Example for Solution of Transportation Problem

EXAMPLE FOR SOLUTION OF TRANSPORTATION PROBLEM  

An organization has four destinations and three sources for supply of goods. The transportation cost per unit is given below. The entire availability is 700 units which exceeds the cumulative demand of 600 units. Decide the optimal transportation scheme for this case.

Solution

Step 1:            Check for balance of supply and demand

                        S Supply = 250 + 200 + 250 = 700 units

                        S Demand = 100 + 150 + 250 + 100 = 600 units

Decision Rul

(i) If S Supply = S Demand then go to next step.

(ii) Else; if S Supply > S Demand then, add a “dummy destination” with zero transportation cost.

(iii) Or else; if S supply < S Demand then, add a “dummy source” with zero transportation cost.

Since, in this problem

S supply > S Demand

Hence; add a “dummy destination” (say D5) with zero transportation cost and balance demand which is difference in supply and demand (= 100 units).

The initial transportation matrix is now formulated with transportation cost in the small box of each route. Note that each cell of the transportation matrix represents a potential route.

                             Introducing dummy column for balancing the supply and demand

Step 2:

(i) Decide the nature of problem : Minimization of transportation-cost

(ii) Make initial assignment

Initial assignment may be done by using any of the following approaches :

(i) Least-cost method

(ii) North-West corner method

(iii) Vogel's approximation method

We would demonstrate all the three methods.

(i) Initial Solution by Least Cost method

Select the lowest transportation (or shipping) cost cell (or route) in the initial matrix. For example: it is route S1D5, S2D5 and S3D5 in our problem with zero shipping cost.

Allocate the minimum of remaining balance of supply (in last column) and demand (in last row).

Let us select S1D5 route. One can also select other route (S2D5 or S3D5) in case of tie. For S1D5, available supply is 250 and available demand is 100 units. The lower is 100 units. Hence, allocate 100 units-through this route (i.e, S1D5).

With this allocation, entire demand of route S1D5 is consumed but supply of corresponding source, S1, is still (250-100) or 150 units left. This is marked in last column of supply. The entire demand of destination, D5, is consumed. We get the following matrix (Fig. 12.6) by crossing out the consumed destination (D5):

Now, we leave the consumed routes (i.e., column D5) and work for allocation of other routes.

Next, least cost route is S1D1, with 13 per unit of shipping cost. For this route, the demand is 100 units and remaining supply is 150 units. We allocate minimum of the two, i.e., 100 units in this route. With this destination, D1 is consumed but source S1 is still left with (150-100) = 50 units of supply. So, now leave the destination D1 and we get the following matrix.

                                                      With 100 units allocation in route S1D5

                                           Assignment for destination D1 and D5 consumed

Now, we work on remaining matrix, which excludes first column (D1) and last column (D5). Next assignment is due in the least cost route, which is route S2D4. For this route, we can allocate 100 units which is lesser of the corresponding demand (100 units) and (200 units). By this allocation in route S2D4, the demand of destination D4 is consumed. So, this column is now crossed out.

                               Assignment with destination D1, D4 and D5 consumed

Now, we work on the remaining matrix which excludes, column, D1, D4 and D5. Next assignment is due in the least cost route of the remaining routes. Note that we have two potential routes: S1D2

and S2D3. Both have 16 units of transportation cost. In case of any tie (such as this), we select any of the routes. Let us select route, S1D2, and allocate 50 units (minimum of demand of 150 and supply of remaining 50 units). With this, all supply of source S1 is consumed. Therefore, cross out row of S1. We get the following matrix:

                                    Destination D1, D4 and D5 source S1 are consumed

Now, remaining allocation is done in route S2D3 (as 100 units). With this source, S2 is consumed. Next allocation of 100 units is done in route S3D2 and 150 units in route S3D3. Final initial assignment is as follows:

Total cost in this assignment is (13 × 100 + 16 × 50 + 100 × 0 + 16 × 100 + 15 × 100 + 17 × 150) or Rs. 9450.

                                             Initial assignment by least cost method

Step 3: Count the number of filled (or allocated) routes.

