NWPGCL Assistant Manager (ICT) Exam 12 January 2024 — Question Solution | নর্থ-ওয়েস্ট পাওয়ার (এনডব্লিউপিজিসিএল) সহকারী ব্যবস্থাপক (আইসিটি) নিয়োগ পরীক্ষা ২০২৪ — প্রশ্ন সমাধান

Given: IP = 192.168.1.50, subnet mask = 255.255.255.240

Step 1: mask in binary and prefix length

255.255.255.240 = 11111111.11111111.11111111.11110000 → 28 network bits → /28, leaving 4 host bits.

Step 2: block size (subnet increment)

Block size = 256 − 240 = 16 (= 2⁴). Subnets in the last octet start at 0, 16, 32, 48, 64, …

Step 3: find the block that holds .50

48 ≤ 50 < 64, so 192.168.1.50 lies in the block 48–63.

Subnet	Network	Host range	Broadcast
previous	192.168.1.32	.33 – .46	192.168.1.47
this one	192.168.1.48	.49 – .62	192.168.1.63
next	192.168.1.64	.65 – .78	192.168.1.79

Step 4: binary check

IP        11000000.10101000.00000001.00110010   192.168.1.50
Mask      11111111.11111111.11111111.11110000   255.255.255.240
AND  ->   11000000.10101000.00000001.00110000   192.168.1.48   (network)

Wildcard  00000000.00000000.00000000.00001111   0.0.0.15
Network OR wildcard
     ->   11000000.10101000.00000001.00111111   192.168.1.63   (broadcast)

The last octet shows it directly: 50 = 0011 0010; keeping the top 4 bits and zeroing the host bits gives 0011 0000 = 48; setting the host bits to 1 gives 0011 1111 = 63.

Step 5: usable hosts

Usable hosts = 2⁴ − 2 = 14 (the network and broadcast addresses cannot be given to a host).

Answer

Item	Value
Prefix	/28 (block size 16)
Network address	192.168.1.48
Broadcast address	192.168.1.63
First valid host	192.168.1.49
Last valid host	192.168.1.62
Valid host range	192.168.1.49 – 192.168.1.62 (14 hosts)

Quick check: 48 + 16 − 1 = 63 for the broadcast, and 62 − 49 + 1 = 14 hosts.

Exam tip: block size = 256 − (mask octet that is not 255); network = the largest multiple of the block size not above the IP's octet; broadcast = next boundary − 1; hosts lie strictly between them.

This is a decision under risk: each option has possible costs with known probabilities, so we compare the expected cost (expected monetary value, EMV) and choose the lowest, because these are costs, not profits.

Expected cost = Σ (probability × cost)

flowchart LR
  D{"Decision: how to get the software?"}
  D -->|"Buy"| B["Cost Tk 50 lakh (certain)"]
  D -->|"Build"| C1(("Chance node: EV 47"))
  D -->|"Modify open-source"| C2(("Chance node: EV 46"))
  C1 -->|"Easy, p = 0.3"| E1["Tk 40 lakh"]
  C1 -->|"Hard, p = 0.7"| E2["Tk 50 lakh"]
  C2 -->|"Small changes, p = 0.2"| E3["Tk 30 lakh"]
  C2 -->|"Large changes, p = 0.8"| E4["Tk 50 lakh"]

Each chance node's probabilities add up to 1 (0.3 + 0.7 = 1; 0.2 + 0.8 = 1), so the data are consistent.

Step-by-step (roll back the tree)

Option	Outcomes	Expected cost
Buy	50 with p = 1	1 × 50 = 50.0 lakh
Build	40 (p = 0.3), 50 (p = 0.7)	0.3 × 40 + 0.7 × 50 = 12 + 35 = 47.0 lakh
Modify open-source	30 (p = 0.2), 50 (p = 0.8)	0.2 × 30 + 0.8 × 50 = 6 + 40 = 46.0 lakh

Decision: modify the open-source software. Its expected cost (Tk 46 lakh) is the lowest: Tk 1 lakh below building and Tk 4 lakh below buying.

Why this is also safe

• Worst case: every option can end at Tk 50 lakh, so choosing to modify adds no extra downside risk.
• Best case: modifying can cost only Tk 30 lakh, the lowest possible outcome.

Sensitivity (how firm is the choice?)

• Let s = probability of small changes. EV(modify) = 30s + 50(1 − s) = 50 − 20s. Modify beats build (47) when 50 − 20s < 47, i.e. s > 0.15. With s = 0.2 the margin is small.
• Let e = probability that building is easy. EV(build) = 50 − 10e. Build would win only if 50 − 10e < 46, i.e. e > 0.4.

So if the estimate of small changes falls below 0.15, or the chance that building is easy rises above 0.4, the choice flips to building.

Other factors a manager should weigh (not in the numbers): delivery time, licence terms of the open-source code, in-house skills to maintain it, vendor support and security updates.

Exam tip: draw the tree, multiply each outcome by its probability, add them at each chance node, and pick the minimum for costs (maximum for profits).