A Stochastic Differential
(Difference) Game Model With an
LP Subroutine for Mixed and
Pure Strategy Optimization
INFORMS International Meeting 2007, Puerto Rico
Peter Lohmander
Professor
SLU
Umea, SE-901 83, Sweden,
http://www.Lohmander.com
Version 2007-09-29
1
Abstract:
This paper presents a stochastic two person differential
(difference) game model with a linear programming subroutine
that is used to optimize pure and/or mixed strategies as a
function of state and time.
In ”classical dynamic models”, ”pure strategies” are often
assumed to be optimal and deterministic continuous path
equations can be derived. In such models, scale and timing
effects in operations are however usually not considered.
When ”strictly convex scale effects” in operations with defence
and attack (”counter air” or ”ground support”) are taken into
consideration, dynamic mixed OR pure strategies are optimal in
different states.
The optimal decision frequences are also functions of the relative
importance of the results from operations in different periods.
The expected cost of forcing one participant in the conflict to use
only pure strategies is determined.
The optimal responses of the opposition in case one participant in
the conflict is forced to use only pure strategies are calculated.
Dynamic models of the presented type, that include mixed
strategies, are highly relevant and must be used to determine
the optimal strategies. Otherwise, considerable and quite
unnecessary losses should be expected.
2
Inspiration:
• Isaacs R. (1965) Differential Games – A
mathematical theory with applications to
warfare and pursuit, control and
optimization, Wiley, 1965 (Also Dover,
1999)
• Washburn, A.R. (2003) Two-person
zero-sum games, 3 ed., INFORMS,
Military applications society, Topics in
operations research
3
Flexibility
Some resources may be used in
different ways.
This is true and important in the air
force, in the army and in the
enterprises in the economy.
4
Our Plan to Win
This plan is focused on continuously increasing McDonald's
relevance to consumers' everyday lives through multiple
initiatives surrounding the five key drivers of great restaurant
experiences.
Our efforts, complemented with financial discipline, have
delivered for shareholders some of the strongest results in our
history.
5
W-lan
Max Hamburger restaurants is the only national restaurant chain to
offer free Internet access for all guests. Connection is offered via WLan.
The most satisfied guests
For the fith year in a row Max Hamburger Restaurants has Sweden's
most satisfied customers among the national hamburger
restaurants according to a new survey by SIFO/ISI Wissing.
6
Boeing
B-52 Stratofortress
Intercontinental Strategic Bomber
7
JAS 39 Gripen
Multirole fighter
8
9
t  T 1
V ( xT 1 , yT 1 )  0
xT 1  0,1,..., N x 
yT 1  0,1,..., N y 
10
t T
V ( xT , yT )  xT  yT
xT  0,1,..., N x 
yT  0,1,..., N y 
11
t
t T




V ( xt , yt )  max min  Rt ()  d  ( xt 1 , yt 1 )V ( xt 1 , yt 1 ) 
GS1t ,CA1t GS2t ,CA2t
xt 1 yt 1




(GS1t , CA1t )  1 ( xt )
(GS 2t , CA2t )   2 ( yt )
t  0,1,..., T  1
xt  0,1,..., N x  t
yt  0,1,..., N y  t
12
Rt ()  Rt (GS1t , CA1t , GS2t , CA2t , xt , yt )
 ( xt 1 , yt 1 )   ( xt 1 , yt 1 GS1 , CA1 , GS2 , CA2 , xt , yt )
t
t
t
t
13
Special case 1:
GS1t  CA1t  xt
GS2t  CA2t  yt
Special case 2:
GS1t  CA1t  xt
GS2t  CA2t  yt
14
In Model 1:
 GS1t
GS2t
Rt ()  

 1  CA2 1  CA1
t
t





15
In Model 2:
5


2
GS
2t
 (GS1t )

3
Rt ()  

2
2 
1

(
CA
)
1

(
CA
)
2
1


t
t


16
In Model 3 and Model 4:
5


2
GS
2
(
GS
)


t
3
1t
3
Rt ()  

