Zach's Math Zone · YouTube · 6:47
This 120-second clip introduces a transportation problem with three factories and four cities, shows the cost/supply/demand table, verifies that total supply equals total demand (), states the goal of minimizing shipping cost, and begins the least cost method by selecting the unique minimum-cost cell ($5 at Factory 3 to City 4). It also gives the tie-breaking rule for equal minimum costs and ends just after identifying the relevant demand (30) and supply (40) for the first allocation, before any allocation value is entered. This 120-second clip demonstrates the least cost method on a 3-by-4 transportation problem. The narrator first satisfies City 4 by allocating 30 units from Factory 3, updates Factory 3's remaining supply from 40 to 10, and excludes the other City 4 routes. He then selects the next cheapest remaining route, Factory 1 to City 2 at $6, allocates 20 units, reduces Factory 1's supply from 35 to 15, and closes the City 2 column. The final seconds begin a third iteration by choosing the $8 route from Factory 1 to City 1 against demand 45, but the clip stops before that allocation is completed. This 120-second clip works through the tail end of a transportation problem using the least cost method on a 3-by-4 tableau. Starting from a partially filled table, the instructor allocates 15 units from Factory 1 to City 1, then 30 units from Factory 2 to City 1 at cost $9, then 20 units from Factory 2 to City 3 at cost $13 after correctly rejecting the exhausted $10 route from Factory 1, and finally 10 units from Factory 3 to City 3. The completed visible solution is , , , , , . The clip ends just as the instructor writes TOTAL COST: and begins explaining how to compute the objective value, but the final numerical total is not reached within the provided duration. This 47-second excerpt shows the final stage of a transportation-problem example presented on a digital whiteboard. A completed 3-by-4 tableau lists factories, cities, remaining supply, remaining demand, unit shipping costs, and six occupied routes marked by circled quantities, while unused routes carry X marks. The narrator converts each occupied route into a product of quantity and unit cost, writing TOTAL COST = 15($8) + 30($9) + 20($6) + 20($13) + 10($16) + 30($5). The expression is then evaluated to a boxed answer of $1,080. The clip explicitly connects this calculation to the least cost method, but it does not show the earlier route-selection process that produced the allocation.
Use the learning inspector for key ideas and moments, or open the reading tabs for the complete notes.
Generated from the video's visuals and explanation; not verbatim speech.
The clip opens on a transportation table. The rows are labeled Factory 1, Factory 2, and Factory 3; the columns are labeled City 1, City 2, City 3, and City 4. A rightmost column labeled Supply lists 35, 50, and 40, and a bottom row labeled Demand lists 45, 20, 30, and 30. The speaker frames this as a transportation problem with three producing factories and four consuming cities.
Next, the video checks whether the model is balanced. The supply entries are summed as , and the demand entries are summed as . Because total supply equals total demand, the speaker classifies the example as a balanced transportation problem.
The objective is then stated explicitly: satisfy each city's demand while minimizing total shipping cost. The interior cells of the table are identified as unit shipping costs from a given factory to a given city, with examples such as $8 from Factory 1 to City 1 and $6 from Factory 1 to City 2.
To begin the least cost method, the speaker instructs the viewer to find the cell with the smallest shipping cost. Scanning the table, the minimum is $5 at Factory 3 to City 4. The speaker also adds a tie-breaking rule: if two cells share the same minimum cost, choose the one with the higher demand. Here the $5 cell is unique, so it becomes the starting cell.
Finally, the chosen cell is linked to its row and column limits. The column is City 4, whose demand is 30 units, and the row is Factory 3, whose supply is 40 units. The clip ends just as the speaker begins asking whether the full demand of City 4 can be met from Factory 3; no allocation number is written before the segment stops.
The clip opens on a transportation tableau with three factory rows, four city columns, unit shipping costs inside the cells, a Supply column on the right, and a Demand row along the bottom. The narrator is already comparing Factory 3's supply with City 4's demand, and the visual emphasis is on the $5 route between them.
Because Factory 3 has 40 units available and City 4 needs 30, the narrator concludes that the demand can be satisfied entirely from this factory. The rule being used is that the shipment on the selected route equals the smaller of remaining supply and remaining demand, so here the allocation is 30.
Once 30 units are written into the Factory 3, City 4 cell, City 4's demand is considered fully met. The demand entry 30 is crossed out, signaling that no additional units should be sent to that destination.
The algorithm then updates the source side of the table. Factory 3's supply is reduced from 40 to 10 by the subtraction , showing that 10 units remain available for other cities.
The narrator explains why the rest of the City 4 column is removed from future consideration. If any more units were sent to City 4, the total received there would exceed its demand of 30, violating the destination constraint.
Accordingly, the unused cells in the City 4 column are marked with X. This visual exclusion encodes the rule that once a demand column is satisfied, all other routes into that column are closed for the remainder of the iteration sequence.
The process restarts on the reduced set of available cells. The narrator asks which remaining cell has the smallest shipping cost, and the answer is the $6 entry in the Factory 1, City 2 cell.
That selected route belongs to City 2, whose demand is 20. The next check is whether Factory 1's remaining supply of 35 can cover this demand. Since 35 is greater than 20, the full demand can again be assigned on one route.
The allocation is written into the Factory 1, City 2 cell. This satisfies City 2 completely, so the demand value 20 is crossed out in the bottom row.
Factory 1's supply is then updated from 35 to 15 using . The table now records that Factory 1 still has 15 units left for future assignments.
As in the first iteration, satisfying a city's demand closes that entire column. The remaining unused cells in the City 2 column are marked with X so they will not be considered in later steps.
The narrator begins a third iteration by asking for the lowest shipping cost among the cells that remain open. The next chosen route is Factory 1 to City 1, with unit cost $8.
City 1 has total demand 45, and the narrator says to meet as much of that demand as possible using this $8 route. The clip ends before the actual allocation number is written, so the third step is only initiated, not completed.
The clip opens on a transportation tableau with factories as rows, cities as columns, unit costs inside the cells, and marginal Supply and Demand totals. Factory 1 still has 15 units left, and City 1 still needs 45. Following the least-cost procedure already underway, the instructor assigns all 15 remaining units from Factory 1 to City 1, writes 15 in that cell, exhausts Factory 1, and reduces City 1's demand from 45 to 30.
Next, the instructor scans only the still-available cells and chooses the smallest remaining unit cost, which is the $9 route from Factory 2 to City 1. Since City 1 needs 30 more units and Factory 2 has 50 available, the whole remaining demand of City 1 is satisfied on this route. The instructor writes 30 in cell (2,1), updates Factory 2's supply from 50 to 20, and eliminates City 1 by crossing out its demand and the unused cell in that column.
The instructor then briefly considers the $10 entry from Factory 1 to City 3, but immediately corrects course: Factory 1's supply is already exhausted, so that cell is not eligible and should have been crossed out earlier. This makes the practical rule explicit: only cells with positive remaining supply and positive remaining demand can be chosen. Among the truly remaining options, the costs are 13 and 16, so the minimum is 13, corresponding to Factory 2 to City 3.
