Zhao Wu
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Answer:
The story is still about me buying a car.
The Player: Me(Zhao Wu), the seller(M)
Sequences of moves: The seller moved first and I responded to her. She said if I could accept the
original price, she would gave me a set of new snow-tyres.
Options for Me: I can either accept the seller’s suggestion or I could choose not to have
snow-tyres so that I could beat the price.
Objectives: The seller wanted to sell her car and the snow-tyre in a relatively high price. I wanted
to spend as little as I could.
Information: Complete Vs Incomplete. I knew that the seller wanted to sell both the car and the
snow-tyres but she wasn’t sure if I would only buy the car or buy both.
Time: One-shot
Equilibrium: I knew that winter in Buffalo is terrible and I am sure I would need a set of
snow-tyres. After comparing the price with other cars and snow-tyres, I thought the original price
would be reasonable for both the car and the snow-tyres. At last, I bought the car without
beating the price and got a new set of tyres and the seller saved her time for selling tyres.
Answer:
I am interested in Game Theory so that I think I like most parts of IE675 class. But I would like to
give one suggestion that more examples should be given when there are some abstract and
complex formulations and conceptions.
Zhao Wu
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Answer:
We can see the feasible area with the constraints from the picture below. We can see that 0  x1
 10 and 0  x2  5.
2
(a) U1= x1 x2
First, from the three constraints for three inequality, we can have x2  5substitute (1) into U1 so that we could get:
2
U1= x1 x2  
1 3
x1  5 x12
2
We can see the graph of 
1 3
x1  5 x12 function as below.
2
1
x1 (1) and we
2
Zhao Wu
d (
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1 3
x1  5 x12 )
3
20
1 3
2000
2
x1  5 x12 =
, 
  x12  10 x1  0  x1 
2
27
dx1
2
3
Because x2  5-
1
5
x1  x2 
2
3
So the maximum point should be(
20 5
2000
, ), the maximum value of U1 should be
.
3 3
27
(b) U2=2 x1 + x2
Similarly, we can have x2  5U2=2 x1 + x2 
1
x1 (1) and we substitute (1) into U1 so that we could get:
2
3
x1  5
2
We can see the graph of
3
x1  5 as below:
2
Zhao Wu
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3
x1  5 gets its maximum value 20 when x1 =10.
2
1
Because x2  5- x1  x2  0
2
Obviously,
So the maximum point should be(10,0), the maximum value of U1 should be 20.
(c) U3=min( x1 ,3 x2 )
Similarly, we have: 3 x2  15 
Let x1  15 
3
x1
2
3
x1  x1  6  3x2  6
2
Now we have U3=min(6,6)=6
If x1 decreases, then obviously U3 would decrease too;
If x1 increases, then 3x2 would decrease and smaller than 6.
So we can draw the conclusion that the maximum point should be(6,2), the maximum value
of U1 should be 6.
Answer:
In order to find the global minimum and global maximum, we can calculate the derivative of
f ( x ) and then we get f '( x) as below:
f '( x)  (
Vx
(Vx) '( x  d )  ( x  d ) '(Vx)
dV
  x) ' 
 

2
xd
(x  d )
( x  d )2
Zhao Wu
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Then we calculate the derivative of f '( x ) and we get f ''( x) as below:
f ''( x)  2dV ( x  d )3 <0
We know that f ( x ) gets its maximum or minimum when f '( x ) =0. So we would have
f '( x)  (
or x  
Vx
(Vx) '( x  d )  ( x  d ) '(Vx)
dV
dV
  x) ' 
 
 0 x 
d
2
2
xd
(x  d )
(x  d )

dV

 d (which obviously contradicts with the given condition)
We can also see from f ''( x)  2dV ( x  d )
3
 0 that when x 
dV

 d , f ( x)
would get its maximum.
(a) Under
no
conditions
would
f ( x ) get
its
global
minimum
f ''( x)  2dV ( x  d )3  0 .
(b) When x 
dV

 d  0 , that means V  d  , we can have global maximum.
*
(c) Global maximum point x 
dV

 d (when V  d  )
Global maximum value f ( x )  V   d  2 dV 
*
because
Zhao Wu
dV
*
(d) 1.For x 
i)

