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Unit IIAssignment and Transportation Models Q1) Define Assignment model.A1) Given n promotion and n jobs, it's clear that there are n allocations during this case. Each facility or worker can perform each job, one at a time. However, certain steps are required to form the allocation so as to maximise profits or minimize costs or time.In this table, Coij is defined because the cost when the jth job is assigned to the ith worker. Note that this is often a special case of transportation problems when the number of rows is adequate to the amount of columns.Mathematical formulation:The basic viable solution to the allocation problem consists of (2n – 1) variables, where the (n – 1) variable is zero and n is that the number of jobs or facilities. thanks to this high degree of degeneracy, solving problems with normal shipping methods are often a posh and time-consuming task. Therefore, another method springs. it's vital to formulate the matter before proceeding to absolutely the law.Suppose xjj may be a variable defined as follows:If the primary job is assigned to the jth machine or facility0 if the i-th job isn't assigned to the j-th machine or facility.
Here, the matter is made one-to-one, or one job is assigned to at least one facility or machine.The total allocation cost is
The above definition is often evolved into a mathematical model as follows:Determine xij> 0 (i, j = 1,2,3… n),
Be constrained
xij is either 0 or 1. Q2) How to solve the matter (Hungarian technique)?A2) Consider a minimization type objective function. Resolving this allocation issue involves the subsequent steps:1. Find the littlest cost element in each row of the required cost table, starting with the primary row. Here, this smallest element is subtracted from each element therein row. Therefore, get a minimum of one zero in each row of this new table.2. After creating the table (as in step 1), get the columns of the table. Start with the primary column and find rock bottom cost think about each column. Then subtract this smallest element from each element therein column. After performing steps 1 and a couple of, you'll see a minimum of one zero in each column of the reduced cost table.3. The reduced table is now allocated within the following way:(I) Rows are continuously inspected until a row with just one zero is found. This assignment to one zero is formed by placing a square □ around it, with all other zeros being strikethrough (x) within the corresponding column. this is often because they're not used for other assignments during this column. The steps are performed for every row.(II) Step 3 (i) am performed on the column as follows: -Columns are continuously inspected until a column with just one zero is found. Now, by placing a square around it, this single zero is assigned, and at an equivalent time all other zeros within the corresponding row are erased (x) A step is performed on each column. I will.(III) Steps 3, (i), and three (ii) are repeated until all zeros are marked or strikethrough is drawn. Here, if the amount of zeros marked or the amount of allocations made is adequate to the amount of rows or columns, the simplest solution is achieved. There’s exactly one assignment for every column or column, with none assignment. During this case, attend step 4.4. At this stage, draw the minimum number of lines (horizontal and vertical) needed to hide all the zeros within the matrix obtained in step 3. Use the subsequent procedure.(i) Put a check on all unassigned lines.(ii) Now mark of these columns that have zeros within the graduated rows (iii) Now check all rows that haven't yet been marked and are assigned to the marked columns.(iv) All steps, namely (4 (i), 4 (ii), 4 (iii), are repeated until no row or column are often marked.(V) Draw a line through all unmarked rows and marked columns. Also note that in an n x n matrix, lines but "n" always cover all zeros if there's no solution between them.5. In step 4, the number of lines drawn is n or the number of lines, it's the best solution. If not, go to step 6.6. Select the smallest element from all uncovered elements. Here, this element is subtracted from all uncovered elements and added to the element at the intersection of the two lines. This is the new allocation matrix.7. Repeat step (3) until the number of allocations is equal to the number of rows or columns. Q3) Explain the Balanced and Unbalanced Problems.A3) Whenever the cost matrix of an allocation problem is not a square matrix, that is, the number of sources is not equal to the number of destinations, the allocation problem is called an imbalanced allocation problem. In such problems, dummy rows (or columns) are added to the matrix to make a matrix. The dummy row or column contains all cost elements as zero. The Hungarian method could also be wont to solve the matter.Example: A company has 5 machines used for 4 jobs. Each job can only be given to one machine. The following table shows the cost of each job on each machine.Example of an unbalanced maximization allocation problem