With City 3 still requiring 30 units and Factory 2 holding exactly 20 units, the instructor assigns all 20 remaining units from Factory 2 to City 3. The allocation 20 is written in cell (2,3), Factory 2's supply drops to zero, and City 3's demand is reduced from 30 to 10. The tableau now has only one unresolved demand left, namely the final 10 units for City 3.
In the last active cell, Factory 3 also has exactly 10 units remaining. The instructor therefore fills those 10 units from Factory 3 to City 3, writes 10 in cell (3,3), and simultaneously drives both the remaining supply of Factory 3 and the remaining demand of City 3 to zero. At this point every row total and column total has been satisfied, so the displayed set of allocations forms a complete feasible shipping plan.
The instructor announces this as the final answer for the allocation stage and writes TOTAL COST: below the table. The next step is to evaluate the objective by multiplying each occupied cell's unit cost by its allocated quantity and summing those products. The clip begins this explanation with the first term involving 15 units from Factory 1, but it ends before the full arithmetic expression and numerical total are completed.
The excerpt opens on a completed transportation tableau. Rows are labeled Factory 1, Factory 2, Factory 3, and Demand; columns are labeled City 1, City 2, City 3, City 4, and Supply. Six interior cells contain circled shipment quantities paired with unit costs, and the remaining interior cells are marked with X. The narrator begins translating the occupied cells into cost terms.
First, the route from Factory 1 to City 1 contributes , because 15 units are shipped at $8 per unit. Next, Factory 2 to City 1 contributes . These two products are written onto the line labeled TOTAL COST.
The process continues route by route: Factory 1 to City 2 adds , Factory 2 to City 3 adds , Factory 3 to City 3 adds , and Factory 3 to City 4 adds . Each new term corresponds to a cell that is visually emphasized just before it is written.
At this point the full objective-value expression is on screen: The mathematical rule being used is that each occupied route contributes quantity times unit cost, and the total cost is the sum over all occupied routes.
Evaluating the products gives , which sums to . The video writes the result as \$1,080 and boxes it in red.
The narrator identifies this as the final answer for the displayed transportation solution and says it illustrates the least cost method. However, within this excerpt only the final cost evaluation is shown; the earlier least-cost selection steps that generated the circled allocation are not included.
A transportation problem is represented as a matrix-like table: rows are supply sources (here factories), columns are demand destinations (here cities), interior entries are unit shipping costs, the rightmost column gives supplies, and the bottom row gives demands.
The problem is balanced when the sum of all supplies equals the sum of all demands. In this example, total supply is and total demand is , so the model is balanced.
The goal is to meet every city's demand while minimizing total shipping cost. The table entries are unit costs, so the optimization concerns the weighted sum of shipments across factory-city pairs.
The least cost method starts by selecting the cell with the smallest unit shipping cost among all available factory-city pairs. In this example, that cell is Factory 3 to City 4 with cost $5.
If two or more cells have the same minimum shipping cost, the video says to choose the cell with the higher demand. This rule resolves ambiguity in the initial selection step.
After selecting the minimum-cost cell, the method inspects the associated destination demand and source supply. Here the chosen cell lies in City 4 (demand 30) and Factory 3 (supply 40), setting up the next allocation decision.
The video organizes the problem as a transportation table. Rows are factories with supply values 35, 50, and 40; columns are cities with demand values 45, 20, 30, and 30; and the interior cells list unit shipping costs such as $8, $6, $10, $9, $12, $13, $7, $14, $16, and $5. All algorithmic updates happen inside this same tableau.
At each iteration, choose the remaining interior cell with the smallest unit shipping cost. In the clip, this rule first leads to the $5 route from Factory 3 to City 4, then to the $6 route from Factory 1 to City 2, and finally to the $8 route from Factory 1 to City 1.
After selecting the cheapest route, ship the smaller of the factory's remaining supply and the city's remaining demand. The narrator applies this as min(40,30)=30 for Factory 3 to City 4 and min(35,20)=20 for Factory 1 to City 2.
When a city's demand is fully met, its demand entry is crossed out and the other unused cells in that column are marked X. This is shown for City 4 after allocating 30 units and for City 2 after allocating 20 units. The reason given is that any extra shipment to that city would exceed demand.
Each allocation subtracts the shipped quantity from the chosen factory's remaining supply. The table shows 40 becoming 10 after sending 30 units, and 35 becoming 15 after sending 20 units. These updated supplies remain available for later routes.
Within the 120 seconds provided, the completed visible allocations are and . After these steps, Factory 3 has 10 units left and Factory 1 has 15 units left. The next selected route is Factory 1 to City 1 at cost $8, but the clip ends before that allocation amount is written.
The printed dollar amounts in the cells are unit shipping costs, not shipment sizes. The handwritten numbers 30 and 20 are the actual allocated quantities. Keeping these two kinds of entries distinct is essential for reading the transportation tableau correctly.
The video demonstrates a greedy rule for building an initial feasible transportation solution: among all cells that are still usable, pick the one with the smallest unit shipping cost, ship min(remaining supply, remaining demand) on that route, then update the row and column totals. Any row or column reduced to zero is eliminated from future consideration.
The problem is organized as a matrix with Factory 1-3 as supply rows, City 1-4 as demand columns, dollar amounts in the interior as unit costs, a Supply column on the right, and a Demand row on the bottom. The method works directly on this tableau by annotating allocations and crossing out exhausted routes.
Starting from a partially completed table, the clip produces the additional allocations , , , and . Together with the already visible and , the final solution shown is , , , , , , with all other routes zero.
A key teaching moment occurs when the instructor notices the $10 cell from Factory 1 to City 3 but rejects it because Factory 1's supply has already been used up. The correction emphasizes that low cost alone is not enough; the cell must still have positive remaining supply and demand to be eligible.
Once the allocation table is complete, the objective value is found by summing cost times quantity over all occupied cells. The clip introduces this final step with the heading TOTAL COST: and begins the explanation, but it stops before the full numerical computation is shown.
The video displays a 3-by-4 transportation table with factories as rows, cities as columns, and marginal supply and demand entries. Interior cells combine a shipped quantity with a unit cost, and unused routes are marked with X. The excerpt begins with the allocation already completed.
To evaluate a given feasible allocation, multiply each occupied route's quantity by that route's unit cost and add all such products. This is the computational rule used throughout the clip.
The occupied routes are Factory 1 to City 1 with 15 units at $8, Factory 2 to City 1 with 30 units at $9, Factory 1 to City 2 with 20 units at $6, Factory 2 to City 3 with 20 units at $13, Factory 3 to City 3 with 10 units at $16, and Factory 3 to City 4 with 30 units at $5.
Adding the six route costs yields , so the total shipping cost of the displayed allocation is $1,080. The video boxes this result in red.
Although the narrator says this demonstrates the least cost method, the clip itself only shows the final objective-value calculation. It does not show how the least-cost algorithm selected the circled routes or updated supply and demand.
