AutoSave On Capacity H5 - Last Modified: 6m ago - Lindsey Littlefield File Home Insert Draw Design Layout References Mailings Review View Help O Search A Share P Comments O Find - S Replace X Cut - A^ A Aa- A AаBЬСcDd AaBbСcDd AaBbC AаBЬСс[ Аа B АавьссD Helvetica 12 AaBbCcDd AaBbCcDd AaBbCcDd B Copy Paste E- 1 Normal 1 No Spac. Heading 1 Heading 2 Subtitle Intense E... Dictate BIU - ab x, x A - l-A- Title Subtle Em... Emphasis A Select - S Format Painter Clipboard Font Paragraph Styles Editing Voice .. . . . . 1. . . |: . . 2.. . | · .. ITE 221 Capacity 04/15/2020 Lindsey Littlefield A system currently accessing 3.8 G bytes of RAM and 10 M bytes of ROM. Compute the number of address lines required to access the designated storage. 2. For a given # of lines how many more storage space is accessible? Currently out of 3.8 G, 3 G is used by the system programs, .4 G by an application. Can we expand the memory by another 890 M to support a new application? show total storage use and available. Show all your work. 3. RAM = 3.8 G ROM = 10 M Total Capacity = (3.8 G + 10 M) = 3.81 G The number of address lines = 32 lines 32 lines allows for 4 G 3.8G – (3G +.4G) = 3.4G 4G – 3.4G = 0.6G left .89 * 1024 = 911.36 M so yes it does support Page 1 of 1 143 words + 85% 7:44 PM O Type here to search 4/15/2020 O Zoom Participants (14) Q Find a participant (5 LL Lindsey Littlefield (Me) ite221lecture - Word File Home Insert Design Layout References Mailings Review View O Tell me what you want to do.. Anwari, M H. 2 Share AH Anwari, Hashem (Host) X Cut P Find - Calibri 11 - A A Aa - АавьСсDc AаBьссDс AaBbC AаBbCc[ Aав давьссС АавЬСсD AaвьссD 暗Copy abc Replace Paste U - abe X, x’ A - aly , A- 1 Normal I No Spac. Heading 1 Heading 2 Subtitle Subtle Em. Emphasis ws Waqas Siddiqui Title Format Painter A Select- Clipboard Font Paragraph Styles Editing A aziz o Example >> ... Raise Hand yes go slower go faster no more A system currently accessing 3.1 G bytes of RAM and 0.3 Gbytes of ROM. Compute the number of address lines required to access the designated storage. For a given # of lines how many more storage space is accessible can we expand the memory by another 600Mbytes to support a program to run Unmute Me Zoom Group Chat From Patrick Nero to Everyone: RAM =3.1 G 3.1G+ 0.3G 3.4 G 3.1 G + 0.3 G ROM=0.3 G total capacity= 3.4 G From Erick Bonilla to Me: (Privately) л The number of address lines= ? 32 lines 32 lines allows 4 G From Erick Bonilla to Everyone: 4G- 3.4G =.6G .6* 1024 =614 M yes it does support From Kasra Kamgar to Everyone: got it thanks Patrick From Patrick Nero to Everyone: np man Page 59 of 70 2076 words 188% To: Erick Bonilla v (Privately) Type message here... 9:23 PM O Type here to search PDF 4/13/2020

Assume you now have 1kB of memory, i.e. the memory address space runs from 0 to 1023. The starting address of the first word is 0, the second word is 4, the third word is 8, and so on. The last word comprising 4 bytes resides in addresses 1020, 1021, 1022, 1023. Thus, the last word starts at 1020, which is a multiple of 4.  Now assume the same 1kB of memory but now, word size is 64 bits. The starting address of the first word is 0, the second, third, and last word starts at? my answer: (correct) the second is equal to  = 8 the third is equal to  = 16       help me  find the last word

Assume that a memory module contains three holes of 10MB each. A sequence of 14 requests for 1MB each will be processed (See the diagram below). For each of the memory allocation methods listed below, draw a diagram representing how memory is allocated and determine the sizes of the remaining holes after all 14 requests have been satisfied. First fit Next fit C. Best fit d. a. b. Worst fit Hint - For the Next fit, the following allocation starts with the hole following the previous allocation 10 Mb 10 Mb 10 Mb

Suppose a program’s 15-th logical instruction (counting starts from zeroth) is at physical address 1234ABDE16 in the RAM. (contents of any logical address fits into any physical address) What is the physical address in the RAM where the program has been loaded?  What is the physical address in the RAM of the last instruction of the program if it has 37 logical instructions altogether?

Please help! Please take a look at the code I have provided as a attachment to this question.  Please comment each line of code of the differences between program 1 and 2. What happens when you subtract one pointer from another? Is it subtracting the two addresses, or something else? How do you know? What happens when you increment a pointer? Compare and contrast how C-strings are laid out in memory compared to arrays of integers. What's one similarity and one difference? Show how the individual bytes of this integer array are laid out in memory.    Thanks!

Draw a picture illustrating the contents of memory, given the following data declarations: (You need to draw a diagram showing the memory addresses and their contents.)    Assume that your data segment starts at 0x1100 in memory. Your answer should fill the table below and explain why. -- photo attached below Thank you for your help!

Case Study Assume a computer system has a main memory of 256 Bytes. The following is a memory byte-access trace history of a program run on this system. For example, as it is shown, the program first accesses memory address 0000 0000, and then it accesses memory address 0000 0001 and so on. Note the memory addresses are represented in binary: 00000000, 00000001, 00000010, 00000011, 00001000, 00010000, 00010001, 00000100, 00000101, 00000110, 00000111, 00001001, 00001010, 00001011, 00001100, 00001000, 00001001, 00001010, 00011100, 00011101. Q3. Assume the system has a 16-Byte direct mapped unified L1 cache with a block size of 2 Bytes. The following table shows how the cache looks like after the first access to the memory is finished. Please show; in the provided table, how it looks like after the 20th access is finished. You could ignore the "Data" Column. (Add or remove Rows/Columns in the provided table for your answer, as you see fit). Cache contents after the 1st access: Cache Index…

Given a computer with 300 memory locations, each 18-bit assessable, how many bits of addressing are needed?  Given a computer with 10-bit addressability and 5 bits for the address, what's the maximum number of memory this computer can have in bits?

Let's double the size of the address field from 32b to 64b.  What's the largest decimal value now?  How might you organize this for readability?

Question 9