Course Classroom

Introduction to deep learning · Which shipment gets tomorrow’s inspection slot?

Lesson summary

At noon a planner knows that a shipment missed two checkpoints. It later arrives late. Follow one training step to explain how that outcome changes tomorrow’s score. Freeze the record at noon, calculate the prediction, and compare two parameter settings before choosing a cutoff.

Lesson 1 / 1 · 4 min · Beginner · English

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Lesson route

  1. Freeze the inputs
  2. Read the neuron
  3. Measure the error
  4. Update one parameter
  5. Test the ranking

The lecture builds from a multi-output perceptron to training a network and later introduces early stopping. In the first Python lab, gradients connect a computed loss to trainable parameters. The shipping calculation below isolates one weight so that change can be audited by hand. Slides · PDF 36 · Companion lab

Original GuideWit example. The figures illustrate the exercise. The linked times open related sections of the MIT lecture.

Freeze the inputs

Let x = 2 be the missed checkpoints by noon and y = 1 the later observation of lateness. Arrival time may establish the label after delivery. Putting it into the input would let the model see the answer during training.

Freeze the inputs · x = 2 → y = 1 → t₀ < t₁

Freeze the inputs · x = 2 → y = 1 → t₀ < t₁. 15:40 · MIT 6.S191

Read the neuron

The input weight is 0.4 and the bias is −1. Multiplying and adding gives z = −0.2. The sigmoid maps this to 0.4502. A score below one half can still belong to a shipment that arrives late.

Read the neuron · 0.4 × 2 − 1 → −0.2 → σ ≈ 0.450

Read the neuron · 0.4 × 2 − 1 → −0.2 → σ ≈ 0.450. 27:36 · MIT 6.S191

Measure the error

Because this shipment was late, its binary cross-entropy is −ln(0.4502), about 0.798. An on-time shipment would use −ln(1−p). Crossing a classification threshold and reducing the training loss are separate events.

Measure the error · y = 1 → −ln(0.450) → ≈ 0.798

Measure the error · y = 1 → −ln(0.450) → ≈ 0.798. 30:44 · MIT 6.S191

Update one parameter

Hold the bias fixed. The gradient of this loss with respect to the weight is (p−y)x ≈ −1.100. Subtract 0.1 times that value. The weight rises to 0.510, giving a probability near 0.505 and loss near 0.683.

Update one parameter · −1.100 → 0.4 + 0.110 → w ≈ 0.510

Update one parameter · −1.100 → 0.4 + 0.110 → w ≈ 0.510. 35:54 · MIT 6.S191

Test the ranking

An on-time shipment with the same input receives the same increase in score. Evaluate the first ten alerts on later shipments, keeping all records of a journey together. Count how many calls catch a genuinely late delivery.

Test the ranking · 0.450 → 0.505 → y = 1: L ↓ → y = 0: L ↑

Test the ranking · 0.450 → 0.505 → y = 1: L ↓ → y = 0: L ↑. 45:59 · MIT 6.S191

Worked comparison

Weight

Probability and loss

0.4

0.450 / 0.798

0.510

0.505 / 0.683

Troubleshooting

Signal

Check

The score never moves

Check x and the learning rate

Test results look perfect

Look for arrival fields dated after noon

Practice

  1. Recalculate with x = 0 and the same bias. Which parameter can this row teach?

Answer · The weight gradient is zero. The bias can still move if it is trainable.

  1. Can a lower loss on this one shipment justify ten calls?

Answer · No. Use later labeled journeys to measure the ranking at the desk’s real call capacity.

Glossary

Weight = learned input multiplier. Bias = input-independent offset. Gradient = local sensitivity of the loss.

Sources