Proportional Allocation
Duration: 4 min
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Inside: a video lesson and guided study material.
Module outline
- Quantitative Ability: Speed Mathematics and Basic Mathematics, Percentage, Ratio and Proportion (Ratios), Average, Profit and Loss, Simple Interest and Compound Interest, Time and Work, Pipe and Cistern, Speed Time and Distance, Number System, HCF LCM, Data Interpretation, Age Problems, Mixture and Alligation, Partnership, Algebra, Sequence and Series (Series), Mensuration and Geometry, Arithmetic
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- Verbal Ability: Comprehension / Reading Comprehension / Unseen Passages (Critical Reasoning) (Paragraph Questions), Cloze Test (Cloze Passage), Sentence Re-arrangements (Para Jumbles) (Narrative Sequencing), Vocabulary, Sentence Completion (Fill in the blanks), Active Passive Voice (Voice), Direct and Indirect Speech, Subject Verb Agreement (Verb Noun Agreement), Tenses, Preposition, Articles, Sentence Correction (Error Correction)
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- HR Interview: MR & HR Interview Preparation
AI summary & chapters
AI Summary
An AI-generated summary of this video lecture.
This educational video lecture explains the concept of Weighted or Proportional Allocation in operating system memory management. The instructor introduces a specific formula used to distribute physical memory frames among processes based on their relative sizes. The core formula presented is a(i) = s(i) / S x m, where allocation is proportional to the process size. A detailed numerical example is worked through on a digital whiteboard, demonstrating how to calculate the exact number of frames assigned to three distinct processes with sizes of 20k, 30k, and 50k, given a total of 30 available frames.
Chapters
0:00 – 2:00 00:00-02:00
The video begins with a slide titled 'Weighted / Proportional Allocation'. The instructor explains that frames are allocated according to the weights of the process. He defines the formula a(i) = s(i) / S x m. He clarifies that a(i) represents the number of allocated frames, s(i) is the size of the specific process, S is the total size of virtual memory (calculated as the sum of all process sizes), and m is the total number of frames available in the system. He sets up an example scenario involving three processes with sizes of 20k, 30k, and 50k, and a total of 30 frames to be allocated.
2:00 – 4:14 02:00-04:14
The instructor moves to the whiteboard to solve the example. He writes down the process identifiers P1, P2, and P3 with their respective sizes: 20k, 30k, and 50k. He calculates the total virtual memory size S as 100k (20+30+50). He then calculates the allocation for P1 as (20/100)*30, resulting in 6 frames. He similarly determines that P2 gets 9 frames and P3 gets 15 frames. He draws a diagram labeled '1GB' with a rectangle divided into blocks to visualize the memory structure, circling the final formula to emphasize its importance in the calculation.
The lecture effectively bridges theoretical formulas with practical application. By defining the variables in the proportional allocation formula and then immediately applying them to a concrete example, the instructor demonstrates how larger processes receive a larger share of memory frames. The visual aids, including the written calculations and the memory block diagram, reinforce the concept that allocation is strictly proportional to the process size relative to the total system memory.