Solution: Add dummy row D5 to convert the 4x5 matrix to a square matrix.Dummy line D5 has been added
Matrix row direction reduction
No per-column reduction is required as every column contains a single zero. Then draw the minimum number of lines that cover all zeros, as shown in the table.All zeros in the matrix of interest
Number of lines drawn ≠ Matrix order. Therefore, it is not optimal.As shown in the table, select the least covered element, that is, 1, subtract it from the other uncovered elements, add it to the element at the intersection of the lines, and cover it with a single line. Do not change the element.Subtract or add to an element
Number of lines drawn ≠ Matrix order. Therefore, it is not optimal.Add or subtract 1 again from the element
Number of lines drawn = degree of matrix. Therefore, optimality is reached. Then assign the job to the machine, as shown in the table.Assigning jobs to machines
Example: In a plant layout, four different machines M1, M2, M3, and M4 are installed in a machine shop. There are five missing areas: A, B, C, D and E. Due to limited space, Machine M2 cannot be installed in Area C and Machine M4 cannot be installed in Area A. The installation cost of the machine is shown in the table. Q4) What steps are used in Hungarian method?A4) The Hungarian method are often summarized within the following steps:Step 1: Create a price table from the given problem.If the number of rows isn't adequate to the number of columns, or the other way around, you would like to feature dummy rows or columns. The value of allocating dummy cells is usually zero.Step 2: Find the chance Cost Table:(A) Find the littlest element in each row of the required cost table and subtract it from each element therein row.(B) Within the reduction matrix obtained from 2 (a), find the littlest element in each column and subtract it from each element. Each row and column has a minimum of one zero value. Step 3: Make an allocation within the cost Matrix:The allocation procedure is as follows:(A) Examine the rows continuously until you get a row with just one unmarked zero. Make the assignment one zero by creating a square around it.(B) For every zero-value assigned, delete (cancel) all other zeros within the same row and / or column.(C) Repeat steps 3 (a) and three (b) for every column, leaving just one unassigned zero value.(D) If there are two or more unmarked zeros within the rows and / or columns and therefore the check cannot select one, select any assigned zero cell.(E) Continue this process until all zeros within the matrix are enclosed (assigned) or deleted (x).Step 4: Optimal criteria:If the assigned cell members are adequate to the row sequence that is the best solution. The entire cost related to this solution is obtained by adding the first cost figure to the occupied cell.If zero cells are arbitrarily selected in step (3), an alternate optimal solution exists. However, if you can't find the simplest solution, attend step (5).Step 5: Modify the chance cost table.Use the subsequent procedure to draw a group of horizontal and vertical lines that cover all zeros within the revised cost table obtained in step (3).(A) Add a check (√) to every row that has not been assigned.(B) Examine the marked line. If zeros occur in these columns, check each row that contains the assigned zeros.(C) Repeat this process until you'll not mark rows or columns.(D) Draw a line through each marked column and every unmarked row.If the number of drawn lines is adequate to the amount (or columns), then the present solution is that the best solution otherwise, attend step 6.Step 6: Create a newly revised cost Table.(A) Select the littlest element from the cells that aren't covered by the road, and call this value K.(B) Draw K from all the weather within the cell that isn’t covered by the road.(C) Add K to the element within the cell covered by the 2 lines, that is, the intersection of the 2 lines.(D) The weather of the cell covered by one row aren't changed.Step 7: Repeat steps 3-6 close up the lights for the simplest solution Q5) Write note on Balanced transportation issues.A5) For shipping issues:Minimize z = 
Be constrained
If the aggregate supply from all sources is equal to the aggregate demand of all destinations, then x11 of all i and j is said to be a balanced transport problem, otherwise the problem is an imbalanced transport problem is said to be.There may be a solution that can only be implemented if the transportation problem is a balanced problem. Imbalanced issues can be balanced by adding dummy supply centers (rows) or dummy demand centers, depending on your requirements. Q6) Explain Prohibited Transportation Problems, Unique or Multiple Optimal Solutions.A6) Sometimes, there may be situations when a particular route cannot be used for transportation, for example, when road is being constructed, accidents, floods, etc. These problems are known as prohibited transportation problems. These can be solved by two methods:By assigning a very high or infinite cost value represented by M to the unavailable route. By blocking allocation to the cell which has the prohibited route, thus helping cancel the cell.
