http://eckstein.rutgers.edu/mis/handouts/storage-calcs.pdf WebSee Answer Question: 1. How many bits are required to address a 2M x 32 main memory if a. The memory is byte-addressable? b. The memory is word-addressable? (assume word size is 32 bits) 2. Suppose that a 2M x 32 memory is build using 256K x 8 RAM chips and is word-addressable a. How many RAM chips are necessary? (X rows by Y columns) b.
Bits (binary digits) (article) Khan Academy
WebFeb 13, 2015 · A bit is either 0 or 1. So it can store 2 values. Two bits can store 2*2 values, 4. Three bits can store 2*2*2 values, 8. And so on. So to store 3 values, you need at least two … Weba. The maximum size of data field in each fragment = 680 (because there are 20 bytes IP header). Thus the number of required fragments 4 0 0 20 » » º « « ª b. Each fragment will have Identification number 422. Each fragment except the last one will be of size 700 bytes (including IP header). The last datagram will be of size 360 bytes ... ct walker augusta ga
How Do Bits, Bytes, Megabytes, Megabits, and Gigabits Differ?
Webbook, T-shirt, history, merchandising 312 views, 13 likes, 2 loves, 0 comments, 1 shares, Facebook Watch Videos from Simple History: Check out our... How many different patterns can be made with 1, 2, or 3 bits? 1. 3 bits vs. 2 bits 2. Consider just the leftmost bit 3. It can only be 0 or 1 4. Leftmost bit is 0, then append 2-bit patterns 5. … See more WebLikewise, you need 20 bits to address every byte in a megabyte, and 30 bits to address every byte in a gigabyte. 2 32 = 4294967296, which is the number of bytes in 4 gigabytes, so you need a 32 bit address for 4 GB of memory. Share Improve this answer edited Dec 10, 2012 at 16:11 answered Dec 10, 2012 at 16:03 Caleb 38.8k 8 94 152 ct walk from obesity