CHAPTER 3 COMMAND AND CONTROL (C2), FUZZY RELATION EQUATION
3.6 An Illustrated Example
3.6.4 The Calculation of The Battlefield Model After Adding Examples
In this battlefield example, we would substitute the battlefield model of Fig. 3.2, which according to the above description battle conditions from Chapter 3.6, to do the calculation. The calculation of the model has to use the average battle condition of Anzio war ( M ) at first. Due to the M is the mean value of the battlefield conditions after the calculation of average during all the procession from begin till the end of quarrel (M ,1 M ,…,2 M ); thus, we have to list out all the battle condition one after 20 another. About the beginning battle condition of the Anzio War, M , we would use 1 the method of Chapter 3.3 and Eq. (3.1) to defined each abilities of the landing army.
After the calculation, the beginning situation of battlefield (M ) as follows： 1
Then, according to the description of the Table 3.15, using Chapter 2.2 to calculateM ; Due to the battle condition of 2 M is formed by each element in 2 change from M ; thus, in here, we would take the attacking ability of the panzer 1 division as the sample. And then, using the method of Chapter 2.2.1 to set the rules
of the fuzzy rule base, for its linguistic variables of the frontward and backward part.
Let’s assume the attacking ability of the panzer division has 6 linguistic variables from the frontward part, defined as x ~1 x . Which are tank amount of allied force, 6 infantry amount of allied force, airborne troop amount of allied force, and tank amount of German force, infantry amount of German force, and blockhouse amount of German force; Each linguistic varies of frontward part from each army ability is according to the army, the cooperate fighting friendly force, all the affection from against enemy army, and its cooperation fighting friendly force, to set up the number of varies. However, the linguistic variables of the backward part remain only one, which is the changing attacking ability of the allied force in tanks, defined as y. This paper assume the linguistic variables of the frontward part into five: VS (Very Small), S (Small), M (Medium), B (Big) and VB (Very Big)；But the sentence varies of the backward part into three: NE (Negative), ZO (Zero), and PE (Positive). Also, define their membership function shape according Fig. 3.5 and Fig. 3.6.
Fig. 3.5 Membership functionx ~x
Fig. 3.6 Membership function y
After defining the membership function of the front part and back part, then you can decide the rule of fuzzy manipulation, line up the fuzzy rules：
Rule :1 IF x1 is VS, and x2 is VS, and x3 is VS, and x4 is VS, and x5 is VS, and
x6 is VS, THEN y is ZO.
Rule :2 IF x1 is S, and x2 is S, and x3 is S, and x4 is S, and x5 is S, and x6 is S, THEN y is ZO.
Rule :3 IF x1 is M, and x2 is M, and x3 is M, and x4 is M, and x5 is M, and x6 is M, THEN y is ZO.
Rule :4 IF x1 is B, and x2 is B, and x3 is B, and x4 is B, and x5 is B, and x6 is B, THEN y is ZO.
Rule :5 IF x1 is VB, and x2 is VB, and x3 is VB, and x4 is VB, and x5 is VB, and x6 is VB, THEN y is ZO.
： ： ： ： ： ：
Rule :10 IF x1 is VB, and x2 is S, and x3 is S, and x4 is S, and x5 is VS, and x6 is VS, THEN y is PE.
Rule :11 IF x1 is B, and x2 is B, and x3 is S, and x4 is VS, and x5 is S, and x6 is VS, THEN y is PE.
： ： ： ： ： ：
Rule :n-1 IF x1 is S, and x2 is S, and x3 is S, and x4 is B, and x5 is B, and x6 is M, THEN y is NE.
Rule :n IF x1 is VS, and x2 is VS, and x3 is VS, and x4 is VB, and x5 is VB, and x6 is VB, THEN y is NE.
Finally, let’s go on the fuzzy inference and its defuzzification. We would use the Minimum inference engine of Eq. (2.17) to be the fuzzy inference engine, and the center average method of Eq. (2.19) for the part of the defuzzification. According to the M battlefield condition from the table 3.15, we assume the allied force owned 2 the amount of tanks in 255, infantry in 36000, airborne troops in 10000, and the German force owned the amount of tanks in 103, infantry in 11000, and blockhouse in 3; thus the y is 0.1. M is the calculation solution according to the each ability in 2 changing of the armies.
As Fig. 3.7 shown, when all the battlefield conditions (M ,1 M ,…,2 M ) of campaign 20 formed, we could get its average M .
