Frames

1. For this problem, use a formula from this chapter, but first state the formula. Frames arrive randomly at a 100-Mbps channel for transmission. If the channel is busy when a frame arrives, it waits its turn in a queue. Frame length is exponentially distributed with a mean of 10,000 bits/frame. For each of the following frame arrival rates, give the delay experienced by the average frame, including both queueing time and transmission time.

T = Time  µ = mean frame length      C  = capacity channel

T = 1 / (µC - ג)

a.  90 frames/sec.

 T = 1 / ( 10000 – 90)

       T =  1,01 . 10^-10 s

b.  900 frames/sec.

   T = 1 / ( 10000 – 900)

      T =  1,1 . 10^-10 s

c.           9000 frames/sec.

    T = 1 / ( 10000 – 900)

    T =  1,1 . 10^-9 s

2.   A group of N stations shares a 56-kbps pure ALOHA channel. Each station outputs a 1000-bit frame on an average of once every 100 sec, even if the previous one has not yet send (e.g., the stations can buffer outgoing frames). What is the maximum value of N?

pure ALOHA memiliki efisiensi sebesar 18.4%.

Bandwith pure ALOHA = 56 kbps × 18.4% = 10.304 kbps = 10304 bps

Masing-masing Station membutuhkan 1000bit/100second= 10 bps

Jumlah maximum Station yang bisa menggunakan pure ALOHA adalah :

10304 / 10 = 1030.4 ~ 1030 station.

19. Question. A 1-km-long 10Mbps CSMA/CD LAN (not 802.3) has a propagation speed of 200m/microsec. Data frames are 256bits long, including 32 bits of header, checksum and other overead. The first bit slot after a successful transmission is reserved for the receiver to capture the channel to send a 32-bit acknowledgement. What is the effecitve data rate, excluding overhead, assuming that there are no collisions?

Waktu yang dibutuhkan untuk sebuah cycle :

- Transmitter menggunakan channel channel: 2 tou

- Transmitter mengirimkan frame : waktu untuk mengirimkan sebuah frame.

- Transmitter menunggu sampai bit terakhir datang (contention period):    contention period (1 tou)

- Receiver menangkap channel: 2 tou

- Receiver mengirimkan ACK frame: waktu untuk mengirimkan ACK frame

- Receiver menunggu bit terakhir dari ACK yang datang: contention period (1 tou)

Untuk mendapatkan :

     Effective Data Rate = (Actual Frame Size) / (Time required for one cycle)

Kita mendapatkan :

PropDelay = Length of link (in metres) / Velocity of propagation (metres per second)

Dengan kata lain :

     Distance = Speed * Time

Jadi propagation delay-nya:

     1000m = 200m/microsec * t

yang akan mengantarkan kita ke :

     tou = 5 microsec.

Jadi Round trip propagation delay :

     2tou = 10 microsec.

Waktu untuk mengirimkan sebuah frame :

     numberOfBits = datarate * time

     256 = 10,000,000bps * t

     t = 25.6 microsec

Waktu untuk mengirimkan sebuah ACK :

     numberOfBits = datarate * time

     32 = 10,000,000bps * t

     t = 3.2 * 10^-6 --> 3.2 microsec

Contension period :

     ContensionPeriod = 1tou -> 5 microsec.

Full cycle:

- Transmitter menangkap channel: 10 microsec

- Transmitter mengirimkan frame: 25.6

- Transmitter menunggu sampai bit terakhir datang (contention period):5

- Receiver menangkap channel : 10

- Receiver mengirimkan Ack frame: 3.2

- Receiver menunggu sampai bit terakhir dari ACK datang: 5

Total waktu  cycle adalah 58.8 microsec.

Actual frame size = 25.6 – 3.2 = 22.4

So that the effective data rate is

     dataRate = bits / time

     dataRate = 22.4 / (58.8 * 10^(-6))sec = 3.81 * 10^6 bps

yaitu = 3.81 Mbps

21. Question. Consider building a CSMA/CD network running at 1Gbps over a 1-km cable with no repeaters. The signal       speed in the cable is 200,000 km/sec. What is the minimum frame size.

Mendapatkan waktu propagasi :

data = speed * time

1 km = 200,000 km/sec * t

t = 1/200,000 sec = 5 microsec.

Jadi waktu untuk bolak-balik adalah  10 microsec.

Mendapatkan ukuran frame minimum :

minNumberOfBits = dataRate * time

minNumberOfBits = 10^9 bit/sec * 2Tou = 10^9 * 10 * 10^(-6) = 10,000bits

24. Some books quote the maximum size of an Ethernet frame as 1518 bytes instead of 1500 bytes. Explain your answer why.

Jika semua byte dari alamat tujuan , alamat sumber,type serta checksum di masukkan/dijumlahkan maka Ethernet frame akan menjadi 1500 byte data di tambah dengan 18 yang hasilnya adalah 1518 byte.

Jadi yang 1500 byte adalah byte untuk data yang dikirmkan oleh frame dalam suatu ethernet..