Monday, 30 October 2017
Sunday, 22 October 2017
CBSE UGC NET NETWORKING IMPORTANT POINTS
Important key points in networking:
- The connection points between the Network Layer and the Transport Layer is the IP number(s) whereas the connection points between the Transport Layer and the upper layers are called Ports which are represented by 16 bits in the TCP segment header.
- Unlike TCP the Transport Layer Protocol UDP is a connectionless protocol.
- The programmatic connection between the Network Layer and the upper (application) layers is usually established by small software running in Transport Layer. These are usually called Sockets.
- In a 802.3 LAN, IP to ethernet address translation is performed by/with ARP which stands for Address Resolution Protocol.
- In an IP address with CIDR standard a quad can range from 0 to 255.
- Bluetooth uses Time Division Multiplexing.
- A Bluetooth network with 7 active slaves is called a Piconet.
- Bluetooth standard specifies 13 applications which are called profiles.
- As modulation technology, Bluetooth employs FSK with 1M bps.
- The number of necessary OSI layers in a bridge is two
- Routers and Gateways are internetworking-connecting devices
- A networked computer must minimally have a host-IP, a netmask and default gateway-IP numbers set up in order to do IP-networking properly.
- An IP-address with all host bits set to 1 is called Broadcast address.
- A Class-B IP-range may have about 65000 hosts in it.
- IEEE-802.3 defines broadcast networks standards.
- The word ‗Base‘ in 10BaseT identifies the media as baseband.
- The letter ‗T‘ in 10BaseT indicates that it is a twisted-pair cable.
- An ethernet address with all bits set to 1 is called ethernet broadcast address.
- CIDR addressing scheme allows us to create subnets for efficient IP use.
- ARP is a process of obtaining ethernet addresses from IP-addresses.
- Performance of a network is usually measured by the amount of data transferred per unit time.
- Reliability of a network is usually measured by the frequency of failure (inverse of it).
- Logical to physical addressing conversion is done in the Network Layer.
- In stop and wait flow control method receiver ACKs each data frame received. The alternative to this is called sliding window.
- CSMA stands for “Carrier Sense Multiple Access” which is standardized in IEEE 802.3.
- The 10 Mbps twisted pair ethernet is usually designated by 10BaseT.
- Ethernet or NIC addresses are 48 bit unique numbers used to identify physical devices in CSMA/CD networks.
- Bridges divide the networks into smaller segments in order to reduce traffic.
- Repeaters and Bridges are networking connecting devices.
- In class-B IP-addresses 16 and 16 bits are reserved for host and network identification respectively.
- Topologically fully connected networks have the highest performance and the highest cost among the possible.
- In full-duplex transmission mode devices can transmit and receive simultaneously.
- For bit-rate to be three times the baud rate we need at least 8 constellations.
- Physical Layer deals with the physical and electrical specifications.
- The term point-to-point indicates the dedicated links between two nodes.
- Bus topology is the simplest and cheapest topology to implement in small networks.
- Full-Duplex transmission mode can be characterized simply as “two way simultaneous transmission/reception”.
- Performance of a network is usually measured by bits per second
- Switching/Routing is the job of Network Layer.
- Sliding Window line discipline where only some of the enquiries are acknowledged is more efficient then Stop-and-Wait.
- 10BaseT can have a maximum segment length of 100 meters.
- All NICs are manufactured having unique Hardware Address.
- Physical addressing, error control and access (to media) control are managed by the Data Link layer.
- An advantage of 10BaseT over 10Base2 is that 10Base2 is maintained easier than 10Base2.
- ‗Preamble‘ field at the beginning of an ethernet frame is used for synchronization.
- Bridges must have the following layers; Physical Layer 2 Data Link Layer.
- ‗Time To Live‘ field in an IP packet determines the number of hops (routers) it passes through before it is discarded.
- A host running in an IP-network on the ethernet must be assigned these three numbers properly in order for the networking software to operate correctly; IP address, 2. Netmask, 3. Gateway address.
- ARP is used on IEEE-802.3 networks in order to obtain ethernet addresses using the IP addresses.
- The field named as ‗Window Size‘ in a TCP segment header indicates the sliding window size.
- Standard TCP services use some standard TSAP numbers known as well known ports numbers.
- TCP is a connection oriented protocol while UDP is not.