Decision rule

(i)             If filled route = (m + n – 1) then go for optimality check (i.e. step 5).

(ii)            If filled route < (m + n – 1) then the solution is degenerate. Hence, remove degeneracy and go to step 4.

Here, m = number of destinations, including dummy column, if any

n = number of source, including dummy, row, if any

For our problem (m + n- l) = 5 + 3-1 = 7.

The number of filled route is equal to 7. Hence, problem is not degenerate. Therefore, proceed to step 5.

       Initial Assignment by North-West Corner Method (an alternative to least cost method)

This approach is also for making initial assignment, as we have done in the least cost method. Therefore, this approach should not be applied if initial assignment has already been made by any other method. In the North-West Corner (NWC) method, we start with the top-left (corner-most) route, which is S:DrIrrespective of cost, allocation is made in this route for the minimum of supply or demand. In our case, demand for this route is 100 and supply is 250. Therefore, allocate 100 units in this route. With this, column corresponding to D1 is consumed.

Now, work on the remaining matrix, which excludes column Dr Again, select the top-left route. Now, it is cell S1D2. Allocate in the same way. Thus, 150 units are allocated in this route. Note that, with this, both D2 and S, are consumed.

Remaining matrix excludes S1, D1 and D2. Hence, allocation in the top-left cell is due in route S2D3. Here, 200 units may be allocated and S2 is now consumed.

Remaining allocations are done in S2D3, S3D4 and S3D5 in sequential order. We get the initial solution by north-west corner method as follows (Fig. 12.11):

                                          Initial assignment by North-West corner method

For this assignment, the total cost is (13 × 100 + 16 × 150 + 16 × 100 + 0 × 100) or Rs. 9350.

Step 3: Check for degeneracy

                        (m + n – 1) = 5 + 3 – 1 = 7

Number of filled cells = 6, which is one less than (m + n – 1). Hence, go to step 4 for removing degeneracy.

Step 4: In case of degeneracy, allocate a very-very small quality,  (which is zero for all calculation purposes), in the least cost of un-filled cells. In the above figures of North-West corner method allocation, the least cost un-filled cells are S1D5 and S2D5. Let us select S1D5 and allocate  in this. We get the following allocation after removing degeneracy.

                  Initial assignment by North-West corner method after removing degeneracy

Initial Assignment by Vogel’s Approximation Method (VAM)

This is the third alternative method for doing initial assignment of a transportation problem.

In this method, we calculate the difference between the two least-cost routes for each row and column. The difference is called as penalty cost for not using the least-cost route.

                                  First calculation of Penalty cost in VAM

Highest of all calculated penalty costs is for S3 and (S2). Therefore, allocation is to made in row of source S3. The route (or cell), which one must select, should be the lowest cost of this row. This route S3D5. Hence, first allocation is as follows.

                                             First calculation in Vogel’s method

Now, with the first allocation, destination D5 is consumed. We exclude this column and work on the remaining matrix for calculating the penalty cost. We get the following matrix.

Now for this, source S1 has highest penalty cost. For this row, the least cost route is S1D1. Hence, next assignment is due in this route:

                                          Second calculation of Penalty cost in VAM

                                              Second allocation in Vogel’s method

After second allocation, since destination D1 is consumed, we leave this column and proceed for calculation of next penalty cost. Allocation is done in route S1D2. Since there is tie between all routes, we break the tie by arbitrarily selecting any route (S1D2 in this case.) 

                                           Third calculation of Penalty cost

                                             Third allocation in Vogel’s method

                                        Fourth calculation of Penalty cost in VAM

                                          Fourth allocation in Vogel’s method

With the fourth allocation, column D4 is consumed. In the only left column D3, the allocations of 100 units and 150 units are done in route S2D3 and S4D3 respectively. Thus, we get the following allocations in the Vogel’s approximation method.

                                        Final allocation through Vogel’s method

The initial cost for this allocation is (13 × 100 + 16 × 150 + 16 × 100 + 15 × 100 + 17 × 150 + 0 × 100) or equal to Rs. 9350:

Step 3:           Check for degeneracy

                        (m + n – 1) = 7

Number of filled cell = 6, which is one less than (m + n + 1). Hence, go to step 4 for removing the degeneracy.

