Gradient accumulation in PyTorch - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to zero the gradients before starting the backward pass.

PyTorch

optimizer.[1]()

Drag options to blanks, or click blank then click option'

Azero_grad

Bstep

Cbackward

Deval

Attempts:

3 left

💡 Hint

Common Mistakes

Using optimizer.step() before zeroing gradients

Calling backward() on optimizer instead of loss

Forgetting to clear gradients causing accumulation unintentionally

Before computing gradients, we clear old gradients by calling optimizer.zero_grad().

2fill in blank

medium

Complete the code to perform a backward pass on the loss.

PyTorch

loss.[1]()

Drag options to blanks, or click blank then click option'

Adetach

Bbackward

Czero_grad

Dstep

Attempts:

3 left

💡 Hint

Common Mistakes

Calling optimizer.step() instead of loss.backward()

Forgetting to call backward() causing no gradient computation

To compute gradients, call loss.backward() which backpropagates the error.

3fill in blank

hard

Fix the error in the code to accumulate gradients over multiple batches before updating weights.

PyTorch

if (batch_idx + 1) % [1] == 0:
    optimizer.step()
    optimizer.zero_grad()

Drag options to blanks, or click blank then click option'

Anum_epochs

Bbatch_size

Clearning_rate

Daccumulation_steps

Attempts:

3 left

💡 Hint

Common Mistakes

Using batch_size instead of accumulation_steps

Updating optimizer every batch without accumulation

Gradient accumulation updates weights every accumulation_steps batches.

4fill in blank

hard

Fill both blanks to correctly scale the loss for gradient accumulation.

PyTorch

loss = loss / [1]
loss.[2]()

Drag options to blanks, or click blank then click option'

Aaccumulation_steps

Bbatch_size

Cbackward

Dstep

Attempts:

3 left

💡 Hint

Common Mistakes

Not scaling loss causing gradients to be too large

Calling optimizer.step() instead of loss.backward() here

We divide loss by accumulation_steps to average gradients, then call backward() to accumulate.

5fill in blank

hard

Fill all three blanks to implement gradient accumulation correctly in a training loop.

PyTorch

for batch_idx, (inputs, targets) in enumerate(dataloader):
    outputs = model(inputs)
    loss = criterion(outputs, targets) / [1]
    loss.[2]()
    if (batch_idx + 1) % [3] == 0:
        optimizer.step()
        optimizer.zero_grad()

Drag options to blanks, or click blank then click option'

Aaccumulation_steps

Bbackward

Dstep

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variables for scaling and update frequency

Forgetting to zero gradients after optimizer step

Divide loss by accumulation_steps, call backward() to accumulate gradients, and update weights every accumulation_steps batches.

Practice

1. What is the main purpose of gradient accumulation in PyTorch training?

easy

A. To reduce the number of model parameters

B. To increase the learning rate automatically

C. To change the model architecture dynamically

D. To split a large batch into smaller steps to save memory

Solution

1. Step 1: Understand gradient accumulation concept

   Gradient accumulation splits a large batch into smaller mini-batches to avoid memory overflow.

2. Step 2: Identify the main benefit

   This technique helps save GPU memory by processing smaller batches sequentially before updating weights.

3. Final Answer:

   To split a large batch into smaller steps to save memory -> Option D

4. Quick Check:

   Gradient accumulation = batch splitting for memory saving [OK]

Hint: Gradient accumulation saves memory by splitting batches [OK]

Common Mistakes:

• Thinking it changes learning rate
• Confusing with model size reduction
• Assuming it alters model layers

2. Which PyTorch code snippet correctly implements gradient accumulation over 4 steps?

easy

A. for i, data in enumerate(dataloader):
outputs = model(data)
loss = criterion(outputs, labels)
optimizer.step()
loss.backward()
optimizer.zero_grad()

B. for i, data in enumerate(dataloader):
outputs = model(data)
loss = criterion(outputs, labels)
loss.backward()
optimizer.step()
if (i + 1) % 4 == 0:
optimizer.zero_grad()

C. for i, data in enumerate(dataloader):
outputs = model(data)
loss = criterion(outputs, labels)
loss.backward()
if (i + 1) % 4 == 0:
optimizer.step()
optimizer.zero_grad()

D. for i, data in enumerate(dataloader):
outputs = model(data)
loss = criterion(outputs, labels)
optimizer.zero_grad()
loss.backward()
if (i + 1) % 4 == 0:
optimizer.step()

Solution

1. Step 1: Identify correct order of operations

   Gradient accumulation requires calling loss.backward() each step, then optimizer.step() and zero_grad() after accumulating gradients.

