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<article xmlns:xlink="http://www.w3.org/1999/xlink">
  <front>
    <journal-meta />
    <article-meta>
      <title-group>
        <article-title>The Project of Information System for Students Knowledge Evaluation</article-title>
      </title-group>
      <contrib-group>
        <contrib contrib-type="author">
          <string-name>Sergiy Chernov</string-name>
          <xref ref-type="aff" rid="aff0">0</xref>
        </contrib>
        <contrib contrib-type="author">
          <string-name>Sergiy Titov</string-name>
          <xref ref-type="aff" rid="aff0">0</xref>
        </contrib>
        <contrib contrib-type="author">
          <string-name>Ludmila Chernova</string-name>
          <xref ref-type="aff" rid="aff0">0</xref>
        </contrib>
        <contrib contrib-type="author">
          <string-name>Nataliia Kunanets</string-name>
          <xref ref-type="aff" rid="aff0">0</xref>
        </contrib>
        <contrib contrib-type="author">
          <string-name>Lubava Chernova</string-name>
          <xref ref-type="aff" rid="aff0">0</xref>
        </contrib>
        <contrib contrib-type="author">
          <string-name>Evgeniy Trushliakov</string-name>
          <email>eugeniy.trushliakov@nuos.edu.ua</email>
          <xref ref-type="aff" rid="aff0">0</xref>
        </contrib>
        <contrib contrib-type="author">
          <string-name>Pavlo Fedorka</string-name>
          <xref ref-type="aff" rid="aff0">0</xref>
        </contrib>
        <aff id="aff0">
          <label>0</label>
          <institution>Admiral Makarov National University of Shipbuilding</institution>
          ,
          <addr-line>Heroiv Ukrainy ave 9, Mykolaiv, 54025</addr-line>
          ,
          <institution>Ukraine Lviv Polytechnic National University</institution>
          ,
          <addr-line>Stepana Bandery Street 32-a, Lviv, 79013</addr-line>
          ,
          <institution>Ukraine Uzhgorod National University</institution>
          ,
          <addr-line>University Street 14, Uzhgorod 88000</addr-line>
          ,
          <country country="UA">Ukraine</country>
        </aff>
      </contrib-group>
      <abstract>
        <p>The paper provides an analysis of using the bimatrix game algorithm in designing an information system for students' knowledge evaluation. The information system function tools are based on the theory of matrix games resting upon analysis of the result of a conflict of two players - a teacher and a student whose interests were opposite to some extent. Comparison of the obtained results of solving matrix games with results of solving a bimatrix game confirms the possibility of justifying not only quantitative results on the players' averagebuwtainlsso their qualitative behavior.</p>
      </abstract>
      <kwd-group>
        <kwd>1 Bimatrix game algorithm</kwd>
        <kwd>information system</kwd>
        <kwd>students</kwd>
        <kwd>knowledge evaluation</kwd>
      </kwd-group>
    </article-meta>
  </front>
  <body>
    <sec id="sec-1">
      <title>-</title>
      <p>Introduction
game</p>
      <p>The researchers note that approbation of the proposed system confirms its efficiency in usage of the
accumulative evaluation model.</p>
      <p>Thus, they have proposed a model of an online evaluation of knowledge consisting of a series of marks
linearly combined in a chain-like structure. This approach is aimed at orientation towards continuous knowledge
level improvement.</p>
      <p>After each automatic evaluation, a student receives a notice on the level of each educational target
achievement, as well as tasks to be completed for successful further evaluation and repeated evaluation of
previous educational topics.</p>
      <p>Thus, usage of the developed system enabling a combination of different types of questions, adaptation
between evaluations and individual feedback after evaluation for each educational target, contributes to
increasing the level of students’ knowledge[7,8].</p>
      <p>The function tools of the system allow imitating the traditional evaluation system, and receiving a positive
mark requires completing 50% of tasks.</p>
      <p>Maier, Wolf und Randler [9] have analyzed the importance of feedback proposing to arrange it with use of
computer multilevel tests, when their utility is acknowledged.</p>
      <p>Researchers Adebayo, E. I. &amp; Wokocha [10] convince that an information system for an educational
institution control requires continuous support after implementation. They have studied industries where
knowledge control information systems were created for maintaining the competitive advantage [11,12].</p>
      <p>Mukhneri Mukhtar, Sudarmi Sudarmi, Mochamad Wahyudi and Burmansah [13] have completed a research