Explore conditions, steps and evidence. Supplementary explanations are labeled separately from content shown in the video.
The leftmost column of the table lists Factory 1, Factory 2, and Factory 3.
The speaker says there are three factories that produce some good.
Factory 1, Factory 2, Factory 3
Supply sources in the transportation table.
Discrete set of three factories.
The top header row lists City 1, City 2, City 3, and City 4.
The speaker says there are four cities that have a demand for that good.
City 1, City 2, City 3, City 4
Demand destinations in the transportation table.
Discrete set of four cities.
The rightmost column is labeled Supply and contains 35, 50, and 40 for Factory 1, Factory 2, and Factory 3 respectively.
The speaker states that Factory 1 has a total supply of 35 units.
Supply
Total number of units available at each factory.
Nonnegative integer quantities for each factory.
The bottom row is labeled Demand and contains 45, 20, 30, and 30 under City 1 through City 4.
The speaker states that City 1 has a demand of 45 units.
Demand
Total number of units required by each city.
Nonnegative integer quantities for each city.
Interior cells contain dollar amounts such as $8, $6, $10, $9, $12, $13, $7, $14, $16, and $5.
The speaker says these are unit shipping costs from each factory to each city.
Unit shipping cost
Cost per unit shipped from a given factory to a given city.
Positive monetary values indexed by factory-city pairs.
The speaker adds and says the total is 125.
The handwritten number 125 appears beside the circled Supply column.
Total supply
Sum of all factory supplies.
Scalar quantity equal to 125 in this example.
The speaker adds and says the total demand is also 125.
The handwritten number 125 appears below the circled Demand row.
Total demand
Sum of all city demands.
Scalar quantity equal to 125 in this example.
The rightmost column is labeled Supply and lists 35, 50, and 40 for Factory 1, Factory 2, and Factory 3.
The narrator compares each factory's supply with a city's demand, saying that supply is greater than demand before allocating units.
Supply
Available shipment quantity at each factory row.
Nonnegative integer quantities shown as 35, 50, and 40 in the table.
The bottom row is labeled Demand and lists 45, 20, 30, and 30 for City 1 through City 4.
The narrator says the total demand for City 4 was 30 and later says City 1 has a total demand of 45.
Demand
Required shipment quantity for each destination column.
Nonnegative integer quantities shown as 45, 20, 30, and 30 in the table.
Interior cells contain dollar amounts such as $8, $6, $10, $9, $12, $13, $7, $14, $16, and $5.
The narrator refers to the smallest shipping cost and names values including $6 per unit and $8.
Unit shipping cost from factory i to city j, displayed inside the transportation table.
Positive dollar amounts attached to individual factory-city routes.
Handwritten 30 is placed in the Factory 3, City 4 cell; handwritten 20 is placed in the Factory 1, City 2 cell.
The narrator says to send 30 units from Factory 3 to City 4 and later to send 20 units through the Factory 1, City 2 route.
Number of units allocated on the route from factory i to city j.
Nonnegative integer quantities written into interior cells during the least-cost procedure.
After writing 30 in Factory 3, City 4, the Supply entry 40 is crossed out and replaced by 10; after writing 20 in Factory 1, City 2, the Supply entry 35 is crossed out and replaced by 15.
The narrator says 40 minus 30 leaves 10 and 35 minus 20 leaves 15.
Remaining supply
Updated factory capacity after subtracting the units already allocated from that factory.
Nonnegative integers updated row by row during the algorithm.
The speaker introduces the topic as solving the transportation problem using the least cost method.
A table with factories as rows, cities as columns, unit costs inside, Supply on the right, and Demand on the bottom is shown.
The video presents a transportation problem as a table whose rows are factories (sources), whose columns are cities (destinations), whose interior entries are unit shipping costs, whose rightmost column gives each factory's supply, and whose bottom row gives each city's demand.
There are multiple supply sources.
There are multiple demand destinations.
Each source-destination pair has an associated unit shipping cost.
Each source has a supply amount and each destination has a demand amount.
The speaker computes total supply as 125 and total demand as 125, then says this is known as a balanced problem because total supply equals total demand.
The Supply column is circled and annotated with 125; the Demand row is circled and annotated with 125.
A transportation problem is called balanced when the sum of all supplies equals the sum of all demands. In this example, both totals are 125 units.
All factory supplies are included in the left-hand sum.
All city demands are included in the right-hand sum.
Equality of the two sums defines the balanced case used here.
The speaker says the goal is to meet the demand of each city while minimizing the shipping cost.
The interior cost cells are highlighted one by one as examples of unit shipping costs.
The stated goal is to satisfy every city's demand while minimizing total shipping cost, using the unit shipping costs listed in the table.
Demand must be met for each city.
Shipping decisions are evaluated by unit costs from factory to city.
The overall aim is cost minimization.
The speaker says the first step is to identify the cell with the smallest shipping cost.
The cell containing $5 at Factory 3 / City 4 is circled.
The least cost method begins by locating the cell with the minimum unit shipping cost among all available factory-city pairs. In this example, that minimum is $5 at Factory 3 to City 4.
Compare all currently available unit shipping costs.
Select the cell with the smallest cost.
If there is a tie, use the tie-breaking rule stated in the video.
The speaker says that if two cells have the same minimum shipping cost, choose the one with the higher demand.
When more than one cell shares the same minimum shipping cost, the video instructs choosing the cell whose destination has the higher demand.
Applies only when there is a tie for the minimum shipping cost.
Comparison is made using the demand associated with the candidate cells.
A 3-by-4 cost matrix is shown with rows Factory 1, Factory 2, Factory 3, columns City 1, City 2, City 3, City 4, a Supply column, and a Demand row.
The narration repeatedly refers to factories, cities, supply, demand, and shipping costs while working within this table.
The video uses a transportation tableau to organize a distribution problem. Rows represent factories with available supply, columns represent cities with required demand, interior cells show unit shipping costs, and the margins list supply and demand totals. The method updates this same table by writing allocations, crossing out satisfied demands, and reducing remaining supplies.
Rows correspond to sources (factories).
Columns correspond to destinations (cities).
Interior entries are unit shipping costs.
Right margin lists supply totals.
Bottom margin lists demand totals.
The narrator says, 'of the remaining cells, now which one has the smallest shipping cost?' and later repeats the same question for the next iteration.
A circle is drawn around $6 in the Factory 1, City 2 cell, and later around $8 in the Factory 1, City 1 cell.
At each iteration, the algorithm chooses the still-available interior cell with the lowest unit shipping cost. Once chosen, that route receives an allocation, and the corresponding row or column is updated or removed from future consideration.
Only remaining unexcluded cells are considered.
The selected cell has the minimum shipping cost among them.
For City 4, the narrator asks whether the demand can be met using Factory 3's supply, answers yes because supply is greater than demand, and sends 30 units. For City 2, he again checks whether supply 35 can meet demand 20 and sends 20 units.
The allocation written into Factory 3, City 4 is 30, matching City 4 demand; the allocation written into Factory 1, City 2 is 20, matching City 2 demand.
