PREVIOUS YEAR SOLVED PAPERS - GATE 2020
- Option : A
- Explanation : No. of host = 1500
No. of host bits = [log2 1500] = 11 bits
∴ Total possible hosts = 211 = 23 × 28
n is the netmask bits,
Range of addresses is = 232 – n
211 = 232-n
Available IP address ⇒ 202.61.0.0/17
So the IP address follow the pattern
0.0 to 7.255
8.0 to 15.255
16.0 to 23.255
:
64.0 to ...
:
104.0 to ...
∴ The possible IP addresses are 202.61.64.0/21 & 202.61.104.0/21
- Option : C
- Explanation : L1 L(m) = ϕ ⇒ emptiness problem of TM.TM is undecidable under emptiness.L2 = where a TM visits a particular state in finite steps in decidable, as we can do this with UTM.L3 = L(m) is non-Recursive,
Clearly from the diagramL(A) ⇒ non recursive language accepted by TML(B) ⇒ non-recursive language not accepted by TM.∴ it is a non-trivial property, hence undecidable.L4 = Undecidable problem using rice-theorem.Hence, L1, L3, and L4 are undecidable.
- Option : C
- Explanation : Given Hash Function ⇒ h1(k) = k mod 23 h2(k) = 1 + (k mod 19) Table size = 23 Key = 90 h1(k) = 90 mod 23 ≡ 21 h2(k) = 1 + 90 mod 19 ≡ 1 + 14 = 15 Double hashing, (h1(k) + i.h2(k)) mod 23 Asked for probe 1, put i = 1 (21 + 1. (15)) mod 23 36 mod 23 = 13
- Option : B
- Explanation : In non-persistent HTTP, each packet takes 2 RTT (Round trip Time): one for TCP connection, one or HTTP Text (Image file As, it is given text and 5 images that totals 6 objects.) So, it takes 12 RTT in total. But, 12 RTT includes 6 HTTP connections + 6TCP connections. So, the minimum number of TCP connections required is 6.
- Option : C
- Explanation : According to Lagrange’s theorem, state that for any finite group G, the order (number of element) of every subgroup t1 of G divides the order of G. therefore, possible subgroup of group of 35 elements. {1, 5, 7, 35}