for determining requirements to a knowledge control information system in a higher school having analyzed the
experience of using information technology for creation of the mentioned systems.</p>
      <p>Janis Grundspenkis [14] has generalized the experience of developing a knowledge evaluation information
system, particularly a system of knowledge evaluation procedures automation.</p>
      <p>The author notes that the knowledge level evaluation can be automated in accordance with Bloom taxonomy
with selecting a format replies and/or solutions submitted by students. The researcher justifies usage of concept
cards for knowledge evaluation and analyzes the specifics of the function tools of the developed adaptive
intellectual knowledge evaluation system.</p>
      <p>Mohamed AF Ragab, Amr Arisha [15] note that the effectiveness of knowledge management is determined
by the possibility of evaluating knowledge based on the constructions of individual knowledge. The information
on the conceptualization of individual knowledge and the characteristics of knowledge carriers has been
analyzed.</p>
      <p>The proposed information system for evaluating knowledge from several points of view, using a
multidimensional system of theoretically based indicators, which creates an Individual Knowledge Index. The
index is calculated using a unique mathematical formula that combines multi-criteria decision analysis methods
to consolidate results.</p>
      <p>The implementation of the technology allows to fully automate the evaluation process and helps to solve
parametric multiplicity and arithmetic complexity. The authors proposed a complex integrated system of
individual assessment of knowledge, in which knowledge is considered as a personal and humanistic concept
[16].
3</p>
      <p>Statement of basic material</p>
      <p>The information system function tools are based on the theory of matrix games resting upon analysis of the
result of a conflict of two players – a teacher and a student whose interests were opposite to some extent.</p>
      <p>For development of an information system function algorithm, let us examine a conflict between two players
С A and D B . Player C may follow his own strategies that we give in the following tuple: .
Player D selects one of the strategies that we give in the following tuple: .</p>
      <p>Both the players are participants of the game with known rules and a combinatory selection receives the
reward. In the case when player C inclines to strategy
. The win of player D shall be . In this case
and player D to strategy , we can present win C as</p>
      <p>, each of the players expects to get a reward.</p>
      <p>Player A can adopt two strategies A1 or A2 to trust the student or to check his statements on the
studied theoretical sections of the game theory discipline. The second player B (the student) has two
strategies too. B1 - to prepare a theoretical course for examination or not to prepare. The teacher may
check or not check the student’s theoretical knowledge directly during the examination. Let’s assum
that the value of win both for the teacher and for the student is conditionally equal to a  0 units.
Therefore, payment matrixes of the players look as follows:
a 
 .</p>
      <p>
        If the teacher is sure that the student has not mastered the theoretical course, he would check the
student’s knowledge(location in payment matrixes (
        <xref ref-type="bibr" rid="ref1 ref2">1,2</xref>
        )). The student would study the course if he is
sure that the teacher would check his knowledge (location (
        <xref ref-type="bibr" rid="ref1 ref1">1,1</xref>
        )). And the other way round, the
teacher does not need to check the student if he has mastered the course (location (
        <xref ref-type="bibr" rid="ref1 ref2">2,1</xref>
        ). Finally, the
student would not master the course if he is sure that the teacher would not check his knowledge
(location (
        <xref ref-type="bibr" rid="ref2 ref2">2,2</xref>
        )).
      </p>
      <p>Model examples of a bimatrix game solution in mixed strategies.</p>
      <p>Let us consider the features of a conceptual bimatrix game if the players’ payment matrixes are:
r
CA  
s
r  s</p>
      <p> ,
0 </p>
      <p> t
CB  
t  s
s </p>
      <p> .
0</p>
      <p>Let us have X A   p1, p2    p,1  p, X B  q1, q2   q,1  q. In this case, the
players’ wins average values will be:</p>
      <p>2 2
M A  X A, X B     aij piq j  (r  s  2qs) p  sq ,</p>
      <p>2 2</p>
      <p>M B  X A, X B    bij piq j  (s  t  2 ps)q  sp .</p>
      <p>
        The game equilibrium description system (
        <xref ref-type="bibr" rid="ref1">1</xref>
        ) looks as follows:
i1 j1
i1 j1
(r  s  2qs)(1  p)  0,