Once the cheapest route is selected, the amount shipped on that route is the smaller of the current remaining supply at the chosen factory and the current remaining demand at the chosen city. In the examples shown, the demand side is smaller, so the full city demand is assigned to that route.
Use the current remaining supply for the selected factory row.
Use the current remaining demand for the selected city column.
Allocate the minimum of these two quantities.
The narrator says that once 30 units are sent, the total demand has been met, so the demand value is crossed out and the other cells in that column are not used. He repeats the same logic for City 2.
Demand 30 for City 4 is crossed out, then the Factory 1 and Factory 2 cells in the City 4 column are marked X. Later demand 20 for City 2 is crossed out, and the Factory 2 and Factory 3 cells in the City 2 column are marked X.
If an allocation exactly satisfies a city's remaining demand, that demand entry is crossed out and all other unused cells in the same column are excluded from later iterations. This prevents over-sending to that destination.
The chosen route allocation equals the city's remaining demand.
After satisfaction, the column is closed to further allocations.
The narrator says to reduce the total supply from Factory 3 by 30, computing 40 minus 30 to leave 10, and later reduces Factory 1's supply by 20, computing 35 minus 20 to leave 15.
Supply 40 is replaced by 10, and supply 35 is replaced by 15.
Whenever units are shipped from a factory, that factory's remaining supply is reduced by the allocated amount. The updated supply stays available for later routes unless it becomes zero.
Subtract only the units actually allocated from that factory row.
The updated supply may remain positive for future iterations.
The speaker repeatedly says to look at the remaining cells and identify the next lowest shipping cost, then allocate as much as possible along that route.
Cells with lower costs are circled before allocations are written, and exhausted rows/columns are crossed out.
The video demonstrates a greedy construction rule for an initial feasible solution: among all still-available routes, choose the cell with the smallest unit shipping cost, assign the maximum feasible quantity equal to min(remaining supply, remaining demand), update both the row supply and column demand, and eliminate any row or column that becomes zero.
Only uneliminated cells are considered when selecting the minimum cost.
Allocation in a chosen cell is limited by the smaller of the remaining supply and remaining demand.
After allocation, any row or column reduced to zero is crossed out and no longer used.
A 3-by-4 grid lists factories as rows, cities as columns, unit costs inside cells, Supply on the right, and Demand on the bottom.
The transportation problem is represented by a tableau whose rows are supply sources, columns are demand destinations, interior entries are unit shipping costs, and marginal totals give available supply and required demand.
Row totals represent supplies.
Column totals represent demands.
Interior cell values represent per-unit transportation costs.
The speaker explicitly says the total supply is 125, the total demand is also 125, and therefore this is a balanced problem.
Both the Supply column and Demand row are annotated with 125.
For the displayed table, total supply equals total demand, so the problem is balanced.
Factory supplies are 35, 50, and 40.
City demands are 45, 20, 30, and 30.
For this specific numerical example.
The speaker says the smallest shipping cost is $5 and identifies that cell as the starting point.
The $5 entry in the Factory 3 / City 4 cell is circled.
Among all unit shipping costs shown, the minimum is $5, located at Factory 3 to City 4.
The table entries are the unit shipping costs.
All interior cost cells are considered.
Over all factory-city cost cells in the displayed table.
The narrator explicitly states that the demand can be met because the supply is greater than the demand, then allocates the full demand amount on the selected route.
Factory 3 supply 40 exceeds City 4 demand 30, and Factory 1 supply 35 exceeds City 2 demand 20; in both cases the allocation equals the demand.
If the selected factory's remaining supply is greater than the selected city's remaining demand, the algorithm assigns the full remaining demand to that route and then closes the city column.
A cheapest remaining route has been selected.
The chosen factory's remaining supply exceeds the chosen city's remaining demand.
For the currently selected factory-city pair in the iterative least-cost procedure.
The narrator gives a counterexample: sending 5 more units on another City 4 route would make total supply 30 plus 5, which exceeds the demand.
The City 4 column is subsequently excluded except for the already filled Factory 3 cell.
After a city's total demand has been met, assigning any additional positive amount on another route to that same city would exceed the demand constraint, so those routes must be excluded.
The city's demand has already been fully satisfied by earlier allocations.
An additional positive shipment is proposed on another route to the same city.
For any unused route leading to an already satisfied destination column.
Initial supplies shown are 35, 50, 40; initial demands shown are 45, 20, 30, 30.
The speaker treats the problem as one where all demand must be met by exhausting supplies through successive allocations.
The video does not explicitly state the balance condition in words; this is inferred from the displayed totals and the successful completion of all demands and supplies.
For the displayed instance, total supply equals total demand: .
The initial marginal totals are those shown in the tableau.
For the specific 3-factory, 4-city instance displayed in the video.
The speaker says to identify the next lowest shipping cost among remaining cells and use that route to meet as much demand as possible.
The chosen cells are always the currently lowest-cost available cells: $9, then $13, then the last remaining feasible cell.
At each iteration of the demonstrated method, the algorithm selects an available cell with minimum unit cost c_{ij} and assigns x_{ij}=(Supply_i,Demand_j).
The cell is still available, meaning its row supply and column demand have not been eliminated.
Supply_i and Demand_j denote the current remaining amounts.
For every allocation step shown in the clip.
The speaker says, "So if you punch all of this into a calculator, the final total shipping cost comes out to $1,080."
The handwritten result below the sum is "$1,080" enclosed in a red box.
The total shipping cost of the displayed transportation allocation is $1,080.
The occupied-route quantities and unit costs are exactly those shown in the table.
The total cost is computed as the sum of quantity times unit cost over all occupied routes.
For the single allocation displayed in the tableau.
The speaker adds the supply values and then the demand values aloud.
Handwritten 125 appears next to the Supply column and below the Demand row.
Add the supplies of Factory 1, Factory 2, and Factory 3.
Direct summation of the values in the Supply column.
Add the demands of City 1, City 2, City 3, and City 4.
Direct summation of the values in the Demand row.
Compare the two totals.
Since total supply equals total demand, the example satisfies the definition of a balanced transportation problem.
The example is balanced because total supply and total demand are both 125 units.
The speaker says to identify the cell with the smallest shipping cost and notes that $5 is the smallest and not tied with any other cell.
The $5 cell at Factory 3 / City 4 is circled.
Scan all unit shipping costs and find the minimum value.
The least cost method starts from the smallest available cost.
Identify the location of the minimum cost cell.
The value 5 appears in the Factory 3 row and City 4 column.
Check whether another cell also has cost 5.
The speaker explicitly says it is not tied with any other cell, so the tie-breaking rule is not needed here.
The algorithm begins with the Factory 3 to City 4 cell because it has the unique minimum unit shipping cost of $5.
The narrator explains that Factory 3 can cover City 4 because supply is greater than demand, writes 30, crosses out demand 30, reduces supply 40 to 10, and excludes the other City 4 cells.