Step 4:           We allocate  in the least-cost un-filled cell. This cell is route S1D5 or S2D5. Let us select route S1D5. Thus, we get following matrix after removing degeneracy.

                                Final allocation after removing degeneracy in Vogel’s method

Optimization of Initial Assignment

The initial feasible assignment is done by using least-cost method or North-West corner method or Vogel's approximation method. However, none of these methods guarantees optimal solution. Hence, next step is to check the optimality of the initial solution.

Step 5:   Check the optimality of the initial solution

For this, we have to calculate the opportunity cost of un-occupied routes.

First, we start with any row (or column). Let us select row 1, i.e., source S1; For this row, let us define row value, u1 = 0. Now consider all filled routes of this row. For these routes, calculate column values v. using following equation:

u1 + v1 = Cij (For any filled route)

where             u1 = row value

vj = column value

Cij = unit cost of assigned route

Once first set of column values (vj is known, locate other routes of filled cells in these columns. Calculate next of ui (or vj values using above equation. In this way, for all rows and columns, ui and vj values are determined for a non- degenerate initial solution.

Step 6:           Check the optimality

Calculate the opportunity of non-allocated orunfilled routes. For this, use the following equation:

                                               Opportunity unassigned route = ui + vj – Cij

where             ui = row value

vj = column value

Cij = unit cost of unassigned route

If the opportunity cost is negative for all unassigned routes, the initial solution is optimal. If in case any of the opportunity costs is positive, then go to next step.

Step 7:           Make a loop of horizontal and vertical lines which joins some filled routes with the unfilled route, which has a positive opportunity cost. Note that all the corner points of the loop are either filled cells or positive opportunity cost un-assigned cells.

Now, transfer the minimal of all allocations at the filled cells to the positive opportunity cost cell. ¥or this, successive corner points from unfilled cell are subtracted with this value. Corresponding addition is done at alternate cells. In this way, the row and column addition of demand and supply is maintained. We show the algorithm with our previous problem.

Let us consider the initial allocation of least-cost method (Fig. 12.10) :

For this, we start with row, S1and take u1 = 0. Now S1DpS1D2,and S1D5are filled cells. Hence, for filled cells; (vj = Cij – ui).

v1 = 13 – 0 = 13

v2 = 16 – 0 = 16

v5 = 0 – 0 = 0

                               Calculation for ui and vj in least cost initial assignment

Now, cell S3D2 is taken, as this has a vj value. For this cell u3 = 17 – 16 = 1

Now, cell S3D3 is selected, as this has a ui value. For this cell v3 = 17 – 1 = 16

Now, cell S2D3 is selected, as it has a vj value. For this cell u2 = 16 – 16 = 0

Now, cell S2D4 is selected, as it has a ui value. For this cell v4 = 15 – 0 = 0

Thus, all ui and vj are known.

Step 6: Calculate opportunity cost of un-assigned routes.

Unassigned route Opportunity cost (ui + vj – Cij)

S1D3

S1D4

S2D1

0 + 16 – 19 = –3

0 + 15 – 17 = –2

0 + 13 – 17 = –4

S2D2

S2D5

S3D1

S3D4

S3D5

0 + 16 – 19 = –3

0 + 0 – 0 = 0

1 + 13 – 15 = –2

1 + 15 – 16 = 0

1 + 0 – 0 = +1

Since route S3D5 has positive opportunity cost, the solution is non-optimal; hence, we go to next step and make a loop as follows.

                                               Closed loop for cell S3D5

The revised allocation involves 100 units transfer from cells S1D5 and S3D2 to cells S3D5 and S1D2.

Thus, revised allocation is as follows:

                                                       Revised allocation in least-cost assignment

Since above solution is degenerate now, we allocate  to the least-cost un-filled cell S1D5. Fresh calculation of ui and vj is also done in the similar way as explained in Step 5.

For this assignment, the opportunity cost of unassigned cells is as follows.

