Soal Deadlock Sistem Operasi

1.  Apa yang dimaksud dengan sumber daya ? Berikan contohnya!

Sumber daya (resources) adalah sesuatu proses yang digunakan untuk proses-proses untuk menyelesaikan task.Sumber daya yang ada pada sistem terdiri dari tipe resource CPU cycle, ruang memori, perangkat I/O yang disebut dengan tipe sumber daya R1, R2, . . ., Rm.

2.  Apa yang dimaksud deadlock ?

Deadlock adalah proses dimana saling terjadinya penungguan antara proses-proses yang ingin menggunakan resources (sumber daya), proses yang sedang menunggu tidak mau saling mengalah maka akan terjadi deadlock.

3.  Sebutkan 4 kondisi yang menyebabkan deadlock?

a.    Mutual Exclusion

hanya satu proses pada satu waktu yang dapat menggunakan sumberdaya.

b.    Genggam dan Tunggu (Hold and Wait)

suatu proses membawa sedikitnya satu sumberdaya menunggu mendapatkan tambahan sumber daya baru yang dibawa oleh proses.

c.    Non-Preemption

sebuah sumber daya dapat dibebaskan dengan sukarela oleh proses yangmemegangnya setelah proses menyelesaikan task.

d.    Menunggu Secara Sirkuler (Circular Wait)

Terdapat sekumpulan proses {P0, P1, …, P0} yang menunggu sumber daya dimana P0 menunggu sumber daya yang dibawa P1, P1 menunggu sumber daya yang dibawa P2, dan seterusnya, Pn–1 menunggu sumber dayayang dibawa oleh Pn, dan Pn menunggu sumber daya yang dibawa P0.

4.  Sebutkan cara mencegah deadlock dari 4 kondisi tersebut pada soal 3?

a. Mencegah Mutual Exclusion

Mutual exclusion benar-benar tak dapat dihindari. Hal ini dikarenakan tidak ada sumber dayayang dapat digunakan bersama-sama, jadi sistem harus membawa sumber daya yang tidak dapat digunakan bersama-sama.

b. Mencegah Hold and Wait

Untuk mencegah hold and wait, sistem harus menjamin bila suatu proses meminta sumberdaya, maka proses tersebut tidak sedang memegang sumber daya yang lain. Proses harus meminta dan dialokasikan semua sumber daya yang diperlukan sebelum proses memulai eksekusi atau mengijinkan proses meminta sumber daya hanya jika proses tidak membawasumber daya lain. Model ini mempunyai utilitas sumber daya yang rendah dan kemungkinan terjadi starvation jika proses membutuhkan sumber daya yang popular sehingga terjadi keadaan menunggu yang tidak terbatas karena setidaknya satu dari sumberdaya yang dibutuhkannya dialokasikan untuk proses yang lain.

c. Mencegah Non Preemption

Peniadaan non preemption mencegah proses-proses lain harus menunggu. Seluruh proses menjadi preemption, sehingga tidak ada tunggu menunggu. Cara mencegah kondisi non preemption : Jika suatu proses yang membawa beberapa sumber daya meminta sumber daya lainyang tidak dapat segera dipenuhi untuk dialokasikan pada proses tersebut, maka semuasumber daya yang sedang dibawa proses tersebut harus dibebaskan. Proses yang sedang dalam keadaan menunggu, sumber daya yang dibawanya ditunda dan ditambahkan pada daftar sumber daya. Proses akan di restart hanya jika dapat memperoleh sumber daya yang lama dan sumberdaya baru yang diminta.

d. Mencegah Kondisi Menunggu Sirkular

Sistem mempunyai total permintaan global untuk semua tipe sumber daya. Proses dapat meminta proses kapanpun menginginkan, tapi permintaan harus dibuat terurut secara numerik. Setiap proses yang membutuhkan sumber daya dan memintanya maka nomor urut akan dinaikkan. Cara ini tidak akan menimbulkan siklus. Masalah yang timbul adalah tidak ada cara pengurutan nomor sumber daya yang memuaskan semua pihak.