The table shows 30 entered at Factory 3, City 4; demand 30 crossed out; supply changed from 40 to 10; X marks placed on Factory 1, City 4 and Factory 2, City 4.
The cheapest remaining route under discussion is the Factory 3, City 4 cell with unit cost $5.
The narrator identifies this route as the one being used and the table displays $5 in that cell.
The allocation size is the smaller of Factory 3's supply 40 and City 4's demand 30.
The narrator says supply is greater than demand and therefore sends 30 units on this route.
Thirty units are assigned from Factory 3 to City 4.
The handwritten 30 appears in the Factory 3, City 4 cell.
Factory 3's remaining supply is updated from 40 to 10.
The narrator performs this subtraction aloud and the table replaces 40 with 10.
Because City 4's demand is fully met, its demand entry is crossed out and the other cells in that column are marked unavailable.
The narrator states that no goods will be sent through those routes and the X marks appear in the City 4 column.
The first completed iteration satisfies City 4 completely, leaves Factory 3 with 10 units of remaining supply, and removes the rest of the City 4 column from future consideration.
The narrator asks which remaining cell has the smallest shipping cost, identifies $6, notes City 2 demand is 20, confirms Factory 1 supply 35 is enough, sends 20 units, crosses out demand 20, reduces supply 35 to 15, and excludes the other City 2 cells.
A circle is drawn around $6 in Factory 1, City 2; 20 is written there; demand 20 is crossed out; supply 35 becomes 15; X marks are added to Factory 2, City 2 and Factory 3, City 2.
Among the cells still available after the first iteration, the Factory 1, City 2 route has the lowest unit shipping cost.
The narrator explicitly asks for the smallest remaining shipping cost and circles $6.
The allocation is the smaller of Factory 1's remaining supply 35 and City 2's demand 20.
The narrator checks whether supply 35 can meet demand 20 and answers yes.
Twenty units are assigned from Factory 1 to City 2.
The handwritten 20 appears in the Factory 1, City 2 cell.
Factory 1's remaining supply is updated from 35 to 15.
The narrator states this subtraction and the table replaces 35 with 15.
Since City 2's demand is fully satisfied, the demand entry is crossed out and the other cells in that column are removed from future use.
The narrator says those cells will no longer be used and X marks appear in the City 2 column.
The second iteration satisfies City 2 completely, leaves Factory 1 with 15 units of remaining supply, and removes the rest of the City 2 column from future consideration.
The narrator asks which remaining cell has the lowest shipping cost, identifies $8, notes City 1 demand is 45, and says to meet as much of that demand as possible using this $8 shipping route.
A circle is drawn around $8 in the Factory 1, City 1 cell, and demand 45 for City 1 is circled.
The final allocation number for the Factory 1, City 1 route is not shown before the clip ends.
After excluding the City 4 and City 2 columns, the next cheapest remaining route identified in the video is Factory 1 to City 1 with cost $8.
The narrator asks for the lowest remaining shipping cost and circles $8.
The destination associated with the selected route has total demand 45.
The narrator states this and the bottom row shows 45 under City 1.
The next step is to allocate on the Factory 1, City 1 route up to the limit imposed by the smaller of remaining supply and remaining demand.
The narrator says to meet as much of the demand as possible using this route.
The clip establishes the next selected route as Factory 1 to City 1 with unit cost $8, but it stops before the actual allocation value is written.
The speaker says the supply left in Factory 1 is 15, uses all of that 15, and reduces City 1 demand from 45 to 30.
Cell (Factory 1, City 1) receives 15; Supply for Factory 1 changes from 35 to 15 and then is exhausted; Demand for City 1 changes from 45 to 30.
At the start of the clip, Factory 1 has 15 units remaining and City 1 still needs 45 units.
Read directly from the tableau and narration.
The method assigns the maximum feasible amount along the selected route.
Least-cost allocation rule demonstrated in the video.
After shipping 15 units, Factory 1 is exhausted and City 1's remaining demand drops to 30.
Row and column updates follow the assigned quantity.
The first step of the clip allocates 15 units from Factory 1 to City 1, exhausting Factory 1 and leaving 30 units of demand at City 1.
The speaker identifies the next lowest shipping cost as $9, notes City 1 still needs 30, says Factory 2 has supply 50, and computes .
Cell (Factory 2, City 1) is marked with 30; Supply for Factory 2 changes from 50 to 20; Demand for City 1 is crossed out.
Among the still-available routes, the cheapest unit cost is the $9 route from Factory 2 to City 1.
Explicitly stated by the speaker and circled on screen.
Factory 2 can cover all of City 1's remaining demand.
Allocation is limited by the smaller of remaining supply and demand.
Factory 2 keeps 20 units, and City 1's demand is fully satisfied.
Marginal totals are updated after assigning 30 units.
Once City 1 demand reaches zero, the remaining unused cell in that column is eliminated.
The speaker says the cell is not needed anymore and crosses it out.
The second step allocates 30 units from Factory 2 to City 1, satisfying City 1 completely and reducing Factory 2's remaining supply to 20.
The speaker initially considers $10 but says Factory 1 supply is already exhausted, so the remaining candidates are 13 and 16; the minimum is 13. Then the speaker says Factory 2 has 20 left and uses all of it for City 3, reducing City 3 demand from 30 to 10.
Cell (Factory 1, City 3) is crossed out; cell (Factory 2, City 3) receives 20; Supply for Factory 2 becomes 0; Demand for City 3 becomes 10.
Although $10 is low, the route from Factory 1 to City 3 cannot be used once Factory 1 is exhausted.
The speaker explicitly corrects this and crosses out the cell.
Among the remaining usable cells, the cheapest route is Factory 2 to City 3 at $13.
Stated verbally and indicated by circling the cell.
Factory 2 has only 20 units left, so all 20 are sent to City 3.
Allocation is capped by remaining supply.
Factory 2 is exhausted and City 3 still needs 10 units.
Row and column updates follow the assigned quantity.
The third step allocates 20 units from Factory 2 to City 3, exhausting Factory 2 and leaving 10 units of demand at City 3.
The speaker says the total supply of Factory 3 is down to 10 and the total demand left in City 3 is also 10, so those 10 units are filled there.
Cell (Factory 3, City 3) receives 10; Supply for Factory 3 changes from 10 to 0; Demand for City 3 changes from 10 to 0.
Only one feasible route remains for the outstanding City 3 demand, and Factory 3 has exactly enough supply.
Read from the updated marginals in the tableau.
All remaining supply is assigned to the remaining demand.
Least-cost method continues until a row or column is exhausted.
Both the row and column become zero simultaneously.
The speaker states supply will be completely gone and demand reduced to zero.
The fourth step allocates 10 units from Factory 3 to City 3, simultaneously exhausting Factory 3's remaining supply and City 3's remaining demand.
The speaker says this is the final answer and begins explaining how to find the total shipping cost by sending 15 units from Factory 1...
The completed allocation table is visible and the heading TOTAL COST: is written below it.
The clip ends before the full arithmetic expression and numerical total are given.
