logo Google

sumber : http://id.berita.yahoo.com/apa-arti-logo-google-hari-ini-051500314.html

INILAH.COM, Jakarta – Sudahkah Anda mengakses Google hari ini? Heran logo Google berubah menjadi sekawanan burung hinggap di ranting-rating? Simak berikut.
John James Audubon lahir pada 26 April 226 tahun lalu. Tak seperti logo Google sebelumnya yang interaktif, kali ini Google membuat logonya tetap sederhana. Huruf pembentuk Google bisa dikenali dengan mudah pada ranting yang ada.
Aububon merupakan orang yang mengembangkan apresiasi mendalam untuk alam sejak awal ia masih anak-anak hingga berpindah ke Amerika dari Prancis pada usia 18 untuk menghindari bergabung dengan tentara Napoleon.
Ia berusaha menunjukkan jati dirinya dengan melukis burung. “Saya merasa intim dengan burung,” katanya.
Pencapaian Audubon mempengaruhi sejarah ornithologi dan alam hingga dikutip Charles Darwin dalam ‘Asal Muasal Spesies’.
Pada Desember 2010, salinan lukisan Audubon berjudul ‘Burung Amerika’ terjual di dengan harga 7,3 juta GBP (Rp1,04 triliun) dalam lelang Sotheby. Harga itu menjadi rekor harga termahal untuk buku cetak tunggal. [mor]

Hati2 dengan Wi-FI

Wi-Fi Jadi Media Untuk Mencuri Data Pribadi

sumber : Metro TV News | Metro TV – Rab, 27 Apr 2011 17.21 WIB


PENGGUNA ponsel pintar dan laptop harus berhati-hati bila sedang berkoneksi memakai Wi-Fi untuk umum. Kabarnya informasi data pribadi, seperti rincian kartu kredit bisa dicuri lewat koneksi Wi-Fi.

Seorang hacker bisa mengatur hotspot internet kabel di tempat umum, dan kemudian mencuri informasi pribadi dari pengguna ponsel pintar dan laptop tanpa disadari pengguna.

British Telecom, operator terbesar di Inggris yang selama ini menyediakan hotspot internet publik telah menyadari kerentanan ini.

British Telecom dalam pernyataannya, scammers dengan memggunakan perangkat berbiaya US$27 dan perangkat lunak yang diunduh gratis, bisa mengumpulkan nama pengguna, password, bahkan rincian kartu kredit, melaluo Wi-Fi palsu yang tampaknya sah.

Dalam sebuah investigasi yang dilakukan The Guardian, relawan menguji betapa mudahnya menghubungkan hotspot Wi-Fi palsu.

Dari hasil investigasi, setelah hacker melakukan penipuan jaringan Wi-Fi di stasiun kereta api, seorang ahli menyaksikannya dengan menggunakan laptop dan ponsel pintar mendapatkan akses koneksi Wi-Fi yang tidak aman, dengan menuliskan nama pengguna dan password.

Tes lain mengundang orang untuk memasukkan rincian kartu kredit mereka, untuk melakukan pembayaran melalui akses koneksi internet publik. Tanpa disadari pengguna sudah masuk dalam perangkat Wi-Fi palsu.

Di Inggris, sejumlah ponsel pintar bisa mengakses ke 2,5 juta BTS zona terbuka, yang tersedia di stasiun kereta api dan hotel. Hacker bisa mendapat label koneksi Wi-Fi pengguna, dengan nama apapun dan tanpa disadari si pengguna.

Padahal semua informasi pribadi bisa masuk ke perangkat orang lain. Bahkan informasi yang dienkripsi dapat diuraikan menggunakan perangkat lunak yang tersedia gratis.(MI/*)

problem dan solusi jaringan komputer

alamat
probelm and solution computer network :
http://www.rose-hulman.edu/class/csse/csse432/201030/Wireshark-Labs/Wireshark_IP_Solution.pdf
http://www.ccs.neu.edu/home/rraj/Courses/U640/F04/ProblemSets/ss3.pdf
http://itee.uq.edu.au/~coms3200/tutorials/COMS3200-2010-Tutorial8-solns.pdf
http://www.comm.utoronto.ca/~jorg/teaching/ece461/handouts/
http://www.comm.utoronto.ca/~jorg/teaching/ece461/handouts/Quiz1_461_F10-sol.pdf
http://web.uettaxila.edu.pk/CMS/coeCCNbsSp09/notes%5CFA08MIDII%20Solution.pdf
http://www.comp.lancs.ac.uk/~adrian/networking-notes/uploads/Main/ans

