Lec-94: Numerical Example on I/O Cost in Indexing | Part-1 | DBMS 

U r ri8

Maybe because your mother tongue is Hindi. As somebody like me whose mother tongue is not Hindi, lectures in English are a lot helpful compared to Hindi. Btw, I can understand Hindi.

Baat to bilkul sahi hai

Yeah, his biggest plus Point is his lectures are in Hindi Still good after 4 years 😊❤

This particular case isn't that hard though. 2 raised to the power 10 is 1024, 11 would be 2048 but still 2500 is greater, so one more power ->12. That's how he calculated. But there's no doubt that he is a legendary teacher. Hats off to him.

@@samarpitshrivastavaofficial Exactly !

Bilkul sahi

Same doubt

in that scenario, the best case will be 4ms and the worst case will be 10000ms. But, in the video sir is only focusing on block access.

When I referred to the worst case being 2500 blocks and the best case being 1 block, I was specifically referring to the number of blocks needed to be processed. Typically, worst-case and best-case scenarios are discussed in terms of the number of iterations or steps required in a process, rather than the absolute record count. Therefore, in the context of block processing, the worst-case scenario involves processing 2500 blocks, and the best-case scenario would require processing just 1 block.

Binary search is indeed an efficient algorithm when applied on sorted data. However, when dealing with records stored on a hard disk (HD), there are additional considerations to take into account. Let's explore how the binary search algorithm can still reduce the number of disk reads in this scenario. When performing binary search on a sorted data set, the algorithm divides the data into halves and compares the search key with the middle value. This process is repeated on the corresponding half until the desired record is found or determined to be absent. In the context of searching on a hard disk, the limitation lies in the fact that accessing data from the disk is significantly slower compared to accessing data from RAM. To mitigate this, the concept of blocks or pages is introduced. A block typically consists of multiple records and is the minimum unit of data transfer between the hard disk and RAM. The block size is often a few kilobytes or more. To apply binary search efficiently in this scenario, we first need to bring a block that potentially contains the desired record into RAM. This step is called fetching or reading a block. Even though bringing a block from the HD into RAM incurs some disk read operations, it reduces the overall number of disk reads required. Initially, we fetch the block that contains the middle element of the sorted data set into RAM. Then, we compare the search key with the middle element to determine if it matches or if it lies in the preceding or subsequent half. If it matches, the search is complete. Otherwise, we repeat the process with the corresponding half, fetching the relevant block from the HD into RAM. With each iteration of the binary search, we reduce the data set to half its size, allowing us to narrow down the search range faster. Since we are only accessing blocks in RAM during these iterations, the number of disk reads required is significantly reduced compared to a linear search, where each record would need to be individually accessed from the disk. In summary, binary search reduces the number of disk reads by leveraging the concept of blocks or pages to bring data into RAM efficiently. It allows us to perform comparisons on the desired data in RAM rather than accessing each record individually from the hard disk, which significantly improves the search time for sorted data stored on a hard disk.

@Shivanan Chetia mind your own business

mujhe bhi yahi lagra hai

arey nahi, mujhe abb samjha, 10K is not in bytes, if 1k is block size and 250 is 1 record size, the in 1 block there can be 4 records, so for 10K such records, we would need 2500 blocks

​@@atharvamarathe7134 muje b pehle 10 blocks laga tha, ache se socha fir..sahi h 2500 blocks😅