5.  Diketahui snapshot dari suatu sistem :

Allocation Max Available

A B C D A B C D A B C D

P0 0 0 1 2 0 0 1 2 1 5 2 0

P1 1 0 0 0 1 7 5 0

P2 1 3 5 4 2 3 5 6

P3 1 6 3 2 1 6 5 2

P4 0 0 1 4 0 6 5 6

Jawablah pertanyaan berikut :

a. Bagaimana isi matrik Need ?

b. Apakah sistem dalam state selamat ?

c. Jika proses P1 meminta (0,4,2,0) dapatkah permintaan dipenuhi segera ?

a. Isi matrik Need didefinisikan dengan Max – Allocation :

Need

A B C

P0 7 4 3

P1 1 2 2

P2 6 0 0

P3 0 1 1

P4 4 3 1

Sistem dalam keadaan state selamat dengan urutan <> yang memenuhi kriteria algoritma safety.

Misalnya proses P1 meminta tambahan anggota tipe sumber daya A dan dua anggota tipesumber daya C sehingga Request1 = (1, 0, 2). Untuk menentukan apakah permintaan dapat segera dipenuhi, pertama harus diperiksa apakah Request1 ≤ Available ((1, 0, 2) ≤ (3, 3, 2)) ternyata benar. Maka akan diperoleh state baru berikut :

Allocation Need Available

A B C A B C A B C

P0 0 1 0 7 4 3 2 3 0

P1 3 0 2 0 2 0

P2 3 0 1 6 0 0

P3 2 1 1 0 1 1

P4 0 0 2 4 3 1

Kemudian harus ditentukan apakah sistem berada dalam state selamat. Setelah mengeksekusi algoritma safety ternyata urutan memenuhi criteria safety. Setelah sistem berada pada state doatas, permintaan (3, 3, 0) oleh P4 tidak dapat dipenuhi karena sumber daya tidak tersedia. Permintaan (0, 2, 0) oleh P1 juga tidak dapat dipenuhi karena meskipun sumber daya tersedia, state hasil tak selamat.

b. Sistem dalam state tidak selamat

c. Dapat

WireShark Lab HTTP

1. The Basic HTTP GET/response interaction

Let’s begin our exploration of HTTP by downloading a very simple HTML file - one that is very short, and contains no embedded objects. Do the following:

1.      Start up your web browser.

2.      Start up the Wireshark packet sniffer, as described in the Introductory lab (but don’t yet begin packet capture). Enter “http” (just the letters, not the quotation marks) in the display-filter-specification window, so that only captured HTTP messages will be displayed later in the packet-listing window. (We’re only interested in the HTTP protocol here, and don’t want to see the clutter of all captured packets).

3.      Wait a bit more than one minute (we’ll see why shortly), and then begin Wireshark packet capture.

4.      Enter the following to your browser http://gaia.cs.umass.edu/wireshark-labs/HTTP-wireshark-file1.html Your browser should display the very simple, one-line HTML file.

5.      Stop Wireshark packet capture.

Your Wireshark window should look similar to the window shown in Figure 1. If you are unable to run Wireshark on a live network connection, you can download a packet trace that was created when the steps above were followed.²

1.      Is your browser running HTTP version 1.0 or 1.1? What version of HTTP is the server running?

Jawab : Browser saya versi http adalah 1.1 , server juga versi http 1.1

2.      What languages (if any) does your browser indicate that it can accept to the server?

Jawab : language = en-us , en

3.      What is the IP address of your computer? Of the gaia.cs.umass.edu server?

Jawab :

Ip saya : 192.168.1.7

IP gaia.cs.umass.edu server : 128.119.245.12

4.      What is the status code returned from the server to your browser?

Jawab : Http /1.1 200 OK (text/html)

5.      When was the HTML file that you are retrieving last modified at the server?

Jawab : Last modified = sun, 21 jan 2018 06:59:01 GMT

6.      How many bytes of content are being returned to your browser?

Jawab : content-length = 128

7.      By inspecting the raw data in the packet content window, do you see any headers within the data that are not displayed in the packet-listing window? If so, name one.

Jawab : Tidak semua header bisa ditemukan dalam data raw

2. The HTTP CONDITIONAL GET/response interaction

Recall from Section 2.2.6 of the text, that most web browsers perform object caching and thus perform a conditional GET when retrieving an HTTP object. Before performing the steps below, make sure your browser’s cache is empty. (To do this under Firefox, select Tools->Clear Recent History and check the Cache box, or for Internet Explorer, select Tools->Internet Options->Delete File; these actions will remove cached files from your browser’s cache.) Now do the following:

•       Start up your web browser, and make sure your browser’s cache is cleared, as discussed above.

•       Start up the Wireshark packet sniffer

•       Enter the following URL into your browser http://gaia.cs.umass.edu/wireshark-labs/HTTP-wireshark-file2.html Your browser should display a very simple five-line HTML file.