2
2 
2  1  (CA2t ) 1  (CA1t ) 




17
Model 1
18
19
Model 1
!TAW_070611;
!Peter Lohmander;
Model:
sets:
xset/1..3/:;
yset/1..3/:;
nset/1..3/:;
D1set/1..3/:;
D2set/1..3/:;
xyset(xset,yset):;
xynset(xset,yset,nset):V;
xyD1set(xset,yset,D1set):GS1, CA1;
xyD2set(xset,yset,D2set):GS2, CA2;
xynD1s(xset,yset,nset,D1set):xdec;
xD1D2s(xset, D1set, D2set):;
xD1D2Xs(xset,D1set, D2set, xset):TPX;
yD2D1s(yset,D2set,D1set):;
yD2D1yset(yset,D2set,D1set,yset):TPY;
endsets
20
max = Z;
Z = @sum(xyset(x,y):V(x,y,3));
!Terminal condition: No more activities occur
when n <= 1 . ;
@for(xyset(x,y):V(x,y,1) = 0);
21
! The last day when activities occur, n=2, all resources should be used for GS. ;
@for(xyset(x,y):V(x,y,2) = x-y);
22
! Day n >= 3, you may use some resources for GS and some for CA.
The rows and columns represent the resources used for GS. ;
@for(xyD1set(x,y,D1)|D1 #LE# x: GS1(x,y,D1) = D1-1 );
@for(xyD1set(x,y,D1)|D1 #GT# x: GS1(x,y,D1) = 0 );
@for(xyD1set(x,y,D1): CA1(x,y,D1) = x - 1 - GS1(x,y,D1) );
@for(xyD2set(x,y,D2)|D2 #LE# y: GS2(x,y,D2) = D2-1 );
@for(xyD2set(x,y,D2)|D2 #GT# y: GS2(x,y,D2) = 0 );
@for(xyD2set(x,y,D2): CA2(x,y,D2) = y - 1 - GS2(x,y,D2));
23
@for(xynset(x,y,n): @sum(D1set(D1):xdec(x,y,n,D1)) = 1);
@for(xynD1s(x,y,n,D1)| D1#GT#x: xdec(x,y,n,D1) = 0);
24
!Sum (Transition probability) = 1;
@for(xD1D2s(x1,GS1,CA2): @sum(xset(x2): TPX(x1,GS1,CA2,x2) ) = 1 );
@for(yD2D1s(y1,GS2,CA1): @sum(yset(y2): TPY(y1,GS2,CA1,y2) ) = 1 );
25
!********************************************;
! Transition probability calculations for x ;
!********************************************;
! Transition probabilities for x if x1 = 0;
@for(xD1D2s(x1,GS1P1,CA2P1)|x1#EQ#1 : TPX(x1,GS1P1,CA2P1,x1) = 1 );
! Transition probabilities for x if CA2 = 0;
@for(xD1D2s(x1,GS1P1,CA2P1)|CA2P1 #EQ# 1 : TPX(x1,GS1P1,CA2P1,x1) = 1 );
! Transition probabilities for x if GS1 = 0;
@for(xD1D2s(x1,GS1P1,CA2P1)|GS1P1 #EQ# 1 : TPX(x1,GS1P1,CA2P1,x1) = 1 );
26
! Transition probabilities for x if GS1 >= 1 and CA2 = 1;
@for(xD1D2s(x1,GS1P1,CA2P1)|x1#GE#2 #AND# GS1P1#GE#2 #AND# CA2P1#EQ#2 :
TPX(x1,GS1P1,CA2P1,x1) = 1/2 );
@for(xD1D2s(x1,GS1P1,CA2P1)|x1#GE#2 #AND# GS1P1#GE#2 #AND# CA2P1#EQ#2 :