- The three problems which limit a communication line are attenuation, distortion and noise.
- In order for a 2400 baud modem to achieve 9.6 kbps the constellation diagram must have at least 16 distinct points.
- The filter with a bandwidth of 300-3400 Hz at the end office for an ordinary voice phone line is removed for DSL operation.
- An ADSL modem and a splitter are the required equipment residing at the customer‘s premises for the ADSL system.
- Three of the BlueTooth profiles are Service Discovery, Serial Port, Cordless Telephony.
- In piconets of BlueTooth, Frequency Hopping Spread Spectrum technique is used with 1600
- hops/sec and hop sequence is dictated by the master.
- Pseudoternary encoding technique is used on T and S interfaces of ISDN system in order to maintain synchronization during the long sequences of zeros.
- PRI ISDN service carries 23 bearer and 1 data channels with a total of 1544 kbps.
- Bus topology usually requires terminators at both ends of the cable.
- In half-duplex transmission mode both station can transmit and receive but not at the same time.
- Logical addressing and routing is the function of network layer.
- Mail services are being made available by application layer.
- Manchester coding is one of the polar encoding techniques which effectively eliminate DC component of the signal.
- In Diff. Manchester, the transition in the middle is used for synchronization.
- The number of signal units per second is called baud-rate.
- In QAM both amplitude and phase of the carrier signal are varied.
- In stop-and-wait flow control technique, every frame is acknowledged.
- Physical Layer deals with the physical and electrical specifications.
- Full-duplex transmission mode can be characterized simply as “two way simultaneous transmission/reception”.
- All NICs are manufactured having unique ethernet address/MAC number.
- 10BaseT can have a maximum segment length of 100 meters.
- The words and numbers ’10’, ‗Base‘ and ‘T’ in 10BaseT respectively indicate 10 Mbps baseband and twisted pair.
- Carrier extension and Frame bursting are features added to 802.3 by the gigabit ethernet standards.
- 1000Base-T uses 4 pairs of Cat-5 UTP.
- Flooding is a packet routing method in which incoming packet is sent to every neighbor except where it came from.
- In Distance Vector Routing a router receives routing information from all of its neighbors and by using the knowledge about its distance to its neighbors it constructs its own routing table which in turn used by the router and distributed to the neighbors.
- Hierarchical Routing reduces the memory requirements at some penalty on the path optimality in large networks with large number of routers.
- If packets from a live audio/video source are to be distributed to multiple destinations we need to talk about Broadcast Routing.
- The 802.11 configuration in which no central coordination is used for is called Distributed Coordination Function
- Multipath Reception is a problem in wireless systems, which deteriorates the received signal at the receiver.
- When there is no central coordination is employed in 802.11, channel access privileges (who transmits when) are determined by a protocol called Carrier Sense Multiple Access / Collision Avoidance.
- In wireless ethernet, when stations directly talk to each other, a station wanting to transmit data when the channel is idle transmits Request To Send (RTS) frame first.
- The small clusters of stations communicating using Bluetooth are called Piconets.
- There can be only seven active slaves in a Bluetooth station cluster.
- The master station in a Bluetooth cluster employs Time Division Multiplexing in order to communicate with slaves and send commands to them.
- The destination address field in a Bluetooth frame is three bits
- Bluetooth operates on 2.4 GHz ISM band and its range is about 10 meters.
- The PSTN term Local Loop refers to the wiring between the customer and end office of the telecom company.
CBSE UGC NET THEORY OF COMPUTATION
THEORY OF AUTOMATA – JUNE 2013 – PAPER III
36. The grammar with production rules S →aSb |SS|λ generates language L given by :
(A) L = {w∈{a, b}* | na(w) = nb(w) and na(v) ≥ nb(v) where v is any prefix of w}
(B) L = {w∈{a, b}* | na(w) = nb(w) and na(v) ≤ nb(v) where v is any prefix of w}
(C) L = {w∈{a, b}* | na(w) ≠ nb(w) and na(v) ≥ nb(v) where v is any prefix of w}
(D)L = {w∈{a, b}* | na(w) ≠ nb(w) and na(v) ≤ nb(v) where v is any prefix of w}
Ans:- A
Explanation:-
Understanding and solving automata problems, will be a challenge. But if we understand the basics, solving these problems will not be a problem at all.
