Mencari Rumus T(n) Dari Procedure Rata-rata

PSEDUCODE RATA-RATA


Procedure Rata_Rata (input Total, Nilai : integer)-> Real

Deklarasi 

 {Tidak ada }

Algoritma

     rata_rata <-- Total / N

endfunction

input (nilai)
Total <-- 0
for i <-- 1 to N do
   input ( mhs(i).Nim , Mhs(i).Nama ,Mhs(i).Nilai )
   Mhs(i).indeks <-- indeks_nilai ( Mhs(i).Nilai )
   Output ( Mhs(i).indeks )
   Total <-- total + Mhs(i).Nilai

endfor

    Output('Rata-rata Nilai     :', rata_rata(Total,N))
    Otuput('Nilai Tertinggi     :', Nilai_max(Mhs,N))

Function Nilai_max (input Mhs : mahasiswa, input N : integer) --> integer

Deklarasi 

  max,i : integer;

Algoritma

max <-- Mhs(i).Nilai
for i <-- 2 to N do
  if (Mhs(i).Nilai > max)
     then
        max<-- Mhs(i).Nilai

  endif
endfor
Nilai_max <-- max
endfunction

Menghitung Tmin, Tmax, Tavg


OPERASI DASAR

            <--
              >
              *
              +
               /

Operasi Paling Dalam
            <--

Tmin(n) = 1+2 = 3
Tmax(n) = N+2
Tavg(n)  = 3+4+5+....(n+2) ~~ n

                           n

O (Big Oh)	Ω (Big Omega)	Θ (Big Theta)
Tmin (n) = 3	Tmin (n) = 3	Tmin (n) = 3
O = t(n) € O (g(n)); t(n) ≤ Cg(n)	Ω = t(n) € O (g(n)); t(n) ≤ Cg(n)	Θ = t(n) € Θ (g(n)); C2g(n) ≤ t(n) ≤ C1g(n)
Tmin(n) = 3 € O (1)	Tmin(n) = 3 € Ω (1)	Batas Atas = 3 € O (1)
no = 0	no = 0	Batas Bawah = 3 € Ω (1)
C = 1	C = 1	Jadi = no = 0
		C1 = 1, C2 = 1

Tmax = n+2

O (Big Oh)

O = t(n) € O (g(n)); t(n) ≤ Cg(n)
       n+2 € O(n)

n+2 ≤ 2n

2   ≤   0       n = 0      (x)
5   ≤   6       n = 3      (√)
11  ≤   18       n =  9        (√)
22  ≤   40       n =  20      (√) 

Jadi no = 3
         C  = 2


Ω (Big Omega)

Ω = t(n) € O (g(n)); t(n) ≤ Cg(n)
       n+2 € Ω (n)
       n+2 > n

2     >   0           n = 0      (√)
6      >    0              n = 4         (√)
10    >    0              n = 8         (√)
102  >    0              n = 100    (√)

Jadi no = 0   
C = 1



O (Big Theta)

Θ = t(n) € Θ (g(n)); C2g(n) ≤ t(n) ≤ C1g(n)
Batas Atas = n+2 < 2n = no = 3, C1 = 2
Batas Bawah = n+2 > n= no = 0, C2 = 1
Jadi no= 3, C1= 2, C2 = 1


Tavg(n)  3+4+5+......(n+2)
                          n
            = 1/2 n (3+(n+2))/n
            = 1/2(n+5)
            = n/2 + 5/2 
            = 1/2 + 5/n


          

O (Big Oh)

O = t(n) € O (g(n)); t(n) ≤ Cg(n)
       1/2n + 5/2 € O(n)

1/2n + 5/2  ≤ 2n

5/2   ≤  0       n = 0      (x)
5/2   ≤ 12      n = 6      (√)
5/2   ≤   18      n =  9        (√)
5/2   ≤   90      n =  30      (√) 

Jadi no =6
         C  = 2



Ω (Big Omega)

Ω = t(n) € O (g(n)); t(n) ≤ Cg(n)
       1/2 + 5/2 € Ω (n)
       1/2 + 5/2 > n

5/2       >   0           n = 0      (√)
6/2        >    1              n = 1        (√)
9/2        >    4              n = 4        (√)
95/2      >    90            n = 90      (x)
105/2    >    100          n = 100   (x)

Jadi no = 0   
C = 1



O (Big Theta)

Θ = t(n) € Θ (g(n)); C2g(n) ≤ t(n) ≤ C1g(n)
Batas Atas = 1/2n + 5/2 < 2n
Jadi no = 6, C1 = 2
Batas Bawah = 1/2n + 5/2 > 2
Jadi no= 0, C2= 1
Jadi no =6, C1 = 2, C2=1