TPX(x1,GS1P1,CA2P1,x1-1) = 1/2 );
27
! Transition probabilities for x if GS1 = 1 and CA2 = 2;
@for(xD1D2s(x1,GS1P1,CA2P1)|x1#GE#2 #AND# GS1P1#EQ#2 #AND# CA2P1 #EQ# 3 :
TPX(x1,GS1P1,CA2P1,x1) = 1/4 );
@for(xD1D2s(x1,GS1P1,CA2P1)|x1#GE#2 #AND# GS1P1#EQ#2 #AND# CA2P1 #EQ# 3 :
TPX(x1,GS1P1,CA2P1,x1-1) = 3/4 );
28
! Transition probabilities for x if GS1 >= 2 and CA2 = 2;
@for(xD1D2s(x1,GS1P1,CA2P1)|x1#GE#3 #AND# GS1P1#GE#3 #AND# CA2P1 #EQ# 3 :
TPX(x1,GS1P1,CA2P1,x1) = 1/4 );
@for(xD1D2s(x1,GS1P1,CA2P1)|x1#GE#3 #AND# GS1P1#GE#3 #AND# CA2P1 #EQ# 3 :
TPX(x1,GS1P1,CA2P1,x1-1) = 1/2 );
@for(xD1D2s(x1,GS1P1,CA2P1)|x1#GE#3 #AND# GS1P1#GE#3 #AND# CA2P1 #EQ# 3 :
TPX(x1,GS1P1,CA2P1,x1-2) = 1/4 );
29
!*******************************************;
! Transition probability calculations for y ;
!*******************************************;
! Transition probabilities for y if y1 = 0;
@for(yD2D1s(y1,GS2P1,CA1P1)|y1#EQ#1 : TPY(y1,GS2P1,CA1P1,y1) = 1 );
! Transition probabilities for y if CA1 = 0;
@for(yD2D1s(y1,GS2P1,CA1P1)|CA1P1 #EQ# 1 : TPY(y1,GS2P1,CA1P1,y1) = 1 );
! Transition probabilities for y if GS2 = 0;
@for(yD2D1s(y1,GS2P1,CA1P1)|GS2P1 #EQ# 1 : TPY(y1,GS2P1,CA1P1,y1) = 1 );
30
! Transition probabilities for y if GS2 >= 1 and CA1 = 1;
@for(yD2D1s(y1,GS2P1,CA1P1)|y1#GE#2 #AND# GS2P1#GE#2 #AND# CA1P1#EQ#2 :
TPY(y1,GS2P1,CA1P1,y1) = 1/2 );
@for(yD2D1s(y1,GS2P1,CA1P1)|y1#GE#2 #AND# GS2P1#GE#2 #AND# CA1P1#EQ#2 :
TPY(y1,GS2P1,CA1P1,y1-1) = 1/2 );
31
! Transition probabilities for y if GS2 = 1 and CA1 = 2;
@for(yD2D1s(y1,GS2P1,CA1P1)|y1#GE#2 #AND# GS2P1#EQ#2 #AND# CA1P1#EQ#3 :
TPY(y1,GS2P1,CA1P1,y1) = 1/4 );
@for(yD2D1s(y1,GS2P1,CA1P1)|y1#GE#2 #AND# GS2P1#EQ#2 #AND# CA1P1#EQ#3 :
TPY(y1,GS2P1,CA1P1,y1-1) = 3/4 );
32
! Transition probabilities for y if GS2 >= 2 and CA1 = 2;
@for(yD2D1s(y1,GS2P1,CA1P1)|y1#GE#3 #AND# GS2P1#GE#3 #AND# CA1P1 #EQ# 3 :
TPY(y1,GS2P1,CA1P1,y1) = 1/4 );
@for(yD2D1s(y1,GS2P1,CA1P1)|y1#GE#3 #AND# GS2P1#GE#3 #AND# CA1P1 #EQ# 3 :
TPY(y1,GS2P1,CA1P1,y1-1) = 1/2 );
@for(yD2D1s(y1,GS2P1,CA1P1)|y1#GE#3 #AND# GS2P1#GE#3 #AND# CA1P1 #EQ# 3 :
TPY(y1,GS2P1,CA1P1,y1-2) = 1/4 );
33
In Model 1:
 GS1t
GS2t
Rt ()  