Solution:As can be seen in the question, we have some restrictions.An initial solution is obtained by the matrix minimum method Final Table
Initial basic feasible solution16 X 50 + 12 X 150 + 14 X 40 + 8 X 120 + 26 X 90 = 6460.The minimum transportation cost is Rs. 6460. Q7) Elaborate and write Simple Formulation of Transportation Problems.A7) While solving a problem, the key part lies in formulating the model by decoding the problem. For a transportation problem, usually, the problem will be given in a tabular form or matrix form called transportation table or cost-effective matrix.Let’s check an example below.
An example for transportation problem with Source {A, B, C} with total supply as 70 and Destinations {1,2,3} with total demand as 70. Here,Source = {A, B, C}They represent sources with supply capacity 10,35,25 units of commodities respectively (given in orange)Hence, total supply=10+35+25=70Destinations= {1,2,3}They represent destinations that require 20,25,25 units of commodities respectively (given in green).Hence, total demand=20+25+25=70The elements represented in the matrix are called cost. i.e., the unit cost involved in moving unit commodity from one origin to another destination.For example,The cost incurred in moving 1 unit of the commodity from source A to destination 1= ₹2/-The cost incurred in moving 1 unit of the commodity from source A to destination 1= ₹2/- (here, we are looking at the cross-section of source A and destination 1)Likewise,The cost incurred in moving one unit of the commodity from source B to destination 2= ₹4 and so on. [Here, we are looking at the cross-section of source and destination] Q8) How will you explain North West Corner Rule (NWCR) A8) Northwest corner ruleIn this part, we'll look at the Northwest Corner rules to find the first basic viable solution. There are three steps. Find the cell in the northwest corner of the transport tableau. Allocate as much as possible to the selected cells and subtract the allocated amount to adjust the associated supply and demand amount. Cancels a row or column with zero supply or demand. If both the row and column have 0s, then the row or column is randomly cancelled. If one cell does not intersect, cancel the cell and stop. Otherwise, go to step 1. ExampleStart with an empty matrix with supply at the bottom and supply at the right.
When you get the cell in the northwest corner, its values are the minimum d1 and s1. Since s₁ is smaller than d₁, cancel the first line.
In the next iteration, do the same steps.
We come up with such a table and a basic viable solution.
Q9) Define Least Cost Method (LCM).A9) There are several methods available to get the first basic viable solution to your shipping problem. Only the following three are discussed here. To find the first basic viable solution, the total supply should be equal to the total demand.Method: Least common multiple method (LCM)The lowest cost method is more economical than the northwest corner rule because it starts with a low starting cost. The various steps involved in this method are summarized below.Step 1: Find the cell with the lowest (minimum) cost in the transport table.Step 2: Assign the maximum amount that can be executed in this cell.Step: 3: Delete the row or column to which the assignment will be made.Step: 4: Repeat the above steps for the reduced transport table until all allocations are complete.Example Use the lowest cost method to get the first basic viable solution to the next shipping problem.
Here, Oi and Dj indicate the i-th starting point and the j-th destination, respectively.Solution:Total supply = total demand = 24The problem given is a balanced transportation problem.Therefore, there is a viable solution to the given problem. Q10) How is initial basic feasible solution is obtained by VAM?A10) The Vogel’s Approximation Method or VAM can be defined as an iterative procedure which is used to find out the initial feasible solution of the transportation problem. The initial basic feasible solution obtained by VAM is either optimal solution or very nearer to the optimal solution. Let’s go through the procedure with an example. Example:
A mobile phone manufacturing company has three branches which are located in three different regions, say Jaipur, Udaipur and Mumbai. The company is supposed to transport mobile phones to three destinations, say Kanpur, Pune and Delhi. The availability from Jaipur, Udaipur and Mumbai is 40, 60 and 70 units respectively. The demand at Kanpur, Pune and Delhi are 70, 40 and 60 respectively. The transportation cost is shown in the matrix below (in Rs). Use the Vogel’s Approximation Method to find a basic feasible solution (BFS).
Solution:Step 1: Balance the problemBalance the problem meaning we need to check that if;∑ Supply=∑ DemandIf this condition is satisfied, then we will consider the given problem as a balanced problem.Now, what if it’s not balanced?i.e., Σ Supply≠ Σ DemandIf such a condition occurs, then we have to add a dummy source or market; whichever makes the problem balanced.Step 2: Find out the Row difference and the Column differenceNow, we will find out the row and the column difference of the provided matrix.For the same we will solve this as follows:
Here as you can see in the first row, we need to find out cell containing smallest value (least cost), and the second least.
