The size of the memory could be given as: Size of the memory =Number of Words× Number of bits per word What will be the number of address lines and data lines required to construct a memory circuit of size 256 X 32 using a decoder and a multiplexer.
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- Suppose r0 = ox300010A0, r2 = 0x00000011, and the memory layout is as follows Address: Data: 0x300010A7 0x72 0x300010A6 0XA5 0x300010A5 0x9F 0x300010A4 0x00 0x300010A3 0x50 0x300010A2 0x2B 0x300010A1 0XA5 0x300010A0 0x01 -What is the Value of r0 and r1 after executing LDR r1, [r0, #2] -What are the values of r0 and r1 after executing the following code? Illustrate your process in a memory map. LDR r1, [r0], #3 ADD r1, r1, r2 STR r1, [r0, r#4]Suppose r0 = ox300010A0, r2 = 0x00000011, and the memory layout is as follows Address: Data: 0x300010A7 0x72 0x300010A6 0XA5 0x300010A5 0x9F 0x300010A4 0x00 0x300010A3 0x50 0x300010A2 0x2B 0x300010A1 0XA5 0x300010A0 0x01 -What are the values of r0 and r1 after executing the following code? Illustrate your process in a memory map. LDR r1, [r0], #3 ADD r1, r1, r2 STR r1, [r0, r#4]A digital computer has a memory unit with 24 bits per word. The instruction set consists of 150 different operations. All instructions have an operation code part (opcode) and an address part (allowing for only one address). Each instruction is stored in one word ofmemory.a. How many bits are needed for the opcode?b. How many bits are left for the address part of the instruction?c. What is the maximum allowable size for memory?d. What is the largest unsigned binary number that can be accommodated in one word of memory?
- The use of transistors in the construction of RAM and ROM leads me to believe that there is no need for cache memory.The term "temporary storage" may also be thought of as "random access memory" (RAM) that is momentarily vacant. Imagine a machine that only had one kind of memory—is it even possible?By assuming that X = 3, and 33 is a two digit number, consider memory storage of a 64-bit word stored at memory word 33 in a byte-addressable memory (a) What is the byte address of memory word 33? (b) What are the byte addresses that memory word 33 spans? (c) Draw the number 0xF1234567890ABCDE stored at word 33 in both big endian and little-endian machines. Clearly label the byte address corresponding to each data byte value.In the memory of Computer, if we say 256 * 8, it means 256 is size and 8 is Data bus size
- Suppose the RAM for a certain computer has 256M words, where each word is 16 bits long. a. What is the capacity of this memory expressed in bytes? b. If this RAM is byte addressable, how many bits must an address contain? c. If this RAM is word addressable, how many bits must an address contain?0001 = Load AC from memory 0010 = Store AC to memory 0101 = Add to AC from memory 0011 = Load AC (the accumulator register) from an I/O device 0111 = Store AC to an I/O device With these instructions, a particular I/O device is identified by replacing the 12-bit address portion with a 12-bit device number. Remember that a number ending with a small ‘h’ means the number is a hexadecimal number. What is the hexadecimal string that expresses the following instructions? Load AC from memory location 62h. Add the contents of memory location 451h to AC. Store AC to memory location 8h. Store AC to I/O device number 8h.For the instruction (0x6479), select all data paths that are used from the beginning of the Decode Instruction phase through the end of the Store Result phase. FYI: Be certain; Canvas deducts points for incorrect choices. OL tol OH to J OK to N OL to E OH to F OM to B OC to N OH to L A to F ON to O OM to N O to M
- Write an 8085 microprocessor program to calculate the number of negative data among the following data (D4, 5E, 14, 93, 80, 11, 21, EE, 5D, 23, F1, C4, 39, 79, 10, OD). Note that D7 bit represents the sign, where 1 is for negative and 0 for positive.A CPU that supports little endian format reads two integer (4-byte) values from address 0x1000 and 0x2000. The values read are 55 and 6850 respectively. Please show the memory contents (byte-wise) at address 0x1000 and 0x2000?Consider the following hexadecimal readout: 000000 8A00 8E00 CFA1 48BF 7900 3202 9015 AD34 000010 0218 6D30 028D 3402 AD35 0288 3102 8D35 000020 0E30 0290 DAEE 3102 4C00 0200 0040 004B Refer to the first byte of memory shown above, address 000000. Assume that this byte is used to store an 8-bit unsigned integer. What is the decimal value stored in this byte? Group of answer choices 138 -27 22,842 66