Transcript Chapters 7 – 9: PowerPoint slides - ECE

Computer Networks
Ivan Marsic
Rutgers University
Chapter 7 – Network Security
Chapter 8 – Network Monitoring
Chapter 9 – Internet Protocols
APPENDIX: Probability Refresher
Network Security
Chapter 7
Topic:
Secure Communication
 Network Security Problem
 Symmetric and Public-Key Cryptosystems
 Cryptographic Algorithms
 Authentication
Network Security Problem
Secure/Confidential Communication ?
Padlock
and shared
key copy
Content
Shared
key copy
Message
Sender
Sender needs:
• receive securely a copy of the shared key
• positively identify the message receiver
Intermediary
Threats posed by intruder/adversary:
• forge the key and view the content
• damage/substitute the padlock
• damage/destroy the message
• observe characteristics of messages
(statistical and/or metric properties)
Receiver
Receiver needs:
• receive securely a shared key copy
• positively identify the message sender
• detect any tampering with messages
4
Objectives of Information Security
• Confidentiality: information not disclosed or
revealed to unauthorized persons
• Integrity: consistency of data—preventing
unauthorized creation, modification, or destruction
• Availability: legitimate users are not unduly
denied access to resources, including information
resources, computing resources, and
communication resources
• Authorized use: resources are not used by
unauthorized persons or in unauthorized ways
Message Encoding and Decoding
• Encoding takes a message M and produces a
coded form f(M)
• Decoding the message requires an inverse
1
 f (M )= M.
f
function , such that
Two Basic Types of Cryptosystems
• Symmetric systems: both parties use the
same (secret) key in encryption and
decryption transformations
• Public-key systems (aka asymmetric
systems): the parties use two related keys,
one of which is secret and the other can be
publicly disclosed
Public-Key Cryptosystem
Sender’s
padlock
1. Sender secures the briefcase
Receiver’s
padlock
with his/her padlock and sends
Sender
2. Receiver additionally secures
Receiver
the briefcase with his/her
padlock and returns
3. Sender removes his/her
padlock and sends again
4. Receiver removes his/her
padlock to access the content
8
Public-Key Cryptosystem - mod
“Public key”
Receiver distributes his/her padlock (unlocked)
to sender ahead of time, but keeps the key
“Private key”
Receiver
Receiver’s
padlock (unlocked)
Receiver’s
key
Sender uses the receiver’s padlock
to secure the briefcase and sends
Sender
Receiver removes his/her
padlock to access the content
9
Public-Key Cryptography
In RSA, receiver does the following:
•
Randomly select two large prime numbers p and q, which always must
be kept secret.
•
Select an integer number E, known as the public exponent, such that
(p  1) and E have no common divisors, and (q  1) and E have no
common divisors.
•
Determine the product n = pq, known as public modulus.
•
Determine the private exponent, D, such that (ED  1) is exactly
divisible by both (p  1) and (q  1). In other words, given E, we
choose D such that the integer remainder when ED is divided by (p
 1)(q  1) is 1.
•
Release publicly the public key, which is the pair of numbers n and E,
K = (n, E). Keep secret the private key, K = (n, D).
Example: send the plaintext “hello world”
• receiver chooses p = 5 and q = 7
• receiver chooses E = 5, because 5 and (5  1)(7
 1) have no common factors. Also, n = pq = 35
• receiver chooses D = 29, because
•
5291
ED1
144


i.e., they are exactly divisible.
( p1)( q1)
46
24
receiver’s public key is K  = (n, E) = (35, 5),
which is made public. The private key K  = (n,
D) = (35, 29) is kept secret.
6
Example, cont’d
Encryption
Plaintext numeric
Plaintext letter
BE
Ciphertext BE % n
representation
h
8
85 = 32768
85 % 35 = 8
e
5
55 = 3125
55 % 35 = 10
l
12
125 = 248832
125 % 35 = 17
l
12
248832
17
o
15
155 = 759375
155 % 35 = 15
Decryption
Ciphertext
CD
B = CD % n
Plaintext letter
8
829 = 154742504910672534362390528
829 % 35 = 8
h
10
100000000000000000000000000000
5
e
17
481968572106750915091411825223071697
12
l
17
481968572106750915091411825223071697
12
l
15
12783403948858939111232757568359375
15
o
Example, cont’d
• While the adversary knows n and E, he or
she does not know p and q, so they cannot
work out (p  1)(q  1) and thereby find D.