Hence, we find 1 and 4 as the two lowest values in the first row.
Take the difference of these two numbers, i.e. (4−1)=3.
In a similar manner, find out differences for all the rows as shown above.
The same procedure is to be done with columns.
So, for the first column, you will get, (4−3)=1.
Find out difference for all columns as done above.Step 3: Select the row/column with highest difference
As we can see above, the highest difference value is 4 which is for the last row.
So we will select that row, and find out the minimum cost in that row.
You will find that, minimum cost is 2.
Now, compare the supply and demand for this row and column respectively.
We find that, we have supply as 70 and demand as 40 for the selected cell.
Compare both and select the minimum of them for the allocation in the cell, as shown below.
Cross out the column whose demand is fulfilled.Step 4: Remove the row/column whose supply or demand is fulfilled and prepare new matrix and repeat the same procedure
As it can be seen, we now have a new matrix with second column removed (as its demand was 40 which was fulfilled)
Now, we will repeat the same procedure until we're done with all allocations.
i.e.,Find row and column difference. Select highest difference. Allocate cell with minimum cost, associated with selected highest row or column difference. Check out row/column, if demand or supply is fulfilled.
Step 5: After all the allocations are completed, write all the allocations in the matrix and find Transportation cost
We have all allocations with their respective cell now.Find transportation cost as follows:Transportation cost= (1x40)+(3x40)+(3x20)+(6x30)+(2x40)=Rs 480. Q11) How do we find Optimal Solution by Modified Distribution (MODI) Method? (u, v and ∆)A11) There are two steps to solving a transportation problem. The first phase requires finding the first basic viable solution, and the second phase optimizes the first basic viable solution obtained in the first phase. There are three ways to find the first basic viable solution.Northwest corner method Minimum cost cell method Vogel approximation This article describes how to optimize your first basic viable solution through the examples provided. Consider the following transportation problem.
Solution:Step 1: Check if the problem is balanced.If the sum of all supplies from sources O1, O2, and O3 is equal to the sum of all demands for destinations D1, D2, D3, and D4, then the shipping problem is a balanced shipping problem.Note: If the problem is not imbalanced, you can follow the concept of dummy rows or columns to transform an imbalanced problem into equilibrium, as described in this article.
Step 2: Find the first basic viable solution.You can use one of the three methods described above to find the first basic viable solution. Here we use the Northwest Corner method. And according to the Northwest Corner Method, this is the final first basic viable solution.
Now the total shipping cost is (200 * 3) + (50 * 1) + (250 * 6) + (100 * 5) + (250 * 3) + (150 * 2) = 3700.Step 3: U-V method to optimize the first basic viable solution.Below is the first basic viable solution.
– For U-V methods, you need to find the row and column values ui and vj, respectively. Since we have 3 rows, we need to find 3 ui values. That is, u1 on the first row, u2 on the second row, and u3 on the third row.Similarly, for four columns, you need to find the four vj values, v1, v2, v3, and v4. Please check the image below.
There is another formula for finding ui and vj.ui + vj = Cij where Cij is the cost value of the assigned cell only. Please check this out for details. Before applying the above formula, we need to make sure that m + n – 1 is equal to the total number of cells allocated. Where m is the total number of rows and n is the total number of columns.In this case, m = 3, n = 4, and the total number of cells allocated is 6, so m + n – 1 = 6. Later post if m + n – 1 is not equal to the total number of allocated cells.Now assign any of the three u or any of the four v as 0 to find the values for u and v. In this case, assign u1 = 0. Then, using the above formula, we get v1 = 3 as u1 + v1 = 3 (that is, C11) and v2 = 1 as u1 + v2 = 1 (that is, C12). Similarly, we got the value of v2 = 1, so we get the value of u2 = 5, which means v3 = 0. From the value of v3 = 0, we get u3 = 3, which means v4 = -1. See the image below
Now use the formula Pij = ui + vj – Cij to calculate the penalty only for unallocated cells. There are two unallocated cells in the first row, two in the second row and two in the third row. Let's calculate this one by one.For C13, P13 = 0 + 0 – 7 = -7 (here C13 = 7, u1 = 0 and v3 = 0)For C14, P14 = 0 + (-1) -4 = -5For C21, P21 = 5 + 3 – 2 = 6For C24, P24 = 5 + (-1) – 9 = -5For C31, P31 = 3 + 3 – 8 = -2For C32, P32 = 3 + 1 – 3 = 1Rule: If all penalty values are taken as zero or negative, it means that optimality has been reached and this answer is the final answer. However, if you get a positive value, it means that you need to continue the sum in the next step.Find the biggest positive penalty here. The maximum value here is 6, which corresponds to cell C21. This cell is now the new base cell. This cell is also included in the solution.