 1  CA2 1  CA1
t
t





34
In Model 3 and Model 4:
5


2
GS
2
(
GS
)


t
3
1t
3
Rt ()  

2
2 
2  1  (CA2t ) 1  (CA1t ) 




35
!**********************;
! The game problems
!**********************;
;
@for(xyset(x,y):
@for(D2set(D2)|D2 #LE# y:
[ydec] V(x,y,3) <= @sum(D1set(D1):
(GS1(x,y,D1)/(1+CA2(x,y,D2)) - GS2(x,y,D2)/(1+CA1(x,y,D1))
+
@sum(xyset(x2,y2):
TPX(x,GS1(x,y,D1)+1,CA2(x,y,D2)+1,x2)*TPY(y,GS2(x,y,D2)+1,CA1(x,y,D1)+1,y2)*
V(x2,y2,2) ))
*
xdec(x,y,3,D1) )
);
);
36
@for(xynset(x,y,n): @FREE(V(x,y,n)));
end
37
In Model 3 and Model 4:
5


2
GS
2
(
GS
)


t
3
1t
3
Rt ()  

2
2 
2  1  (CA2t ) 1  (CA1t ) 




38
Model 3; (x,y,t) = (2,2,T-1)
GS2 = 0
CA2 = 2
GS2 = 1
CA2 = 1
GS2 = 2
CA2 = 0
GS1 = 0
CA1 = 2
3/2*(0-0)
+
0
=0
3/2*(0 – (5/3)/(1+4)) +
0.25*(0) + 0.75*(+1)
= 0.25
3/2*(0 – (5/3)*2/(1+4))
+
0.25*(0) + 0.5*(1) + 0.25*(2)
=0
GS1 = 1
CA1 = 1
3/2*(1/(1+4)-0)
+
0.25*(0) + 0.75*(-1)
= -0.45
3/2*(1/(1+1) – (5/3)/(1+1))
+
0.5*(0)+0.25*(1) + 0.25*(-1)
= -0.5
3/2*(1/(1+0) – (5/3)*2/(1+1))
+
0.5*(0) + 0.5*(1)
= -0.5
3/2*(4/(1+4) – 0)
+
0.25*(0) + 0.5*(-1) +
0.25*(-2)
= 0.2
3/2*(4/(1+1) – (5/3)/(1+0))
+
0.5*(0) + 0.5*(-1)
=0
3/2*(4/(1+0)-(5/3)*2/(1+0))
+0
=1
GS1 = 2
CA1 = 0
39
Model 3; (x,y,t) = (2,2,T-1)
GS2 = 0
CA2 = 2
GS2 = 1
CA2 = 1
GS2 = 2
CA2 = 0
GS1 = 0
CA1 = 2
0
0.25
0
GS1 = 1
CA1 = 1
-0.45
-0.5
-0.5
GS1 = 2
CA1 = 0
0.2
0
1
40
 11 12


22
 21
 .
.

 m1  m 2
. 1n 

. 2n 
. . 

.  mn 
41
 x1 y1 . x1 yn 
 .

.
.


 xm y1 . xm yn 
42
max E
s.t.
E  11 x1  ...   m1 xm
(against 1 )
E  12 x1  ...   m 2 xm
(against  2 )
.
E  1n x1  ...   mn xm
1
(against  n )
x1  ........  xm
43
min E
s.t.
E  11 y1  ...  1n yn
(against 1 )
E   21 y1  ...   2 n yn
(against  2 )
.
E   m1 y1  ...   mn yn
1
y1  ........  yn
(against  m )
44
Results from Model 1
45
Table 1.1
V(x,y,3) with Model 1
(This (n=3) means that t =T-1 since we use backward recursion
and period T+1 is defined as n=1. )
-4
-1.5
0
-2
0
1.5
0
2
4
46
47
Table 1.2
(GS1*, GS2*) at n=3 with Model 1
(0,2)
(0,2)
(0,0)
(0,1)
(1,1)
(2,0)
(0,0)
(1,0)
(2,0)
48
Table 1.2
(GS1*, GS2*) at n=3 with Model 1
(0,2)
(0,2)
(0,0)
(0,1)
(1,1)
(2,0)
(0,0)
(1,0)
(2,0)
49
50
Table 1.2
(GS1*, GS2*) at n=3 with Model 1
(0,2)
(0,2)
(0,0)
(0,1)
(1,1)
(2,0)
(0,0)
(1,0)
(2,0)
51
52
Table 1.2
(GS1*, GS2*) at n=3 with Model 1
(0,2)
(0,2)
(0,0)
(0,1)
(1,1)
(2,0)
(0,0)
(1,0)
(2,0)
53
54
Table 1.2
(GS1*, GS2*) at n=3 with Model 1
(0,2)
(0,2)
(0,0)
(0,1)
(1,1)
(2,0)
(0,0)
(1,0)
(2,0)
55
56
Model 2
57
In Model 2:
5