The final feasible solution contains six positive allocations.
Visible in the completed tableau.
The speaker introduces the standard way to evaluate the solution once allocations are known.
Explicitly announced as the next step.
The explanation starts with the allocation , but the multiplication and summation are cut off.
Audio ends mid-sentence.
The clip establishes the completed allocation table and starts the total-cost computation, but does not reach the final numerical cost within the provided duration.
The written sum accumulates term by term until it reads 15($8) + 30($9) + 20($6) + 20($13) + 10($16) + 30($5).
The final boxed value is $1,080.
The speaker verbally identifies each route's quantity and unit cost before adding the next term.
Start with the first occupied route, using 15 units at a unit cost of $8.
Directly read from the circled entry in Factory 1 / City 1 and stated in the audio.
Add the second occupied route, 30 units at $9 per unit.
Directly read from the circled entry in Factory 2 / City 1 and stated in the audio.
Add the third occupied route, 20 units at $6 per unit.
Directly read from the circled entry in Factory 1 / City 2 and stated in the audio.
Add the fourth occupied route, 20 units at $13 per unit.
Directly read from the circled entry in Factory 2 / City 3 and stated in the audio.
Add the fifth occupied route, 10 units at $16 per unit.
Directly read from the circled entry in Factory 3 / City 3 and stated in the audio.
Add the sixth occupied route, 30 units at $5 per unit.
Directly read from the circled entry in Factory 3 / City 4 and stated in the audio.
Evaluate each product separately.
Standard arithmetic expansion of the displayed sum; these intermediate products are not written on screen.
Sum the six products to obtain the total cost.
Arithmetic check of the displayed final answer.
The displayed allocation has total shipping cost $1,080.
A full transportation table is shown with three factories, four cities, unit costs, supplies, and demands.
The speaker walks through the setup, checks balance, explains the objective, and begins the least cost method.
The clip ends before the actual allocation amount for the chosen cell is written into the table.
Given three factories and four cities with unit shipping costs, supplies, and demands, explain how to begin solving the transportation problem by the least cost method.
Factories: Factory 1, Factory 2, Factory 3.
Cities: City 1, City 2, City 3, City 4.
Supplies: 35, 50, 40.
Demands: 45, 20, 30, 30.
Unit costs shown in the table include $8, $6, $10, $9, $9, $12, $13, $7, $14, $9, $16, and $5.
Set up the problem, verify whether it is balanced, state the minimization objective, and identify the first cell selected by the least cost method.
Read the table structure.
The visual layout directly assigns factories to rows, cities to columns, and costs to intersections.
Compute total supply and total demand.
The speaker performs these sums to test balance.
Conclude the problem is balanced.
By the definition stated in the video, equality of totals makes the problem balanced.
Find the smallest shipping cost cell.
The least cost method begins by selecting the minimum-cost cell.
Inspect the row and column associated with the chosen cell.
The speaker identifies the destination demand and source supply relevant to the next allocation decision.
The example is a balanced transportation problem, and the least cost method starts at the Factory 3 to City 4 cell with unit cost $5.
Balance is verified by matching totals of 125 and 125; the starting cell is verified by comparing all listed unit costs and finding $5 to be the unique minimum.
The full table shows costs: Factory 1 = $8, $6, $10, $9; Factory 2 = $9, $12, $13, $7; Factory 3 = $14, $9, $16, $5; supplies 35, 50, 40; demands 45, 20, 30, 30.
The narrator works through this instance using the least cost method, referring to specific factories, cities, supplies, demands, and shipping costs.
The example is only partially solved within the clip; the final complete allocation plan is not reached.
Given three factories and four cities with the displayed unit shipping costs, supplies, and demands, construct an initial feasible shipping plan by repeatedly choosing the lowest-cost remaining route.
Factory supplies: Factory , Factory , Factory .
City demands: City , City , City , City .
Unit costs shown in the table include $8, $6, $10, $9 for Factory 1; $9, $12, $13, $7 for Factory 2; and $14, $9, $16, $5 for Factory 3.
Apply the least cost method step by step until the next allocation is determined.
The cheapest route discussed is Factory 3 to City 4, and 30 units are allocated there.
The narrator compares supply 40 with demand 30 and writes 30 in the cell.
Factory 3's remaining supply becomes 10, and the other City 4 cells are excluded.
The table replaces 40 with 10 and marks X in the remaining City 4 cells.
Among remaining cells, Factory 1 to City 2 has the lowest cost, so 20 units are allocated there.
The narrator circles $6 and writes 20 in the Factory 1, City 2 cell.
Factory 1's remaining supply becomes 15, and the other City 2 cells are excluded.
The table replaces 35 with 15 and marks X in the remaining City 2 cells.
The next cheapest remaining route identified is Factory 1 to City 1, and the narrator starts allocating against City 1's demand of 45.
The narrator circles $8 and says to meet as much demand as possible using this route.
Within this clip, the visible partial solution is and , with updated supplies Factory and Factory ; the next selected route is Factory 1 to City 1 at cost $8, but its allocation is not completed on screen.
Each shown allocation matches the smaller of the relevant remaining supply and demand, satisfied demand entries are crossed out, and the corresponding column cells are excluded from later use.
A full worked transportation tableau is shown with costs, supplies, demands, and step-by-step handwritten allocations.
The speaker narrates each selection, subtraction, crossing-out, and update until declaring the final answer.
The final total shipping cost is not computed within the clip.
Construct an initial feasible shipping plan for a 3-factory, 4-city transportation problem using the least cost method, starting from the partially filled tableau shown on screen.
Factories: Factory 1, Factory 2, Factory 3.
Cities: City 1, City 2, City 3, City 4.
Initial supplies: 35, 50, 40.
Initial demands: 45, 20, 30, 30.
Unit costs visible in the table include $8, $6, $10, $9, $12, $13, $7, $14, $16, $5.
Some allocations were already present before this clip begins: , , and the remaining supply/demand values shown at the start.
Determine the remaining allocations produced by the least cost method and set up the total shipping cost calculation.
Use the remaining 15 units of Factory 1 to satisfy part of City 1's demand.
Minimum of remaining supply 15 and demand 45.
Update the marginals after the first allocation and eliminate Factory 1.
Row/column update rule of the method.
Choose the cheapest available route, Factory 2 to City 1 at $9, and send 30 units.
Minimum of remaining supply 50 and demand 30.
Update marginals and cross out City 1 once satisfied.
Column elimination after demand reaches zero.
After rejecting the exhausted Factory 1 route at $10, choose Factory 2 to City 3 at $13 and send all remaining 20 units.
Minimum of remaining supply 20 and demand 30.
Update marginals and eliminate Factory 2.
Row elimination after supply reaches zero.
Send the last 10 units from Factory 3 to City 3.
Minimum of remaining supply 10 and demand 10.
Both the last row remainder and last column remainder become zero.
Simultaneous exhaustion shown in the final update.
Begin evaluating the completed solution by multiplying each allocation by its unit cost and summing.
Standard objective evaluation for a transportation solution.