daftar alamat solusi latihan jaringan komputer

www.cs.cmu.edu/~srini/15-441/F06/hw1.pdf
http://www.inf.uni-konstanz.de/dbis/teaching/ss06/os/os_pw_4_sol.pdf
http://www.cs.cmu.edu/~srini/15-441/F06/hw1.pdf
http://www.cs.uiuc.edu/class/fa07/cs438/exams/CS438-midterm-sol.pdf
http://www1.ju.edu.jo/ecourse/netcourse/Exams/Spr08NetMidSolU.pdf
http://www.systems.ethz.ch/education/past-courses/fs09/vs/exercises/solution1.pdf
http://users.crhc.illinois.edu/vraman3/438.fall08/solutions/HW1-soln.pdf
http://www.cs.uiuc.edu/class/fa06/cs438/
http://web.uettaxila.edu.pk/CMS/coeCCNbsSp09/notes%5CFA08FINAL%20Solution.pdf

Soal Latihan dan Jawaban dari computer network edisi 4 - peterson

bab 1 no 6
Calculate the total time required to transfer a 1.5 MB file in the following cases, assuming RTT
of 80ms, a packet size of 1KB and an initial 2XRTT of “handshaking” before it is sent.
a) The b/w is 10Mbps, and the data packets can be sent continuously.
b) The b/w is 10Mbps, but after we finish sending each data packet, we must wait one RTT
before sending the next
c) The link allows infinitely fast transmits, but limits bandwidth such that only 20 packets can be
sent per RTT zero transmit time as in (c), but during the first RTT, we can send one packet
during the 2nd RTT we can send 2 packets, during the 3rs we can send 4 = 23-1 and so on

Solution:
We will count the transfer as completed when the last data bit arrives at its destination
a) 1.5MB = 12,582,912 bits, 2 initial RTTs (160ms) + (12,582,912)/(10,000,000) bps (transmit) +
RTT/2 (propagation) = 1.458 secs (approx)
b) To the above, we add the time for 1499 RTTs for a total of 1.46 + 119.92 = 121.38 secs
c) This is 74.5 RTTs, plus the two initial RTTS, for 6.12 secs
d) Right after the handshaking is done, we send one packet. One RTT after handshaking, we send 2
packets. At n RTTs past the handshaking, we have sent 1+2+4+... + 2n = 2(n+1) – 1. At n =10, we have
thus been able to send all 1500 pkts, the last batch arrives 0.5 RTT later. Total time is 2 + 10.5 RTTs or
1 sec.

bab 1 no 16
Suppose a 128 kbps p2p link is set up between earth and a rover on mars. The distance from the
earth to mars (when they are the closest together) is approximately 55Gm, and data travels over
the link at the speed of light 3X108 m/s
a) Calculate the minimum RTT for the link
b) Calculate the delay X bandwidth product of the link
c) A camera on the rover takes pictures of its surroundings and sends these to the earth. How
quickly can it reach Mission Control on Earth? Assume that each image is 5Mb in size

Solution:
a) Propagation Delay of the link is 55X109 / (3X108) = 184 secs, Thus RTT = 368 secs
b) The delay X bandwidth product for the link is the RTT X bandwidth = 23.5Mb
c) After a picture is taken, it must be transmitted on the link and completely propagated, before Mission
Control can interpret it. Transmit delay for 5Mb of data is 29 secs
Hence, total time = transmit delay + propagation delay = 223 secs

bab 1 no 19
Calculate the latency (from first bit sent to the lat bit received) for the following:
a) 1 Gbps Ethernet with a single store and forward switch in the path, and a packet size of
5000bits. Assume that each link introduces a propagation delay of 10 micro second and
that the switch begins retransmitting immediately after it has finished receiving the
packet.
b) Same as a) but with 3 switches
c) Same as b) but assume the switch implements 􀊠cut-through􀊡 switching. It is also able to
begin retransmitting the packet after the first 128 bits have been received.