In order to derive the language L given a grammar with production rules, start deriving some strings starting with the start symbol S.
Example:-
Given the production rules,
S-> a S b | SS | λ
Let us start a derivation,
S -> a S b
-> a S S b ( After applying the production rule S -> SS)
-> a a S b S b ( After applying the production rule S -> a S b )
-> a a b S b (After applying the rule S -> λ)
->a a b b ( After applying the rule S -> λ)
So, we get the string, aabb which is accepted by the language.
Let us look at another derivation. (Always start with the start symbol)
S-> SS
-> a S b S ( After applying the rule S->aSb)
->a S b a S b (After applying the rule S->a S b)
->a b a S b (After applying the rule S -> λ)
->a b a b ( After applying the rule S-> λ)
So, we get the string abab, which is accepted by the language.
You can apply the production rules in any order, any number of times, you will get strings which are of the above two kinds. That is the number of ‘a’ and ‘b’ in the string will be the same.
L = {w∈{a, b}* | na(w) = nb(w) and na(v) ≥ nb(v) where v is any prefix of w}
In the above expression, L stands for the language. w is string of terminals. w is made up of zero or more occurrence of ‘a’ and ‘b’s. na(w) = nb(w) means the number of ‘a’ in the string w is equal to the number of ‘b’s which is true. Now, for understanding the second part of the expression, v is any prefix of w.
A prefix is a string of any number of leading symbols. For example, the string xyz has prefix (empty string), x, xy, xyz. When we consider the prefix v of w, the number of ‘a’s in it is either greater than or equal to ‘b’s in the string w.
For example, aab is prefix of the string aabb. Here, number of ‘a’s is more than the number of ‘b’s.
ab is a prefix of the string abab. The number of ‘a’ and ‘b’ are equal.
So, when we consider the prefix v, the number of ‘a’s and ‘b’s are equal. So, the correct answer is option A. In option B, the first part is correct, but the prefix part is wrong. Option C and D says that the string w will have unequal number of ‘a’ and ‘b’s which is not possible with the production rules given.
So, try to get some strings by applying the production rules, starting with the start symbol. Look for a pattern emerging in all those strings and try to answer these kind of questions.
37. A pushdown automation M = (Q,Σ, Γ, δ, q0, z, F) is set to be deterministic subject to which of the following condition(s), for every q ∈Q, a ∈Σ∪{λ} and b ∈Γ
(s1) δ(q, a, b) contains at most one element
(s2) if δ(q, λ, b) is not empty then δ(q, c, b) must be empty for every c ∈Σ
(A) Only s1
(B) Only s2
(C) Both s1 and s2
(D) Neither s1 nor s2
Ans:- C
Explanation:-
A DFA or NFA recognizes only regular language. They do not recognize context free languages. The automaton to recognize the CFL may require additional amount of storage which will be used to store the data. Since the DFA’s or NFA’s cannot count and cannot store the input for future reference, we are forced to have a new machine called Pushdown Automaton(PDA). PDA is an NFA with the exception that PDA has an extra stack. PDA is of two types, Deterministic and non deterministic. Unless otherwise explicitly mentioned, a PDA is non deterministic. The definition will be easier for you to understand if you know NFA better. Please refer to a good book on Theory of automata for understanding the basics well(I recommend Padma Reddy for the beginners).
The definition is understood in the following manner.
M = (Q,Σ, Γ, δ, q0, z, F)
M stands for the pushdown automaton.
Q is set of finite states
Σ – set of input alphabets
Γ – set of stack alphabets
δ – Transition function
q0 is the start state of the machine
z is the initial symbol on the stack
F is the set of final states.
In order for the PDA to be deterministic,both the conditions s1 and s2 must be satisfied. So, the correct answer is option C.
36. The grammar with production rules S →aSb |SS|λ generates language L given by :
(A) L = {w∈{a, b}* | na(w) = nb(w) and na(v) ≥ nb(v) where v is any prefix of w}
(B) L = {w∈{a, b}* | na(w) = nb(w) and na(v) ≤ nb(v) where v is any prefix of w}
(C) L = {w∈{a, b}* | na(w) ≠ nb(w) and na(v) ≥ nb(v) where v is any prefix of w}
(D)L = {w∈{a, b}* | na(w) ≠ nb(w) and na(v) ≤ nb(v) where v is any prefix of w}
Ans:- A
Explanation:-
Understanding and solving automata problems, will be a challenge. But if we understand the basics, solving these problems will not be a problem at all.
