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# 数学代写|运筹学代写Operations Research代考|MATH3202 North West Corner Rule

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## 数学代写|运筹学代写Operations Research代考|North West Corner Rule

1. Consider the top left hand corner (the North West Corner). The supply available is 40 and the requirement is 20 . The maximum possible for allocation is the minimum of the two, which is 20 . We allocate 20 . The demand of the first point is met and we draw a vertical line indicating this. The available supply in the first supply point now reduces to 40 .
2. We now identify first row-second column as the north west corner. The available supply is 20 and the requirement is 30 . The maximum possible allocation is 20 , and is made. The supply in Row 1 is exhausted and we indicate this with a horizontal line. The demand (unfulfilled) in the second demand point reduces to 10.
3. We now identify second row-second column as the north west corner. The available supply is 60 and the requirement is 10 . The maximum possible allocation is 10 , and is made. The demand in Row 2 is met completely and we indicate this with a vertical line. The available supply in the second row reduces to 50 .
4. We now identify first second row-third column as the north west corner. The available supply is 50 and the requirement is 50 . The maximum possible allocation is 50 , and is made. The supply in Row 1 is exhausted and the demand in third column is entirely met. We indicate this with a horizontal line and vertical line, respectively.
5. The only available position for allocation is the fourth row-third column. The supply is 50 and the demand is 50 . We make the allocation and observe that all allocations have been made. The supplies from all the rows are used and the requirements of all columns have been completely met.
6. We observe that we have made five allocations and the total cost is:
$$20 \times 4+20 \times 6+10 \times 8+50 \times 6+50 \times 8=980$$

## 数学代写|运筹学代写Operations Research代考|Minimum Cost Method

1. From the cost coefficients, we observe that Row 1 -Column 1 has the minimum cost of 4. The supply is 40 and the demand is 20 . We allocate the maximum possible, which is the minimum of 40 and 20 . The requirement of Column 1 is met and we indicate this with a vertical line. The available supply in Row 1 is now 20 .
2. From the cost coefficients of the remaining positions, we observe that Row 1-Column 2 has the minimum cost of 6 . The available supply is 20 and the demand is 30 . We allocate the maximum possible, which is 20 . The supply from Row 1 is used fully and we indicate this with a horizontal line. The unfulfilled demand of Column 2 is now 10 .
3. From the cost coefficients of the remaining positions, we observe that Row 2-Column 3 has the minimum cost of 6 . The supply is 60 and the demand is 50 . We allocate the maximum possible, which is 50 . The requirement of Column 3 is met and we indicate this with a vertical line. The available supply in Row 2 is now 10.
4. From the cost coefficients of the remaining positions, we observe that Row 2-Column 4 as well as Row 3-Column 2 have the minimum cost of 7 . We break the tie arbitrarily and choose Row 2-Column 4. The available supply is 10 and the demand is 50 . We allocate the maximum possible, which is 10 . The supply from Row 2 is used fully and we indicate this with a horizontal line. The unfulfilled demand of Column 4 is now 40 .
5. Only Row 3 has available supply and the unfulfilled requirements are in Row 3-Column 2 is 10 and Row 3-Column 4 is 40 . We make the allocations. Now, all the requirements have been met and we have a basic feasible solution. The cost is:
$$20 \times 4+20 \times 6+50 \times 6+10 \times 7+10 \times 7+40 \times 8=960$$

## 数学代写|运筹学代写Operations Research代考|North West Corner Rule

$$20 \times 4+20 \times 6+10 \times 8+50 \times 6+50 \times 8=980$$

## 数学代写|运筹学代写Operations Research代考|Minimum Cost Method

$$20 \times 4+20 \times 6+50 \times 6+10 \times 7+10 \times 7+40 \times 8=960$$

## MATLAB代写

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