2
GS
2t
 (GS1t )

3
Rt ()  

2
2 
1

(
CA
)
1

(
CA
)
2
1


t
t


58
Model 2
!**********************;
! The game problems
!**********************;
;
@for(xyset(x,y):
@for(D2set(D2)|D2 #LE# y:
[ydec] V(x,y,3) <= @sum(D1set(D1):
(GS1(x,y,D1)*GS1(x,y,D1)/(1+CA2(x,y,D2)*CA2(x,y,D2))
- 5/3*GS2(x,y,D2)/(1+CA1(x,y,D1)*CA1(x,y,D1))
+
@sum(xyset(x2,y2):
TPX(x,GS1(x,y,D1)+1,CA2(x,y,D2)+1,x2)*TPY(y,GS2(x,y,D2)+1,CA1(x,y,D1)+1,y2)*
V(x2,y2,2) ))
*
xdec(x,y,3,D1) )
);
);
59
Results from Model 2
60
Table 2.1
V(x,y,3) with Model 2
-5.333
-2.167
0
-2.667
-0.333
2.5
0
2
6
61
Table 2.2
(GS1*, GS2*) at n=3 with Model 2
(0,2) (0,2)
(0,0)
(0,1) (0,1)
(2,0)
(0,0) (1,0)
(2,0)
62
Table 2.2
(GS1*, GS2*) at n=3 with Model 2
(0,2) (0,2)
(0,0)
(0,1) (0,1)
(2,0)
(0,0) (1,0)
(2,0)
63
64
Model 3
65
In Model 3 and Model 4:
5


2
GS
2
(
GS
)


t
3
1t
3
Rt ()  

2
2 
2  1  (CA2t ) 1  (CA1t ) 