In order to derive the language L given a grammar with production rules, start deriving some strings starting with the start symbol S.
Example:-
Given the production rules,
S-> a S b | SS | λ
Let us start a derivation,
S -> a S b
-> a S S b ( After applying the production rule S -> SS)
-> a a S b S b ( After applying the production rule S -> a S b )
-> a a b S b (After applying the rule S -> λ)
->a a b b ( After applying the rule S -> λ)
So, we get the string, aabb which is accepted by the language.
Let us look at another derivation. (Always start with the start symbol)
S-> SS
-> a S b S ( After applying the rule S->aSb)
->a S b a S b (After applying the rule S->a S b)
->a b a S b (After applying the rule S -> λ)
->a b a b ( After applying the rule S-> λ)
So, we get the string abab, which is accepted by the language.
You can apply the production rules in any order, any number of times, you will get strings which are of the above two kinds. That is the number of ‘a’ and ‘b’ in the string will be the same.
L = {w∈{a, b}* | na(w) = nb(w) and na(v) ≥ nb(v) where v is any prefix of w}
In the above expression, L stands for the language. w is string of terminals. w is made up of zero or more occurrence of ‘a’ and ‘b’s. na(w) = nb(w) means the number of ‘a’ in the string w is equal to the number of ‘b’s which is true. Now, for understanding the second part of the expression, v is any prefix of w.
A prefix is a string of any number of leading symbols. For example, the string xyz has prefix (empty string), x, xy, xyz. When we consider the prefix v of w, the number of ‘a’s in it is either greater than or equal to ‘b’s in the string w.
For example, aab is prefix of the string aabb. Here, number of ‘a’s is more than the number of ‘b’s.
ab is a prefix of the string abab. The number of ‘a’ and ‘b’ are equal.
So, when we consider the prefix v, the number of ‘a’s and ‘b’s are equal. So, the correct answer is option A. In option B, the first part is correct, but the prefix part is wrong. Option C and D says that the string w will have unequal number of ‘a’ and ‘b’s which is not possible with the production rules given.
So, try to get some strings by applying the production rules, starting with the start symbol. Look for a pattern emerging in all those strings and try to answer these kind of questions.
37. A pushdown automation M = (Q,Σ, Γ, δ, q0, z, F) is set to be deterministic subject to which of the following condition(s), for every q ∈Q, a ∈Σ∪{λ} and b ∈Γ
(s1) δ(q, a, b) contains at most one element
(s2) if δ(q, λ, b) is not empty then δ(q, c, b) must be empty for every c ∈Σ
(A) Only s1
(B) Only s2
(C) Both s1 and s2
(D) Neither s1 nor s2
Ans:- C
Explanation:-
A DFA or NFA recognizes only regular language. They do not recognize context free languages. The automaton to recognize the CFL may require additional amount of storage which will be used to store the data. Since the DFA’s or NFA’s cannot count and cannot store the input for future reference, we are forced to have a new machine called Pushdown Automaton(PDA). PDA is an NFA with the exception that PDA has an extra stack. PDA is of two types, Deterministic and non deterministic. Unless otherwise explicitly mentioned, a PDA is non deterministic. The definition will be easier for you to understand if you know NFA better. Please refer to a good book on Theory of automata for understanding the basics well(I recommend Padma Reddy for the beginners).
The definition is understood in the following manner.
M = (Q,Σ, Γ, δ, q0, z, F)
M stands for the pushdown automaton.
Q is set of finite states
Σ – set of input alphabets
Γ – set of stack alphabets
δ – Transition function
q0 is the start state of the machine
z is the initial symbol on the stack
F is the set of final states.
In order for the PDA to be deterministic,both the conditions s1 and s2 must be satisfied. So, the correct answer is option C.
