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50.005 Computer System Engineering
Information Systems Technology and Design
Singapore University of Technology and Design
Natalie Agus (Summer 2024)

Banker’s Algorithm

You need Python 3.10.x or above to complete this lab.

In deadlock avoidance, before granting a resource request (even if the request is valid and the requested resources are now available), we need to check that the request will not cause the system to enter a deadlock state in the future or immediately.

The Banker’s Algorithm is a resource allocation and deadlock avoidance algorithm. It can be used to predict and compute whether or not the current request will lead to a deadlock.

  • If yes, the request for the resources will be rejected,
  • Otherwise, it will be granted.

There are two parts of the Banker’s Algorithm:

  1. Resource Allocation Algorithm
  2. Safety Algorithm

You will be implementing the Safety Algorithm and then calling it in Resource Allocation Algorithm in this lab.

Starter Code

Install Python 3.10.x or above

If you don’t have Python 3.10.x installed already, here’s a list of commands you can enter to install:

sudo apt update && sudo apt upgrade
sudo apt install software-properties-common -y
sudo add-apt-repository ppa:deadsnakes/ppa -y
sudo apt update
sudo apt install python3.10 -y

When done, check that it is installed, and install pip as well:

python3.10 --version
sudo apt install python3-pip

Clone

Download this repository:

git clone https://github.com/natalieagus/lab_banker

You should have the following files:

lab_banker/
  |- test_files/
      |- q0_answer.txt
      |- q0.txt
      |- q1_answer.txt
      |- q1.txt
      |- q2_answer.txt
      |- q2.txt
      |- q3_answer.txt
      |- q3.txt
      |- q4_answer.txt
      |- q4.txt
      |- q5_answer.txt
      |- q5.txt
  |- .gitignore
  |- banker_test.py
  |- banker.py
  |- README.md

You will be required to modify only certain sections in banker.py.

Leave ALL other files untouched. Also, DO NOT print anything else in banker.py. Only type your answers in the given space labeled in the starter code as TASK 1 and TASK 2. Remember, DO NOT print anything else, and DO NOT import any other modules, and DO NOT modify any other instructions.

Banker’s Algorithm

Prerequisite

Suppose we have N processes competing for M different types of resources. We call these processes “customers” (to these available resources).

In order for the Banker’s Algorithm to work, need to know the maximum demand of process i for resource j instances.

System State

We also need to represent the system STATE, such as the amount of resources available per-type-per-process using the following data structures:

  1. available: 1 by M vector
    • available[i]: the available instances of resource i
  2. max: N by M matrix
    • max[i][j]: maximum demand of process i for resource j instances
  3. allocation: N by M matrix
    • allocation[i][j]: current allocation of resource j instances for process i
  4. need: N by M matrix
    • need[i][j]: how much more of resource j instances might be needed by process i

For example, the following arrays display the available, max, allocation, and need values of a current state of a system with 5 processes: P0, P1, P2, P3, and P4, and 3 types of resources: A, B, C;

Banker.py

Open Banker.py and give it a quick read.

There are a few methods defined in Class Banker:

  1. set_maximum_demand: implemented for you, this populates the corresponding row of max matrix
  2. request_resources: request resource algorithm implementation. Task 1 is here.
  3. release_resources: releases resources borrowed by a customer. Assume release is valid for simplicity
  4. check_safe: safety check algorithm implementation. Task 2 is here.
  5. print_state and run_file: functions to load the input files and printing the states of the system: available, max, allocation, need values.

Input Format

The inputs to banker.py are inside test_files/. All .txt files named as qi.txt are various input files.

The format are as follows:

  1. The first three lines contain values of N (number of processes), M (number of resource type), and the contents of the available vector.
  2. For lines starting with c, it signifies the max demand of a process for ALL resource types.
    • The format is c,pid,rid_1 rid_2 rid_3 ...
  3. For lines starting with r, it signifies a request by a process.
    • The format is: r,pid,rid_1 rid_2 rid_3 ...
  4. For lines starting with f, it signifies a release of resources by a process.
    • The format is: f,pid,rid_1 rid_2 rid_3 ...
  5. For lines with just p, we print the state of the system

The lines in the input are read from top to bottom, and are treated as incoming requests or releases by each process as time progresses.

