CSE360_HW2

# CSE360_HW2 - CSE 360 Homework#2 Summer11 Abdulwahab Alkharashi Hany Rashwan CSE 360 Homework#2 HW2-1 19 23d 26 27 19.a 00110011 = 51 19.b 00010111

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CSE 360 Homework #2 Summer11 Abdulwahab Alkharashi & Hany Rashwan CSE 360 Homework #2 HW2-1. ( 19, 23d, 26, 27) 19.a : 00110011 = 51 19.b : 00010111 = 23 19.c: 10000010 = -126 19.d : 10101010 = -86 19.e : 11111111 = -1 23.d : E= 1, T = 00, Z= 01011, Q = 01010, U = 011, M = 0100 .2769 .1464 .0902 = (T) .0023 .0277 = (U) .0262 = (M) .0562 .0009=(Z) .00014= (Q) .0285 .1305 = (E)

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CSE 360 Homework #2 Summer11 26.a: { 0001, 1001, 0101, 1110} = 1 26.b: { 11100, 00011, 01010, 10101, 11011} = 2 26.c: { 0011, 1010, 1100, 0000, 1111, 01101} = 2 27a: { 0110, 1010, 0011, 1111 } 27.b: { 01010, 10111, 11110, 11011, 00011 } 27.c: { 00011, 01010, 01100, 00000, 01111, 00101 } HW2-2. A: 2 15 = 32768 B: (32768 address) * (1 byte / address) = 32768 C: (32768 address) * (16 bits / address) * (1 byte / 8 bits) = ( 32768) * (16/ 8 bytes) = 32768 * 2 = 65536 bytes. HW2-3. A: It is 16 bits, because the diagram shows the data bus to be 16 bits wide. Transferring 16 bits is easer between the CPU at the memory subsystem. B-i:
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## This note was uploaded on 10/08/2011 for the course CSE 360 taught by Professor Sam during the Spring '11 term at Ohio State.

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CSE360_HW2 - CSE 360 Homework#2 Summer11 Abdulwahab Alkharashi Hany Rashwan CSE 360 Homework#2 HW2-1 19 23d 26 27 19.a 00110011 = 51 19.b 00010111

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