The completed allocation shown in the clip is , , , , , , with all other x_{ij}=0. The total shipping cost calculation is introduced but not finished within the clip.
Row sums match the initial supplies 35, 50, 40 and column sums match the initial demands 45, 20, 30, 30 using the allocations visible in the final tableau.
The full worked example is a 3-by-4 transportation table with circled allocations and a handwritten total-cost calculation underneath.
The speaker closes by saying, "That's how you can use the least cost method as it relates to the transportation problem."
The clip does not show the earlier least-cost selection steps that produced the circled allocation.
The source of the crossed-out supply and demand updates is not demonstrated within this excerpt.
Given a transportation tableau with occupied routes (Factory 1 to City 1: 15 at $8; Factory 2 to City 1: 30 at $9; Factory 1 to City 2: 20 at $6; Factory 2 to City 3: 20 at $13; Factory 3 to City 3: 10 at $16; Factory 3 to City 4: 30 at $5), find the total shipping cost.
Factory 1 to City 1 ships 15 units at $8 per unit.
Factory 2 to City 1 ships 30 units at $9 per unit.
Factory 1 to City 2 ships 20 units at $6 per unit.
Factory 2 to City 3 ships 20 units at $13 per unit.
Factory 3 to City 3 ships 10 units at $16 per unit.
Factory 3 to City 4 ships 30 units at $5 per unit.
Unselected cells are marked with X.
Compute the total shipping cost for the displayed allocation.
Write one product for each occupied route, multiplying shipped units by that route's unit cost.
Uses the total-cost rule for a transportation solution.
Evaluate the six products.
Direct multiplication; these intermediate values are derived rather than shown on screen.
Add the six route costs to get the final total.
Arithmetic summation of the expanded expression.
$1,080
The boxed handwritten result matches the arithmetic check: . The circled quantities also satisfy the displayed remaining supply and demand totals.
A static black-background table appears with headers City 1 through City 4, row labels Factory 1 through Factory 3, a Supply column, and a Demand row.
A teal oval is drawn around the factory labels, then another around the city labels.
Transportation table
Factory labels
City labels
Supply column
Demand row
Teal highlight ovals
First the factory row labels are circled.
Then the city column labels are circled.
The numerical contents of the table remain unchanged during this interval.
The overall table layout remains fixed.
The highlighting distinguishes sources (factories) from destinations (cities) before discussing supplies, demands, and costs.
The Supply column is circled and the handwritten number 125 is added beside it; then the Demand row is circled and another handwritten 125 is added below it.
Supply values 35, 50, 40 and demand values 45, 20, 30, 30 remain visible.
Supply column
Demand row
Handwritten 125 annotations
Teal circles
The Supply column is emphasized and annotated with 125.
The Demand row is emphasized and annotated with 125.
The individual supply and demand entries do not change.
The cost matrix is not yet modified.
The annotations visually demonstrate that total supply equals total demand, supporting the claim that the problem is balanced.
Individual cost cells are circled in sequence, including $8 at Factory 1 / City 1 and $6 at Factory 1 / City 2.
Cost cells
Teal circles
Specific interior cost entries are circled one after another.
The table values themselves remain unchanged.
No allocation numbers are written yet.
The circling links spoken examples to exact matrix positions, clarifying that interior entries represent unit shipping costs from a factory to a city.
The $5 cell at Factory 3 / City 4 is circled; later the City 4 header, the demand value 30, the Factory 3 label, and the supply value 40 are also circled.
The clip ends before any allocation quantity is entered in the chosen cell.
Factory 3 / City 4 cost cell
City 4 header
Demand value 30
Factory 3 row label
Supply value 40
Teal circles
The minimum-cost cell is marked first.
Then the associated destination column and demand are marked.
Then the associated source row and supply are marked.
The cost matrix values remain unchanged.
No new numeric allocation is written before the clip ends.
The sequence shows how the least cost method identifies a starting cell and then inspects the corresponding demand and supply limits that will constrain the next allocation.
At the beginning, the Factory 3, City 4 cost $5, the Factory 3 supply 40, and the City 4 demand 30 are circled in light blue.
Transportation table
Factory 3 row
City 4 column
Cost cell $5
Supply value 40
Demand value 30
Circles emphasize the route, supply, and demand that will be compared first.
The underlying cost matrix and marginal totals remain unchanged at this moment.
The visual emphasis introduces the first least-cost comparison: whether Factory 3's supply can satisfy City 4's demand.
A handwritten 30 appears in the Factory 3, City 4 cell.
Factory 3, City 4 cell
Handwritten number 30
The selected route receives an explicit allocation of 30 units.
Other table entries remain unchanged during this writing step.
This visual action records the decision to ship 30 units from Factory 3 to City 4.
The demand value 30 under City 4 is crossed out with a diagonal slash.
City 4 demand entry 30
Diagonal strike-through
The City 4 demand is visually marked as fulfilled.
The cost entries in the City 4 column are still visible at this instant.
The strike-through signals that no more units should be assigned to City 4.
The Factory 3 supply value 40 is crossed out and 10 is written beside it.
Factory 3 supply entry 40
New handwritten 10
Factory 3's remaining supply is updated from 40 to 10.
The allocation 30 in Factory 3, City 4 remains visible.
The table now reflects that 30 of Factory 3's 40 units have already been shipped.
X marks are drawn in the Factory 1, City 4 and Factory 2, City 4 cells.
Factory 1, City 4 cell
Factory 2, City 4 cell
X marks
Two additional routes to City 4 are excluded from future consideration.
The already chosen allocation remains active.
Because City 4's demand is already met, the remaining cells in that column are removed from the search space.
A circle is drawn around $6 in the Factory 1, City 2 cell, and later around the demand value 20 under City 2.
Factory 1, City 2 cost $6
City 2 demand 20
Circular highlights
Attention shifts to the next minimum-cost cell and its associated destination demand.
The previous allocations and exclusions remain in place.
The highlights introduce the second iteration of the least cost method.
A handwritten 20 appears in the Factory 1, City 2 cell; the demand 20 is crossed out; the supply 35 is replaced by 15.
Factory 1, City 2 cell
Handwritten 20
City 2 demand 20
Factory 1 supply 35
New handwritten 15
City 2 receives 20 units, its demand is marked satisfied, and Factory 1's remaining supply decreases to 15.
Earlier City 4 allocations and exclusions remain unchanged.
This sequence records the full effect of the second least-cost assignment.
X marks are drawn in the Factory 2, City 2 and Factory 3, City 2 cells.
Factory 2, City 2 cell
Factory 3, City 2 cell
X marks
The remaining unused routes to City 2 are excluded.
The allocation stays fixed.
Once City 2's demand is satisfied, the rest of that column is removed from later iterations.
The speaker explicitly addresses the case where two cells have the same minimum shipping cost.
One might think the least cost method simply picks any minimum-cost cell without a rule when several cells tie.
The video states that if two cells share the same minimum shipping cost, the correct choice is the one with the higher demand.