CBSE UGC NET FUZZY LOGIC NUMERICAL QUESTIONS
JUNE 2012 – PAPER III Q.No 6
6. If two fuzzy sets A and B are given with membership functions μA(x) = {0.2, 0.4, 0.8, 0.5, 0.1} μB(x) = {0.1, 0.3, 0.6, 0.3, 0.2} Then the value of μ ––– will be A∩B
(A) {0.9, 0.7, 0.4, 0.8, 0.9}
(B) {0.2, 0.4, 0.8, 0.5, 0.2}
(C) {0.1, 0.3, 0.6, 0.3, 0.1}
(D) {0.7, 0.3, 0.4, 0.2, 0.7}
Ans:-A
Explanation:- The fuzzy intersection of two fuzzy sets A and B on universe of discourse X: μA∩B(x) = min [μA(x), μB(x)] , where x∈X But here in the question, they are asking for complement of A intersection B and so the answer would be 1-min[A(x),B(x)].
The minimum of 0.2 and 0.1 will be 0.1, and 1-0.1 will be 0.9
The second value is min(0.4,0.3)=0.3 and 1-0.3=0.7
The third value is min(0.8,0.6)=0.6 and 1-0.6=0.4
The fourth value is min(0.5,0.3)=0.3 and 1-0.3=0.7
The last value is min(0.1,0.2)=0.1 and 1-0.1=0.9
The only option which has got the values 0.9,0.7,0.4,0.7 and 0.9, although the fourth value is given as 0.8 instead of 0.7 is option A.
So the answer is option A.
DECEMBER 2012 – PAPER III Q.No 13
13. Consider a fuzzy set A defined on the interval x=[0,10] of integers by the membership function.
µA(x) = x / x+ 2
α cut corresponding to α = 0.5 will be
(A) { 0,1,2,3,4,5,6,7,8,9,10}
(B) {1,2,3,4,5,6,7,8,9,10}
(C) {2,3,4,5,6,7,8,9,10}
(D) { }
Ans:- C
Explanation:-
In the fundamentals, refer to the answer given for question no. 6 regarding α-cut.
α-cut of a fuzzy set A denoted as Aα, is the crisp set comprised of the elements x of a universe of discourse X for which the membership function of A is greater than or equal to α.
Given, x = In the range [0,10]
Membership function = x/x+2
Calculate the value of membership function for the interval from 0 to 10, substituting in the formula x/x+2.
µA(0) = 0 / 0+ 2 = 0
µA(1) = 1 / 1+ 2 = 0.33
µA(2) = 2 / 2+ 2 = 0.5
µA(3) = 3 / 3+ 2 = 0.6
µA(4) = 4 / 4+ 2 = 0.66
µA(5) = 5 / 5+ 2 = 0.71
µA(6) = 6 / 6+ 2 = 0.75
µA(7) = 7 / 7+ 2 = 0.77
µA(8) = 8 / 8+ 2 = 0.8
µA(9) = 9 / 9+ 2 = 0.81
µA(10) = 10 / 10+ 2 = 0.83
α= 0.5. We have to find the corresponding α-cut,
That will be a crisp set, having those values of x, for which the membership function is returning a value of 0.5 or above.
µA(2) = 0.5 and all the values of x above 2 is getting a value greater than 0.5. So the crisp set will contain the following values.
{ 2,3,4,5,6,7,8,9,10}.
So the correct answer is C.
DECEMBER 2013 – PAPER III Q.No 28
28. If A and B are two fuzzy sets with membership functions μA(x) = {0.2, 0.5, 0.6, 0.1, 0.9} μB(x) = {0.1, 0.5, 0.2, 0.7, 0.8} Then the value of μA ∩B
will be
(A) {0.2, 0.5, 0.6, 0.7, 0.9}
(B) {0.2, 0.5, 0.2, 0.1, 0.8}
(C) {0.1, 0.5, 0.6, 0.1, 0.8}
(D) {0.1, 0.5, 0.2, 0.1, 0.8}
Ans:-D
Explanation:-
Intersection of two fuzzy sets
µA ∩B (x) = µA(x) ^ µB(x) = min(µA(x), µB(x))
μA(x) = {0.2, 0.5, 0.6, 0.1, 0.9}
μB(x) = {0.1, 0.5, 0.2, 0.7, 0.8}
μA ∩B={0.1,0.5,0.2,0.1,0.8}
So, the correct answer is D.
