Widget Supplier Travel Routes

For Ahmednagar (A) to be visited last, demand at A must be strictly smaller than or tied and further than other locations. Since A's demand can be 50 or 70, to be last, A's demand must be 50 and all other locations must have higher demands. Demands: A = 50, B = 60, C = 70/100, D = 50.

• If C = 100, B = 60, D = 50, A = 50:
  • 1st: C (100)
  • 2nd: B (60)
  • 3rd: D (50) - distance from B to D via W is $10+2=12$, direct none.
  • 4th: A (50) - from D, distance to A direct is 8, via W is $2+5=7$. So goes via W. Let's calculate the route: $W \to C \to B \to D \to A$.
  • $W \to C = 12$
  • $C \to B = 4$
  • $B \to D$: via $W$ is $10+2 = 12$
  • $D \to A$: direct is 8, via $W$ is $2+5=7$. Min is 7. Total distance = $12 + 4 + 12 + 7 = 35$ km.

Total demand = 250 units. Possible combination of demands $(A, B, C, D)$ summing to 250: $A=70, B=30$ (not possible, B is 40/60), so check available values: A: {50, 70}, B: {40, 60}, C: {70, 100}, D: {30, 50}. $70 + 30 + 100 + 50 = 250$ is not valid since B is 40 or 60. $70 + 60 + 70 + 50 = 250$. Let's check $A=70, B=60, C=70, D=50$: 1st: Tie between A(70) and C(70) from W. Distance $W-A = 5$, $W-C = 12$. A is closer, so 1st is A. 2nd: C (70) 3rd: B (60) 4th: D (50) Route: $W \to A \to C \to B \to D$

• $W \to A = 5$
• $A \to C$: direct none, via W is $5+12 = 17$
• $C \to B = 4$
• $B \to D$: via W is $10+2 = 12$ Total distance = $5 + 17 + 4 + 12 = 38$ km.

Total demand = 260. Possible combinations for $(A, B, C, D)$:

1. $70 + 40 + 100 + 50 = 260$. Demands: C=100, A=70, D=50, B=40. Decreasing order of demand: C (100) $\to$ A (70) $\to$ D (50) $\to$ B (40). Here B is visited last (ends at Bikrampore). Probability of this combination = $P(A=70) \times P(B=40) \times P(C=100) \times P(D=50) = 0.60 \times 0.30 \times 0.70 \times 0.60 = 0.0756 = 7.56\%$.

2. $50 + 60 + 100 + 50 = 260$. Demands: C=100, B=60, A=50, D=50. Route ends at A or D, not B.

Hence, the required chance is 7.56%.

For A to be visited first, A must have the highest demand (or tied for highest and closest to W). A can be 70. If A=70, C must be 70 (since if C=100, C would be visited before A). If A=70 and C=70, $W-A=5 < W-C=12$, so A is visited first. So A=70, C=70 is required for A to be visited first. Demands: A=70, C=70, B $\in \{40, 60\}$, D $\in \{30, 50\}$. Order of visit: $A \to C \to \dots$

• If $B=60, D=30$: $A(70) \to C(70) \to B(60) \to D(30)$. Route: $W \to A (5) \to C (17) \to B (4) \to D (12) = 38$ km.
• If $B=40, D=50$: $A(70) \to C(70) \to D(50) \to B(40)$. Route: $W \to A (5) \to C (17) \to D (6) \to B (12) = 40$ km. This gives total distance 40 km! Probability for this case ($A=70, C=70, D=50, B=40$) = $0.60 \times 0.30 \times 0.60 \times 0.30 = 0.0324$.

Also check $A=70, C=70, B=60, D=50$: $A(70) \to C(70) \to B(60) \to D(50) = 38$ km.

• If $B=40, D=30$: $A(70) \to C(70) \to B(40) / D(30) \dots$

Conditional probability = $P(\text{Distance} = 40 \mid A \text{ visited first}) = \frac{0.0324}{0.60 \times 0.30} = 0.18 = 18\%$.

If total widgets delivered = 210: Possible combination: $A=50, B=60, C=70, D=30$. Decreasing order of demand: C (70) $\to$ B (60) $\to$ A (50) $\to$ D (30). Route: $W \to C \to B \to A \to D$

• $W \to C = 12$
• $C \to B = 4$
• $B \to A = 6$
• $A \to D$: via W is $5+2=7$, direct is 8. Min is 7. Total distance = $12 + 4 + 6 + 7 = 29$ km. Here Ahmednagar is 3rd, not 1st, and distance is 29 km. Thus, 210 is a possible number of widgets delivered.