(2qs  r  s) p  0,

(s  t  2 ps)(1  q)  0,
(2 ps  t  s)q  0.
      </p>
      <p> t  s s  t 
X A   2s , 2s </p>
      <p>,
 r  s s  r </p>
      <p>X B   2s , 2s  ,</p>
      <p>Solving the system we obtain frequencies of using strategies by the players and their respective
wins:</p>
      <p>The existence of equilibrium in bimatrix game pure strategies does not exclude the existence of
equilibrium in mixed strategies.</p>
      <p>Let the added utility of joint using computers of the same hardware platform be higher than the
utility of acquiring the known platform, i.e. s  r and s  t .</p>
      <p> xA  M A  X A, X B  
 xB  M B  X A, X B  </p>
      <p>
        (
        <xref ref-type="bibr" rid="ref2">2</xref>
        )
r  s
      </p>
      <p>2
s  t
2
In this case, we the first player’s feedback
The second player would have the reaction

0, q 


p  0,1, q 


1, q 

t  s</p>
      <p>2s
t  s</p>
      <p>2s
t  s</p>
      <p>2s

0, p 


q  0,1, p 


1, p 

r  s
2s
r  s
2s
,
,
square  p, q | p [0,1], q [0,1] (Fig. 1). Fig. 1 shows the
available three equilibrium points two of which are realized in pure strategies  A1, B2  with game
values 1A  r  s and 1B  s і  A2, B1  with values  2A  s and  2B  t  s plus one point
in
mixed
strategies
for
the
first
player
with
frequency
vector
 t  s s  t 
X A   2s , 2s  and value  xA 
r  s
2</p>
      <p> r  s s  r 
and for the second one - X B   2s , 2s  ,
2
No. 2. Let us consider example No. 2 of a conceptual bimatrix game.</p>
      <p>Payment matrixes of the players will be:</p>
      <p>
        i1 j1
The game equilibrium description system (
        <xref ref-type="bibr" rid="ref1">1</xref>
        ) looks as follows:
      </p>
      <p>2 2
M A  X A, X B     aij piq j  ap(1  2q) ,</p>
      <p>i1 j1
2 2
M B  X A, X B    bij piq j  a(1  q)(1  2 p) .</p>
      <p>a(1  p)(1  2q)  0,

ap(1  2q)  0,

a(1  q)(1  2 p)  0,
aq(1  2 p)  0.</p>
      <p>Having solved the system, we obtain the frequencies of using strategies by the players and their
respective wins:</p>
      <p> 1 1 
X A   2 , 2  ,  xA  M A  X A, X B   0 ,</p>
      <p> 1 1 </p>
      <p>X B   2 , 2  ,  xB  M B  X A, X B   0 .</p>
      <p>We have the first player’s feedback
0, q  1 , q
 2
p  0,1, q  12 , 1
1, q  1 .</p>
      <p> 2  1 , 1 
The second player will have the following reaction: 1  2 2 
0, p  1 , 2
 2
q  0,1, p  1 ,
1, p  12 . 12 1
 2 Fig. 2 One equilibrium
Let’s represent graphically the players’ reaction on squarpeoint in mixed strategies.
 p, q | p [0,1], q [0,1] Fig. 2, which shows the presence</p>
      <p> 1 1 
of equilibrium point  ,  in mixed strategies for the first
 2 2 </p>
      <p> 1 1 
player with frequency vector X A   ,  and value  xA  0 and for the second player
 2 2 
 1 1 
X B   ,  ,  xB  0 .</p>
      <p> 2 2 
No. 3 The bimatrix game is given by payment matrixes</p>
      <p>8 8 5 4 7 2
CA  8 1 8 , CB  9 5 1 .</p>
      <p>2 6 6 1 4 8
Let the first and the second players’ mixed solutions frequency vectors be:</p>
      <p>X A   p1, p2 , p3    p1, p2 ,1  p1  p2 , X B  q1, q2 , q3    q1, q2,1 q1  q2 .</p>
      <p>In this case, the game value for player A :
p
(7q1  3q2  1) p1  (4q1  7q2  2) p2  0,

