A TPZSGD will now be analyzed in the most general way. BLUE is the maximizer, who selects the row, i. RED is the minimizer, who selects the column, j. The decision of BLUE is not known by RED before RED takes a decision and the decision of RED is not known by BLUE before BLUE takes the decision. The game matrix, A(i,j), is diagonal. All diagonal elements cij=A(i,j) are strictly positive and represent the reward that BLUE obtains from RED in case i=j. "('The reward that BLUE obtains is equal to the loss that RED gets")". In case i≠j, the reward is zero. Equations (2.1) and (2.2) define these conditions.
(2.1)
(2.2)
A concrete example is the following: RED should move an army convoy from one city to another. One road, among the existing n available roads, should be selected. BLUE wants to destroy as many RED trucks as possible. RED sends the convoy via road j and BLUE moves the equipment and troops to road i and prepares an attack there. If i=j, BLUE attacks RED and destroys the number of RED trucks found in a diagonal element of the game matrix where i=j.
. If BLUE and RED select different roads, no attack takes place and no trucks are destroyed.
.
Different roads usually have different properties with respect to slope, curvature, protection, options to hide close to the road and so on. As a consequence, the values of the diagonal elements of the game matrix,
, are usually not the same for different values of i.
The maximization problem of BLUE
The maximization problem of BLUE is defined here. The expected reward, X0, is the objective function, which is found in (2.1.1). The number of possible decisions is n and the probability of a particular decision, i, is Xi. The total probability can not exceed 1, which is shown in (2.1.2.).gi is defined in (2.2). Since RED can select any decision , j,x0 is constrained via (2.1.3). Furthermore, no probability can be negative, which is seen in (2.1.4).
(2.1.1)
s.t.
(2.1.2)
(2.1.3)
(2.1.4)
Let
denote dual variables. The following Lagrange function is defined:
(2.1.5)
The following derivatives will be needed in the proceeding analysis:
(2.1.6)
(2.1.7)
(2.1.8)
(2.1.9)
Karush Kuhn Tucker conditions in general problems
In general problems, we may have different numbers of decision variables and constraints. Furthermore, the elements
are not necessarily zero (Table 1).
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Table 1 Karush Kuhn Tucker conditions in general maximization problems
Particular conditions in problems that satisfy (2.1) and (2.2)
Note that in these problems, i=j in all relevant constraints.
(2.1.10)
(2.1.11)
(2.1.12)
(2.1.13)
(2.1.14)
(2.1.15)
Proof 1: Proof that
:
(2.1.2) and (2.1.4) make it feasible to let
.
(2.2) says that
.
When
, (2.1.3) makes it feasible to let .
(2.1.1) states that we want to maximize x0. Let stars indicate optimal values.
Hence, when optimal decisions are taken,
.
Proof 2: Proof that
:
(2.1.7) says that
Proof 1 states that
. (2.2) says that
.
.
Hence,
.
Proof 3: Proof that
can be determined from a linear equation system.
(2.1.15)
.
(2.1.16)
(2.1.17)
Proof 4: Proof that
.
(2.1.16)
.
Hence, at least for one strictly positive value i,
is strictly greater than zero.
(2.1.17)
.
(2.1.18)
(2.1.17)
(2.1.18)
(2.1.19)
(2.1.18)
(2.1.19)
(2.1.20)
Proof 5: Proof that
, can be determined from a linear equation system.
(2.1.12)
.
(2.1.21)
(2.1.22)
Determination of explicit equations that give all values:
:
(2.1.22)
(2.1.23).
(2.1.23)
(2.1.21)
(2.1.24).
(2.1.24)
(2.1.25)
(2.1.26)
(2.1.27)
(2.1.28)
(2.1.29)
Determination of explicit equations that give all values:
:
(2.1.17)
(2.1.30).
(2.1.30)
(2.1.16)
(2.1.31)
(2.1.31)
(2.1.32)
(2.1.33)
(2.1.34)
(2.1.35)
(2.1.36)
Observations:
(2.1.37)
(2.1.38)
The minimization problem of RED
We are interested in the solution to
. The objective function is formulated as
. The frequences of the different decisions, i are
.
(2.2.1)
s.t.
(2.2.2)
(2.2.3)
(2.2.4)
Proof that
(2.2.2)
(2.2.5).
(2.2.5)
(2.2.6)
(2.2.3)
(2.2.5)
(2.2.6)
(2.2.7).
(2.2.7)
Let
denote dual variables. The following Lagrange function is defined for RED:
(2.2.8)
These derivatives will be needed in the analysis:
(2.2.9)
(2.2.10)
(2.2.11)
(2.2.12)
Proof that
According to (2.2.1), we want to maximize
, which implies that we minimize
.
(2.2.2)
(2.2.4)
Let us start from an infeasible point, origo, and move to a feasible point in the way that keeps
as low as possible. Initially, let
. According to (2.2.2), this point is not feasible.
(2.2.3)
.
Now, we have to move away from the infeasible point
. We have to reach a point that satisfies
without increasing y0 more than necessary. To find a point that satisfies (2.2.2), we have to increase the value of at least one of the
. Select one arbitrary index
. To simplify the exposition, we let k=1. According to (2.2.3): If we increase y1 by dy1, increases by g1dy1, as long as
. Hence,
. Let
.