•       Quickly enter the same URL into your browser again (or simply select the refresh button on your browser)

•       Stop Wireshark packet capture, and enter “http” in the display-filter-specification window, so that only captured HTTP messages will be displayed later in the packet-listing window.

•       (Note: If you are unable to run Wireshark on a live network connection, you can use the http-ethereal-trace-2 packet trace to answer the questions below; see footnote 1. This trace file was gathered while performing the steps above on one of the author’s computers.)

Answer the following questions:

8.      Inspect the contents of the first HTTP GET request from your browser to the server. Do you see an “IF-MODIFIED-SINCE” line in the HTTP GET?

Jawab : Tidak

9.      Inspect the contents of the server response. Did the server explicitly return the contents of the file? How can you tell?

Jawab : Ya karena kita bisa melihat isinya di field Line-based text data

10.  Now inspect the contents of the second HTTP GET request from your browser to the server. Do you see an “IF-MODIFIED-SINCE:” line in the HTTP GET? If so, what information follows the “IF-MODIFIED-SINCE:” header?

Jawab : Ya , informasinya adalah sun, 21 jan 2018 06:59:01 GMT yang merupakan tanggal modifikasi terakhir file dari GET request sebelumnya

11.  What is the HTTP status code and phrase returned from the server in response to this second HTTP GET? Did the server explicitly return the contents of the file? Explain.

Jawab : status code = 304 dan phrase = Not Modified. Server tidak mengembalikan isi file karena browser memuatnya dari cache-nya

3. Retrieving Long Documents

In our examples thus far, the documents retrieved have been simple and short HTML files. Let’s next see what happens when we download a long HTML file. Do the following:

•       Start up your web browser, and make sure your browser’s cache is cleared, as discussed above.

•       Start up the Wireshark packet sniffer

•       Enter the following URL into your browser http://gaia.cs.umass.edu/wireshark-labs/HTTP-wireshark-file3.html Your browser should display the rather lengthy US Bill of Rights.

•       Stop Wireshark packet capture, and enter “http” in the display-filter-specification window, so that only captured HTTP messages will be displayed.

•       (Note: If you are unable to run Wireshark on a live network connection, you can use the http-ethereal-trace-3 packet trace to answer the questions below; see footnote 1. This trace file was gathered while performing the steps above on one of the author’s computers.)

Answer the following questions:

12.  How many HTTP GET request messages did your browser send? Which packet number in the trace contains the GET message for the Bill or Rights?

Jawab : ada 1 HTTP GET

13.  Which packet number in the trace contains the status code and phrase associated with the response to the HTTP GET request?

Jawab : di trace saya paket nomor 15

14.  What is the status code and phrase in the response?

Jawab : Status code 200 ,phrase OK

15.  How many data-containing TCP segments were needed to carry the single HTTP response and the text of the Bill of Rights?

Jawab : Ada 5 data TCP segment yaitu 1440 , 1440, 1440 dan 541 totalnya 4861

4.  HTML Documents with Embedded Objects

Now that we’ve seen how Wireshark displays the captured packet traffic for large HTML files, we can look at what happens when your browser downloads a file with embedded objects, i.e., a file that includes other objects (in the example below, image files) that are stored on another server(s).

Do the following:

•       Start up your web browser, and make sure your browser’s cache is cleared, as discussed above.

•       Start up the Wireshark packet sniffer

•       Enter the following URL into your browser http://gaia.cs.umass.edu/wireshark-labs/HTTP-wireshark-file4.html

Your browser should display a short HTML file with two images. These two images are referenced in the base HTML file. That is, the images themselves are not contained in the HTML; instead the URLs for the images are contained in the downloaded HTML file. As discussed in the textbook, your browser will have to retrieve these logos from the indicated web sites. Our publisher’s logo is retrieved from the www.aw-bc.com web site. The image of the cover for our 5^th edition (one of our favorite covers) is stored at the manic.cs.umass.edu server.

•       Stop Wireshark packet capture, and enter “http” in the display-filter-specification window, so that only captured HTTP messages will be displayed.

Answer the following questions:

16.  How many HTTP GET request messages did your browser send? To which Internet addresses were these GET requests sent?

Jawab : ada 4 , alamat ip yang menjadi Get request adalah 128.119.245.12 ada 2 dan 128.119.245.90 ada 2

17.  Can you tell whether your browser downloaded the two images serially, or whether they were downloaded from the two web sites in parallel? Explain.

