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If 5/6< x and x < 7/4, THEN X COULD be?? - Computers & Internet

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5/6 is the same as 20/24. 7/4 is the same as 42/24. This means that X could be any number between 21/24 and 41/24. Hope this helps!

Posted on Apr 30, 2010

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Banker's Algorithm


* multiple instances of resource types IMPLIES cannot use resource-allocation graph

* banks do not allocate cash unless they can satisfy customer needs when a new process enters the system

* declare in advance maximum need for each resource type

* cannot exceed the total resources of that type

* later, processes make actual request for some resources

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Banker: Data Structures define MAXN 10 /* maximum number of processes */
#define MAXM 10 /* maximum number of resource types */
int Available[MAXM]; /* Available[j] = current # of unused resource j */
int Max[MAXN][MAXM]; /* Max[i][j] = max demand of i for resource j */
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Notation:

X <= Y iff X[i] <= Y[i] for all i

(0,3,2,1) is less than (1,7,3,2)

(1,7,3,2) is NOT less than (0,8,2,1)

Each row of Allocation and Need are vectors: Allocation_i and Need_i



Banker: Example

Initially:

Available
A B C
10 5 7

Later Snapshot:

Max - Allocation = Need Available
A B C A B C A B C A B C
P0 7 5 3 0 1 0 7 4 3 3 3 2
P1 3 2 2 2 0 0 1 2 2
P2 9 0 2 3 0 2 6 0 0
P3 2 2 2 2 1 1 0 1 1
P4 4 3 3 0 0 2 4 3 1



Banker: Safety Algorithm

* consider some sequence of processes

* if the first process has Need less than Available

* it can run until done

* then release all of its allocated resources

* allocation is increased for next process

* if the second process has Need less than Available

* ...

* then all of the processes will be able to run eventually

* IMPLIES system is in a safe state



Banker: Safety Algorithm

STEP 1: initialize
Work := Available;
for i = 1,2,...,n
Finish[i] = false
STEP 2: find i such that both
a. Finish[i] is false
b. Need_i <= Work
if no such i, goto STEP 4
STEP 3:
Work := Work + Allocation_i
Finish[i] = true
goto STEP 2
STEP 4:
if Finish[i] = true for all i, system is in safe state



Banker: Safety Example

Using the previous example, P1,P3,P4,P2,P0 satisfies criteria.

Max - Allocation = Need <= Work Available
A B C A B C A B C A B C
P1 3 2 2 2 0 0 1 2 2 3 3 2 3 3 2
P3 2 2 2 2 1 1 0 1 1 5 3 2
P4 4 3 3 0 0 2 4 3 1 7 4 3
P2 9 0 2 3 0 2 6 0 0 7 4 5
P0 7 5 3 0 1 0 7 4 3 10 4 7
10 5 7<<< initial system



Banker: Resource-Request Algorithm

STEP 0: P_i makes Request_i for resources, say (1,0,2)
STEP 1: if Request_i <= Need_i
goto STEP 2
else ERROR
STEP 2: if Request_i <= Available
goto STEP 3
else suspend P_i
STEP 3: pretend to allocate requested resources
Available := Available - Request_i
Allocation_i := Allocation_i + Request_i;
Need_i := Need_i - Request_i
STEP 4: if pretend state is SAFE
then do a real allocation and P_i proceeds
else
restore the original state and suspend P_i



Banker: Resource-Request Algorithm [129]

Say P1 requests (1,0,2)

Compare to Need_1: (1,0,2) <= (1,2,2)

Compare to Available: (1,0,2) <= (3 3 2)

Pretend to allocate resources:

Max - Allocation = Need Available
A B C A B C A B C A B C
P0 7 5 3 0 1 0 7 4 3 2 3 0<<<
P1 3 2 2 3 0 2<<< 0 2 0<<<
P2 9 0 2 3 0 2 6 0 0
P3 2 2 2 2 1 1 0 1 1
P4 4 3 3 0 0 2 4 3 1

Is this safe? Yes: P1, P3, P4, P0, P2

Can P4 get (3,3,0)? No, (3,3,0) > (2,3,0) Available

Can P0 get (0,2,0)? (0,2,0) < (2,3,0) Available

Pretend: Available goes to (2,1,0)

Thanks And Regards

May 13, 2009 | Microsoft Windows XP Professional

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