exxamer

[exxamer]

Maxmin princ.

#include<iostream.h>

#include<conio.h>

#include <math.h>

#include<iomanip.h> void main()

signed int a (5)(5) = (1750, 1125, 315, - 1125) ,

(0, 1500, 3000, 2250, 1500),

(-750.750,2250, 4125),

(-1500, 0, 1500, 3000, 4500) 17

inti,j.1, 1, mi, r[3], k

clrscr();

cout<<"\n The summery of DALLY Net Profit:-\n";

for (i=0;i<4;i++)

for(j=0; j<5; j++)

cout<<setw

cout<<endl;

cout<<"\n Rowwise Maximum: \n"

for (i=0;i<4;14+)

1 = a[1][0]

for(j=0:15; j++)

if(1 < a(1)(j)) 1=a(i11j\ )

x(1)^ =1

for (i=0;i<4 (71++) cout<<r[i]<<endl;

cout<<"\n Minimax value:

m=r[0];

for (i=1;i<4;i++)

if (m>r[i])

m=r[i];

cout<<m;

getch();

[exxamer]

program to find best alternative when the intrestrate is 09 //mean and variance of each alternative

//alternative j (Rs)

1

2

//expected annual revenue (x5) 500000 //variance of annual revenue 1000000

400000

10000

#include<iostream.h>

#include<conio.h> #include<math.h>

void main()

clrscr();

double x1-500000, i2=400000. double v1=100000,v2=1000. cv1.cv2;

int i;

cout<<"mean and vantriance of each alternative":

cout<<"\n

\n":

cout<<"\n alternative 5(rs) backslash n^ prime prime 7

cout<<"\n

cout<<"\n 12

cout<<"\n

cout <<^ m 1 n\nexpected annual revenue (x!)"<x1 <<" "<<x2: cout<<"\n\n variance of annual revenue ( j)"<<v l<<"<w2z

cout<<"\n\n

cvi-sqrt(v1)/x1; Cv2-sqrt (v2)/x2;

if (cvl>cv2)

cout<<"\n\n the coefficient of variation of alternative

cout<<"\n is lesser than that of alternative 1:"<<<cv2;

An":

else cout<<"\n the coefficient of variation of alternative1:"<<cv2:cout<<"\n is lesser than that of alternative 2:"<<cv1:

getch();

[exxamer]

WAP FOR MINMAX

#include<iostream.h> #include<conio.h>

#include <math.h>

void main()

signed int a 15115]=1750,1125 , 375,-375,-1125), 10, 1500, 10, 225epsilon , 1500),

(-750,750,2250,412^ .

(-1500,0,1500,300,4^ 2 89\ 12

inti,s,j,1 ,m,ml,r151,k; clrscr();

cout<<"\n The summery of DALLY Net Profit:-\n";

for (1-0;1<4;1++)

for(j=0; j<5; j++)

cout <a[i]\ j]<<^ prime prime ; cout<<endl;

cout<<"\n Rowwise Maximum: \n":

for (1m0:1<4/+*)

swa[i][0];

for (3=0; j<5;3+*

if (g<a(111j)

s-a[1] (317

1

for (i=0;i<4;3+*

cout <<r[1] <<endl; cout<<" in Hammin valum:

101 2

for (11;i<;**if (m>r[i])

m=r[i];

}

}

cout<<m;

getch();

[exxamer]

B1 PROGRAM FOR HURWICZ CRITERIA.

#include<iostream.h>

#include<conio.h>

#include<iomanip.h>

class hurviCE

private :

int A[100] [100] .1,1,m,n,r,r1, intsmall .1,L[100) , large, a[100], h[100];

public :

void getdata (void);

void disp(void);

void hurwicz :: getdata(void)

cout<<"\n\n\t Enter the size of Matrix. ";

cout<<"\n\n\t Number of row's: ";

cin>>m;

cout<<"\n\n\t Number of col's: "

|n>>0;

for i=0:i<m:i++)

for (j=0;j<n;j++)

cout<<"\n\n\t Enter the value for row "<<+<<" : "

cin>>A[1 1(31;

void hurwicz :: disp(void)

int min-0, max=0;

float alpha 0.5;

large = 0;

small 9999:

for (i=0;i<n;i++)

large 0:

small 9999;

for(j=0; j<n; j++)

if(large-A[1151

large large;

L[1] large;

else

large A[i](31:

L[i] = large;

if (small<-A[4] [])

small small;

s[i] = small;

else

small=A[i][j];

s[i] small;

for(i=0; i<m; i++)

h[i] (float(alpha) *L[1])+(float (1-alpha) s[1]);

min = h[01:

max=h[0];

for (i=0:1<m:1++)

Lf(min<=h(11)

min minz

else

min hlil;

if(max > - h * (1))

max max;

else

max - hlil;

for(i=0;i<m:1++)

if(h(1) = min)

r = i + 1

for (i=0;i<m;i++)

if(h[i] = max)

r1=l+1.

cout<<"\n\n\t Your entered matrix is :\tit Maximum Minimum": cout<<" hurwicz";

cout<<"\n\n\t\t\t";

for (i=0;i<m;i++)

cout<<" A"<<i+l<<" ->";

for(j=0; j<n; j++)

cout<<" "<<setw r(4)<<A1 perp1/j] :

cout<<" "<<setw(4)<<L[1]<<^ prime prime "<<setw(4) <<s[1]; cout<<" " <setw(4)<<h(1):

cout<<"\n\ n|t|t|t^ prime prime

cout<<"\n\n\t Maximum is coaxi

cout<<"\ |n|+30. for profit the answer is : A"<<rice"

A1xeraxive=.

