Giải pháp điều khiển nghẽn trong mạng OBS bằng phương pháp làm lệch hướng đi - 10

set(handles.NUT1,'Value',0); set(handles.NUT2,'Value',0); set(handles.NUT3,'Value',0); set(handles.NUT4,'Value',0); set(handles.NUT5,'Value',0); set(handles.NUT6,'Value',0); set(handles.NUT7,'Value',0); set(handles.NUT8,'Value',0); set(handles.NUT9,'Value',0); set(handles.NUT10,'Value',1); set(handles.NUT11,'Value',0); set(handles.NUT12,'Value',0);

%THONG SO C = 6;

L = 1;

lamda = 1000;

%KHI SU DUNG FDL = 50 ms B7=50;

n0 = 10^3.*C./(lamda.*L);

%CHON i THOA DIEU KIEN i>n0+1 i=7:12;

Ti7=B7.*C./(lamda.*L.*(i-n0).*(i-n0+1)); a7=3.5714;

b7=1.3889;

%CHON n THOA DIEU KIEN n>i n=25;

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p=factorial(n)./(factorial(i).*factorial(n-i)).*((b7./(a7+b7)).^i).*((b7./(a7+b7)).^(n-i)); Pl7=p.*exp(-i.*a7.*Ti7);

%VE HAM Pl2 THEO i

Giải pháp điều khiển nghẽn trong mạng OBS bằng phương pháp làm lệch hướng đi - 10

x=plot(i,Pl7,'bo:');

title('khong FDL voi 1/a=280 va 1/b=720'); xlabel('luu luong tai');

ylabel('xac suat chum suy hao');

%HIEN THI TREN HE TOA DO

axes(x,'square')

function NUT11_Callback(hObject, eventdata, handles) set(handles.NUT1,'Value',0); set(handles.NUT2,'Value',0); set(handles.NUT3,'Value',0); set(handles.NUT4,'Value',0); set(handles.NUT5,'Value',0); set(handles.NUT6,'Value',0); set(handles.NUT7,'Value',0); set(handles.NUT8,'Value',0); set(handles.NUT9,'Value',0); set(handles.NUT10,'Value',0); set(handles.NUT11,'Value',1); set(handles.NUT12,'Value',0);

C = 6;

L = 1;

lamda = 1000;

%KHI SU DUNG FDL = 50 ms B6=0;

n0 = 10^3.*C./(lamda.*L);

%CHON i THOA DIEU KIEN i>n0+1 i=7:12;

Ti6=B6.*C./(lamda.*L.*(i-n0).*(i-n0+1)); a5=3.1250;

b5=1.4706; a=3.333; b=1.4268; a7=3.5714; b7=1.3889;

%CHON n THOA DIEU KIEN n>i

n=25;

p=factorial(n)./(factorial(i).*factorial(n-i)).*((b./(a+b)).^i).*((b./(a+b)).^(n-i));

p5=factorial(n)./(factorial(i).*factorial(n-i)).*((b5./(a5+b5)).^i).*((b5./(a5+b5)).^(n-i));

p7=factorial(n)./(factorial(i).*factorial(n-i)).*((b7./(a7+b7)).^i).*((b7./(a7+b7)).^(n-i)); Pl6=p.*exp(-i.*a.*Ti6);

Pl5=p5.*exp(-i.*a5.*Ti6); Pl7=p7.*exp(-i.*a7.*Ti6);

%VE HAM Pl2 THEO i

q=plot(i,Pl6,'r+:',i,Pl5,'k*:',i,Pl7,'bo:'); title('so sanh ket qua');

xlabel('luu luong tai'); ylabel('xac suat chum suy hao');

%HIEN THI TREN HE TOA DO

axes(q,'square')

function NUT12_Callback(hObject, eventdata, handles) set(handles.NUT1,'Value',0); set(handles.NUT2,'Value',0); set(handles.NUT3,'Value',0); set(handles.NUT4,'Value',0); set(handles.NUT5,'Value',0); set(handles.NUT6,'Value',0); set(handles.NUT7,'Value',0); set(handles.NUT8,'Value',0); set(handles.NUT9,'Value',0); set(handles.NUT10,'Value',0); set(handles.NUT11,'Value',0); set(handles.NUT12,'Value',1);

C = 6;

L = 1;

lamda = 1000;

%KHI SU DUNG FDL = 50 ms B6=50;

n0 = 10^3.*C./(lamda.*L);

%CHON i THOA DIEU KIEN i>n0+1 i=7:12;

Ti6=B6.*C./(lamda.*L.*(i-n0).*(i-n0+1)); a5=3.1250;

b5=1.4706; a=3.333; b=1.4268; a7=3.5714; b7=1.3889;

%CHON n THOA DIEU KIEN n>i n=25;

p=factorial(n)./(factorial(i).*factorial(n-i)).*((b./(a+b)).^i).*((b./(a+b)).^(n-i));

p5=factorial(n)./(factorial(i).*factorial(n-i)).*((b5./(a5+b5)).^i).*((b5./(a5+b5)).^(n-i));

p7=factorial(n)./(factorial(i).*factorial(n-i)).*((b7./(a7+b7)).^i).*((b7./(a7+b7)).^(n-i)); Pl6=p.*exp(-i.*a.*Ti6);

Pl5=p5.*exp(-i.*a5.*Ti6); Pl7=p7.*exp(-i.*a7.*Ti6);

%VE HAM Pl2 THEO i

q=plot(i,Pl6,'r+:',i,Pl5,'k*:',i,Pl7,'bo:'); title('so sanh ket qua');

xlabel('luu luong tai'); ylabel('xac suat chum suy hao');

%HIEN THI TREN HE TOA DO

axes(q,'square')

NHÂN XÉT CỦA CÁN BỘ HƯỚNG DẪN

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Ngày đăng: 11/02/2023