Now, since un-allocated routes have negative (or zero) opportunity cost, the present assignment is the optimal one. Thus, optimal allocation of route is given in Figure.

Note that total cost is less than the initial assignment cost of least-cost method (= Rs. 9450).

Similarly, optimality of North-West corner method solution is done.

Unassigned route Opportunity cost (ui + vj – Cij)

S1D3

S1D4

S2D1

S2D2

0 + 17 – 19 = –2

0 + 16 – 17 = –1

–1 + 13 – 17 = –5

–1 + 16 – 19 = –4

S2D5

S3D1

S3D2

S3D4

–1 + 0 – 0 = –1

0 + 13 – 15 = –0

0 + 16 – 17 = –1

0 + 16 – 16 = 0

                                                                  Opportunity cost

Route Unit Cost in this route

S1D1

S1D2

S2D3

S2D4

S3D3

S3D5

100

150

100

100

150

100

13 × 100 = 1300

16 × 150 = 2400

16 × 100 = 1600

15 × 100 = 1500

17 × 150 = 2550

0 × 100 = 0

Total cost = Rs.9350

                                        Optimal allocation in different routes

                         Calculation of ui and vj for N-W corner method’s initial solutions

Opportunity cost of above assignment is as follows:

Since all opportunity costs are negative or zero, the initial assignment is optimal one with total cost of Rs. 9350.

The optimal assignment of routes is 100 units is S1D1, 150 units in S1D2, 200 units in S2D3, 50 units in S3D3, 100 units in S3D4.

Similarly, the optimality of Vogel’s method’s initial solution is done.

Opportunity cost of above N-W corner assignment is as follows

Unassigned route Opportunity cost (ui + vj – Cij)

S1D3

S1D4

S2D1

S2D2

S2D4

S2D5

S3D1

S3D2

0 + 17 – 19 = –2

0 + 16 – 17 = –1

–1 + 13 – 17 = –5

–1 + 16 – 19 = –4

–1 + 16 – 15 = 0

–1 + 0 – 0 = –1

0 + 13 – 15 = –2

0 + 16 – 17 = –1

                             Calculation of ui and vj for Vogel method’s initial solutions

Opportunity cost of above assignment is as follows:

Unassigned route Opportunity cost (ui + vj – Cij)

S1D3

S1D4

S2D1

S2D2

S2D5

S3D1

S3D2

S3D4

0 + 17 – 19 = –2

0 + 16 – 17 = –1

–1 + 13 – 17 = –5

–1 + 16 – 19 = –4

–1 + 0 – 0 = –1

0 + 13 – 15 = –2

0 + 16 – 17 = –1

0 + 16 – 16 = 0

Since all opportunity costs are negative or zero, the initial assignment of Vogel’s solution is optimal with total cost of Rs. 9350.

The optimal assignment of routes is 100 units in S1D2, 100 units in S2D3, 100 units in S2D4, and 150 units in S3D3. Note that this solution is different from North-West corner solution but total cost is same and minimum.

                                            The transportation problem solution approach

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MathStat: Nikhil M. JoshiTransportation   Problem

Transportation Method

A transportation tableau is given below. Each cell represents a shipping route (which is an arc on the network and a decision variable in the LP formulation), and the unit shipping costs are given in an upper right hand box in the cell.

To solve the transportation problem by its special purpose algorithm, the sum of the supplies at the origins must equal the sum of the demands at the destinations (Balanced transportation problem).

• If the total supply is greater than the total demand, a dummy destination is added with demand equal to the excess supply, and shipping costs from all origins are zero.

• Similarly, if total supply is less than total demand, a dummy origin is added. The supply at the dummy origin is equal to the difference of the total supply and the total demand. The costs associated with the dummy origin are equal to zero.

When solving a transportation problem by its special purpose algorithm, unacceptable shipping routes are given a cost of +M (a large number).

Develop an Initial Solution

Two methods to get the initial solution:

• Northwest Corner Rule

• Minimum Cell-Cost Method

Northwest Corner Rule

1. Starting from the northwest corner of the transportation tableau, allocate as much quantity as possible to cell (1,1) from Origin 1 to Destination 1, within the supply constraint of source 1 and the demand constraint of destination 1.

