Question 11
In the context of parsing techniques, LL(1) refers to:
Step-by-Step Solution
Detailed Explanation:
In LL(1):
- First L: Left-to-right scan of input
- Second L: Leftmost derivation
- 1: 1 symbol lookaheadMatches Option B.
In the context of parsing techniques, LL(1) refers to:
Detailed Explanation:
In LL(1):
- First L: Left-to-right scan of input
- Second L: Leftmost derivation
- 1: 1 symbol lookaheadMatches Option B.
A host in an organization is connected to a host in another organization using 1 Gbps link with RTT = 25 ms. The network administrator tuned the network for 1000 Bytes packet size. To achieve channel utilization > 96%, what should be the window size (in number of packets)?
Detailed Explanation:
$Transmission time T_t = (1000 * 8 bits) / (1 * 10^9 bps) = 8 * 10^{-6} s = 0.008 ms.$
RTT = 25 ms.
$Utilization U = W * T_t / (RTT + T_t) >= 0.96.$
$W >= 0.96 * (25 + 0.008) / 0.008 \approx 3000.96 / 0.96 -> W = (25 / 0.008) = 3125 packets.$
In a computer network, the minimum frame size is 1500 bytes. What will be the data transfer speed in this network over 1 Km cable, if the signal speed in the cable is 250000 Km/sec and no repeaters are used?
Detailed Explanation:
Distance $d = 1 km$, speed $v = 250,000 km/s$.
Propagation delay $T_p$ = 1 / 250,000 = 4 microseconds.
For CSMA/CD, minimum frame size $T_t$ >= 2 * $T_p$ = 8 microseconds.
Frame size = 1500 bytes = 12,000 bits.
$Bandwidth B = 12,000 bits / 8 microseconds = 1.5 * 10^9 bps = 1.5 Gbps.$
For internal data communication through computer network, a 5 layer protocol hierarchy is proposed. Applications generate 270 bytes long messages. At each layer, a 6-byte header is added. What fraction of network bandwidth (in %) is wasted due to headers?
Detailed Explanation:
Total headers added = 5 layers * 6 bytes = 30 bytes.
Total packet size = 270 bytes data + 30 bytes headers = 300 bytes.
Bandwidth fraction wasted = (30 / 300) * 100% = 10%.
Match the OSI layers with their primary functions:
(i) Physical Layer -> (P) Bit transmission over physical medium
(ii) Data Link Layer -> (Q) Framing, error detection, node-to-node delivery
(iii) Network Layer -> (R) Routing packets across networks
(iv) Transport Layer -> (S) End-to-end process communication & flow control
Detailed Explanation:
Physical layer transmits raw bits.
Data Link layer manages frames and node-to-node delivery.
Network layer routes packets using IP addresses.
Transport layer provides process-to-process reliable transport.