However, when
, we may also partly increase
without increasing
above z. This follows from (2.2.3) and (2.2.10). Since we want to satisfy , we want to increase as much as possible, without increasing dyemabove . Hence, we select:
(2.2.13)
(2.2.14)
(2.2.15)
Since we started in origo, we have
(2.2.16)
We already know that
. Hence,.
(2.2.17)
Observation: The following direct method can be used to solve the optimization problem of RED.
First, remember that
. We may directly determine the optimal values of
without using the Lagrange function and KKT conditions, in this way:
(2.2.18)
(2.2.19)
(2.2.20)
(2.2.21)
(2.2.22)
(2.2.23)
(2.2.24)
(2.2.25)
Proof that
can be solved via a linear equation system and that
.
Since
, we may determine that
via a linear equation system.
(2.2.26)
(2.2.27)
(2.2.26)
(2.2.28)
(2.2.27)
(2.2.28)
(2.2.29)
(2.2.27)
(2.2.29)
(2.2.30)
(2.2.29)
(2.2.30)
(2.2.31)
Proof that
can be solved via a linear equation system and that
.
Since
, we may determine that
via a linear equation system.
(2.2.32)
(2.2.33)
(2.2.32)
(2.2.34)
(2.2.33)
(2.2.35)
(2.2.35)
(2.2.36)
(2.2.35)
(2.2.36)
(2.2.37)
Determination of explicit equations that give all values:
:
(2.2.33)
(2.2.38).
(2.2.38)
(2.2.32)
(2.2.39).
(2.2.39)
(2.2.40)
(2.2.41)
(2.2.42)
(2.2.43)
(2.2.44)
Determination of explicit equations that give all values:
:
(2.2.27)
(2.2.45).
(2.2.45)
(2.2.26)
(2.2.46)
(2.2.46)
(2.2.47)
(2.2.48)
(2.2.49)
(2.2.50)
(2.2.51)
Observations:
(2.2.52)
(2.2.53)
Generalized Observations:
(2.2.54)
(2.2.55)
Sensitivity analyses
First, the sensitivity analyses will concern these variables:
. How do these variables change under the influence of changing elements in the game matrix?
Observation:
Proof that
.
(2.3.1)
(2.3.2)
(2.3.3)
(2.3.4)
(2.3.5)
(2.3.6)
(2.3.7)
Observation:
is a strictly increasing and strictly concave function of each gi. From the Jensen inequality, it follows that increasing risk in gi will reduce the expected value of
. Compare Figure 1.
Figure 1 In this graph, the horizontal axes represents E(g1), the expected value of g1. Here, g1 is a stochastic variable. There are two possible outcomes, namely
and
, with probabilities ½ and ½ respectively. The vertical axes shows
, the optimal objective function value as a function of the expected value of g1, and
, the expected value of the optimal objective function value of
as a function of the value of g1. The graph also includes a linear approximation of
based on the values of
for
and for
. This linear approximation is equal to
for
. According to the Jensen inequality,
<
, when
is a strictly concave function and g1 is a stochastic variable. This graph illustrates that the Jensen inequality is correct. The graph also illustrates the general conclusion that the expected optimal objective function value
is a strictly decreasing function of the level of risk in g1.
Second, the sensitivity analyses will concern these variables:
. How do these variables change under the influence of changing elements in the game matrix?
Observation:
Proof that
.
(2.3.8)
(2.3.9)
(2.3.10)
(2.3.11)
(2.3.12)
(2.3.13)
(2.3.14)
(2.3.15)
(2.3.16)
(2.3.17)
(2.3.18)
(2.3.19)
(2.3.20)
Observation:
is a strictly decreasing and strictly convex function of gi. From the Jensen inequality, it follows that increasing risk in gi will increase the expected value of
. Compare Figure 2.
Figure 2 In this graph, the horizontal axes represents
, the expected value of g1. Here, g1 is a stochastic variable. There are two possible outcomes, namely
and
, with probabilities ½ and ½ respectively. The vertical axes shows the optimal decision frequency
as a function of the expected value of g1, and
, the expected value of the optimal frequency
as a function of the value of . The graph also includes a linear approximation of
based on the values of
for
and for
. This linear approximation is equal to
for
. According to the Jensen inequality, >
, when
is a strictly convex function and g1 is a stochastic variable. This graph illustrates that the Jensen inequality is correct. The graph also illustrates the general conclusion that the expected optimal decision frequency
is a strictly increasing function of the level of risk in g1.
Proof that
.
(2.3.21)
(2.3.22)
(2.3.23)
(2.3.24)
(2.3.25)
(2.3.26)
(2.3.27)
(2.3.28)
Observation:
is a strictly increasing and strictly concave function of gi. From the Jensen inequality, it follows that increasing risk in gi will decrease the expected value of
. Compare Figure 3.
Figure 3 In this graph, the horizontal axes represents
, the expected value of g1. Here, g1 is a stochastic variable. There are two possible outcomes, namely
and
, with probabilities ½ and ½ respectively. The vertical axes shows the optimal decision frequency
as a function of the expected value of g1, and
, the expected value of the optimal frequency
as a function of the value of g1. The graph also includes a linear approximation of
based on the values of
for
and for
. This linear approximation is equal to
for
. According to the Jensen inequality, <
, when
is a strictly concave function and g1 is a stochastic variable. This graph illustrates that the Jensen inequality is correct. The graph also illustrates the general conclusion that the expected optimal decision frequency
is a strictly decreasing function of the level of risk in g1.