For instance, open test_files/q0.txt:

n,3
m,3
a,0 0 0
c,0,2 2 4
c,1,2 1 3
c,2,3 4 1
r,0,1 2 1
r,1,2 0 1
r,2,2 2 1
p

From the first two lines, we know that there are 3 distinct processes (N) competing for 3 different types of resources (M).

  • Let’s name them as P0, P1, and P2 so that it is easier for us to address them.
  • Also label each of the 3 resources as A, B, and C.

From the third line, we know that NONE of the resources are available: available = [0,0,0].

In the fourth to sixth line, we can initialise the max matrix as follows:

max = [ [2,2,4]
        [2,1,3]
        [3,4,1] ]

This means P0 will need at most 2 instances of Resource A, 2 instances of Resource B, and 4 instances of Resource C during its lifetime of execution. Similar logic for P1 and P2.

In the seventh line r,0,1 2 1, P0 starts requesting for 1 instance of Resource A, 2 instances of Resource B, and 1 instance of Resource C simultanously.

We will need to run the Banker’s algorithm now to determine whether this request is safe, and to be granted.

This is already done for you inside run_file function. Whenever line with r is encountered, the function request_resources is called.

  • If the request is granted, we need to update available, allocation and need. This is already done for you inside request_resources function.
  • If the request is not granted, we simply ignore that request. The values of available, allocation and need remains the same. This is already done for you inside request_resources function.

The same is done for all lines starting with r. We can then print the state of the system at any point in time with p.

In this example, no resources are available. So obviously we are met with the same state of available, allocation and need after running the file because none of the resource requests can be fulfilled.

You can run the algorithm using the command:

python3.10 banker.py test_files/q0.txt

The output is as such as expected, with allocation matrix remaining at 0.

Customer 0 requesting
[1, 2, 1]
Customer 1 requesting
[2, 0, 1]
Customer 2 requesting
[2, 2, 1]

Current state:
Available:
[0, 0, 0]

Maximum:
[2, 2, 4]
[2, 1, 3]
[3, 4, 1]

Allocation:
[0, 0, 0]
[0, 0, 0]
[0, 0, 0]

Need:
[2, 2, 4]
[2, 1, 3]
[3, 4, 1]

Experiment with the other 5 files as well. Try them one by one and study the result.

python3.10 banker.py test_files/q1.txt
python3.10 banker.py test_files/q2.txt
python3.10 banker.py test_files/q3.txt
python3.10 banker.py test_files/q4.txt
python3.10 banker.py test_files/q5.txt

Resource Request Algorithm

This algorithm is already implemented for you inside request_resources function. This algorithm is called each time we encounter a resource request (line starting with r). The function receives two parameters and returns a bool:

def request_resources(self, customer_index, request):
    """
    Request resources for a customer loan.

    Parameters
    ----------
    customer_index : int
        the customer's index (0-indexed)
    request : list[int]
        an array of the requested count for each resource

    Returns
    -------
    True : if the requested resources can be loaned
    False : otherwise
    """

The resource allocation algorithm goes as follows:

  1. If request[i] <= need[customer_index][i] for all i < M, go to step 2. Otherwise, return False (request rejected) since the process has exceeded its maximum claim.

  2. If request[i] <= available[i] for all i < M, go to step 3. Otherwise, return False.

    • Process i must wait since its requested resources are not immediately available (request rejected).

  3. At this point, the requested resources are available, but we check if granting the request is safe (will not lead to deadlock).

    • Create a deepcopy of available, allocation, and need
    • Pass these new data structures to the safety_check algorithm, which will return True (safe) or False (unsafe)

  4. If the outcome of Step(3) is:

    • True: UPDATE all system states concerning Process i (request granted):
      • available[i] = available[i] - request[i] for all i<M
      • need[customer_index][i] = need[customer_index][i] - request[i] for all i<M
      • allocation[customer_index][i] = allocation[customer_index][i] + request[i] for all i<M
    • False: This means request rejected, and process i has to try again in the future. This is because granting the request results in deadlock in the future.

Example 1

You may run: python3.10 banker.py test_files/q1.txt and study the output.