Rules for drawing closed paths or loops. Start with a new base cell and draw a closed path so that only the assigned cell or the new base cell has a right-angle rotation. See the image below.
Assign an alternative plus or minus sign to all cells that bend (or corner) at right angles in the loop, and assign a plus sign to the new base cell.
Consider a cell with a negative sign. Compare the assigned values (200 and 250 in this case) and select the minimum value (that is, select 200 in this case). Then subtract 200 from the cells with the minus sign and add 200 to the cells with the plus sign. Then draw a new iteration. Now that we're done with the loop, the new solution looks like this:
Make sure that the total number of allocated cells is equal to (m + n – 1). Again, use the expression ui + vj = Cij to find the u and v values. Where Cij is the cost value for the assigned cell only. Assigning u1 = 0 yields v2 = 1. Similarly, for ui and vj you get the following values:
Use the formula Pij = ui + vj – Cij to find penalties for all unassigned cells.For C11, P11 = 0 + (-3) – 3 = -6For C13, P13 = 0 + 0 – 7 = -7For C14, P14 = 0 + (-1) – 4 = -5For C24, P24 = 5 + (-1) – 9 = -5For C31, P31 = 0 + (-3) – 8 = -11For C32, P32 = 3 + 1 – 3 = 1There is one positive value. That is, 1 for C Now this cell becomes new basic cell.
Then draw a loop starting from the new base cell. Assign alternative plus and minus signs using the new base cell assigned as a plus sign.
Select the minimum value from the values assigned to the cells marked with a minus sign. Subtract this value from the cell with the minus sign and add it to the cell with the plus sign. The solution now looks like the following image. Check if the total number of allocated cells is equal to (m + n – 1). Find the values of u and v as above.
Now find the penalty for the unallocated cell again, as described above.When P11 = 0 + (-2) – 3 = -5When P13 = 0 + 1 – 7 = -6When P14 = 0 + 0 – 4 = -4When P22 = 4 + 1 – 6 = -1When P24 = 4 + 0 – 9 = -5When P31 = 2 + (-2) – 8 = -8All penalty values are negative. Therefore, optimality is reached.Here you find the total cost, that is, (250 * 1) + (200 * 2) + (150 * 5) + (50 * 3) + (200 * 3) + (150 * 2) = 2450. Q12) Explain Maximum Two Iterations (i.e., Maximum Two Loops) after IFS.A12) Use Vogel’s method to find an initial basic feasible solution and then find the minimum transportation cost for the following transportation problem by Modi method.
Solution: Step I: Find the initial basic feasible solution The initial basic feasible solution obtained by applying Vogel’s approximation method is shown below.
Step II Perform Optimality Test Required number of allocations = m+n-1= 3+4-1 =6=Actual number of allocations. Therefore, optimality test can be performed. Now, we use Modi for the optimality test.
Now we find the value uj and vj by the relation uj+vj=Cj For the basic cell (1,1): u1+ v1=1 For the basic cell (1,3): u1+ v3=1 For the basic cell (2,2): u2+ v2=3 For the basic cell (2,3): u2+ v3=2 For the basic cell (2,4): u2+ v4=1 For the basic cell (3,2): u3+ v2=2 Then, let v1=0, then we get V2 =1; v3=0; v4=-1; u1=1; u2=2; u3=1.Now displaying these values in the table and calculate the cell evaluation.
Iteration 1 Now, subtracting the cell values of the above matrix from the original cost matrix;
Iteration 2Step III: Since all cell evaluations are positive, the initial basic feasible solution is optimal. The minimum transportation cost is given by=1x20+1x10+3x20+2x20+1x10+2x20=180.