29. The height h(A) of a fuzzy set A is defined as h(A) =sup A(x) where x belongs to A. Then the fuzzy set A is called normal when
(A)h(A)=0
(B)h(A)<0
(C)h(A)=1
(D)h(A)<1
Ans:- C
Explanation:-
Explanation:- The height of a fuzzy set is the highest membership value of the membership function: Height(A) = max µA(xi)
A fuzzy set with height 1 is called a normal fuzzy set.
In contrast, a fuzzy set whose height is less than 1 is called a subnormal fuzzy set. So, according to the above rule, the fuzzy set A is called normal when h(A)=1.
So, the correct answer is 1.
JUNE 2013 – PAPER III Q.No 74
74. If A and B are two fuzzy sets with membership functions μA(x) = {0.6, 0.5, 0.1, 0.7, 0.8} μB(x) = {0.9, 0.2, 0.6, 0.8, 0.5}
Then the value of μ Complement A∪B(x) will be
(A) {0.9, 0.5, 0.6, 0.8, 0.8}
(B) {0.6, 0.2, 0.1, 0.7, 0.5}
(C) {0.1, 0.5, 0.4, 0.2, 0.2}
(D){0.1,0.5,0.4,0.2,0.3}
Ans:- C
Union of two fuzzy sets
µAUB(x) = µA(x) V µB(x) = max(µA(x), µB(x))
μA(x) = {0.6, 0.5, 0.1, 0.7, 0.8}
μB(x) = {0.9, 0.2, 0.6, 0.8, 0.5}
µAUB(x) = {0.9,0.5,0.6,0.8,0.8}
Complement of µAUB(x)={0.1,0.5,0.4,0.2,0.2}
So, the correct answer is C.
JUNE 2014 – PAPER III Q.No 7,8 7.
Given U = {1, 2, 3, 4, 5, 6, 7} A = {(3, 0.7), (5, 1), (6, 0.8)} then
~ A will be : (where ~ →complement)
(A) {(4, 0.7), (2, 1), (1, 0.8)}
(B) {(4, 0.3), (5, 0), (6, 0.2) }
(C) {(1, 1), (2, 1), (3, 0.3), (4, 1), (6, 0.2), (7, 1)}
(D) {(3, 0.3), (6.0.2)}
Ans:- C
Explanation:-
Complement of a fuzzy set
The complement of a fuzzy set A is a new fuzzy set A Complement, containing all the elements which are in the universe of discourse but not in A, with the membership function
Complement of µA(x) = 1 – µA(x)
Complement of a fuzzy set A is a new fuzzy set A complement. Since it is a fuzzy set, there will be two members in a singleton. The first member will be all the elements which are in the universe of discourse but not in A. The membership function will be 1- µA(x).
So, the complement of A will be
{(1,1),(2,1),(3,0.3),(4,1),(6,0.2),(7,1)}
The first is (1,1). The first 1 is in U but not in A, so it should be added in the complement. The second 1 is because the membership function is 1- µA(x). 1-0=1.
The same reason why you get (2,1).
The third one (3,0.3) because it is (3,1-0.7)=(3,0.3).
Same reason why you have (4,1) and (7,1).
(6,1-0.8)=(6,0.2).
The member (5,0) is not included because , a singleton whose membership to a fuzzy set is 0, can be excluded .
8. Consider a fuzzy set old as defined below
old={(20,0),(30,0.2),(40,0.4),(50,0.6),(60,0.8),(70,1),(80,1)}. Then the alpha-cut for alpha=0.4 for the set old will be (A){(40,0.3)}
(B){50,60,70,80}
(C){(20,0.1),(30,0.2)}
(D){(20,0),(30,0),(40,1),(50,1),(60,1),(70,1),(80,1)}
Ans:-D
Explanation:-
alpha-cut of a fuzzy set A will contain those elements where the membership function value is equal to or greater than alpha.
Here, alpha is given a value 0.4. Starting from (40,0.4) all the members have membership function equal or greater than 0.4. so, except
(20,0) and (30,0.2) all the menbers are included in the alpha-cut of the fuzzy set. The only option which has 40,50,60,70, and 80 included is option D. It has
(20,0) and (30,0) too. But it is already noted that any singleton where the membership function is 0 can be considered not included. So basically these two members are not part of the alpha-cut of the fuzzy set A. So the correct option is D.
