(RAND(),0,13.53)
Generate Normal Errors
Creating the fixed error matrix
• Copy all the generated and do a paste
special(values only) in a new worksheet
• Once the values are copied YOU MUST
DELETE ALL THE NORMINV
FORMULAS IN THE ORIGINAL
WORKSHEET
• IF YOU DON’T DO THIS THE FILE WILL
BE TOO LARGE TO HANDLE
How to calculate Winner’s CurseE(C)
• Let C be the continuous random variable
which gives the largest number in a
sample of 19(class project has 19
companies) observations of R.
• Assuming that each company bids its
signal, E(C) will be the expected value of
the winner’s curse.
Winner’s Curse-E(C)
sample mean for the 10,000 observations of
C is $25.00 M. If a company bids its
signal and wins the auction, it can
expect, on average, to fall $25M below
its needed fair return on the lease.
How to calculate Winner’s
Blessing- E(B)
• Let B be the continuous random variable
which gives the difference between the
largest and second largest errors in a
random set of 19 observations of R
• E(B) will be the expected value of the
winner’s blessing
• sample mean for the 10,000 observations
of B is 6.38 million
E(C) & E(B)
Extra profit
• Extra profit is the amount by which the
winning bid is below the fair value of a
lease.
Expected Value of an adjustment
(for all other companies-combined)
The best measure of success is the expected
value of an adjustment. Let Xi be the
continuous random variable giving the extra
profit, in millions of dollars, for Company i. Xi
can be positive or negative, if Company i wins,
but it will always be 0 if Company i does not
win. Signal Adjustment uses the 10,000
simulated auctions to approximate E(X1) and the
average of E(X2), E(X3), , and E(X19).
Extra Profit
Extra Profit
Identifying the
Random variables
Let V be the continuous random variable
that gives the fair profit value, in millions of
dollars, for an oil lease that is similar to the
20 tracts in the data. The 20 proven
values form a random sample for V.
the proven leases have a sample mean of
229.8million dollars
Random variable
V- fair profit value
Continuous random variable Sv
• Recall from the project description that each signal is an
observation of a continuous random variable Sv, where v
is the actual fair profit value of the given lease. We have
assumed that E(Sv) = v, for every lease.
• To test the reasonableness of this assumption we have
computed the sample mean for the 19 signals on each
of the proven leases.
• For example historical lease number 1 –>Mean of the
signals of 19 companies is $233.7M S1, and proven
value is $237.2M
• Even with the small sample sizes of 19, there is relatively
good agreement between the sample means of the
signals and the proven values of the leases.
Continuous random variable Sv
Random variable, Rv
• For each lease value, v, we define a new
random variable, Rv, that gives the error in a
company’s signal. This is given by the signal
minus the actual fair profit value of the lease, Rv
= Sv  v.
• By project Assumption 2,
Mean of Sv  v ( actual value of lease )
• the long term average of the signals for a fixed
lease will be the actual value of the lease.
Thus, for each value of v, the expected value
of Rv is assumed to be 0.
Assumption 3
• According to project Assumption 3, the
signals for each lease have similar
distributions about their averages. This
means that, for all values of v, the signal
errors have the same distributions
about 0. In terms of our random
variables, we conclude that the Rv’s are all
identical random variables.
Error random variable, R
We can assume that all of the individual
error random variables represent a
common error random variable, R. This
gives
error = signal  proven value.
Bidding on
Probability, Mathematics,
Tests, Homework, Computers
an Oil Lease
on the project
Integration provides expressions for the expected values of V and
R.

E (V ) 
 v  fV (v) dv


E ( R) 
 r  f R (r) dr

Unfortunately, we have no way to know exact formulas for the
p.d.f.’s, fV, for V. and fR for R. Thus, we cannot compute these means.
The only thing that we can do is to replace the parameters E(V)
and E(R) with their estimating statistics, the sample means. Hence, we
assume that E(V) is 229.82 million dollars. Likewise, we have used
project assumptions to conclude that E(R) = 0.
Distributions,
Focus
Auction
Focus.xls
Class Project
(material continues)

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Bidding on
Tests, Homework, Computers
an Oil Lease
Probability, Mathematics,
on the project
From the histogram, it appears that the distribution of V is reasonably
close to being normal. From now on, we will assume that V is a normal random
variable, with V = 229.82 and V = 38.73. With this assumption the p.d.f. for V
has the following form.
1
fV (v ) 
38.73  2  
Normal, Focus
Auction
Focus.xls
Class Project
1  v 229.82 2
 

2
38
.
73


e
(material continues)

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Height
ERRORS AND NORMAL
on the project
0.035
0.030
Sample
0.025
Normal
Probability,
Tests, Homework, Computers
0.020 Mathematics,
0.015
0.010
0.005
0.000
-40 -32 -24 -16 -8 0 8 16 24 32 40
Bidding on
an Oil Lease
Millions of $
In the sheet Normal
we have also computed values for the normal density, with a mean of 0 and a
standard deviation of 13.53, at the midpoints of the bars in the histogram for our
sample of R. We see that the distribution of R appears to be very close to that of a
normal random variable.
We will assume that R is a normal random variable, with R = 0 and R =
13.53. The p.d.f. for R has the following form.
f R (r) 
Normal, Focus
Auction
Focus.xls
1
13.53  2  
Class Project
1  r 2
 

2
13
.
53


e
(material continues)

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