66
!**********************;
! The game problems
!**********************;
;
@for(xyset(x,y):
@for(D2set(D2)|D2 #LE# y:
[ydec] V(x,y,3) <= @sum(D1set(D1):
(
3/2*(GS1(x,y,D1)*GS1(x,y,D1)/(1+CA2(x,y,D2)*CA2(x,y,D2))
- 5/3*GS2(x,y,D2)/(1+CA1(x,y,D1)*CA1(x,y,D1)) )
!A constant is added and later removed;
+ (1 +)
@sum(xyset(x2,y2):
TPX(x,GS1(x,y,D1)+1,CA2(x,y,D2)+1,x2)*TPY(y,GS2(x,y,D2)+1,CA1(x,y,D1)+1,y2)*
V(x2,y2,2) ))
*
xdec(x,y,3,D1) )
);
);
67
Results from Model 3
68
Table 3.1
V(x,y,3) with Model 3
-7
-3
0.111
-3.5
-0.75
3.5
0
2.5
8
69
Table 3.2
(GS1*, GS2*) at n=3 with Model 3
(0,2)
(0,2)
(0,0)
(2,0)
(0,1)
(2,1)
(0,1)
(0,1)
(2,0)
(0,0)
(1,0)
(2,0)
Prob = .4444444 * .5555556 =
Prob = .5555556 * .5555556 =
Prob = .4444444 * .4444444 =
Prob = .5555556 * .4444444 =
Sum =
.246914
.308642
.197531
.246914
1.000001
70
Table 3.2
(GS1*, GS2*) at n=3 with Model 3
(0,2)
(0,2)
(0,0)
(2,0)
(0,1)
(2,1)
(0,1)
(0,1)
(2,0)
(0,0)
(1,0)
(2,0)
Prob = .4444444 * .5555556 =
Prob = .5555556 * .5555556 =
Prob = .4444444 * .4444444 =
Prob = .5555556 * .4444444 =
Sum =
.246914
.308642
.197531
.246914
1.000001
71
Table 3.2
(GS1*, GS2*) at n=3 with Model 3
(0,2)
(0,2)
(0,0)
(2,0)
(0,1)
(2,1)
(0,1)
(0,1)
(2,0)
(0,0)
(1,0)
(2,0)
Prob = .4444444 * .5555556 =
Prob = .5555556 * .5555556 =
Prob = .4444444 * .4444444 =
Prob = .5555556 * .4444444 =
Sum =
.246914
.308642
.197531
.246914
1.000001
72
73
Table 3.2
(GS1*, GS2*) at n=3 with Model 3
(0,2)
(0,2)
(0,0)
(2,0)
(0,1)
(2,1)
(0,1)
(0,1)
(2,0)
(0,0)
(1,0)
(2,0)
Prob = .4444444 * .5555556 =
Prob = .5555556 * .5555556 =
Prob = .4444444 * .4444444 =
Prob = .5555556 * .4444444 =
Sum =
.246914
.308642
.197531
.246914
1.000001
74
75
Table 3.2
(GS1*, GS2*) at n=3 with Model 3
(0,2)
(0,2)
(0,0)
(2,0)
(0,1)
(2,1)
(0,1)
(0,1)
(2,0)
(0,0)
(1,0)
(2,0)
Prob = .4444444 * .5555556 =
Prob = .5555556 * .5555556 =
Prob = .4444444 * .4444444 =
Prob = .5555556 * .4444444 =
Sum =
.246914
.308642
.197531
.246914
1.000001
76
77
Table 3.2
(GS1*, GS2*) at n=3 with Model 3
(0,2)
(0,2)
(0,0)
(2,0)
(0,1)
(2,1)
(0,1)
(0,1)
(2,0)
(0,0)
(1,0)
(2,0)
Prob = .4444444 * .5555556 =
Prob = .5555556 * .5555556 =
Prob = .4444444 * .4444444 =
Prob = .5555556 * .4444444 =
Sum =
.246914
.308642
.197531
.246914
1.000001
78
79
Model 4
80
In Model 3 and Model 4:
5


2
GS
2
(
GS
)


t
3
1t
3
Rt ()  

2
2 
2  1  (CA2t ) 1  (CA1t ) 