(7q1  3q2  1)( p1  1)  (4q1  7q2  2) p2  0,
(7q1  3q2  1) p1  (4q1  7q2  2)( p2  1)  0,

(9 p1  15 p2  7)q1  (9 p1  8 p2  4)q2  0,
(9 p1  15 p2  7)(q1  1)  (9 p1  8 p2  4)q2  0,

(9 p1  15 p2  7)q1  (9 p1  8 p2  4)(q2  1)  0.</p>
      <p>
        The solution of (
        <xref ref-type="bibr" rid="ref4">4</xref>
        ) is:
(
        <xref ref-type="bibr" rid="ref4">4</xref>
        )
For calculation of the problem solution, we compose a system of conditions (
        <xref ref-type="bibr" rid="ref1">1</xref>
        )
      </p>
      <p>Each point gives its win to the players. With account taken of the presence of equilibrium in the
mixed strategies, we need to ground our selection of either of the equilibrium points – the game has
equilibrium both in pure and in mixed strategies (when one doesn’t know where to look first).</p>
      <p>When an information system is designed, an algorithm is used, which interprets a bimatrix game as
two matrix ones.</p>
      <p>Bimatrix games are used when it is necessary to ground the choice of the optimum players’
behavior. For making a thorough analysis, we conditionally divide a bimatrix game into two matrix
ones.</p>
      <p>Let us consider a bimatrix game and divide it into two matrix antagonistic games with zero sums
(Fig. 2). In turn, we’ll consider andeacsholovfethem dually, as a primal and then as a dual linear
optimization problem.</p>
    </sec>
    <sec id="sec-2">
      <title>Matrix game</title>
      <p>CA  aij mn</p>
    </sec>
    <sec id="sec-3">
      <title>Bimatrix game</title>
      <p>CA  aij mn
CB  bij mn</p>
    </sec>
    <sec id="sec-4">
      <title>Matrix game</title>
      <p>CB  bij mn</p>
    </sec>
    <sec id="sec-5">
      <title>Primal</title>
      <sec id="sec-5-1">
        <title>WIBprobBxlemmax,</title>
        <p>IB : X BmCB  VBx ,</p>
      </sec>
    </sec>
    <sec id="sec-6">
      <title>Primal</title>
      <sec id="sec-6-1">
        <title>WIAprobAxlemmax,</title>
        <p>IA : X AmCA  VAx ,</p>
      </sec>
    </sec>
    <sec id="sec-7">
      <title>Dual problem</title>
      <sec id="sec-7-1">
        <title>WIAI  Ay  min,</title>
        <p>IAI : CA YAm T  V y ,
A</p>
      </sec>
    </sec>
    <sec id="sec-8">
      <title>Dual problem</title>
      <sec id="sec-8-1">
        <title>WIBI  By  min,</title>
        <p>IBI : CB YBm T  V y ,</p>
        <p>B</p>
        <p>Thus, we have considered a bimatrix game given by payment matrixes CA , CB of two players A
and B .</p>
        <p>CA  aij nn , CB  bij nn .</p>
        <p>
          First, let’s consider the matrix game for the payment matrix Ao.fWpelafyinedr the lower
 A  max min CA and the upper value of the game  A  min max CA . We’ll solve the matrix
game by reducing it to linear optimization problems. The primal problem is recorded as follows:
WIA  Ax  max,
IA : X AmCA  VAx ,
where X Am   x1A, x2A, , xnA  is a vector of frequencies used by the first player A for
(
          <xref ref-type="bibr" rid="ref5">5</xref>
          )
application
of
his
own
        </p>
        <p>strategies
x1A  x2A 
 xnA  1)
A1, A2 ,
, An ,( xiA 0,1 ,
,n ,
VAx   Ax n1 - the right-hand members column vector,  Ax - the game value.</p>
        <p>
          For convenience of calculation, we reduce problem (
