fa09-sl-ef - ECE 3035 5 problems 9 pages Computing Mechanisms Final Exam Fall 2009 10 December 2009 Instructions This is a closed book closed note exam

fa09-sl-ef - ECE 3035 5 problems 9 pages Computing...

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ECE 3035 Computing Mechanisms Fall 2009 5 problems, 9 pages Final Exam 10 December 2009 Instructions: This is a closed book, closed note exam. Calculators are not permitted. If you have a question, raise your hand and I will come to you. Please work the exam in pencil and do not separate the pages of the exam. For maximum credit, show your work. Good Luck! Your Name ( please print ) ________________________________________________ 1 2 3 4 5 Total 20 30 22 26 22 120 1
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ECE 3035 Computing Mechanisms Fall 2009 5 problems, 9 pages Final Exam 10 December 2009 Problem 1 (3 parts, 20 points) Storage Part A (6 points) Suppose a disk has 2 platters, 4 surfaces, 128K tracks, and 2K (avg.) sectors/track. Compute the following sizes ( in bytes ). Express all answers in decimal. storage/surface storage/cylinder storage/disk Part B (6 points) A clever engineer triples the number of tracks per surface (all other parameters remain the same). Describe the storage and performance effects of this change. Storage effect: Performance effect: Part C (8 points) The table below lists memory hierarchy sizes (in bytes) and access times (in seconds) for a system with a 2.5 GHz processor. For each memory type, determine how many times larger and slower it is versus registers. 128 B register file 1x larger access time = 400 pS 1x slower 4 MB L2 cache access time = 8 nS 4 GB main memory access time = 40 nS 512 GB disk access time = 80 mS Suppose an average of two instructions are completed in each processor cycle. How many instructions would be completed in the time it takes to perform one disk access? 2
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ECE 3035 Computing Mechanisms Fall 2009 5 problems, 9 pages Final Exam 10 December 2009 Problem 2 (2 parts, 30 points) Activation Frames Consider the following C code fragment: int Bar() { int A = 256; int *P = &A; int W[] = {3, 5, 8, 13}; int *Q = &(W[2]); int Foo(int, int *, int *, int *); A = Foo(A, W, P, Q-1); return(A); } int Foo(int M, int *R, int *S, int *T) { int X = 100; int Y; Y = *S << 2; *T = R[2] + X; return(M); } Part A (15 points) Describe the current state of the stack before Foo returns to Bar . Fill in the unshaded
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