81
!**********************;
! The game problems
!**********************;
;
@for(xyset(x,y):
@for(D2set(D2)|D2 #LE# y:
[ydec] V(x,y,3) <= @sum(D1set(D1):
(
3/2*(GS1(x,y,D1)*GS1(x,y,D1)/(1+CA2(x,y,D2)*CA2(x,y,D2))
- 5/3*GS2(x,y,D2)/(1+CA1(x,y,D1)*CA1(x,y,D1)) )
+ 1 + .85 *
@sum(xyset(x2,y2):
TPX(x,GS1(x,y,D1)+1,CA2(x,y,D2)+1,x2)*TPY(y,GS2(x,y,D2)+1,CA1(x,y,D1)+1,y2)*
V(x2,y2,2) ))
*
xdec(x,y,3,D1) )
);
);
Note that “discounting” has been introduced!
82
Results from Model 4
83
Table 4.1
V(x,y,3) with Model 4
-6.7
-2.925
0.117
-3.35
-0.825
3.425
0
2.35
7.7
84
Table 4.2
(GS1*, GS2*) at n=3 with Model 4
(0,2)
(0,2)
(0,0)
(2,0)
(0,1)
(2,1)
(0,1)
(0,1)
(2,0)
(0,0)
(1,0)
(2,0)
Prob = .6666667 * .1515152 = .101010
Prob = .3333333 * .1515152 = .050505
Prob = .6666667 * .8484849 = .565657
Prob = .3333333 * .8484849 = .282828
Sum = 1.000000
85
Detailed study of optimal strategies:
( xt , yt , t )  (2, 2, T  1)
Optimal strategies with Model 3
(In Model 3, future results are not discounted. d = 1)
P1:
P2:
Strategy
Probability
Strategy
Probability
Full CA
0.4444444
Full CA
0.5555556
Full GS
0.5555556
50% CA &
50% GS
0.4444444
86
Detailed study of optimal strategies:
( xt , yt , t )  (2, 2, T  1)
Optimal strategies with Model 4
(In Model 3, future results are not discounted. d = 0.85)
P1:
P2:
Strategy
Probability
Strategy
Probability
Full CA
2/3
Full CA
0.1515152
Full GS
1/3
50% CA &
50% GS
0.8484848
87
When the importance of instant results in relation to future
results increases (= when the discounting of future results
increases from 0 to 15%):
P1 increases the probability of instant full CA.
P1 decreases the probability of instant GS.
P2 decreases the probability of instant full CA.
P2 increases the probability of partial CA and partial GS.
88
Observations:
• Even in case there is just one more period of
conflict ahead of us and even if the
participants only have two units available per
participant, we find that the optimal
strategies are mixed.
• The optimal decision frequencies are
affected by the result discounting.
• The different participants should optimally
change the strategies in qualitatively
different ways when the degree of
discounting changes.
89
Observations cont.:
• Differential games in continuous time can not
describe these mixed strategies.
• Furthermore, even if we would replace deterministic
differential games in continuous time by “stochastic
differential games” based on stochastic differential
equations in continuous time, this would not capture
the real situation, since the optimal frequencies are
results of the scales of missions.
• Two resource units used during one time interval will
not give the same result as one resource unit during
two time intervals.
90
Isaacs (1965), in section 5.4, The
war of attrition and attack:
“The realistic execution of the strategies
would comprise of a series of discrete
decisions. But we shall smooth matters
into a continuous process. Such is
certainly no farther from truth than our
assumptions and therefore can be
expected to be as reliable as a stepped
model. It is also more facile to handle
and yields general results more
readily.”
91
Isaacs (1965), page 66:
• ”Similarily, except possibly on singular surfaces, the
optimal strategies will be continuous functions
denoted by ... when they are unique; when they are
not, we shall often assume such continuous
functions can be selected.”
Observation (Lohmander):
• With economies of scale in operations, the optimal
strategies will usually not be continuous functions!
92
V
X=2
4
3
2
1
X=1
0
X=0
0
u
1
2
93
V*
4
3
2
1
0
x
0
1
2
94
u*(x)
4
3
2
1
0
x
0
1
2
95
Isaacs (1965) obtains differential equations that describe the
optimal decisions over time by the two participants:
The system moves according to these equations:

x1  m1  c1 x2

x2  m2  c2 x1
96
The objective function in Isaacs (1965) is linear in the
decision variables and time is continuous:
T
(1