6. If two fuzzy sets A and B are given with membership functions μA(x) = {0.2, 0.4, 0.8, 0.5, 0.1} μB(x) = {0.1, 0.3, 0.6, 0.3, 0.2} Then the value of μ ––– will be A∩B
(A) {0.9, 0.7, 0.4, 0.8, 0.9}
(B) {0.2, 0.4, 0.8, 0.5, 0.2}
(C) {0.1, 0.3, 0.6, 0.3, 0.1}
(D) {0.7, 0.3, 0.4, 0.2, 0.7}
Ans:-A
Explanation:- The fuzzy intersection of two fuzzy sets A and B on universe of discourse X: μA∩B(x) = min [μA(x), μB(x)] , where x∈X But here in the question, they are asking for complement of A intersection B and so the answer would be 1-min[A(x),B(x)].
The minimum of 0.2 and 0.1 will be 0.1, and 1-0.1 will be 0.9
The second value is min(0.4,0.3)=0.3 and 1-0.3=0.7
The third value is min(0.8,0.6)=0.6 and 1-0.6=0.4
The fourth value is min(0.5,0.3)=0.3 and 1-0.3=0.7
The last value is min(0.1,0.2)=0.1 and 1-0.1=0.9
The only option which has got the values 0.9,0.7,0.4,0.7 and 0.9, although the fourth value is given as 0.8 instead of 0.7 is option A.
So the answer is option A.
DECEMBER 2012 – PAPER III Q.No 13
13. Consider a fuzzy set A defined on the interval x=[0,10] of integers by the membership function.
µA(x) = x / x+ 2
α cut corresponding to α = 0.5 will be
(A) { 0,1,2,3,4,5,6,7,8,9,10}
(B) {1,2,3,4,5,6,7,8,9,10}
(C) {2,3,4,5,6,7,8,9,10}
(D) { }
Ans:- C
Explanation:-
In the fundamentals, refer to the answer given for question no. 6 regarding α-cut.
α-cut of a fuzzy set A denoted as Aα, is the crisp set comprised of the elements x of a universe of discourse X for which the membership function of A is greater than or equal to α.
Given, x = In the range [0,10]
Membership function = x/x+2
Calculate the value of membership function for the interval from 0 to 10, substituting in the formula x/x+2.
µA(0) = 0 / 0+ 2 = 0
µA(1) = 1 / 1+ 2 = 0.33
µA(2) = 2 / 2+ 2 = 0.5
µA(3) = 3 / 3+ 2 = 0.6
µA(4) = 4 / 4+ 2 = 0.66
µA(5) = 5 / 5+ 2 = 0.71
µA(6) = 6 / 6+ 2 = 0.75
µA(7) = 7 / 7+ 2 = 0.77
µA(8) = 8 / 8+ 2 = 0.8
µA(9) = 9 / 9+ 2 = 0.81
µA(10) = 10 / 10+ 2 = 0.83
α= 0.5. We have to find the corresponding α-cut,
That will be a crisp set, having those values of x, for which the membership function is returning a value of 0.5 or above.
µA(2) = 0.5 and all the values of x above 2 is getting a value greater than 0.5. So the crisp set will contain the following values.
{ 2,3,4,5,6,7,8,9,10}.
So the correct answer is C.
DECEMBER 2013 – PAPER III Q.No 28
28. If A and B are two fuzzy sets with membership functions μA(x) = {0.2, 0.5, 0.6, 0.1, 0.9} μB(x) = {0.1, 0.5, 0.2, 0.7, 0.8} Then the value of μA ∩B
will be
(A) {0.2, 0.5, 0.6, 0.7, 0.9}
(B) {0.2, 0.5, 0.2, 0.1, 0.8}
(C) {0.1, 0.5, 0.6, 0.1, 0.8}
(D) {0.1, 0.5, 0.2, 0.1, 0.8}
Ans:-D
Explanation:-
Intersection of two fuzzy sets
µA ∩B (x) = µA(x) ^ µB(x) = min(µA(x), µB(x))
μA(x) = {0.2, 0.5, 0.6, 0.1, 0.9}
μB(x) = {0.1, 0.5, 0.2, 0.7, 0.8}
μA ∩B={0.1,0.5,0.2,0.1,0.8}
So, the correct answer is D.