Topic:
Authentication
 Network Security Problem
 Symmetric and Public-Key Cryptosystems
 Cryptographic Algorithms
 Authentication
Authentication Protocol (1)
Assumption: Only Sender needs to be authenticated to Receiver, not mutually.
Secure communication is not enough … playback attack:
Sender
Receiver’s
public key
Adversary
Receiver
EncryptPK-receiversender-ID, password
record
ACK
Replayed
message
replay
EncryptPK-receiversender-ID, password
ACK
Authentication Protocol (2)
Solution to playback attack:
Sender
Receiver’s
public key
Adversary
number used
once and
never again
EncryptPK-receiversender-ID, password
Sender’s
public key
record
Receiver’s
public key
EncryptPK-sendernonce1
EncryptPK-receivernonce1
ACK
replay
EncryptPK-receiversender-ID, password
EncryptPK-sendernonce2
Does not know how to reply!
Receiver
Impersonation Attack
PROBLEM: Public key distribution … Adversary impersonates Bank
Customer
Adversary
I am Bank and here is my public key
Bank
Assumption:
Adversary obtained Bank’s public key
Adversary’s
public key
EncryptPK-adversarycustomer-ID, password, PK-customer
Customer’s
public key
Decrypt Customer’s message
and obtain ID & password
Bank’s
public key
Adversary’s
public key
EncryptPK-bankcustomer-ID, password, PK-adversary
Customer’s
public key
Decrypt Bank’s message and
obtain Customer account info
EncryptPK-adversarycustomer-account-info
EncryptPK-customercustomer-account-info
PROBLEM: Customer unaware that Adversary obtained his account info!
Network Monitoring
Chapter 8
Packet-pair Dispersion
t4 t3
t2 t1
Pkt-2
t4 t3
Pkt-2
Pkt-1
P2 P1
Send
packet
pair
Link 1
Receive
packet
pair
Link 2
Router 1
Pkt-1
Router 2
Link 3
Same spacing is preserved on higher speed links
= Time to process P bytes packet
Minimum packet spacing
at bottleneck link
P
Packets
Flow direction
t1 t2
t3
Link speed estimation = P/
t4
Internet Protocols
Chapter 9
The Internet Reference Model
Visit http://en.wikipedia.org/wiki/Internet_reference_model for more details on the Internet reference model
http://en.wikipedia.org/wiki/OSI_model
IPv6 Header
0
3 4
version
number
11 12
8-bit traffic class
16-bit payload length
15 16
31
20-bit flow label
next header
8-bit hop limit
128-bit (16-byte) source IP address
40
bytes
128-bit (16-byte) destination IP address
IPv6 Address Prefix Assignments
0
7 8
127
00000000
Reserved
Anything
0
127
00000000
Unspecified
...
00000000
0
127
00000000
Loopback within this network
0
...
7 8
0
Link-local use unicast
Anything
9 10
127
11111110 10
0
Site-local use unicast
127
11111111
Multicast addresses
00000001
Anything
9 10
127
11111110 11
Anything
0
127
Everything else
Global unicast
IPv4 compatible address
(Node supports IPv6 & IPv4)
IPv4 mapped address
(Node does not support IPv6)
0
95 96
00000000
0
...
