================ Start Lecture #13 ================
Is the figure on the right safe or not?
You can NOT tell until I give you the initial claims of the
Please do not make the unfortunately common exam mistake to give
an example involving safe states without giving the claims.
For the figure on the right, if the initial claims are:
P: 1 unit of R and 2 units of S (written (1,2))
Q: 2 units of R and 1 units of S (written (2,1))
the state is NOT safe.
But if the initial claims are instead:
P: 2 units of R and 1 units of S (written (2,1))
Q: 1 unit of R and 2 units of S (written (1,2))
the state IS safe.
Explain why this is so.
A manager can determine if a state is safe.
Since the manager know all the claims, it can determine the maximum
amount of additional resources each process can request.
The manager knows how many units of each resource it has left.
The manager then follows the following procedure, which is part of
Banker's Algorithms discovered by Dijkstra, to
determine if the state is safe.
If there are no processes remaining, the state is
Seek a process P whose max additional requests is less than
what remains (for each resource type).
- If no such process can be found, then the state is
- The banker (manager) knows that if it refuses all requests
excepts those from P, then it will be able to satisfy all
of P's requests. Why?
Ans: Look at how P was chosen.
The banker now pretends that P has terminated (since the banker
knows that it can guarantee this will happen). Hence the banker
pretends that all of P's currently held resources are returned. This
makes the banker richer and hence perhaps a process that was not
eligible to be chosen as P previously, can now be chosen.
Repeat these steps.
A safe state with 22 units of one resource
|process||initial claim||current alloc||max add'l|
One resource type R with 22 unit
Three processes X, Y, and Z with initial claims 3, 11, and 19
Currently the processes have 1, 5, and 10 units respectively.
Hence the manager currently has 6 units left.
Also note that the max additional needs for the processes are 2,
6, 9 respectively.
So the manager cannot assure (with its current
remaining supply of 6 units) that Z can terminate. But that is
not the question.
This state is safe
Use 2 units to satisfy X; now the manager has 7 units.
Use 6 units to satisfy Y; now the manager has 12 units.
Use 9 units to satisfy Z; done!
An unsafe state with 22 units of one resource
|process||initial claim||current alloc||max add'l|
Start with example 1 and assume that Z now requests 2 units and the
manager grants them.
Remark: An unsafe state is not
necessarily a deadlocked state.
Indeed, if one gets lucky all processes in an unsafe state may
A safe state means that the manager can
guarantee that no deadlock will occur.
Currently the processes have 1, 5, and 12 units respectively.
The manager has 4 units.
The max additional needs are 2, 6, and 7.
This state is unsafe
Use 2 unit to satisfy X; now the manager has 5 units.
Y needs 6 and Z needs 7 so we can't guarantee satisfying either
Note that we were able to find a process that can terminate (X)
but then we were stuck. So it is not enough to find one process.
We must find a sequence of all the processes.
3.5.3: The Banker's Algorithm (Dijkstra) for a Single Resource
The algorithm is simple: Stay in safe states. Initially, we
assume all the processes are present before execution begins and that
all initial claims are given before execution begins.
We will relax these assumptions very soon.
Before execution begins, check that the system is safe.
That is, check that no process claims more than the manager has).
If not, then the offending process is trying to claim more of
some resource than exist in
the system has and hence cannot be guaranteed to complete even if
run by itself.
You might say that it can become deadlocked all by itself.
When the manager receives a request, it pretends to grant it and
checks if the resulting state is safe. If it is safe the request is
granted, if not the process is blocked.
When a resource is returned, the manager (politely thanks the
process and then) checks to see if “the first” pending
requests can be granted (i.e., if the result would now be
safe). If so the request is granted. The manager checks to see if
the next pending request can be granted, etc..
3.5.4: The Banker's Algorithm for Multiple Resources
At a high level the algorithm is identical: Stay in safe states.
What is a safe state?
The same definition (if processes are run in a certain order they
will all terminate).
Checking for safety is the same idea as above. The difference is
that to tell if there enough free resources for a processes to
terminate, the manager must check that for all
resources, the number of free units is at least equal to the max
additional need of the process.
Limitations of the banker's algorithm
Often users don't know the maximum requests a process will make.
They can estimate conservatively (i.e., use big numbers for the claim)
but then the manager becomes very conservative.