The speaker checks the sums of supply and demand before calling the example balanced.
A transportation table may be assumed automatically balanced without checking totals.
In this video, balance is established only after computing total supply = 125 and total demand = 125.
The narrator warns that sending 5 more units on another City 4 route would make total supply 30 plus 5, exceeding the demand.
The other City 4 cells are then crossed out.
One might think additional routes to the same city can still be used after that city's demand has already been satisfied.
Once a destination's demand is fully met, all other unused routes to that destination must be excluded; otherwise the total shipped to that city would exceed its demand.
Printed dollar amounts such as $5, $6, and $8 remain in cells while separate handwritten numbers 30 and 20 are added as allocations.
The narrator distinguishes shipping cost from the number of units to send.
A learner may confuse the printed unit cost in a cell with the number of units allocated on that route.
The printed dollar value is the cost per unit; the handwritten number placed in the cell is the quantity shipped on that route.
The speaker says the $10 route looks lowest, then corrects that Factory 1 supply has already been exhausted and that the cell should have been crossed out earlier.
Cell (Factory 1, City 3) is subsequently marked with an X.
One may mistakenly choose the globally smallest visible cost even if its row or column has already been reduced to zero.
In the least cost method, only cells with positive remaining supply and positive remaining demand are eligible; exhausted rows and satisfied columns must be removed from consideration.
The speaker mentions the least cost method only at the end, after the allocation is already visible.
The excerpt begins with a fully populated table rather than an empty one.
One might infer from this excerpt alone that the least cost method consists only of multiplying the circled quantities by their costs.
The clip demonstrates only the final evaluation stage. The actual least-cost construction—choosing routes in order of increasing unit cost and updating supply and demand—is not shown here.
After introducing the table, the speaker immediately sums supplies and demands to classify the example.
Understanding the table layout is necessary before summing the Supply column and Demand row to test balance.
The speaker first states the minimization objective and then introduces the least cost method as the way to achieve it.
The least cost method is presented as a procedure for pursuing the stated cost-minimization objective.
The tie-breaking instruction is given immediately after explaining how to find the smallest shipping cost.
The tie-breaking rule is part of the selection procedure used when applying the least cost method.
The worked numerical table is used throughout to illustrate definitions and the first algorithmic step.
The numerical example is the concrete instance used to verify that the problem is balanced.
All algorithmic steps are performed directly on the transportation tableau.
The narrator repeatedly refers to remaining cells, supplies, and demands inside the table while applying the method.
The least cost method is carried out by operating on the transportation table's costs, supplies, and demands.
After identifying the smallest remaining shipping cost, the narrator immediately determines how many units to send on that route.
The allocation-size rule is applied only after the cheapest remaining route has been selected.
The narrator says that once the total demand is met, the other routes in that column will not be used.
Crossed-out demand values are followed by X marks in the same column.
Column elimination follows when the allocated amount equals the city's remaining demand.
After each allocation, the narrator subtracts the shipped amount from the factory's supply.
Supply values 40 and 35 are replaced by 10 and 15 after allocations of 30 and 20.
Every allocation triggers a corresponding reduction in the chosen factory's remaining supply.
The narrator justifies excluding the remaining City 4 cells by noting that extra shipments would exceed demand.
The proposition about over-sending explains why the algorithm removes the rest of a satisfied demand column.
The method is carried out entirely on the transportation tableau.
The least-cost allocation procedure is applied directly to the transportation tableau's costs, supplies, and demands.
After declaring the final allocations, the speaker immediately moves to computing total shipping cost.
The full cost expression is not completed in the clip.
The allocated quantities x_{ij} are needed before the objective value _{ij}x_{ij} can be evaluated.
Each step is justified by choosing the next lowest shipping cost among remaining cells.
The demonstrated method depends on the greedy minimum-cost selection rule at each iteration.
The speaker introduces factories, cities, supply, demand, and unit shipping costs using the table.
The speaker adds supplies and demands and calls the problem balanced when both totals are 125.
The speaker says the aim is to meet each city's demand while minimizing shipping cost.
The speaker says the first step is to identify the cell with the smallest shipping cost.
The speaker explains what to do if two cells have the same minimum shipping cost.
The $5 cell at Factory 3 / City 4 is circled as the starting point.
The whole clip is organized around a transportation tableau with factories, cities, costs, supply, and demand.
The narrator asks which remaining cell has the smallest shipping cost and circles $6, then later $8.
The narrator compares supply and demand and sends the full demand amount when supply is larger.
The narrator says the other routes will not be used because the total demand is already met.
X marks are placed in the remaining cells of the satisfied columns.
The narrator computes 40 minus 30 and 35 minus 20 to update remaining supplies.
Supply entries are replaced by 10 and 15.
Printed dollar costs remain in cells while handwritten quantities 30 and 20 are added separately.
Covered · Introduction of the transportation table, factories, cities, supply, and demand.
Covered · Computation of total supply and total demand and identification of the example as balanced.
Covered · Statement of the cost-minimization objective and explanation of unit shipping costs.
Covered · Explanation of the first step of the least cost method, including tie handling and selection of the $5 cell.
Covered · Identification of the chosen cell's associated demand (30 for City 4) and supply (40 for Factory 3); the clip ends before an allocation is written.
Covered · The opening interval introduces the transportation tableau and completes the first least-cost iteration for Factory 3 to City 4, including allocation, demand satisfaction, supply update, and column exclusion.
Covered · This interval covers the second iteration: selecting the $6 route, allocating 20 units to City 2, updating Factory 1 supply to 15, and excluding the rest of the City 2 column.
Covered · The final interval begins the third iteration by selecting the $8 route to City 1 and identifying demand 45, but the clip ends before the allocation amount is written.
Covered · First allocation from Factory 1 to City 1 and marginal updates.
Covered · Selection of the $9 route, allocation from Factory 2 to City 1, and elimination of City 1.
Covered · Rejection of exhausted Factory 1 route, selection of $13, allocation from Factory 2 to City 3.
Covered · Final allocation from Factory 3 to City 3, exhausting both remaining supply and demand.
Covered · Presentation of the completed solution and beginning of the total-cost calculation.
Covered · The speaker identifies each occupied route in turn while the corresponding table cell is circled and the total-cost expression is built term by term.
Covered · The completed sum is evaluated to $1,080, boxed in red, and the speaker states that this is the result for the least-cost transportation solution shown.
Reviewed English material v1: card/concept_card "Transportation problem table" starting at 0:00 — A transportation problem is represented as a matrix-like table: rows are supply sources (here factories), columns are demand destinations (here cities), interior entries are unit shipping costs, the rightmost column gives supplies, and the bottom row gives demands.
YouTube8:20Another explanation of Transportation and assignment problems.
Transportation and assignment problems
YouTube10:40Another explanation of Transportation and assignment problems.
Transportation and assignment problems
YouTube6:11Another explanation of Transportation and assignment problems.
Transportation and assignment problemsYouTube8:20Lecture for Transportation and assignment problems, compared with the current whiteboard.
Transportation and assignment problems