5 10 15 20
0 0.5 1
Situation of battle field (event)
degree
Panzer division
5 10 15 20
0 0.5 1
Situation of battle field (event)
degree
Battleship
5 10 15 20
0 0.5 1
Situation of battle field (event)
degree
Bomber
5 10 15 20
0 0.5 1
Situation of battle field (event)
degree
Infantry division
5 10 15 20
0 0.5 1
Situation of battle field (event)
degree
Airborne division
5 10 15 20
0 0.5 1
Situation of battle field (event)
degree
Battleplane
Fig. 3.7 All situations of battle field of Anzio
The chief commander of the allied force, Dwight David Eisenhower could use the average battlefield condition ( M ) from the past campaign, with the battle result ( R ) as he except, to decide the sending military amount ( P ) to Normandy campaign.
We could count out R according to Chapter 3.4 and Eq. (3.1).
After the solution of M and R, we could substitute P Mo =R into Chapter 2.3, to count out the sending military amount ( P ) that commander should put into the battlefield.
[ ]
max 1 0.6 0.6 0.5 1 0.7 P =
[ ]
min1 0.7 0.6 0 0 0 0.7 P =
[ ]
min 2 0 0.5 0 0 0 0.6
P =
M
[ ]
min 7 0 0 0 0 0.5 0.7
P =
The above solution of Pmax and Pmin represent what the commander wish to happen in such battle result, the most and the fewest sending amount of the military. It means in between the most and the fewest amount of military force sending into the battlefield, the commander could all achieve his exceptive battle result. The advantage is, since the general campaign broke out, it usually not only has one battlefield, and
to count out the range amount for his sending military force due to his exceptive result.
As the result, he could arrange his own supporting amount in the well organize, further to remain the battle results of all battlefields could all achieve what the commander wish, and avoid to affect the whole war situation due to the improper arrangement of the support to the wrong battlefield. According to the historical record, the chief commander of allied force, Dwight David Eisenhower, sent the military amount for going on the landing battle includes: 10 panzer divisions, 3 airborne divisions, 23 infantry divisions, 1200 battleships, 5800 bombers, and 4900 battle planes [42]. These armies place the degree in each among all the amount of military force:
It’s sure that, P is in between Pmax and Pmin after checking; thus, it means the landing military amount that Dwight David Eisenhower sent, could actually satisfy what he wish as the battle result.
After that, we have to adjust if the allied force has to input support or not. If the support army (C) is a 6-dimension unit matrix, it doesn’t need any support. By using max-min composition, and then the calculation of Eq. (3.3). Finally, according to
'
P Go =R , we could come out the actual battle result ( 'R ) of Normandy Battlefield.
The solution of G and 'R are shown as following:
When we come out the solution R as the actual battle result, and compare with the ' battle result ( R ) that Dwight David Eisenhower except, and then get the value of error ( E ):
' [ 1 2 3 4 5 6] [0.3 0.3 0.2 0.2 0.4 0.0]
E= −R R = e e e e e e =
From the exceptive battle results of the landing war, the commander must requires all the value from E is necessary within the acceptance range of error (ε ). And if it excess the acceptance range of error (ε ), it could caused the failure due to the gap in large in between actual battle result and the commander’s exceptive battle result. If we assume ε =0.2 from this example of battlefield, and check to see if all the values of E ≤ 0.2。And the move error (e3), hit error (e4), supply error (e6), are all within the value of acceptance range of error; But the attacking error (e1), defense error (e2), and hide error (e5), are excess the value of acceptance range of error. It means if the commander didn’t send any support (C), could lead to the failure as the battle result; thus, the commander must input support (C), to ensure the battle result would match the commander’s exceptive battle result.
For ensure the actual battle result is within the value of acceptance range of error (ε ), we have to assume the error ( E ) to be 0. Then, according to P and 'R is known from P Go =R', therefore, use the Eq. (2.27) and Eq. (2.28) to find the range of G : T
max
1 1 1 1 0.7 1 1 1 1 1 0.6 0.6 1 1 1 1 0.7 1 1 1 1 1 0.5 0.5 1 1 1 1 0.7 1 1 1 1 1 0.5 0.5 GT
⎡ ⎤
⎢ ⎥
⎢ ⎥
⎢ ⎥
= ⎢ ⎥
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎣ ⎦
，
min1
0 0 0 0 0.7 0
0 0 0 0 0 0.6
0 0 0 0 0.7 0
0 0.5 0 0 0 0
0 0 0 0 0 0.7
0 0 0.5 0 0 0
GT
⎡ ⎤
⎢ ⎥
⎢ ⎥
⎢ ⎥
= ⎢ ⎥
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎣ ⎦
，………， min
0 0 0 0 0 0.7
0 0 0 0 0 0.6
0 0 0 0 0.7 0
0 0 0.5 0 0 0
0 0 0 0 0 0.7
0 0 0 0.5 0 0
T
G n
⎡ ⎤
⎢ ⎥
⎢ ⎥
⎢ ⎥
= ⎢ ⎥
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎣ ⎦
，
As long as the value of G is in between the range will do. The value of G from this paper is shown as following:
After that, use Eq. (3.2) and Eq.(3.3) to come out Cmax and Cmin, and then to pick up the support amount (C) that Dwight David Eisenhower sent for Normandy War, according to the historical record [40].