2. The first allocation will satisfy either the supply capacity of Source 1 or the destination requirement of Destination 1.

ƒ If the demand requirement for destination 1 is satisfied but the supply capacity for Source 1 is not exhausted, move on to cell (1,2) for next allocation.

ƒ If the demand requirement for destination 1 is not satisfied but the supply capacity for Source 1 is exhausted, move to cell (2,1)

ƒ If the demand requirement for Destination 1 is satisfied and the supply capacity for Source 1 is also exhausted, move on to cell (2,2).

3. Continue the allocation in the same manner toward the southeast corner of the transportation tableau until the supply capacities of all sources are exhausted and the demands of all destinations are satisfied.

Initial tableau developed using Northwest Corner Method

Total Cost = 12(400)+13(100)+4(700)+9(100)+12(200)+4(500)= 142,000

Minimum Cell-Cost Method

Although the North-west Corner Rule is the easiest, it is not the most attractive because our objective is not included in the process.

Steps of Minimum Cell-Cost Method

1. Select the cell with the minimum cell cost in the tableau and allocate as much to this cell as possible, but within the supply and demand constraints.

2. Select the cell with the next minimum cell-cost and allocate as much to this cell as possible within the demand and supply constraints.

3. Continue the procedure until all of the supply and demand requirements are satisfied. In a case of tied minimum cell-costs between two or more cells, the tie can be broken by selecting the cell that can accommodate the greater quantity.

Initial tableau developed using Minimum Cell-Cost Method

Total Cost = 12(300)+4(200)+4(700)+10(100)+9(200)+4(500)= 120,000

MODI Method (for obtaining reduced costs)

Associate a number, ui, with each row and vj with each column.

• Step 1: Set u1 = 0.

• Step 2: Calculate the remaining ui’s and vj’s by solving the relationship cij

= ui + vj for occupied cells.

• Step 3: For unoccupied cells (i,j), the reduced cost = cij – ui – vj.

Step 1: For each unoccupied cell, calculate the reduced cost by the MODI method. Select the unoccupied cell with the most negative reduced cost. (For maximization problems select the unoccupied cell with the largest reduced cost.) If none, STOP.

Step 2: For this unoccupied cell, generate a stepping stone path by forming a closed loop with this cell and occupied cells by drawing connecting alternating horizontal and vertical lines between them. Determine the minimum allocation where a subtraction is to be made along this path.

Step 3: Add this allocation to all cells where additions are to be made, and subtract this allocation to all cells where subtractions are to be made along the stepping stone path. (Note: An occupied cell on the stepping stone path now

becomes 0 (unoccupied). If more than one cell becomes 0, make only one unoccupied; make the others occupied with 0′s.)

GO TO STEP 1.

Example: Acme Block Co. (ABC)

Acme Block Company has orders for 80 tons of concrete blocks at three suburban locations as follows: Northwood — 25 tons, Westwood — 45 tons, and Eastwood — 10 tons. Acme has two plants, each of which can produce 50 tons per week. Delivery cost per ton from each plant to each suburban location is shown below.

How should end of week shipments be made to fill the above orders?

Since total supply = 100 and total demand = 80, a dummy destination is created with demand of 20 and 0 unit costs.

Iteration 1: Tie for least cost (0), arbitrarily select x14. Allocate 20. Reduce s1 by 20 to 30 and delete the Dummy column.

Iteration 2: Of the remaining cells the least cost is 24 for x11. Allocate 25. Reduce s1 by 25 to 5 and eliminate the Northwood column.

Iteration 3: Of the remaining cells the least cost is 30 for x12. Allocate 5. Reduce the Westwood column to 40 and eliminate the Plant 1 row. Iteration 4: Since there is only one row with two cells left, make the final allocations of 40 and 10 to x22 and x23, respectively.