00000000
79 80
95 96
000000 ... 000000 111...11
127
IPv4 Address
127
IPv4 Address
IPv6 Global Unicast Address
0
IPv6 global unicast address
general format
(n bits)
(m bits)
(128nm bits)
global routing prefix
subnet ID
interface ID
127
(a)
0
IPv6 global unicast address
format for prefix not “000”
(n bits)
(64n bits)
(64 bits)
global routing prefix
subnet ID
interface ID
(b)
127
Example IPv6 Extension Headers
Mandatory
IPv6 header
IPv6 main header
40 bytes
= Next Header field
Optional
extension
headers
Hop-by-hop
options header
variable
Routing header
variable
Fragment header
8 bytes
Destination
options header
variable
TCP header
IPv6 packet
payload
application data payload
20 bytes (default)
variable
Format of IPv6 Extension Headers
0
7 8
Next header
15 16
31
0
7 8
15 16
Next header Hdr ext len
Hdr ext len
23 24
0
Segments left
Reserved
One or more options
Address[1]
(a) Hop-by-Hop Options header;
Destinations Options header
0
7 8
Next header
15 16
Reserved
28 29 31
Fragment offset
Res M
Address[2]
Identification
(b) Fragment header
0
7 8
Next header
15 16
23 24
31
31
Hdr ext len Routing type Segments left
Type-specific data
Address[n]
(c) Generic Routing header
(d) Type 0 Routing header
RIP Header (for IPv4)
0
7 8
command
15 16
version
31
unused (must be zero)
RIP header
address family identifier
route tag
8
bytes
IPv4 address
subnet mask
RIP route entry
next hop
distance metric
Total up to 25 route entries
16
bytes
OSPF Directed Graph of an AS
AS
N3
AS
(a)
7
4
H1
1
D
6
5
N2
B
2
N1
1
1
1
A
3
H2
C
8
6
N2
D
5
7
N1
(b)
2
1
1
B
1
4
1
A
3
C
8
H2
N3
OSPF Header (for IPv4)
0
7 8
version
15 16
31
type
packet length
source router address (IPv4 )
area ID
OSPF header
checksum
authentication type
authentication
OSPF packet payload
24
bytes
OSPF - LSA Header
0
7 8
15 16
LS age
31
options
type
link state ID
20
bytes
LS sequence number
LSA header
authentication
LS checksum
0
flags
length
0
number of links
link ID
Link description
for LSA type = 1
link data
link type
num_TOS
optional TOS information
(more link descriptions)
metric
16
bytes
eBGP and iBGP Sessions
AS 
AS 
B
E
G
C
J
A
H
F
D
I
AS 
AS 
P
L
O
Q
Key:
K
N
M
Link-layer connection
eBGP TCP session
iBGP TCP session
BGP Finite State Machine
ManualStart OR
AutomaticStart /
Idle
ConnectRetryTimer
expired / retry
Setting up
TCP connection
DelayOpenTimer expires OR
DelayOpen attribute == FALSE /
{Connect, Active}
OPEN or msg recvd /
send KEEPALIVE msg
ManualStop /
TcpConnectionFails /
Opening
BGP session
KEEPALIVE
msg recvd /
{OpenSent, OpenConfirm}
KEEPALIVE or
UPDATE msg recvd /
ManualStop OR AutomaticStop OR
HoldTimer expired /
send NOTIFICATION cease
Established
ManualStop OR AutomaticStop OR Error in msg detected OR
NOTIFICATION error recvd /
Detail from Figure 1-49:
AS 
N
}
S
, A
{AS
AS 
AS 
S
R
t}
{Cust
BGP Header & Message Formats
15 16
0
23 24
31
Marker
Length
Type
(a) BGP header format
0
7 8
15 16
23 24
31
0
15 16
Marker
Length
31
Marker
Type: OPEN
My autonomous system
23 24
Version
Length
Type:
KEEPALIVE
(c) BGP KEEPALIVE message format
Hold time
BGP identifier
Optional params
length
0
7 8
15 16
23 24
31
Optional parameters (variable)
(b) BGP OPEN message format
Marker
Length
Type:
NOTIFICATION
Error code
Error subcode
Data (variable)
(d) BGP NOTIFICATION message format
BGP UPDATE Message
0
15 16
7 8
23 24
31
Marker
Attribute type (2 bytes)
Length
Type: UPDATE
Withdrawn routes length
Attrib. length (1 or 2 bytes)
Attribute value (variable)
(b) Path attribute format
Withdrawn routes (variable)
Total path attribute length
Path attributes (variable)
Network layer reachability information (variable)
(a) BGP UPDATE message format
Attribute flags
OTPE
0
Attribute
type code
Extended Length
Partial
Transitive
Optional
(c) Attribute type format
Example BGP UPDATE Message
192.12.69.2
Prefix
128.34.10.0/24 192.12.62.1
AS 
AS 
192.12.62.2
L
AS BGP routing table:
192.12.69.1
Prefix
Path
O
.5
.5 0
2
2.1
19
K
Next Hop
128.34.10.0/24 {AS} 192.12.69.2
+ 
K’s IGP routing table:
Destination
Cost
Next Hop
128.34.10.0/24
0
K’s BGP
M
K’s forwarding table:
Prefix
Next Hop
Next Hop
128.34.10.0/24 192.12.69.2
N
2.1
2.6
1
.
