Computer Organization And Architecture MCQ - Memory Hierarchy

61 The total size of address space in a virtual memory system is limited by

Virtual memory concept is independent of size of main memory and depends only on the availability of the secondary storage.

62 A dynamic RAM has a memory cycle time of 64 nsec. It has to be refreshed 100 times per nsec and each refresh takes 100 nsec. What percentage of the memory cycle time is used for refreshing?

Memory cycle time = 64 ns
Memory is refreshed 100 times per msec.
Number of refreshes in 1 memory cycle (i.e in 64 ns)
= (100 * 64 * 10 – 9)/10 – 3 = 64 * 10 – 4.
Time taken for each refresh = 100 ns
Time taken for 64 * 10 – 4 refreshes
= 64 * 10 – 4 * 100 * 10-9 sec
= 64 * 10 – 11 sec.
Percentage of the memory cycle time used for refreshing :
= (Time taken to refresh in 1 memory cycle/ Total time) * 100
= (64 * 10 – 11/64 * 10 – 9) * 100
= 1 %

63 Increasing the RAM of a computer typically improves performance because

As size increase, page frame increase. If we apply any replacement rule which is not cause ‘blady anamoly’ is always reduce page fault.

64 How many 32K × 1 RAM chips are needed to provide a memory capacity of 256 K-bytes?

Number of chips required = 256*1024*8/32*1024*1 = 64

65 A main memory unit with a capacity 4 megabytes is built using 1M × 1-bit DRAM chips. Each DRAM chip has 1K rows of cells with 1K cells in each row. The time taken for a single refresh operation is 100 nanoseconds. The time required to perform one refresh operation on all the cells in the memory unit is

Main memory unit has capacity = 4 MB
Number of DRAM chips = 4MB/ 1M*1bit = 32
1 DRAM chip has 1 K (2¹⁰) rows and 1 K(2¹⁰) cells
In each row = 32 * 2¹⁰ * 2¹⁰ = 32 * 2²⁰
Time taken in one refresh opn
= 100 ns
Time required to perform one refresh operation on all the cells in MM unit
= 32 * 2²⁰ * 100 = 3200 * 2²⁰ ns