1. Statement of the problem: Distribute the trips among four zones using a production-constrained gravity model with K_ij = 1.0.\n2. Formula used and important rules: The production-constrained gravity model formula is:\n$$T_{ij} = P_i \frac{A_j f_{ij}}{\sum_k A_k f_{ik}}$$\nHere $P_i$ is trip production for origin i, $A_j$ is trip attraction for destination j, $f_{ij}$ is the friction factor corresponding to travel time between i and j, and we ensure $\sum_j T_{ij}=P_i$ for each origin i.\n3. Friction factors from Table 4 (travel time -> friction): 3 -> 62, 4 -> 51, 6 -> 36, 8 -> 22, 9 -> 15, 11 -> 12, 12 -> 8, 13 -> 4.\n4. Travel time matrix (rows origin, columns destination):\n$$T = \begin{bmatrix}3 & 6 & 8 & 11 \\ 6 & 4 & 9 & 13 \\ 8 & 9 & 3 & 12 \\ 11 & 13 & 12 & 4 \end{bmatrix}$$\n5. Friction factor matrix corresponding to the travel times above:\n$$F = \begin{bmatrix}62 & 36 & 22 & 12 \\ 36 & 51 & 15 & 4 \\ 22 & 15 & 62 & 8 \\ 12 & 4 & 8 & 51 \end{bmatrix}$$\n6. Productions $P = [650, 850, 550, 950]$ and attractions $A = [650, 600, 450, 300]$.\n7. Compute $A_j f_{ij}$ and denominator $D_i = \sum_j A_j f_{ij}$ for each origin.\nFor origin 1 the products are $650\cdot62=40300$, $600\cdot36=21600$, $450\cdot22=9900$, $300\cdot12=3600$ and\n$D_1=40300+21600+9900+3600=75400$.\n8. Compute $T_{1j}$ using the formula.\nExample for $T_{11}$:\n$$T_{11}=650\frac{40300}{75400}$$\nShow cancellation of common factor 100:\n$$T_{11}=650\frac{\cancel{40300}}{\cancel{75400}}=650\frac{403}{754}$$\nNumeric evaluation:\n$$T_{11}=\frac{26195000}{75400}=347.41$$\n9. Remaining row 1 values by the same steps are:\n$$T_{12}=650\frac{21600}{75400}=186.21$$\n$$T_{13}=650\frac{9900}{75400}=85.34$$\n$$T_{14}=650\frac{3600}{75400}=31.04$$\n10. Origin 2: the $A_j f_{2j}$ contributions are $650\cdot36=23400$, $600\cdot51=30600$, $450\cdot15=6750$, $300\cdot4=1200$ and\n$D_2=23400+30600+6750+1200=61950$.\nExample cancellation for $T_{21}$ dividing numerator and denominator by 50:\n$$T_{21}=850\frac{23400}{61950}=850\frac{\cancel{23400}}{\cancel{61950}}=850\frac{468}{1239}$$\nNumeric evaluation:\n$$T_{21}=321.07$$\nRemaining origin 2 values:\n$$T_{22}=850\frac{30600}{61950}=419.85$$\n$$T_{23}=850\frac{6750}{61950}=92.62$$\n$$T_{24}=850\frac{1200}{61950}=16.47$$\n11. Origin 3: $A_j f$ are $650\cdot22=14300$, $600\cdot15=9000$, $450\cdot62=27900$, $300\cdot8=2400$ and\n$D_3=14300+9000+27900+2400=53600$.\nExample cancellation for $T_{31}$ dividing by 100:\n$$T_{31}=550\frac{14300}{53600}=550\frac{\cancel{14300}}{\cancel{53600}}=550\frac{143}{536}$$\nNumeric evaluation:\n$$T_{31}=146.73$$\nRemaining origin 3 values:\n$$T_{32}=550\frac{9000}{53600}=92.35$$\n$$T_{33}=550\frac{27900}{53600}=286.29$$\n$$T_{34}=550\frac{2400}{53600}=24.63$$\n12. Origin 4: $A_j f$ are $650\cdot12=7800$, $600\cdot4=2400$, $450\cdot8=3600$, $300\cdot51=15300$ and\n$D_4=7800+2400+3600+15300=29100$.\nExample cancellation for $T_{41}$ dividing by 100:\n$$T_{41}=950\frac{7800}{29100}=950\frac{\cancel{7800}}{\cancel{29100}}=950\frac{78}{291}$$\nNumeric evaluation:\n$$T_{41}=254.64$$\nRemaining origin 4 values:\n$$T_{42}=950\frac{2400}{29100}=78.35$$\n$$T_{43}=950\frac{3600}{29100}=117.53$$\n$$T_{44}=950\frac{15300}{29100}=499.48$$\n13. Final OD matrix (rows origins 1..4, columns destinations 1..4) rounded to two decimals:\n$$\begin{bmatrix}347.41 & 186.21 & 85.34 & 31.04 \\ 321.07 & 419.85 & 92.62 & 16.47 \\ 146.73 & 92.35 & 286.29 & 24.63 \\ 254.64 & 78.35 & 117.53 & 499.48 \end{bmatrix}$$\n14. Check: row sums equal productions within rounding error: 650, 850, 550, 950 respectively.\n15. Final answer: the OD matrix above is the trip distribution using the production-constrained gravity model with K_ij=1.0.\n