Question

In a system using the relocatable dynamic partitions scheme, given the following situation (and using decimal form): Job Q is loaded into memory starting at memory location 42K.

a. Calculate the exact starting address for Job Q in bytes. 42K*1024=43008

b. If the memory block has 3K in fragmentation, calculate the size of the memory block.

3k*1024=3072

42k*1024 =43008

3072+46080=46080

c. Is the resulting fragmentation internal or external? Explain your reasoning.

fragmentation internal or external

Answer 1

a) Job Q starting address = 42 k = 42*1024 = 43008.

b)memory block has 3 k has fragmentation and the size of the memory block you calculated as 3 k = 3*1024 = 3072.

so this is a block means we have end address = 3072 + 43008 = 46080.

c) External Fragmentation. it will come in a system which is using the relocatable dynamic partitions scheme. these are free spaced holes that are generated in your memory or your disk.