2. Step 2: Check modulo condition for accumulation

   Only after 4 mini-batches (i+1 % 4 == 0) should optimizer.step() and zero_grad() be called to update weights and reset gradients.

3. Final Answer:

   Code that calls loss.backward() every step and optimizer.step() plus zero_grad() every 4 steps -> Option C

4. Quick Check:

   Backward every step, step and zero_grad every 4 steps [OK]

Hint: Call backward every step, step and zero_grad every N steps [OK]

Common Mistakes:

• Calling optimizer.step() before loss.backward()
• Resetting gradients too early
• Not using modulo condition correctly

3. Given this PyTorch code with gradient accumulation steps=2, what is the printed loss after 4 batches?

optimizer.zero_grad()
for i, (inputs, labels) in enumerate(dataloader):
    outputs = model(inputs)
    loss = criterion(outputs, labels) / 2
    loss.backward()
    if (i + 1) % 2 == 0:
        optimizer.step()
        optimizer.zero_grad()
    print(f"Batch {i+1} loss: {loss.item()}")

medium

A. Loss values cause runtime error due to division

B. Loss values divided by 2 printed for each batch

C. Loss values accumulated and printed only every 2 batches

D. Loss values printed without division

Solution

1. Step 1: Understand loss scaling in code

   Loss is divided by 2 before backward to average gradients over 2 accumulation steps.

2. Step 2: Check print statement inside loop

   Loss.item() prints the scaled loss for each batch, so printed values are divided by 2 every batch.

3. Final Answer:

   Loss values divided by 2 printed for each batch -> Option B

4. Quick Check:

   Loss scaled by 2 and printed each batch [OK]

Hint: Loss is scaled before backward, printed as scaled every batch [OK]

Common Mistakes:

• Thinking loss prints only every 2 batches
• Ignoring loss division by 2
• Assuming runtime error from division

4. Identify the error in this gradient accumulation code snippet:

optimizer.zero_grad()
for i, (inputs, labels) in enumerate(dataloader):
    outputs = model(inputs)
    loss = criterion(outputs, labels)
    loss.backward()
    if i % 4 == 0:
        optimizer.step()
        optimizer.zero_grad()

medium

A. optimizer.step() and zero_grad() are called too early at i=0

B. loss.backward() is missing gradient scaling

C. optimizer.zero_grad() should be called before loss.backward()

D. The loop should use (i + 1) % 4 == 0 instead of i % 4 == 0

Solution

1. Step 1: Analyze the modulo condition

   Condition i % 4 == 0 triggers optimizer.step() and zero_grad() at i=0, which is before any gradients accumulate.

2. Step 2: Understand correct timing for optimizer step

   Optimizer step should happen after 4 batches, so condition should be (i + 1) % 4 == 0 to avoid early update.

3. Final Answer:

   optimizer.step() and zero_grad() are called too early at i=0 -> Option A

4. Quick Check:

   Modulo condition triggers step too early [OK]

Hint: Use (i + 1) % N == 0 to delay step until after N batches [OK]

Common Mistakes:

• Calling optimizer.step() at first batch
• Confusing zero_grad() timing
• Ignoring off-by-one in modulo condition

5. You want to train a large model on a GPU with limited memory using gradient accumulation with batch size 64 split into 8 steps. Which approach correctly adjusts the learning rate and loss scaling to keep training stable?

hard

A. Divide the loss by 8 before backward and keep the original learning rate

B. Multiply the learning rate by 8 and do not scale the loss

C. Divide the learning rate by 8 and multiply the loss by 8

D. Keep loss and learning rate unchanged, just accumulate gradients

Solution

1. Step 1: Understand gradient accumulation effect on loss

   Splitting batch into 8 steps means gradients accumulate over 8 mini-batches, so loss must be divided by 8 to average gradients correctly.

2. Step 2: Adjust learning rate accordingly

   Learning rate stays the same because gradient accumulation simulates a larger batch size without changing optimization dynamics.

3. Final Answer:

   Divide the loss by 8 before backward and keep the original learning rate -> Option A

4. Quick Check:

   Loss scaled down, learning rate unchanged for stable accumulation [OK]

Hint: Scale loss down by steps, keep learning rate same [OK]

Common Mistakes:

• Scaling learning rate instead of loss
• Not scaling loss causing gradient explosion
• Assuming no changes needed for accumulation