Initially, we know that allocation was initialised as 0 and that available=[5,5,5] from the third line in test_files/q1.txt. The first 3 requests are granted:

Customer 0 requesting
[1, 2, 1]
Customer 1 requesting
[2, 0, 1]
Customer 2 requesting
[2, 2, 1]

After granting the three requests above, we have the system state:

Current state:
Available:
[0, 1, 2]

Maximum:
[2, 2, 4]
[2, 1, 3]
[3, 4, 1]

Allocation:
[1, 2, 1]
[2, 0, 1]
[2, 2, 1]

Need:
[1, 0, 3]
[0, 1, 2]
[1, 2, 0]

Further request by Process 0: request=[0,1,0] is rejected because Process 0 requests MORE than what’s been declared at maximum.

  • It declared that it needs at maximum of 2 resources B (the second type of resource)
  • It already held 2 types of resources B –> Allocation[0][1]: 2
Customer 0 requesting
[0, 1, 0]

Current state: <--- request above is rejected, hence state remains the same.
Available:
[0, 1, 2]

Maximum:
[2, 2, 4]
[2, 1, 3]
[3, 4, 1]

Allocation:
[1, 2, 1]
[2, 0, 1]
[2, 2, 1]

Need:
[1, 0, 3]
[0, 1, 2]
[1, 2, 0]

Example 2

You may run: python3.10 banker.py test_files/q2.txt and study the output.

This time round, we have N=5 processes and M=3 different types of resources, and available=[10,5,7].

After such requests from all customers,

Customer 0 requesting
[0, 1, 0]
Customer 1 requesting
[2, 0, 0]
Customer 2 requesting
[3, 0, 2]
Customer 3 requesting
[2, 1, 1]
Customer 4 requesting
[0, 0, 2]

We should have only 3 counts of Resource A available (the first type of resource).

However here we witness Process 0 releasing 1 instance of Resource A, and hence the updated system state:

Customer 1 releasing
[1, 0, 0]

Current state:
Available:
[4, 3, 2]

Maximum:
[7, 5, 3]
[3, 2, 2]
[9, 0, 2]
[2, 2, 2]
[4, 3, 3]

Allocation:
[0, 1, 0]
[1, 0, 0]
[3, 0, 2]
[2, 1, 1]
[0, 0, 2]

Need:
[7, 4, 3]
[2, 2, 2]
[6, 0, 0]
[0, 1, 1]
[4, 3, 1]

Task 1

TASK 1: Call check_safe algorithm inside request_resources in banker.py.

Right now the variable safe is always set to True, hence we always grant the request as long as the resources are available and the processes do not violate their max. For input q0, q1, q2, we are lucky because we never had any requests that might result in deadlock.

However, if you try running the program with q3, you will notice that the printed output is not the same as the answer: test_files/q3_answer.txt. We need to implement and call the check_safe algorithm to ensure that we get the right answer.

python3.10 banker.py test_files/q3.txt

Let’s start easy. Since check_safe function is already defined (although not yet implemented), call check_safe method and assign their return value to safe:

  # Task 1
  # TODO: Check if the state is safe or not by calling check_safe, right now it is hardcoded to True
  # 1. Perform a deepcopy of available, need, and allocation
  # 2. Call the check_safe method with new data in (1)
  # 3. Store the return value of (2) in variable safe
  # DO NOT PRINT ANYTHING ELSE

  ### BEGIN ANSWER HERE ###
  safe = True  # Change this line

  ### END OF TASK 1 ###

  if safe:
      # If it is safe, allocate the resources to customer customer_number
      for idx, req_val in enumerate(request):
          self.allocation[customer_index][idx] += req_val
          self.need[customer_index][idx] -= req_val
          self.available[idx] -= req_val

      bank_lock.release()
      return True
  else:
      bank_lock.release()
      return False

DO NOT PRINT ANYTHING ELSE for submission. Please remove your debug messages.

The Safety Algorithm

This algorithm is also called each time we encounter a resource request (line starting with r), but only if we managed to enter Step 3 of the Resource Request Algorithm. The function receives five parameters and returns a bool:

def check_safe(self, customer_index, request, work, need, allocation):
    """
    Checks if the request will leave the bank in a safe state.