cout<<"\n\n\n\n\t And Minimin is "<<mini cout<<"\n\n\t So, for cost the answer is: A"<< r<"Altarmativen."

void main()

cirser();

hurwicz h;

h.getdata(); clrscr();

h.disp();

getch();

[exxamer]

10) Write a program to implement queue using linked list

#include<iostream.h>#include<conio.h#include<stdio.h> //using namespace std: struct node

int data:

node*next:

1*front NULL, "rear-NULL, "p-NULL, "np-HULL;

void push(int x)

np new node:

n_{0} = 2 data-x:

mp->next-NULL; if (front==NULL)

front rear =np; rear->next-NULL;

else{rear-> \{t = n\} :

rear np:

rear->next 9015L_{1}

int remove()

int x;

if(fron t==||0Lt)

cout <^ prime prime empty queue 1n^ prime prime ;

else {p=front: front-front->next;

X= p->data;

delete (p):return (x);}}

int main()

int n, c 0.x, true;

cout<<"enter the value to be pushed in to queuein";cin>>n;while (c<n)cout<<"enter the value to be entered in to queue\n";

push (x);cout<<"\n\nremoved value\n\n"; while(true)

if(front !=NULL)

cout<<remove () <<endl;

else

break;

getch()1

[exxamer]

Write a program for simple Inventory.

#include<iostream.h>

#include<iomanip.h>

#include<conio.h>

define MAX PEC 10

class Inventory

{

char itemname (151;

int code:

double cost:

public:

void getdata();

void showdata();

};

void Inventory::getdata()

{

cout<<"\nenter item name:"; cin>>itemname;

cout<<"enter code:";

cin>>code;

cout<<"enter cost :":

cin>>cost;

}

void Inventory :: showdata() (

cout<<endl;

cout.width (15);

cout.setf(los::left, ios::adjustfield):

cout<<citemname;

cout.width (8) ;

cout<<code;

cout.width (15);

cout.setf(ios::right, los::adjustfieldi;

cout.setf(los::showpoint); cout.setf(los::fixed, ios::floatfield;

cout.precision (2);

cout<<cost;

}

Void main()

{

Inventory record [MAX_REC];

inti, n

clrscr();

cout<<"\n......inventory management....\n";

cout<<"\n how many record to be created:";

cin>>n;

cout<<"enter "<<n<<"recordin":

for(i=0; 1<n;1++

record[i].getdata();

cout<<"\n\n......stock information...... ln^ prime prime :

cout<<"\n"<<setw (8) <<"Itemname"

<<setw (10)<<"cost"

<<setw (19)<<"cost"<<end1;

cout <<^ 11 .........."

for (i= 0;i<n;i++)

record[1].showdata();

getch():

}

[exxamer]

Write a program for minimal principal

#include<iostream.h>

#include<conio.h>

#include<math.h>

void main()

signed int a[5][5]={{750,1125,375,-375,-1125},

{0,1500,3000,2250,1500},

{-750,750,2250,4125},

{-1500,0,1500,3000,4500}};

int i,s,j,l,m,mi,r[5],k;

clrscr();

cout<<"the summary of daily net profit:=";

for(i=0;i<4;i++)

{

for(j=0;j<5;j++)

cout<<a[i][j];

cout<<endl;

}

cout<<"rowwise maximum:";

for(i=0;i<4;i++)

{

s=a[i][0];

for(j=0;j<5;j++)

{

if(s<a[i][j])

s=a[i][j];

r[i]=s;

}

}

for(i=0;i<4;i++)

cout<<r[i]<<endl;

cout<<"maxmin value:-";

m=r[0];

for(i=1;i<4;i++)

{

if(m>r[i])

{

m=r[i];

}

}

cout<<m;

getch();

}

[exxamer]

Q.1) Wap to demonstrate the determination of mixed strategies (2*2 matrix).

#include<iostream.h>

#include<conio.h>

#include <math.h>

void main

{

lnt a[ 2][ 2]={ (6,9),(8,4) } ; float rl,cl, i, j, n,m, s,l, c[2], r[2],rp[2],cp[2], V;

clrscr():

cout <<" prime prime backslash r the element in a given array: \n";

for ( i=0:i<2:i++)

{

for(j=0; j<2;j++)

{

Cout<<a[i][j]<<" ";

cout<<endl;

}

n = 0;

for(i=1;i>=0;i--)

{

for(j=0; j<2;j++)

{

r[n]=abs(a [i][0]-a[i][j]);

n++;

cout<<"\n Rowwise Oddments: \n";

for (int p=0;p<=n-1;p++)

{

cout<<r [p]<<endl;

}

m = 0 ;

for(j=1;j>=0;j--)

{

for(i=0;1<2;1++)

{

c[m]=abs(a[0][j]-a[1][j]) ;

m++;

}

cout<<"\n columnwise oddments: \n";

for (int q=0;q<m;q++)

{

cout<<c [q]<<endl;

}

s = 0;

for(i=0;i<2; i++)

{

s=s+r[1];

}

for(i=0;i<2;i++)

{

rp[i] = r[i] / s ;

}

s = 0;

for (i=0;i<2; i++)

{

s = s + c[i] ;

}

for (i=0;i<2; i++)

{

cp[i] = c[i] / s;

}

cout<<"\n the strategies of player A is:";

for (i=0;i<2;i++) cout<<rp[i]<<" ";

cout <"\ n the strategies of player B is:";

for(i=0;i<2;i++)

cout<<cp[i]<<" ";

V=((a[0][0]*rp[0])+(a[1][0]*rp[1]))/(rp[1]+rp[0]);

cout<<"\n\n The value of the game; "<<V;

getch();

}