Jawab : Dengan memeriksa port TCP kita dapat melihat apakah file kita diunduh secara serial atau paralel. Dalam kasus ini, 2 gambar dikirimkan melalui 2 koneksi TCP sehingga mereka diunduh secara serial.

5 HTTP Authentication

Finally, let’s try visiting a web site that is password-protected and examine the sequence of HTTP message exchanged for such a site. The URL http://gaia.cs.umass.edu/wireshark-labs/protected_pages/HTTP-wireshark-file5.html is password protected. The username is “wireshark-students” (without the quotes), and the password is “network” (again, without the quotes). So let’s access this “secure” password-protected site. Do the following:

•       Make sure your browser’s cache is cleared, as discussed above, and close down your browser. Then, start up your browser

•       Start up the Wireshark packet sniffer

•       Enter the following URL into your browser http://gaia.cs.umass.edu/wireshark-labs/protected_pages/HTTP-wireshark-file5.html

Type the requested user name and password into the pop up box.

•       Stop Wireshark packet capture, and enter “http” in the display-filter-specification window, so that only captured HTTP messages will be displayed later in the packet-listing window.

•       (Note: If you are unable to run Wireshark on a live network connection, you can use the http-ethereal-trace-5 packet trace to answer the questions below; see footnote 2. This trace file was gathered while performing the steps above on one of the author’s computers.)

Answer the following questions:

18.  What is the server’s response (status code and phrase) in response to the initial HTTP GET message from your browser?

Jawab : status code 401, phrase = Unauthorized

19.  When your browser’s sends the HTTP GET message for the second time, what new field is included in the HTTP GET message?

Jawab : field baru di pesan HTTP GET yang kedua adalah

Authorization: Basic d2lyZXNoYXJrLXN0dWRlbnRzOm5ldHdvcms=\r\n

Download Laporan WireShark Lab HTTP 

Praktikum WireShark Lab:UDP

WireShark Lab UDP

1.      Select one UDP packet from your trace. From this packet, determine how many fields there are in the UDP header. (You shouldn’t look in the textbook! Answer these questions directly from what you observe in the packet trace.) Name these fields.

Jawab :

Ada 4 kolom : source port , destination port, length dan checksum

2.      By consulting the displayed information in Wireshark’s packet content field for this packet, determine the length (in bytes) of each of the UDP header fields.

Jawab :

Header UDP memiliki panjang 8 byte. Masing-masing dari 4 kolom header ini berukuran 2 byte

3.      The value in the Length field is the length of what? (You can consult the text for this answer). Verify your claim with your captured UDP packet.

Jawab :

Panjang filed menentukan jumlah byte di segmen UDP (header plus data). Nilai panjang eksplisit diperlukan karena ukuran field data mungkin berbeda dari satu segmen UDP ke yang berikutnya.

Panjang muatan UDP untuk paket yang dipilih adalah 43 bytes.51 bytes - 8 bytes = 43 bytes.

4.      What is the maximum number of bytes that can be included in a UDP payload? (Hint: the answer to this question can be determined by your answer to 2. above)

Jumlah byte maksimum yang dapat disertakan dalam muatan UDP adalah (2 ^ 16 - 1) byte ditambah byte header.

(2^16-1) = ( 65536 – 1)  = 65535 bytes

65535 – panjang header UDP =  65535 bytes – 8 bytes = 65527 bytes

5.      What is the largest possible source port number? (Hint: see the hint in 4.)

Jawab : (2^16 – 1) = 65535.

6.      What is the protocol number for UDP? Give your answer in both hexadecimal and decimal notation. To answer this question, you’ll need to look into the Protocol field of the IP datagram containing this UDP segment (see Figure 4.13 in the text, and the discussion of IP header fields).

Jawab :

Protocol number UDP decimal adalah 17 , sedangkan hexadecimal adalah  0x11 hex

     

7.      Examine a pair of UDP packets in which your host sends the first UDP packet and the second UDP packet is a reply to this first UDP packet. (Hint: for a second packet to be sent in response to a first packet, the sender of the first packet should be the destination of the second packet). Describe the relationship between the port numbers in the two packets.

Jawab :

Source Port dari paket UDP yang dikirim oleh host sama dengan destination port dari paket reply, dan sebaliknya, destination port dari paket UDP yang dikirim oleh host sama dengan source port dari paket reply.

                                Paket Reply