【OS】Bankers Algorithm

data structure.

• n: integer, the number of processes;
• m: integer, the number of resource types;
• Available[m]: A one-dimensional array that stores the number of available instances of each resource in [1...m];
• Max[n][m]: n×m matrix, Max[i][j] represents the total demand for resource j by process i;
• Allocation[n][m]: n×m matrix, Allocation[i][j] represents the number of resources j already held by the process i;
• Need[n][m]: n×m matrix, Need[i][j] represents the number of resources j needed by the process i.

vector.

• For two vectors XX of length n$X$和$Y$，当且仅当$X[i]<Y\left[i\right]$ for all$i\in [1..The\; n-]$ When are true, we say that$X<And.$

System security testing algorithm.

• Used to test whether the system is in a safe state at this time, the algorithm is as follows:

TestSafety()
{
    
    
	//1.
	Work[1..m]=Available[1..m];
	//向量Work初始化为当前可用资源实例数向量.

	Finish[1..n]=false;
	//向量Finish每个分量都为false,表示初态都是未完成.

	//2.
	change=true;	//循环控制变量.
	while(change)
	{
    
    
		//Need[i]表示矩阵的第i行,也就是进程Pi所需资源数向量,
		//若进程Pi未完成,并且需求能够得到满足,就执行Pi,后续
		//释放其中的资源,等价于Work获得Pi已经持有的资源数向量
		//Allocation[i].
		change=false;
		for(int i=1;i<=n;++i)
		{
    
    
			if(Finish[i]==false && Need[i]<Work) 
			{
    
    
				Work=Work+Allocation[i];
				Finish[i]=true;
				change=true;
			}
		}
	}
	//不动点算法:只有当Finish和Allocation不再发生改变时,
	//也就意味着所有的进程都已经完成或者剩余进程都无法运行
	//这两种情况之一发生时，才会跳出while循环.

	//3.
	for(int i=1;i<=n;++i)
	{
    
    
		if(Finish[i]==false)
		{
    
    
			return false;
		}
	}
	return true;
	//全真为真,一假俱假.
}

Resource request algorithm.

• Request[n][m]: n×m matrix, Request[i][j] represents the number of resource j requested by the process i at this time.

if(Request[i]<Need[i])
{
    
    
	if(Request[i]<Available)
	{
    
    
		//Simulate allocation.
		Available=Available-Request[i];
		Allocation[i]=Allocation[i]+Request[i];
		Need[i]=Need[i]-Request[i];
		
		if(TestSafety())
		{
    
    
			Agree Allocation;
			//System safe,just allocate.
		}
		else
		{
    
    
			Rollback;
			//System unsafe,rollback.
		}
	}
	else
	{
    
    
		Wait;
		//Cause resource insufficient.
	}
}
else
{
    
    
	Error;
	//Cause request too many resources.
}

Example 1.

• Is the system in a safe state at this time?
• If the process $P_1$Make a resource request Request $Reques{t}_1=(1,0,2)$ Can it be satisfied?
• If the process $P_0$Make a resource request Request $Reques{t}_0=(0,2,0)$ , can it be satisfied?
• If process $P_4$Make a resource request Request $Reques{t}_4=(3,3,0)$ , can it be satisfied?

answer.

Example two.

2. Consider the state of the system at T0 as follows:

	Allocation	Max	Available
	A B C D	A B C D	A B C D
$P_0$	0 0 1 2	0 0 1 2	1 5 2 0
$P_1$	1 0 0 0	1 7 5 0
$P_2$	1 3 5 4	2 3 5 6
$P_3$	0 6 3 2	0 6 5 2
$P_4$	0 0 1 4	0 6 5 6

• What is the content of Need?
• Is the system in a safe state at this moment?
• If $P_1$Request resource Request $Reques{t}_1=(0,4,2,0)$ , does the system satisfy its request? why?

answer.