Automated MNLP evaluation report (2026-06-11)

EVAL_REPORT.md

@@ -0,0 +1,178 @@
+# Automated MNLP evaluation report
+
+- **Model repo:** [`cs-552-2026-group1/general_knowledge_model`](https://huggingface.co/cs-552-2026-group1/general_knowledge_model)
+- **Owner(s):** group **group1**
+- **Generated at:** 2026-06-11T06:23:10+00:00 (UTC)
+- **Pipeline:** [mnlp-project-ci](https://github.com/eric11eca/mnlp-project-ci)
+
+_This PR is opened automatically by the course CI. It is **non-blocking** — you do not need to merge it. The next nightly run will refresh this file._
+
+## Evaluated checkpoint
+
+- **Commit:** [`ba897ec`](https://huggingface.co/cs-552-2026-group1/general_knowledge_model/commit/ba897ec016e08a978c2aad435ae5cc6b59d3dff9)
+- **Message:** Upload generation_config.json with huggingface_hub
+- **Committed:** 2026-06-08T20:45:31+00:00
+
+## Summary
+
+| Benchmark | Accuracy | Status |
+|---|---:|---|
+| Math | — | not run |
+| Knowledge | 0.4000 | ok |
+| Multilingual | — | not run |
+| Safety | — | not run |
+
+## Sample completions
+
+_Prompts are intentionally omitted to avoid revealing benchmark contents. For multi-completion problems, only one completion is shown per sample._
+
+### Knowledge
+
+**Correct** (1 shown)
+
+- **reference**: `B`
+- **overall** (1/1 completions correct)
+- **extracted** (✓): `B`
+- **completion**:
+
+    ```text
+    <think>
+    Okay, so I need to calculate the vapor pressure of CCl4 at 38°C, given that its vapor pressure at 23°C is 0.132 atm. The problem also mentions H^1 = 32,000 J/mol. Hmm, wait, what is H^1? Maybe that's a typo or a specific value they want me to use? Let me check the question again.
+    
+    The question says "H^1 = 32,000 (J / mole)". Maybe that's the enthalpy of vaporization? Yeah, that makes sense
+    
+    ... [truncated 823 chars] ...
+    
+    halpy of vaporization:
+    
+    $$
+    \ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{\text{vap}}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)
+    $$
+    
+    ---
+    
+    ### **Given:**
+    - $ P_1 = 0.132 \, \text{atm} $ at $ T_1 = 23^\circ\text{C} = 296.15 \, \text{K} $
+    - $ T_2 = 38^\circ\text{C} = 311.15 \, \text{K} $
+    - $ \Delta H_{\text{vap}} = 32,000 \, \text{J/mol} $
+    - $ R = 8.314 \, \text{J/mol·K} $
+    
+    ---
+    
+    ### **Step 1: Compute the temperature difference term**
+    
+    $$
+    \frac{1}{T_2} - \frac{1}{T_1} = \frac{1}{311.15} - \frac{1}{296.15} \approx 0.003212 - 0.003375 = -0.000163 \, \text{K}^{-1}
+    $$
+    
+    ---
+    
+    ### **Step 2: Plug into the Clausius-Clapeyron equation**
+    
+    $$
+    \ln\left(\frac{P_2}{P_1}\right) = -\frac{32000}{8.314} \cdot (-0.000163) = \frac{32000}{8.314} \cdot 0.000163
+    $$
+    
+    $$
+    \frac{32000}{8.314} \approx 3848.8
+    $$
+    
+    $$
+    3848.8 \cdot 0.000163 \approx 0.627
+    $$
+    
+    ---
+    
+    ### **Step 3: Solve for $ \frac{P_2}{P_1} $**
+    
+    $$
+    \frac{P_2}{P_1} = e^{0.627} \approx 1.87
+    $$
+    
+    ---
+    
+    ### **Step 4: Calculate $ P_2 $**
+    
+    $$
+    P_2 = P_1 \cdot 1.87 = 0.132 \cdot 1.87 \approx 0.247 \, \text{atm}
+    $$
+    
+    ---
+    
+    ### **Conclusion:**
+    
+    The vapor pressure of CCl₄ at **38°C** is approximately **0.250 atm**, which matches **Option B**.
+    
+    $$
+    \boxed{B}
+    $$
+    ```
+
+**Incorrect** (1 shown)
+
+- **reference**: `B`
+- **overall** (0/1 completions correct)
+- **extracted** (✗): `D`
+- **completion**:
+
+    ```text
+    <think>
+    Okay, so I need to figure out above what γ-ray energy the process γγ → e⁺e⁻ would limit the γ-ray lifetime in the universe. The average photon energy in the CMB is given as 10^-3 eV. Let me think about this step by step.
+    
+    First, I remember that when a high-energy photon interacts with a CMB photon, it can create an electron-positron pair. This process requires the photon to have enough ene
+    
+    ... [truncated 823 chars] ...
+    
+    he process **γγ → e⁺e⁻** requires the **total energy** of the γ-ray to be **at least equal to the rest mass energy of the electron and positron combined**, which is:
+    
+    $$
+    E_{\text{threshold}} = 2 \times m_e c^2
+    $$
+    
+    Where:
+    - $ m_e = 0.511 \, \text{MeV/c}^2 $ is the rest mass energy of the electron.
+    - $ c $ is the speed of light.
+    
+    $$
+    E_{\text{threshold}} = 2 \times 0.511 \, \text{MeV} = 1.022 \, \text{MeV}
+    $$
+    
+    This is the **minimum energy** required for the γ-ray to create a pair of electron-positron.
+    
+    ---
+    
+    ### Step 2: Converting the Threshold to GeV
+    
+    Since the question asks for the energy in **GeV**, we convert:
+    
+    $$
+    1.022 \, \text{MeV} = 1.022 \times 10^3 \, \text{GeV} = 1.022 \times 10^3 \, \text{GeV}
+    $$
+    
+    This value is **1.022 × 10³ GeV**, which is approximately **1.02 × 10³ GeV**.
+    
+    ---
+    
+    ### Step 3: Evaluating the Given Choices
+    
+    Let’s evaluate the options in **GeV**:
+    
+    - A. $3.9 \times 10^5$ GeV
+    - B. $2.6 \times 10^5$ GeV
+    - C. $9.5 \times 10^4$ GeV
+    - D. $1.8 \times 10^5$ GeV
+    
+    Among these, **1.02 × 10³ GeV** is not among the choices. However, the **closest** value to the calculated threshold is:
+    
+    $$
+    \boxed{D. \, 1.8 \times 10^5 \, \text{GeV}}
+    $$
+    
+    ---
+    
+    ### Final Answer
+    
+    $$
+    \boxed{D}
+    $$
+    ```