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d
About d
V
x* 
dV
 d  ( d 

*
When x 
dV


2
V
)2 

4
d 0 d 
V

V
So we can know that when
d increases from (0 ,

2
] , x* will increase;
V
When
d increases from (

2
,
V

] , x* will decrease.
The picture below shows an example.
Assuming that V, β don’t change and let V/ β =10. We can see the graph of function
x* 
ii)
dV

d
About V
x* 
dV

d
*
It is obviously that when V increases, x will increase too.
Zhao Wu
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About β
iii)
x* 
dV

d
It is obviously that when βincreases, x* will increase too.
2.For f ( x )  V   d  2 dV 
*
i)
About d
Let t= d , then
f ( x* )  V   t 2  2 V  t  V   (t 2  2
V

t
V

)  V   (t 
V

)2   ( d 
And We know that to make sure the global maximum exists, we must have
means
d
V

V
d 

Then we can draw the conclusion that:
When d increases from (0,
When d gets bigger than
V

V

] , f * ( x ) will decrease;
, f * ( x ) won’t exist.
ii) About V
Let t= V , then
f ( x* )  V   d  2 Vd   (t   d )2  (V   d )2
Because
V   d ,so
Decrease when V increase from
When  d  1 , then f * ( x ) will
( d ,  d ] ;
Increase when V increase from ( (
 d , )
When  d  1 ,then f * ( x ) will increase when V increase from [  d , )
iii)
About

V

which
)2
Zhao Wu
Let t 
 ,then
f ( x* )  V  dt 2  2 Vdt  V  d (t 2  2
V V
V 2
V 2
t  )  V  d (t 
)  d(  
)
d d
d
d
Similar to problem d-i,
To make sure the global maximum exists, we must have

Then we can draw the conclusion that:
When  increases from (0,
V
] , f * ( x ) will decrease;
d
When  gets bigger than
V
, f * ( x ) won’t exist.
d
Answer:
(a) T={1,2,3,4}
St  {S1 , S2 , S3 , S4 , S5 , S6 , S7 , S8}
A1,1  {(1, 2), (1,3), (1, 4)}
A2,2  {(2,5), (2, 6)}; A3,2  {(3,5), (3, 6), (3, 7)}; A4,2  {(4, 7)};
A5,3  {(5,8)}; A6,3  {(6,8)}; A7,3  {(7,8)};
A8,4  
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V
d
Zhao Wu
50024849
r1 (1,1)  a
r1 (1, 2)  3
r1 (1,3)  4
r2 (2,1)  3
r2 (2, 2)  5
r2 (3,1)  5
r2 (3, 2)  6
r2 (3,3)  2
r2 (4,1)  2
r3 (5,1)  1
r3 (6,1)  2
r3 (7,1)  6
r4 (8, 0)  0
Pt ( j / st , a) is not given, but we know that

jS
Pt ( j / st , a ) =1
(b)i.Set t=4 and u *(4,8)  0
Since t  1, set t=4-1=3
u3 *(5)  1  u4 *(8)  1 , u3 *(6)  2  u 4 *(8)  2 , u3 *(7)  6  u4 *(8)  6
A5,3 *  (5,8) , A6,3 *  (6,8) , A7,3 *  (7,8)
ii. Since t  1, set t=3-1=2
u2 *(2)  max{3  u3 *(5),5  u3 *(6)}  max{4,7}  7 ,
u2 *(3)  max{5  u3 *(5),6  u3 *(6), 2  u3 *(7)}  max{6,8,8}  8 ,
u2 *(4)  max{2  u3 *(7)}  max{8}  8
Optimal Action Sets:
A2,2 *  (2, 6) , A3,2 *  {(3, 6),{3, 7)} , A2,2 *  (4, 7)
iii. Since t  1, set t=2-1=1
u1 *(1)  max{a  u2 *(2),3  u2 *(3), 4  u2 *(4)}  max{a  7,11,12}
When a  7  12  a  5 , u1 *(1)  a  7 ,Optimal Action Sets: A1,1*  (1, 2)
When a  7  12  a  5 , u1 *(1)  12 ,Optimal Action Sets: A1,1*  (1, 4)
Since t=1, Stop.
Conclusion: The optimal solution would be
1-2-6-8
1-4-7-8
if a  5
if 0  a  5