Bisection Methods (Binary Search Methods)

Rumusan akar :

Evaluasi :

Misalnya : tentukan nilai nol (memotong sumbu x) dari persamaan y = x³ - 3x - 5
Pertama kita gunakan nilai awal x₁ = 2 dan x₂ = 3, dengan demikian kita mendapatkan x_mid = 2.5 Selanjutnya, perhatikan flowchart di bawah ini :

i	x1	x2	xmid	f(x1)	f(x2)	f(xmid)
1	2	3	2.5	-3	13	3.125
2	2	2.5	2.25	-3	3.125	-0.35938
3	2.25	2.5	2.375	-0.359375	3.125	1.271484
4	2.25	2.375	2.3125	-0.359375	1.271484	0.428955
5	2.25	2.3125	2.28125	-0.359375	0.428955	0.028107
6	2.25	2.28125	2.26563	-0.359375	0.028107	-0.16729
7	2.26563	2.28125	2.27344	-0.1672935	0.028107	-0.07001
8	2.27344	2.28125	2.27734	-0.07000971	0.028107	-0.02106
9	2.27734	2.28125	2.2793	-0.02105576	0.028107	0.003499
10	2.27734	2.2793	2.27832	-0.02105576	0.003499	-0.00878
11	2.27832	2.2793	2.27881	-0.008784707	0.003499	-0.00264
12	2.27881	2.2793	2.27905	-0.002644293	0.003499	0.000427
13	2.27881	2.27905	2.27893	-0.002644293	0.000427	-0.00111

Keterangan :

• Pada saat iterasi (i) ke-1, harga f(x₂) dan f(x_mid) memiliki tanda yang sama (+), sehingga x₂ = x_mid
• Untuk iterasi (i) ke-2, harga f(x₁) dan f(x_mid) memiliki tanda yang sama (-), sehingga x₁ = x_mid
• Begitu juga untuk iterasi ke-3 dan seterusnya. Sedangkan x_mid adalah nilai x yang dicari.

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Delays

Propagation Delay
With all forms of transmission medium (electrical or optical) there is a short time delay for the signal to travel (propagate) through the medium. For examples :

In free space speed approximates to 3*10⁸ m/s
In cables speed approximates to 2*10⁸ m/s 

Transmission Delay
Transmission delay is the time taken to transmit a block or frame at the specified bit rate.

For examples :

1000 bit frame at a transmission rate of 10kb per second.
T_x = 1000/(10 * 10³) secs = 0.1 secs

Or for Ethernet -
Max packet size of 1500 bytes
R = 10Mbps (approx)
T_x = (1500 * 8)/(10 * 10^-6) = 1200 * 10^-6 = 1.2 * 10^-3 = 1ms (approx)

Satellite Problem
A satellite lies in a geostationary orbit 35800km above the Earths surface. A data link makes use of the satellite to transmit information at a data rate of 50Mbps.

1. Estimate the time taken for the signal to propagate from one station on Earth to another via the satellite.
   v = d/t
   3 * 10⁸ = 35800000/t
   t = 358/3*10⁴ = 0.117
   The signal will take 0.234 seconds to travel from station A to the satellite and back to station B.
2. Estimate the quantity of data (in transit) on the link at any instance of time. (Assume propagation velocity is 3*10⁸ ms^-1.)
   R = N/T_x
   N = R * T_x = 50 * 10⁶ * 0.23
       = 11.7Mbit

Link Efficiency (Utilization)

Idle RQ:
If seach frame was simply transmitted immediately after the previous one then utilization would be 100%. However, Idle RQ also uses various control signals such as ACKnowledgements.

1. Assuming no errors
   The time before a new frame is transmitted is the time taken to transmit the data plus the propagation delay, then the time to receive the acknowledgement and its propogation delay.
   i.e. T_Ix + T_P + T_ackX + T_P
   The time taken to acknowledge the data is insignificant compared to the time taken to transmit it, so we can disregard it in our calculations. (We also ignore the time taken to check each frame as its received, perhaps for CRC errors.)

   

    means the propagation delay is dominant.
   means the transmission delay is dominant.

2. With errors

   

   Where N_r is the number of attempts to transmit each frame. The value of N_r can be obtained from the bit error rate P of the link. If the link operates with frames containing N_i bits, then for values of N_iP < 1 the probability that frame contains an error is :

   N_iP

   Hence the probability a frame has no error is :

   1 - N_iP

   Hence the number of transmission attempts required to transmit a good frame is on average :

   

   For example, if error ratio P is 1 in 10000 then for 1000 bit frames, the probability of an error in frame is 0.1.
   Hence 1 frame in 10 is, on average, corrupted. Average number of transmission attempts required to achieve successful transfer is : 1/(1-0.1) = 1.1

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Pemecahan Persamaan Diferensial (Cara 3)

Persamaan diferensial homogen mempunyai sifat dimana pangkat x dan y yang terlibat dalam masing-masing suku memiliki derajat yang sama. Kunci untuk memecahkan persamaan homogen adalah dengan substitusi y = vx, dengan v adalah fungsi x. Substitusi ini akan mengubah persamaannya menjadi bentuk yang dapat dipecahkan dengan pemisahan variabel. Perhatikan persamaan di bawah ini :

Substitusikan y=vx ke dalam persamaan di atas, sehingga :

Jika fungsi y=vx dideferensiasikan terhadap x, maka :

Dengan substitusi kedua persamaan di atas, maka diperoleh :

Untuk menyelesaikan persamaan di atas, digunakan pemisahan variabel, sehingga :

Persamaan di atas masih terdapat variabel v. Oleh karena itu, kita harus menghilangkan variabel v tersebut agar diperoleh persamaan yang hanya memiliki unsur x dan y saja. Ingat bahwa y=vx, dengan demikian . Jika persamaan v ini disubstitusikan ke dalam persamaan di atas, maka :

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