2
19
/16
}
8.34 S, AS 62.1
2
1
.12.
= {A
fix =
Pre PATH = 192
P
AS_ T_HO
X
NE
UP
TE
DA
AS BGP routing table
N’s forwarding table:
Prefix
Next Hop
128.34.10.0/24 192.12.50.5

AS 
O’s forwarding table:
6
9.2
4/1 } 2.6
E
8.3 S 2.1
12 = {A 19
AT
PD fix = TH P =
U
e A O
Pr _ P _ H
AS EXT
N
Subnet Prefix =
128.34.10.0/24
+
N’s IGP routing table:
Destination
Cost Next Hop
128.34.10.0/24
2
Router M
Router K
2
Router M
AS 
M
AS 
K
N
AT
P
AS refix E
M _P = so
ED AT m
= H = e pre
10 {A fix
0 S in
,A A
S S 
}
D
L
UP
re
AS fix= so
_PA me
ME
T
D = H = { prefix
30 AS, in AS
0

AS
}
UP
DA
TE
P
BGP MULTI_EXIT_DISC (MED)
Attribute
AS 
AS 
B
AS 
E
A
AS 
AS 
G
C
H
D
F
AS 
Address Resolution Protocol (ARP)
Need for multiple addresses, hierarchical vs. non-hierarchical
Vehicle identification number (VIN)
1P3BP49K7JF111966
Registration plate
Address Resolution Protocol (ARP)
IP: 192.200.96.21
MAC: 01-23-45-67-89-AB
IP: 192.200.96.22
MAC: 00-01-03-1D-CC-F7
IP: 192.200.96.20
MAC: 49-BD-2F-54-1A-0F
Sender
ARP Request: to FF-FF-FF-FF-FF-FF
Target
Sender MAC: 0101-2323-4545-6767-8989-AB
Sender IP: 192.200.96.21
Target IP: 192.200.96.23
ARP Reply
Sender MAC: A3A3-B0B0-2121-A1A1-6060-35
Sender IP: 192.200.96.23
Target MAC: 0101-2323-4545-6767-8989-AB
Target IP: 192.200.96.21
IP: 192.200.96.23
MAC: A3-B0-21-A1-60-35
ARP Packet Format (for IPv4)
0
7 8
15 16
Hardware type = 1
Hardware addr len = 6
31
Protocol type = 0x0800
Protocol addr len = 4
Operation
Sender hardware address (6 bytes)
Sender protocol address (first 2 bytes)
Sender protocol address (last 2 bytes)
Target hardware address (6 bytes)
Target protocol address
28
bytes
Mobile IP
Home Agent (HA)
Mobile node (MN)
1
2
4
3
Foreign Agent (FA)
Correspondent node (CN)
SNMP
Network
MIB
(a)
Messages
Agent
Managed
objects
Managed device
Network management system (NMS)
GetRequest
Response
GetNextRequest
Response
GetBulkRequest
(b)
SNMP
SNMP
manager
manager
(client)
(client)
Response
SetRequest
Response
Trap
NMS
InformRequest
SNMP
SNMP
agent
agent
(server)
(server)
Probability Refresher
Appendix
Jar with Black & White Balls
Random Events
Possible outcomes
of two coin tosses:
“Tree diagram” of possible
outcomes of two coin tosses:
Second toss
H
Second toss
H
T
H
½
H
HH
HT
First toss
T
H
First toss
½
T
TH
TT
½
½
T
½
½
T
(a)
(b)
Outcome
HH
HT
TH
TT
Drawing from Jar/Urn Decided by
Rolling a Die
EXPERIMENT 1:
Roll a die; if outcome
is 1 or 2, select Jar;
else, select Urn