    Parameters
    ----------
    work, need, allocation : list[int], list[list[int]], list[list[int]]
        deep copy of available, need, and allocation matrices
    customer_index : int
        the customer's index (0-indexed)
    request : list[int]
        an array of the requested count for each resource

    Returns
    -------
    True : if the request resources will leave the bank in a safe state
    False : otherwise
    """

The algorithm goes as follows:

  1. Create a vector finish: list[int] of size N, initialised to be False for all elements in finish.

    • Then, hypothetically grant the current request by customer customer_index by updating:
      • work[i] = work[i] - request[i] for all i<M
      • need[customer_index][i] = need[customer_index][i] - request[i] for all i<M
      • allocation[customer_index][i] = allocation[customer_index][] + request[i] for all i<M
    • This request granting is hypothetical because work is a copy of available (not the actual available). Similar argument with need, allocation. In reality, we haven’t granted the request yet, we simply compute this hypothetical situation and decide whether it will be safe or unsafe.

  2. Find an index i (which is a customer) such that:

    • finish[i] == False and
    • need[i][j] <= work[j] for all j<M.
    • The two above condition signifies that an incomplete Customer i can complete even after this request by customer_index is granted

  3. If such index i from Step 2 exists do the following, else go to Step 4.

    • Update: work[j] = work[j] + allocation[i][j] for all j<M.
      • This signifies that a Customer i that can complete will free its currently allocated resources.
    • Update: finish[i] = True
    • Then, REPEAT step 2

      You might want to store the values of i each time you execute this Step 3 elsewhere to backtrack a possible safe execution sequence, although it is not explicitly required for this lab. Head to the lecture notes for a summary.

  4. If no such index i in Step 2 exists:

    • If finish[i] == True for all i<N, then it means the system is in a safe state. Return True.
    • Else, the system is not in a safe state. Return False.

Careless Mistake

Common careless mistake: A lot of people missed the “REPEAT step 2” instruction in step 3. Step 2 and 3 must be implemented in a while loop, as you might NOT necessarily obtain i in sequential (increasing) order.

Think!

Why is that so? Does it mean we can have a safe execution sequence e.g: 1,0,2 for a 3-process system? Yes of course! That simply means P1 can be executed first, then P0, then P2. Think about what i represents (just arbitrary naming of consumer processes), of course a safe execution sequence has nothing to do with their naming!

Example

Run the Banker algorithm with q3:

python3.10 banker.py test_files/q3.txt

At first, P0 is making the request [1,0,3] and it is granted as shown in the allocation matrix:

Customer 0 requesting
[1, 0, 3]

Current state:
Available:
[4, 2, 1]

Maximum:
[2, 2, 4]
[2, 1, 3]
[3, 1, 1]

Allocation:
[1, 0, 3]
[0, 0, 0]
[0, 0, 0]

Need:
[1, 2, 1]
[2, 1, 3]
[3, 1, 1]

You can verify whether any requests so far is granted by checking that the allocation matrix value adds up to the requests made.

Then, P1 requests for [1,1,1]. This is a legal request, since if the request is granted, then P1 allocation (allocation[1] is initially [0,0,0]) will still be not more than its maximum: max[1]: [2,1,3].

If your implementation is correct, this should however lead to an unsafe state, and hence the request is rejected and the state of the system remain the same:

Customer 1 requesting
[1, 1, 1]

Current state:
Available:
[4, 2, 1]

Maximum:
[2, 2, 4]
[2, 1, 3]
[3, 1, 1]

Allocation:
[1, 0, 3]
[0, 0, 0]
[0, 0, 0]

Need:
[1, 2, 1]
[2, 1, 3]
[3, 1, 1]

The same logic applies to samples in q4 and q5 as well:

  • In q4, the last request by P0: [0,2,0] leads to an unsafe state and therefore rejected.
  • In q5, the request by P1: [2,0,1] is rejected because it leads to an unsafe state. However, after P0 release the resources it held: [1,3,4], a repeated request by P1 for the same set of resources [2,0,1] is granted.

Task 2

TASK 2: Implement the safety check algorithm inside check_safe function in banker.py:

def check_safe(self, customer_index, request, available, need, allocation):
  """
  Checks if the request will leave the bank in a safe state.