max
1 0.6 0.4 0.5 1 0.4 0.5 0.7 0.5 0.5 1 0.5 0.4 0.4 0.6 0.4 0.4 0.5 0.5 0.5 0.5 0.5 1 0.5 , 0.4 0.4 0.4 0.4 0.4 0.4 1 0.6 0.7 0.5 1 0.7 C
⎡ ⎤
⎢ ⎥
⎢ ⎥
⎢ ⎥
= ⎢ ⎥
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎣ ⎦
min1
0.6 0 0 0.5 0 0
0 0.5 0 0 0.5 0
0 0.4 0.6 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0.6 0 0.7 0 0 0
C
⎡ ⎤
⎢ ⎥
⎢ ⎥
⎢ ⎥
= ⎢ ⎥
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎣ ⎦
，……， min
0.6 0 0 0.5 0 0
0 0.5 0 0 0.5 0
0 0.4 0.6 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0.6 0 0.7 0 0 0
C n
⎡ ⎤
⎢ ⎥
⎢ ⎥
⎢ ⎥
= ⎢ ⎥
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎣ ⎦
Finally, use the each battlefield events in changing (N ,1 N ,…,2 N ) in Normandy 20 campaign from Fig. 3.8, to judge if the support (C) is robust enough or not to remain all the errors (E ,1 E ,…,2 E ) from the actual battle result (20 R ,1' R ,…,2' R ) within the 20' value of the acceptance range of error. As the creative errors (E ,1 E ,…,2 E ) from the 20 actual battle results (R ,1' R ,…,2' R ) as shown in Fig. 3.9 ~ Fig. 3.14 and Fig. 3.15 ~ 20' Fig. 3.20.
5 10 15 20
0 0.5 1
Situation of battle field (event)
degree
Fuel
5 10 15 20
0 0.5 1
Situation of battle field (event)
degree
Equipment
5 10 15 20
0 0.5 1
Situation of battle field (event)
degree
Battle Machinery
5 10 15 20
0 0.5 1
Situation of battle field (event)
degree
Provision
5 10 15 20
0 0.5 1
Situation of battle field (event)
degree
Medical Treatment
5 10 15 20
0 0.5 1
Situation of battle field (event)
degree
Staff
Fig. 3.8 All support situations of battle field of Normandy
2 4 6 8 10 12 14 16 18 20
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Situation of battle field (event)
degree
Fig. 3.9 Combat results of attack of Normandy
2 4 6 8 10 12 14 16 18 20
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Situation of battle field (event)
degree
Fig. 3.10 Combat results of defense of Normandy
2 4 6 8 10 12 14 16 18 20 0
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Situation of battle field (event)
degree
Fig. 3.11 Combat results of move of Normandy
2 4 6 8 10 12 14 16 18 20
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Situation of battle field (event)
degree
Fig. 3.12 Combat results of hit of Normandy
2 4 6 8 10 12 14 16 18 20 0
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Situation of battle field (event)
degree
Fig. 3.13 Combat results of hide of Normandy
2 4 6 8 10 12 14 16 18 20
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Situation of battle field (event)
degree
Fig. 3.14 Combat results of supply of Normandy
2 4 6 8 10 12 14 16 18 20 -0.5
-0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5
Situation of battle field (event)
value
Fig. 3.15 Attack error of combat results
2 4 6 8 10 12 14 16 18 20
-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5
Situation of battle field (event)
value
Fig. 3.16 Defense error of combat results
2 4 6 8 10 12 14 16 18 20 -0.5
-0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5
Situation of battle field (event)
value
Fig. 3.17 Move error of combat results
2 4 6 8 10 12 14 16 18 20
-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5
Situation of battle field (event)
value
Fig. 3.18 Hit error of combat results
2 4 6 8 10 12 14 16 18 20 -0.5
-0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5
Situation of battle field (event)
value
Fig. 3.19 Hide error of combat results
2 4 6 8 10 12 14 16 18 20
-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5
Situation of battle field (event)
value
Fig. 3.20 Supply error of combat results
As we could find from the creative results from Fig. 3.9 till Fig. 3.20, the commander would just need to send the C amount of support of matrix into the battlefield, no matter how the battle condition changing in Normandy, it could always make the error
( E ) from the actual battle result ( 'R ) and the commander’s exceptive battle result ( R ) to remain in the value of acceptance range of error (ε ).