Pubali bank Recruitment Test Short Question Answerfor Junior officer
1(a) Cache Memory Address Fields
Given:
- Main memory = 4 GB
- Byte-addressable
- Cache line (block) size = 64 bytes
- Number of cache lines = 1024
- Direct-mapped cache
Step 1: Address size
Final Answer
| Field | Number of Bits |
|---|---|
| Tag | 16 bits |
| Line Index | 10 bits |
| Block Offset | 6 bits |
1(b) FCFS Scheduling
Given
| Process | Arrival Time (AT) | Burst Time (BT) |
|---|---|---|
| P1 | 0 | 7 |
| P2 | 2 | 4 |
| P3 | 4 | 1 |
| P4 | 5 | 4 |
Gantt Chart
0 7 11 12 16
|---P1---|---P2---|--P3--|---P4---|
Completion Time (CT)
| Process | AT | BT | CT |
|---|---|---|---|
| P1 | 0 | 7 | 7 |
| P2 | 2 | 4 | 11 |
| P3 | 4 | 1 | 12 |
| P4 | 5 | 4 | 16 |
Turnaround Time (TAT)
[ TAT = CT – AT]
| Process | CT | AT | TAT |
|---|---|---|---|
| P1 | 7 | 0 | 7 |
| P2 | 11 | 2 | 9 |
| P3 | 12 | 4 | 8 |
| P4 | 16 | 5 | 11 |
Waiting Time (WT)
[WT = TAT – BT]
| Process | TAT | BT | WT |
|---|---|---|---|
| P1 | 7 | 7 | 0 |
| P2 | 9 | 4 | 5 |
| P3 | 8 | 1 | 7 |
| P4 | 11 | 4 | 7 |
Average Waiting Time
Final Answers
1(a)
- Tag = 16 bits
- Line Index = 10 bits
- Block Offset = 6 bits
1(b)
| Process | CT | TAT | WT |
|---|---|---|---|
| P1 | 7 | 7 | 0 |
| P2 | 11 | 9 | 5 |
| P3 | 12 | 8 | 7 |
| P4 | 16 | 11 | 7 |
- Average Waiting Time = 4.75
- Average Turnaround Time = 8.75