EXPERIMENT 2:
Draw a ball from the
selected container
Jar
Urn
Probability Matrix for Ball Drawing
Random variable X: Color of the ball
c2
Random variable Y:
Identity of the vessel
that will be chosen
x1 = Black
x2 = White
y1 = Jar
n11
n12
y2 = Urn
n21
n22
r1
Illustration for Bayes Theorem
EXPERIMENT 1:
Roll a die; if outcome
is 1 or 2, select Jar;
else, select Urn
EXPERIMENT 2:
Draw a ball from the
selected container
Guess whether the
ball was drawn from
Jar or from Urn
Jar
Urn
Poisson Process
percent of occurrences (%)
average arrival rate  = 5
20
15
10
5
0
0
1
2
3
4 5 6 7 8 9 10 11 12 13 14 15
arrivals per time unit (n)
Partitioning of Areas Under
Normal Curve
34.13%
0.4
Areas between
selected points under
the normal curve
34.13%
0.3
13.59%
0.2
0.1
0.13%
13.59%
2.15%
2.15%
0.13%
0.0
4
3
2

0

2
3
4
How to Read Table A-1
Area between mean and z
(from Column B)
0.4
(A)
z
0.3
Area beyond z
(from Column C)
0.2
43.32%
0.1
6.68%
0.0
4
3
2
1
0
1
2
z = 1.50
(in Column A)
(A)
3
4
z
(B)
area
between
mean and z
(C)
area
beyond
z
(A)
z
(B)
area
between
mean and z
(C)
area
beyond
z
(A)
z
(B)
area
between
mean and z
(C)
area
beyond
z
0.00
0.01
0.02
.0000
.0040
.0080
.5000
.4960
.4920
0.55
0.56
0.57
.2088
.2123
.2157
.2912
.2877
.2843
1.10
1.11
1.12
.3643
.3665
.3686
.1357
.1335
.1314
0.34
0.35
0.36
0.37
0.38
0.39
0.40
0.41
0.42
0.43
0.44
0.45
0.46
0.47
0.48
0.49
0.50
0.51
0.52
0.53
0.54
.1331
.1368
.1406
.1443
.1480
.1517
.1554
.1591
.1628
.1664
.1700
.1736
.1772
.1808
.1844
.1879
.1915
.1950
.1985
.2019
.2054
.3669
.3632
.3594
.3557
.3520
.3483
.3446
.3409
.3372
.3336
.3300
.3264
.3228
.3192
.3156
.3121
.3085
.3050
.3015
.2981
.2946
0.89
0.90
0.91
0.92
0.93
0.94
0.95
0.96
0.97
0.98
0.99
1.00
1.01
1.02
1.03
1.04
1.05
1.06
1.07
1.08
1.09
.3133
.3159
.3186
.3212
.3238
.3264
.3289
.3315
.3340
.3365
.3389
.3413
.3438
.3461
.3485
.3508
.3531
.3554
.3577
.3599
.3621
.1867
.1841
.1814
.1788
.1762
.1736
.1711
.1685
.1660
.1635
.1611
.1587
.1562
.1539
.1515
.1492
.1469
.1446
.1423
.1401
.1379
1.44
1.45
1.46
1.47
1.48
1.49
1.50
1.51
1.52
1.53
1.54
1.55
1.56
1.57
1.58
1.59
1.60
1.61
1.62
1.63
1.64
.4251
.4265
.4279
.4292
.4306
.4319
.4332
.4345
.4357
.4370
.4382
.4394
.4406
.4418
.4429
.4441
.4452
.4463
.4474
.4484
.4495
.0749
.0735
.0721
.0708
.0694
.0681
.0668
.0655
.0643
.0630
.0618
.0606
.0594
.0582
.0571
.0559
.0548
.0537
.0526
.0516
.0505
1.50
(B)
area
between
mean and z
.4332
(C)
area
beyond
z
.0668