)
x

(1


)
x
dt


2
1

0
97
There is no reason to expect that the
Isaacs (1965) model would lead to
mixed strategies.
• The objective function is linear.
• Furthermore there are no scale effects
in the “dynamic transition”.
98
Washburh (2003), Tactical air
war:
”- It is a curious fact that in no case is a mixed
strategy ever needed; in fact, in all cases
each player assigns either all or none of his
air force to GS.”
(The model analysed by Washburn has very
strong similarities to the model analysed
here. The Washburn model was even the
starting point in the development of this
model. However, the objective function used
by Washburn is linear. Hence, the results
become quite different.)
99
An analogy to optimal control:
The following observations are typical when optimal management
of natural resources is investigated:
• If the product price is a constant or a decreasing function of the
production and sales volume and if the cost of extraction is a
strictly convex function of the extraction volume, the objective
function is strictly concave in the production volume. Then we
obtain a smooth differentiable optimal time path of the resource
stock.
• However, if the extraction cost is concave, which is sometimes
the case if you have considerable set up costs or other scale
advantages, then the optimal strategy is usually “pulse
harvesting”. In this case, the objective function that we are
maximizing is convex in the decision variable. The optimal
stock level path then becomes non differentiable.
100
Technology differences:
• The reason why the two participants should make
different decisions with different strategies is that
they have different “GS-technologies”. These
technologies have different degree of convexity of
results with respect to the amount of resources
used.
• Such differences are typical in real conflicts. They
are caused by differences in equipment and applied
methods. Of course, in most cases we can expect
that the applied methods are adapted to the
equipment and other relevant conditions.
101
• For P1, the instant GS result is strictly
convex in the number of units devoted
to the GS mission.
• For P2, the instant GS result is
proportional to the number of units
devoted to the GS mission.
102
Model 5
103
Model 5 corresponds to Model 3
but the following line is added
just before “end”.
Case 1:
xdec(3,3,3,1) = 1;
Case 2:
xdec(3,3,3,2) = 1;
Case 3:
xdec(3,3,3,3) = 1;
104
Results from Model 5
105
Results if the decisions of P1 are not constrained in
(2,2) when n = 3 (Model 3):
Case 0:
xdec(3,3,3,1) = 0.4444444
(GS1 = 0 and CA1 = 2
or
xdec(3,3,3,3) = 0.5555556 ;
GS1 = 2 and CA1 = 0 )
Effects:
The value of the game
: V(3,3)
= 0.111
The opposition should use a mixed strategy:
GS2 = 0, CA2 = 2 with probability 0.5555556
GS2=1, CA2 = 1 with probability 0.4444444.
106
Results if the decisions of P1 are constrained in
(2,2) when n = 3:
Case 1:
xdec(3,3,3,1) = 1;
(GS1 = 0 and CA1 = 2 )
Effects:
The value of the game decreases: V(3,3)
=0
The decision of the opposition changes: GS2
= 2, CA2 = 0
107
Results if the decisions of P1 are constrained in
(2,2) when n = 3:
Case 2:
xdec(3,3,3,2) = 1;
(GS1 = 1 and CA1 =1 )
Effects:
The value of the game decreases: V(3,3)
= -0.5
The decision of the opposition changes: GS2
= 2, CA2 = 0
108
Results if the decisions of P1 are constrained in
(2,2) when n = 3:
Case 3:
xdec(3,3,3,3) = 1;
(GS1 = 2 and CA1 = 0 )
Effects:
The value of the game decreases: V(3,3)
=0
The decision of the opposition changes: GS2
= 1, CA2 = 1
109
Conclusions from Model 5:
Since we are interested in the best
attainable expected outcome,
we should not restrict our decisions to a
pure strategy and let the opposition
select strategy without such
restrictions!
You should avoid being predictable!
110
Conclusions:
This paper presents a stochastic two person differential
(difference) game model with a linear programming subroutine
that is used to optimize pure and/or mixed strategies as a
function of state and time.
In ”classical dynamic models”, ”pure strategies” are often
assumed to be optimal and deterministic continuous path
equations can be derived. In such models, scale and timing
effects in operations are however usually not considered.
When ”strictly convex scale effects” in operations with defence
and attack (”counter air” or ”ground support”) are taken into
consideration, dynamic mixed OR pure strategies are optimal in
different states.
The optimal decision frequences are also functions of the relative
importance of the results from operations in different periods.
The expected cost of forcing one participant in the conflict to use
only pure strategies is determined.
The optimal responses of the opposition in case one participant in
the conflict is forced to use only pure strategies are calculated.
Dynamic models of the presented type, that include mixed
strategies, are highly relevant and must be used to determine
the optimal strategies. Otherwise, considerable and quite
unnecessary losses should be expected.
111
Contact:
Peter Lohmander
Professor
SLU
SE-901 83 Umea, Sweden
e-mail:
[email protected]
Home page:
http://www.Lohmander.com
112
Source:
Lohmander, P., TAW models 070612.doc
113