New processes arriving cause a problem (but not so bad as
The process's claim must be less than the total number of
units of the resource in the system. If not, the process is not
accepted by the manager.
Since the state without the new process is safe, so is the
state with the new process! Just use the order you had originally
and put the new process at the end.
Insuring fairness (starvation freedom) needs a little more
work, but isn't too hard either (once an hour stop taking new
processes until all current processes finish).
A resource can become unavailable (e.g., a tape drive might
This can result in an unsafe state.
3.7: Other Issues
3.7.1: Two-phase locking
This is covered (MUCH better) in a database text. We will skip it.
3.7.2: Non-resource deadlocks
You can get deadlock from semaphores as well as resources. This is
trivial. Semaphores can be considered resources. P(S) is request S
and V(S) is release S. The manager is the module implementing P and
V. When the manager returns from P(S), it has granted the resource S.
As usual FCFS is a good cure. Often this is done by priority aging
and picking the highest priority process to get the resource. Also
can periodically stop accepting new processes until all old ones
get their resources.
3.8: Research on Deadlocks
Chapter 4: Memory Management
Also called storage management or
Memory management must deal with the storage
hierarchy present in modern machines.
Registers, cache, central memory, disk, tape (backup)
Move data from level to level of the hierarchy.
How should we decide when to move data up to a higher level?
- Fetch on demand (e.g. demand paging, which is dominant now).
- Read-ahead for file I/O.
- Large cache lines and pages.
- Extreme example. Entire job present whenever running.
We will see in the next few weeks that there are three independent
Segmentation (or no segmentation)
Paging (or no paging)
Fetch on demand (or no fetching on demand)
Memory management implements address translation.
Convert virtual addresses to physical addresses
- Also called logical to real address translation.
- A virtual address is the address expressed in
- A physical address is the address understood
by the computer hardware.
The translation from virtual to physical addresses is performed by
the Memory Management Unit or (MMU).
Another example of address translation is the conversion of
relative addresses to absolute addresses
by the linker.
The translation might be trivial (e.g., the identity) but not in a modern
general purpose OS.
The translation might be difficult (i.e., slow).
- Often includes addition/shifts/mask--not too bad.
- Often includes memory references.
- VERY serious.
- Solution is to cache translations in a Translation
Lookaside Buffer (TLB). Sometimes called a
translation buffer (TB).
When is address translation performed?
At compile time
- Compiler generates physical addresses.
- Requires knowledge of where the compilation unit will be loaded.
- No linker.
- Loader is trivial.
- Rarely used (MSDOS .COM files).
At link-edit time (the “linker lab”)
Generates relative (a.k.a. relocatable) addresses for each
References external addresses.
- Linkage editor
- Converts the relocatable addr to absolute.
- Resolves external references.
Misnamed ld by unix.
Must also converts virtual to physical addresses by
knowing where the linked program will be loaded. Linker
lab “does” this, but it is trivial since we
assume the linked program will be loaded at 0.
Loader is still trivial.
Hardware requirements are small.
A program can be loaded only where specified and
cannot move once loaded.
Not used much any more.
At load time
Similar to at link-edit time, but do not fix
the starting address.
Program can be loaded anywhere.
Program can move but cannot be split.
Need modest hardware: base/limit registers.
Loader sets the base/limit registers.
No longer common.
At execution time
- Addresses translated dynamically during execution.
- Hardware needed to perform the virtual to physical address
- Currently dominates.
- Much more information later.
- When executing a call, check if module is loaded.
- If not loaded, call linking loader to load it and update
- Slows down calls (indirection) unless you rewrite code dynamically.
- Not used much.
- The traditional linking described above is today often called
- With dynamic linking, frequently used routines are not linked
into the program. Instead, just a stub is linked.
- When the routine is called, the stub checks to see if the
real routine is loaded (it may have been loaded by
- If not loaded, load it.
- If already loaded, share it. This needs some OS
help so that different jobs sharing the library don't
overwrite each other's private memory.
- Advantages of dynamic linking.
- Saves space: Routine only in memory once even when used
- Bug fix to dynamically linked library fixes all applications
that use that library, without having to
relink the application.
- Disadvantages of dynamic linking.
- New bugs in dynamically linked library infect all
- Applications “change” even when they haven't changed.