29. The height h(A) of a fuzzy set A is defined as h(A) =sup A(x) where x belongs to A. Then the fuzzy set A is called normal when
(A)h(A)=0
(B)h(A)<0
(C)h(A)=1
(D)h(A)<1
Ans:- C
Explanation:-
Explanation:- The height of a fuzzy set is the highest membership value of the membership function: Height(A) = max µA(xi)
A fuzzy set with height 1 is called a normal fuzzy set.
In contrast, a fuzzy set whose height is less than 1 is called a subnormal fuzzy set. So, according to the above rule, the fuzzy set A is called normal when h(A)=1.
So, the correct answer is 1.
JUNE 2013 – PAPER III Q.No 74
74. If A and B are two fuzzy sets with membership functions μA(x) = {0.6, 0.5, 0.1, 0.7, 0.8} μB(x) = {0.9, 0.2, 0.6, 0.8, 0.5}
Then the value of μ Complement A∪B(x) will be
(A) {0.9, 0.5, 0.6, 0.8, 0.8}
(B) {0.6, 0.2, 0.1, 0.7, 0.5}
(C) {0.1, 0.5, 0.4, 0.2, 0.2}
(D){0.1,0.5,0.4,0.2,0.3}
Ans:- C
Union of two fuzzy sets
µAUB(x) = µA(x) V µB(x) = max(µA(x), µB(x))
μA(x) = {0.6, 0.5, 0.1, 0.7, 0.8}
μB(x) = {0.9, 0.2, 0.6, 0.8, 0.5}
µAUB(x) = {0.9,0.5,0.6,0.8,0.8}
Complement of µAUB(x)={0.1,0.5,0.4,0.2,0.2}
So, the correct answer is C.
JUNE 2014 – PAPER III Q.No 7,8 7.
Given U = {1, 2, 3, 4, 5, 6, 7} A = {(3, 0.7), (5, 1), (6, 0.8)} then
~ A will be : (where ~ →complement)
(A) {(4, 0.7), (2, 1), (1, 0.8)}
(B) {(4, 0.3), (5, 0), (6, 0.2) }
(C) {(1, 1), (2, 1), (3, 0.3), (4, 1), (6, 0.2), (7, 1)}
(D) {(3, 0.3), (6.0.2)}
Ans:- C
Explanation:-
Complement of a fuzzy set
The complement of a fuzzy set A is a new fuzzy set A Complement, containing all the elements which are in the universe of discourse but not in A, with the membership function
Complement of µA(x) = 1 – µA(x)
Complement of a fuzzy set A is a new fuzzy set A complement. Since it is a fuzzy set, there will be two members in a singleton. The first member will be all the elements which are in the universe of discourse but not in A. The membership function will be 1- µA(x).
So, the complement of A will be
{(1,1),(2,1),(3,0.3),(4,1),(6,0.2),(7,1)}
The first is (1,1). The first 1 is in U but not in A, so it should be added in the complement. The second 1 is because the membership function is 1- µA(x). 1-0=1.
The same reason why you get (2,1).
The third one (3,0.3) because it is (3,1-0.7)=(3,0.3).
Same reason why you have (4,1) and (7,1).
(6,1-0.8)=(6,0.2).
The member (5,0) is not included because , a singleton whose membership to a fuzzy set is 0, can be excluded .
8. Consider a fuzzy set old as defined below
old={(20,0),(30,0.2),(40,0.4),(50,0.6),(60,0.8),(70,1),(80,1)}. Then the alpha-cut for alpha=0.4 for the set old will be (A){(40,0.3)}
(B){50,60,70,80}
(C){(20,0.1),(30,0.2)}
(D){(20,0),(30,0),(40,1),(50,1),(60,1),(70,1),(80,1)}
Ans:-D
Explanation:-
alpha-cut of a fuzzy set A will contain those elements where the membership function value is equal to or greater than alpha.
Here, alpha is given a value 0.4. Starting from (40,0.4) all the members have membership function equal or greater than 0.4. so, except
(20,0) and (30,0.2) all the menbers are included in the alpha-cut of the fuzzy set. The only option which has 40,50,60,70, and 80 included is option D. It has
(20,0) and (30,0) too. But it is already noted that any singleton where the membership function is 0 can be considered not included. So basically these two members are not part of the alpha-cut of the fuzzy set A. So the correct option is D.
Wednesday, 4 October 2017
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