          <xref ref-type="bibr" rid="ref5">5</xref>
          ) to the standard linear optimization problem
form
        </p>
        <p>W*A  x*A  x*A </p>
        <p>I 1 2
 x*A  min,</p>
        <p>n
*A : X *mC</p>
        <p>I A</p>
        <p>A</p>
        <p> 1,
x*A 
i
x A</p>
        <p>i , i  1, 2,
 x</p>
        <p>A</p>
        <p>, n.
where</p>
        <p>X *m   x*A, x*A,</p>
        <p>A  1 2
, x*A </p>
        <p>n 
( x1*A  x*A 
2
 x*A 
n</p>
        <p>1
 x</p>
        <p>A</p>
        <p>)</p>
        <p>In the result of calculating the primal linear optimization problem, we obtain frequency vector
X Am   x1A, x2A,
, x A  and game value  x .</p>
        <p>
          n  A
The following problem will be a dual linear optimization problem to (
          <xref ref-type="bibr" rid="ref5">5</xref>
          ):
his
own
        </p>
        <p>strategies
y A  y A 
1 2
 y A  1)</p>
        <p>n
where YAm   y1A, y2A,
, y A  is a vector of frequencies used by the second player B for
n 
B , B , , B  ,
1 2 n
( y A 0,1 ,
i
i  1, 2,
V y   Ay n1 - the right-hand members column vector,  Ay - game value.</p>
        <p>
          A
We reduce problem (
          <xref ref-type="bibr" rid="ref6">6</xref>
          ) to the standard presentation form
        </p>
        <p>W*A  y A  y A </p>
        <p>II 1 2
 y A  min,</p>
        <p>n
*A : C Y *m T  1,</p>
        <p>I I A A
y*A 
i
y A</p>
        <p>i , i  1, 2,
 y</p>
        <p>A</p>
        <p>, n,
where</p>
        <p>Y *m   y*A, y*A,</p>
        <p>A  1 2
, y*A </p>
        <p>n 
( y1A  y2A 
 y A 
n</p>
        <p>1
 y</p>
        <p>A</p>
        <p>
          )
The calculation of problem (
          <xref ref-type="bibr" rid="ref7">7</xref>
          ) has resulted in frequency vector YAm   y1A, y2A,
, y A  and
n 
value  y . We proceed to consideration of a matrix game for payment matrix CB of the second player
        </p>
        <p>A
B . Let’s determine the game value by calculahteiolnowoefr t B  max min CB and the upper
game value  B
 min max C . We solve the matrix game by reducing it to linear optimization</p>
        <p>B
problems. The primal problem for the second player payment matrix is recorded as follows:
is
a
normalized
frequency
vector.</p>
        <p>W B  x  max,</p>
        <p>I B
B : X mC</p>
        <p>I B
 V x ,</p>
        <p>
          B
(
          <xref ref-type="bibr" rid="ref6">6</xref>
          )
(
          <xref ref-type="bibr" rid="ref7">7</xref>
          )
(
          <xref ref-type="bibr" rid="ref8">8</xref>
          )
where
        </p>
        <p>X *m   x1*B , x2*B ,</p>
        <p>B</p>
        <p>, xn*B 
( x1*B  x*B 
2
 x*B </p>
        <p>n</p>
        <p>The primal linear problem
X Bm   x1B , x2B , , xnB  and the game value  Bx .</p>
        <p>
          The following problem will be a dual linear optimization problem to (
          <xref ref-type="bibr" rid="ref8">8</xref>
          )
calculation resulted in
is
a
normalized
frequency
vector,
his
own
        </p>
        <p>strategies
y1B  y2B </p>
        <p> ynB  1)
VBy   By n1 - the right-hand members column vector,  By - the game value.</p>
        <p>
          We reduce (
          <xref ref-type="bibr" rid="ref10">10</xref>
          ) to the standard form of a linear optimization problem presentation