Solution:
a) For each link it takes 1 Gbps/5 Kb = 5 μs to transmit the packet on the link, after which it takes an
additional 10 μs for the last bit to propagate across the link. Thus for a LAN with only with only one
switch that starts forwarding only after receiving the whole packet, the total transfer delay is the
transmit delays + two propagate delays = 30 μs.
b) For 3 switched and thus 4 links, the total delay is 4 transmit + 4 prop. Delay = 60 μs
c) For 􀊠cut-through􀊡, a switch needs to only decode the 128 bits before it begins to forward. This takes 128 ns. This delay replaces the switch transmit delays in the previous answer for a total delay of one transmit delay + 3 cut through decoding delays + 4 propagation delays = 45.384 μs

soal no 29
For the following, assume that no data compression is done. Calculate the bandwidth necessary
for transmitting in real time.
a) HDTV high-resolution video at resolution of 1920*1080, 24 bits/pixel, 30 frames / second.
b) POTS (Plain Old Telephone Service) voice audio of 8-bits samples at 8 KHz
c) GSM mobile voice audio of 260-bit samples at 50 Hz.
d) HDCD high-definition audio of 24-bit samples at 88.2 KHz
a) 1920 * 1080 * 24 * 30 ~= 115 Gbps
b) 8 * 8000 = 61 Kbps
c) 260 * 50 = 13 Kbps
d) 24 * 882000 ~= 2.1Mbps

Consider an application that transmits data at a steady rate (for example, the sender generates
an N-bit unit of data every k time units, where k is small and fixed). Also, when such an
application starts, it will continue running for relatively longer period of time. Answer the
following questions briefly justifying your answer:
a) Would a packet switched network or a circuit switched network more appropriate for this
application? Why?
b) Suppose that a packet switched network is used and the only traffic in this network comes
from such applications described above. Further assume that the sum of the application data
rates is less than the capacities of each and every link. Is some form of congestion control
needed? Why?
a) A circuit-switched network would be well suited to the application described, because
the application involves long sessions with predictable smooth bandwidth requirements.
Since the transmission rate is known and not bursty, bandwidth can be reserved for each
application session circuit with no significant waste. In addition, we need not worry
greatly about the overhead costs of setting up and tearing down a circuit connection,
which are amortized over the lengthy duration of a typical application session.
b) Given such generous link capacities, the network needs no congestion control mechanism. In the
worst (most potentially congested) case, all the applications simultaneously transmit over one or more
particular network links. However, since each link offers sufficient bandwidth to handle the sum of all
of the applications' data rates, no congestion (very little queuing) will occur.

Pertanyaan seputar jaringan komputer

Pertanyaan seputar PROXY...

Part 1

How does the proxy works?
The Proxy works between the client and the web server. It receives the requests from the clients, communicates with the web server on behalf of the client, receives the request back from the web server and passes it back to the client.

How does the client communicate with the Proxy?
Modify your browser proxy settings so that it isset to the proxy. The proxy listens to the port for incoming connections. On receipt of a connection from a client, it opens a socket for communication

How does the proxy reply to the Client?
After the proxy gets the reply from the proxy server, it writes back on the socket it had opened with the client.

How do I configure my Browser for sonnecting to the proxy?
You will also need to configure your web browser to use your proxy. This depends on your browser. In Internet Explorer, you can set the proxy in "Internet Options" in the Connections tab under LAN Settings. In Netscape (and derived browsers, such as Mozilla), you can set the proxy in Edit->Preferences and then select Advanced and Proxies.

In both cases you need to give the address of the proxy and the port number which you gave when you started the proxy. You can run the proxy and browser on the same computer without any problems.

I am confused on what is question #2 in Ethereal Lab is asking, "Within the IP packet header, what is the value in the upper layer protocol field?". Which field is it asking to look for in ethereal, I can't find something that says upper layer protocol field?

If you go throught the IP Header of a packet, it contains a field which says which transport protocol to deliver the packet to. The IP header in etheral too contains this information and is very much visible. Just see the packets closely.

sumber

http://www.cs.ucsb.edu/~sudipto/TA/cs176a/faq-extra.html