1. **Problem Statement:** A) Prove that the statement pattern $(p \to q) \leftrightarrow (\sim q \to \sim p)$ is a tautology using a truth table. B) Find the initial basic feasible solution for the transportation problem using the North West Corner Method. --- ### Part A: Prove Tautology 2. **Recall:** A tautology is a statement that is true for all truth values of its variables. 3. **Formula:** The biconditional $A \leftrightarrow B$ is true when both $A$ and $B$ have the same truth value. 4. **Construct truth table for $p$, $q$, $p \to q$, $\sim q$, $\sim p$, $\sim q \to \sim p$, and $(p \to q) \leftrightarrow (\sim q \to \sim p)$:** | $p$ | $q$ | $p \to q$ | $\sim q$ | $\sim p$ | $\sim q \to \sim p$ | $(p \to q) \leftrightarrow (\sim q \to \sim p)$ | |-----|-----|-----------|----------|----------|---------------------|----------------------------------------------| | T | T | T | F | F | T | T | | T | F | F | T | F | F | T | | F | T | T | F | T | T | T | | F | F | T | T | T | T | T | 5. **Explanation:** The biconditional column is true for all rows, so the statement is a tautology. --- ### Part B: North West Corner Method 6. **Problem:** Find initial basic feasible solution for the transportation table: | | W1 | W2 | W3 | W4 | Supply | |-----|----|----|----|----|--------| | F1 | 42 | 32 | 50 | 26 | 11 | | F2 | 34 | 36 | 28 | 46 | 13 | | F3 | 64 | 54 | 36 | 82 | 19 | | Demand | 6 | 10 | 12 | 15 | | 7. **Method:** Start allocating from top-left (north-west) corner cell and move right/down adjusting supply and demand. 8. **Stepwise allocation:** - Allocate min(11,6)=6 to F1-W1; update supply F1=5, demand W1=0 - Move right to W2; allocate min(5,10)=5 to F1-W2; update supply F1=0, demand W2=5 - Move down to F2-W2; allocate min(13,5)=5; update supply F2=8, demand W2=0 - Move right to W3; allocate min(8,12)=8; update supply F2=0, demand W3=4 - Move down to F3-W3; allocate min(19,4)=4; update supply F3=15, demand W3=0 - Move right to W4; allocate min(15,15)=15; update supply F3=0, demand W4=0 9. **Final allocations:** | | W1 | W2 | W3 | W4 | |-----|----|----|----|----| | F1 | 6 | 5 | 0 | 0 | | F2 | 0 | 5 | 8 | 0 | | F3 | 0 | 0 | 4 | 15 | --- **Final answers:** - The statement $(p \to q) \leftrightarrow (\sim q \to \sim p)$ is a tautology. - The initial basic feasible solution by North West Corner Method is as above.