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Department of Electrical Engineering and Computer Science
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
6.893 Fall 2009
Quiz II
All problems are open-ended questions. In order to receive credit you must answer the question as precisely as possible. You have 80 minutes to finish this quiz. Write your name on this cover sheet. Some questions may be harder than others. Read them all through first and attack them in the order that allows you to make the most progress. If you find a question ambiguous, be sure to write down any assumptions you make. Be neat. If we can’t understand your answer, we can’t give you credit!
THIS IS AN OPEN BOOK, OPEN NOTES EXAM.
Please do not write in the boxes below.
I (xx/10) II (xx/30) III (xx/10) IV (xx/10)
V (xx/10) VI (xx/20) VII (xx/10) Total (xx/100)
Name: Solutions.
I KeyKOS
- [10 points]: Bob is running the privilege-separated Zoobar web site on a KeyNIX system, using code from lab 3. Suggest a way in which Bob can modify the Zoobar server-side code to take advantage of KeyKOS capabilities to improve the security of his site, in a way that he wouldn’t be able to do on Linux. Answer: Two important advantages of KeyKOS are that a process doesn’t have to be root to do various things (e.g. for zookld to create a service in a chroot jail, by only granting it a capability to that specific directory, and not granting a capability to the top-level root directory), and that a process can protect itself from the Unix root user (e.g. by creating a separate KeyNIX universe, which will be secure even if the root account in the original KeyNIX universe is compromised.)
Now, Alice wants to read Bob’s email, and intercepts all network packets ever sent and received by Bob’s workstation (which is the only computer that Bob uses). However, Alice does not know Bob’s password to access Bob’s post office server, and Bob’s packets to and from the post office server are protected by Kerberos.
- [10 points]: Suppose that after Bob reads and deletes all of his mail, Alice learns what Bob’s password was. Describe how Alice can obtain Bob’s past messages. Answer: Alice can use the password to decrypt Bob’s TGT (which she captured in the past), then use the key in the TGT to decrypt Bob’s service tickets, and then use the session key in the service ticket to decrypt Bob’s mail traffic.
- [10 points]: To prevent Alice from reading any more messages, Bob ensures that Alice cannot intercept any subsequent network traffic, and changes his Kerberos password. Could Alice still read Bob’s mail after this? Explain why not or explain how.
Answer: Alice can continue to read Bob’s mail (by connecting to Bob’s post office server) until her ticket for Bob’s principal expires. After that point, she will not be able to connect to the post office server as Bob.
Also, she can exploit slow slave replication and obtain fresh tickets for Bob’s principal from a slave KDC.
IV BitLocker
- [10 points]: Alice wants to make BitLocker run faster. She decides that computing a different IVs for each sector (pg. 13 in the BitLocker paper) is needlessly expensive, and replaces it with the fixed value E(KAES, e(0)) instead. Explain how an attacker may be able to leverage this change to obtain data from a stolen laptop that uses BitLocker in TPM-only mode. Answer: BitLocker’s sector encryption algorithm is still sector-specific—namely, the sector plaintext is XORed with a sector key Ks, as shown in Figure 1 on page 13 and described in Section 4.3 on page 14. However, since the AES-CBC key is the same for each sector, an attacker may be able to compute the XOR of two sector keys, by swapping the two encrypted sectors on disk. If the attacker swaps encrypted sectors s 1 and s 2 , whose original plaintexts were p 1 and p 2 , then the new decryption of s 2 will be p^ i = p 1 ⊕ Ks i 2 1 ⊕^ Ks 2. If the attacker knows^ p^1 , and can read the new value^ p^2 , then he can deduce Ks 1 ⊕ Ks 2. At this point, if the attacker wants to place malicious data m 1 in sector s 1 , he can place the value m 1 ⊕ Ks 1 ⊕ Ks 2 in sector s 2 and swap the two sectors again. To be able to set a particular range of bytes in some sector to a given value, the attacker must be able to read and write the same range of byte offsets in some other sector inside the OS (e.g. being able to read and write a file is sufficient). Thanks to Stephen Woodrow for pointing out a problem with our previous solution.
V Tor
- [10 points]: Bob is running a hidden service on top of Tor, and wants to know how frequently he should choose new introduction points. Bob cares about his identity not being exposed, and about the availability of his service. Help Bob make an informed choice by explaining the costs and benefits of rotating introduction points either more or less frequently. Answer: Rotating introduction points more frequently helps avoid DoS attacks on a fixed set of in troduction points. Rotation also helps prevent a single introduction point from gaining long-term statistics on how often the service is accessed. Rotation does not improve Bob’s anonymity, because Bob can keep building new circuits to the same introduction point. More frequent rotation places additional load on directory services that provide lookup functionality. However, this does not com promise anonymity either, since lookups and updates happen via anonymous Tor circuits as well.
- [10 points]: The VM-based BackTracker system requires no modifications to the guest OS, but nonetheless makes assumptions about the guest OS. List assumptions that are critical to back-tracking attacks that used high-control events.
Answer: Backtracker assumes that it can observe all system calls (i.e. it depends on a specific system call mechanism), that it can observe system call arguments, and that it can access kernel data structures for things like process IDs, inodes, etc. Backtracker also assumes the guest kernel is not compromised; one could think of this as a special case of the attacker changing the system call format.
VII 6.
We’d like to hear your opinions about 6.893, so please answer the following questions. (Any answer, except no answer, will receive full credit.)
- [10 points]: If you could change one thing in 6.893, what would it be? Answer: Homework questions for each paper. More labs....
End of Quiz
