Lab5 Explanation:.TXT

Lab5 Explanation:

This lab is about simulation of different memory allocation techniques we
learned in week4 such as (first fit, next fit and other techniques).

Select and simulate one of these techniques
for lab4 and select another technique for lab5.


Memory is 256K.

Each unit is 2K. Therefore, there are 128
units that can be allocated to different processes.

process may request between 3 and 10 units of memory

Number of requests: 10,000

Based on the above assumptions, three
constants are defined in the code as follows:

#define NUM_REQUESTS 10000

#define MIN_NUM_UNITS 3

#define MAX_NUM_UNITS 10

A class is defined called Node which has
two member variables called process_id and num_units.

Also a list of nodes needs to be defined.

At start all the units are available. We
can use -1 as the process_id to represent availability (hole). The following diagram shows the list at the
start where the 128 units is available for allocation. Note however, a process
can only ask for up to 10 units of allocation.



The program uses a random generator
function to randomly select between allocation and deallocations. The deallocations can happen if there is at
least one process that memory is allocated for it.

You can keep track of the list of processes
that the memory is currently being allocated to them. Then during the deallocation, you may
randomly select one process from this list and deallocate memory for that

Each allocation or deallocation is
considered as one request. You must also
keep track of number of requests since we want to have 10,000 requests in this

Each process may request 3 to 10 units of
memory. Therefore, number of unit requested by each process is a random number
in the range 3 to 10. You may use a
variable called nextProcessId and at the end of each allocation request, you
may increment that variable for the next process ID request.

You may use a for loop to check the list to
see if there is a hole big enough for the given request as follows:

for (list<Node::iterator cur =
memory_list.begin(); cur != memory_list.end(); ++cur) {

As you check each node, if a node has a
process id of -1 and the size of it is equal or greater than the requested
number of units then allocation is possible.

You need to modify the list by inserting a
node with the given process ID and given number of units requested.


If process 0 requests 5 units, then the
above list becomes:





Next time we do an allocation, we must
traverse the list more to find a free location.

As we do deallocations, we will get more
than one hole. If the size of a hole is
1 or2 units, then we have fragments.

You can also define two constants:

#define HOLE_PROCESS_ID -1


There are three performance parameters that your simulation should
calculate for the chosen two techniques: average
number of external fragments, average
allocation time in terms of the average number of nodes traversed in
allocation, and the
percentage of times an allocation request is denied.

Each time allocation is successful, the
number of traversals is returned. You should have a variable to keep track of
the total number of traversals. Also you should have a variable to keep track
of the number of allocation successes.
These two values can be used to calculate the average of number of nodes
traversed. In addition, when there is a successful allocation, your code should
find the number of fragments. A variable should keep track of total fragments.

Find attached solution
Powered by