where YBm   y1B , y2B ,
, ynB  is a vector of frequencies used by the second player B for
B1, B ,
2
, Bn ,
( yiB 0,1 ,
i  1, 2,
,n ,
own
, xnB  is a vector of frequencies used by the second player B for
VBx   Bx n1 - the right-hand member column vector,  Bx - the game value.
        </p>
        <p>
          For convenience of calculation, we reduce problem (
          <xref ref-type="bibr" rid="ref8">8</xref>
          ) to the standard linear optimization problem
form
        </p>
        <p>W*B  x*B  x*B </p>
        <p>I 1 2
 x*B  min,</p>
        <p>n
*IB : X B*mCB  1,
x*B 
i
xB</p>
        <p>i , i  1, 2,
 x</p>
        <p>B
1</p>
        <p>)
 x</p>
        <p>B
optimization
WIBI  y  min,</p>
        <p>B
IBI : C Y m T  V y ,</p>
        <p>B B B
, n.</p>
        <p> ynB  min,
*IIB : C Y *m T  1,</p>
        <p>B B
y*B 
i
y B</p>
        <p>i , i  1, 2,
 y</p>
        <p>B
, n,
frequency</p>
        <p>
          vector
(
          <xref ref-type="bibr" rid="ref10">10</xref>
          )
(
          <xref ref-type="bibr" rid="ref9">9</xref>
          )
(
          <xref ref-type="bibr" rid="ref11">11</xref>
          )
where
        </p>
        <p>Y *m   y1*B , y2*B ,</p>
        <p>B</p>
        <p>, yn*B 
( y1B  y2B 
 ynB </p>
        <p>1
 y</p>
        <p>B
)
is
a
normalized
frequency
vector.</p>
        <p>
          The calculation of problem (
          <xref ref-type="bibr" rid="ref11">11</xref>
          ) resulted in vector YB*m   y1A, y2A,
, ynA  and value  y .
        </p>
        <p>B</p>
        <p>Comparing the obtained results of matrix games solution with solution of the bimatrix game, we
can state about the possibility of grounding both the quantitative results on the players’ average wins
and their qualitative behavior. The bimatrix game of two players are given by matrixes
Therefore,
6 3
1 CB  7
6 8
the
7 8
8 1
4 4
solution
 Bx 
the second player B
344</p>
        <p>The optimum mixed solution of the first player
in the bimatrix game
coincides with mixed solution</p>
        <p>in the matrix game given by payment matrix CB of
the second player B . Thus, the optimum behavior of the first player is a result of solving the primal
optimization problem for the second player, i.e. the first player will not take account of his own
payment matrix – all the “attention” is paid to the payment matrix of the second player. But at
same time, the value of reward</p>
        <p>is a result of solving dual optimization problem
for the matrix game given by his own payment matrix CA . Using the indicated
algorithm, we can state that the mixed strategy of the second player
coincides with
calculation of dual optimization problem
for the first player. Therefore, the
optimum behavior of the second player is fully conditional on payment matrix CA of the first player –
the values of own wins are ignored. The value of reward
is equal to the value of
bimatrix game</p>
        <p>of the dual problem for the second player.</p>
        <p>Thus, it is reasonable, in designing the information system of students’ knowledge evaluation, t
use an algorithm that is based on solving two separate matrix games. A player shall find the win of the
other player.</p>
      </sec>
    </sec>
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