  Parameters
  ----------
  available, need, allocation : list[list[int]]
      deep copy of available, need, and allocation matrices
  customer_index : int
      the customer's index (0-indexed)
  request : list[int]
      an array of the requested count for each resource

  Returns
  -------
  True : if the request resources will leave the bank in a safe state
  False : otherwise
  """
  bank_lock.acquire()

  # TASK 2
  # TODO: Check if the state is safe
  # 1. Create finish list[int] of length self.N
  #    Then, hypothetically grant the current request by updating:
  #       1. work[i] = work[i] - request[i] for all i<M
  #       2. need[customer_index][i] = need[customer_index][i] - request[i] for all i<M
  #       3. allocation[customer_index][i] = allocation[customer_index][i] + request[i] for all i<M
  # 2. Find index i such that both finish[i] == False, need[i] <= work
  # 3. If such index in (2) exists, update work += allocation[i], finish[i] = True
  # 4. REPEAT step (2 & 3) until no such i exists
  # 5. If no such i exists anymore, and finish[i] == True for all i, set safe = True
  # 6. Otherwise, set safe = False
  # DO NOT PRINT ANYTHING ELSE

  ### BEGIN ANSWER HERE ###
  safe = True  # Change this line according to whether the request will be safe or not

  ### END OF TASK 2 ###

  bank_lock.release()
  return safe

Again, do NOT print any other stuffs as part of your answer when submitting the files.

Further Notes

Rejecting Requests

If a resource request is rejected, dont panic, it’s not the end of the world. The process/consumer just need to try to request it again in the future.

How can this be implemented? Usually schedulers are programmed to tackle this kind of cases; e.g: they can be placed to a special queue and will periodically prompt for resource request until granted, or there exists some kind of event-driven solution – it’s free-for-all to implement.

Resource Release Caveat

We also assume that (for simplicity of this lab) a process will not release more than what has been allocated to them (that the value of release is valid). We wrote this detail under release_resources function:

def release_resources(self, customer_index, release):
    """
    Releases resources borrowed by a customer. Assume release is valid for simplicity.

    Parameters
    ----------
    customer_index : int
        the customer's index (0-indexed)
    release : list[int]
        an array of the release count for each resource

    Returns
    -------
    None
    """
    print(f"Customer {customer_index} releasing\n{release}")

    bank_lock.acquire()
    # Release the resources from customer customer_number
    for idx, val in enumerate(release):
        self.allocation[customer_index][idx] -= val
        self.need[customer_index][idx] += val
        self.available[idx] += val
    bank_lock.release()

Synchronisation

Finally, since max, allocation, need and available are shared data structures among all these methods, we protect each method that modifies these values using a reentrant lock: bank_lock = threading.RLock().

We guard the critical sections with bank_lock.acquire() and bank_lock.release() to make it thread safe.

When you’re confident that your Task 1 and Task 2 are correct, you can run the tester file:

python3.10 banker_test.py

This will run your banker.py against all 6 test cases: q0 to q5. If all goes well, the following message will be printed:

The tester file runs on Windows as well as of 14/06/2023.

It performs a simple output string matching (which is how our autograder works). Therefore it is crucial for you not to print anything else. If you need to debug, use a debugger. Be a proper programmer, it’s never too late to start today 🥳.

Submission

Once you have completed Banker.py, zip that file and submit your answer to our bot. Just type /start and follow the instruction there.

Zip can be done easily via the CLI (POSIX-compliant OS). Call this command inside lab_banker/ directory:

zip lab3_submit.zip banker.py

If you don’t have zip, you shall install it:

sudo apt update
sudo apt install zip

You need to submit the code by the due date stipulated in the course handout. Remember: only type your answers in the given space. DO NOT print anything else, and DO NOT import any other modules, and DO NOT modify any other instructions.

Appendix

How to migrate the .zip file?

Working on a VM and don’t know how to migrate your .zip file out to your host OS? You can choose the hard way like login to your email account in your VM and then email yourself the zipfile. Or, login to Telegram in your VM.

If you’re using VirtualBox, you can also enable shared clipboard:

But why do that? Why do that if you’re a CS student? Use ssh!

ssh

You can enable ssh at your VM, here is a guide if you’re using VirtualBox. Or, you can simply Google “ssh to your-virtual-machine” and follow simple steps there. Then, enable ssh on VSCode: install the ssh extension and connect! You can follow the guide here just this part:

Then, type in your VM address when prompted in this format: ssh -p <port_number> <username>@<local_ip>, for example:

ssh -p 3022 ubuntu@192.168.55.2

You will need to then enter your password. Then click Open Folder and select your Home folder. You will have a nice interface with VSCode now, connected to your VM. Drag and drop to transfer files between your host OS and your VM. No lag, nice terminal, can utilise VSCode functionalities.. Doesn’t it spark joy?