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Be constrained
If the aggregate supply from all sources is equal to the aggregate demand of all destinations, then x11 of all i and j is said to be a balanced transport problem, otherwise the problem is an imbalanced transport problem is said to be.There may be a solution that can only be implemented if the transportation problem is a balanced problem. Imbalanced issues can be balanced by adding dummy supply centers (rows) or dummy demand centers, depending on your requirements. Q6) Explain Prohibited Transportation Problems, Unique or Multiple Optimal Solutions.A6) Sometimes, there may be situations when a particular route cannot be used for transportation, for example, when road is being constructed, accidents, floods, etc. These problems are known as prohibited transportation problems. These can be solved by two methods:It is possible that a transportation problem to have multiple optimal solutions. This happens when one or more of the improvement indices are zero in the optimal solution. This means that there is a possibility for designing alternative shipping routes with the same total shipping cost. The alternate optimal solution can be found by shipping the most to this unused square using a stepping-stone path. In the real world, alternate optimal solutions help with better management and greater flexibility in selecting and using resources.
Consider the following transportation problem.Factory | Warehouse | Supply | ||
W1 | W2 | W3 | ||
F1 | 16 | ∞ | 12 | 200 |
F2 | 14 | 8 | 18 | 160 |
F3 | 26 | ∞ | 16 | 90 |
Demand | 180 | 120 | 150 | 450 |
Factory | Warehouse | Supply | ||
W1 | W2 | W3 | ||
F1 |
| ∞ |
| 200 |
F2 |
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| 18 | 160 |
F3 |
| ∞ | 16 | 90 |
Demand | 180 | 120 | 150 | 450 |
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| DESTINATIONS |
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| SUPPLY |
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| 1 | 2 | 3 |
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| A | 2 | 3 | 1 | 10 |
SOURCE | B | 5 | 4 | 8 | 35 |
| C | 5 | 6 | 8 | 25 |
DEMAND |
| 20 | 25 | 25 |
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A mobile phone manufacturing company has three branches which are located in three different regions, say Jaipur, Udaipur and Mumbai. The company is supposed to transport mobile phones to three destinations, say Kanpur, Pune and Delhi. The availability from Jaipur, Udaipur and Mumbai is 40, 60 and 70 units respectively. The demand at Kanpur, Pune and Delhi are 70, 40 and 60 respectively. The transportation cost is shown in the matrix below (in Rs). Use the Vogel’s Approximation Method to find a basic feasible solution (BFS).
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Hence, we find 1 and 4 as the two lowest values in the first row.
Take the difference of these two numbers, i.e. (4−1)=3.
In a similar manner, find out differences for all the rows as shown above.
The same procedure is to be done with columns.
So, for the first column, you will get, (4−3)=1.
Find out difference for all columns as done above.Step 3: Select the row/column with highest difference
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So we will select that row, and find out the minimum cost in that row.
You will find that, minimum cost is 2.
Now, compare the supply and demand for this row and column respectively.
We find that, we have supply as 70 and demand as 40 for the selected cell.
Compare both and select the minimum of them for the allocation in the cell, as shown below.
Cross out the column whose demand is fulfilled.Step 4: Remove the row/column whose supply or demand is fulfilled and prepare new matrix and repeat the same procedure
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Now, we will repeat the same procedure until we're done with all allocations.
i.e.,
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| D1 | D2 | D3 | D4 | Availability |
O1 | 1 | 2 | 1 | 4 | 30 |
O2 | 3 | 3 | 2 | 1 | 50 |
O3 | 4 | 2 | 5 | 9 | 20 |
Requirement | 4 | 40 | 30 | 10 | 100 |
| D1 | D2 | D3 | D4 | Availability | Row Differences |
01 | 1
(20) | 2 | 1
(10) | 4 | 30/10/0 | [0] [0] [1] |
O2 | 3 | 3
(20) | 2
(20) | 1 (10) | 50/40/20/0 | [1] [1] [1] |
03 | 4 | 2
(20) | 5 | 9 | 20/0 | [2] [2] |
Requirement | 20/0 | 40/20/0 | 30/20/0 | 10/0 | 100/100 |
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Column Differences | [2] [2] | [0] [0] [1] | [1] [1] [1] | [3] |
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