1. Set u1 = 0

2. Since u1 + vj = c1j for occupied cells in row 1, then v1 = 24, v2 = 30, v4 = 0.

3. Since ui + v2 = ci2 for occupied cells in column 2, then u2 + 30 = 40, hence u2 = 10.

4. Since u2 + vj = c2j for occupied cells in row 2, then 10 + v3 = 42, hence v3 = 32.

Calculate the reduced costs (circled numbers on the previous slide) by cij – ui – vj.

Unoccupied Cell Reduced Cost (1,3) 40 – 0 – 32 = 8 (2,1) 30 – 24 -10 = -4 (2,4) 0 – 10 – 0 = -10

Iteration 1:

The stepping stone path for cell (2,4) is (2,4), (1,4), (1,2), (2,2). The allocations in the subtraction cells are 20 and 40, respectively. The minimum is 20, and hence reallocate 20 along this path. Thus for the next tableau:

x24 = 0 + 20 = 20 (0 is its current allocation) x14 = 20 – 20 = 0 (blank for the next tableau) x12 = 5 + 20 = 25

x22 = 40 – 20 = 20

The other occupied cells remain the same.

1. Set u1 = 0.

2. Since u1 + vj = cij for occupied cells in row 1, then v1 = 24, v2 = 30.

3. Since ui + v2 = ci2 for occupied cells in column 2, then u2 + 30 = 40, or u2 =

10.

4. Since u2 + vj = c2j for occupied cells in row 2, then 10 + v3 = 42 or v3 = 32;

and, 10 + v4 = 0 or v4 = -10.

Iteration 2

Calculate the reduced costs (circled numbers on the previous slide) by cij – ui – vj.

Unoccupied Cell Reduced Cost (1,3) 40 – 0 – 32 = 8 (1,4) 0 – 0 – (-10) = 10 (2,1) 30 – 10 – 24 = -4

The most negative reduced cost is = -4 determined by x21. The stepping stone path for this cell is (2,1),(1,1),(1,2),(2,2). The allocations in the subtraction cells are 25 and 20 respectively. Thus the new solution is obtained by reallocating 20 on the stepping stone path. Thus for the next tableau:

x21 = 0 + 20 = 20 (0 is its current allocation)

x11 = 25 – 20 = 5

x12 = 25 + 20 = 45

x22 = 20 – 20 = 0 (blank for the next tableau)

The other occupied cells remain the same.

1. Set u1 = 0

2. Since u1 + vj = c1j for occupied cells in row 1, then v1 = 24 and v2 = 30.

3. Since ui + v1 = ci1 for occupied cells in column 2, then u2 + 24 = 30 or u2 =6.

4. Since u2 + vj = c2j for occupied cells in row 2, then 6 + v3 = 42 or v3 = 36, and 6 + v4 = 0 or v4 = -6.

Iteration 3

Calculate the reduced costs (circled numbers on the previous slide) by cij – ui – vj.

Unoccupied Cell Reduced Cost

(1,3) 40 – 0 – 36 = 4(1,4) 0 – 0 – (-6) = 6(2,2) 40 – 6 – 30 = 4

Since all the reduced costs are non-negative, this is the optimal tableau.

Definition :   Suppose that one is given a linear function of n real variables

z = f (x1, x2,…, xn) = c1x1 + c2x2 +…+ cnxn

and a set of linear inequalities and/or equations, called constraints

(1.1)

where in each line either , = or occurs. The problem of finding  (x1, x2,…, xn), that satisfies the constraints (1.1) and makes z a maximum (or minimum) is called a Linear Programming Problem .

We shall assume that every Linear Programming Problem has included in its constraints, the non-negativity restrictions

xj 0    (1.2)

and these will be written separately from the other constraints. These non-negativity constraints are sometimes known as reality constraints. In examples they typically represent quantities that for physical reasons are non-negative.

Any set of values of x1, x2,…, xn satisfying the constraints (1.1) and inequalities (1.2) is called a feasible solution. The set of feasible solutions is the feasible region. Somewhat perversely, any set of values of x1, x2,…, xn satisfying the constraints (1.1) but not (1.2) is called a non-feasible solution.

The function f is called the objective function and z the objective variable. If set of values of x1, x2,…, xn  is a feasible solution that makes f (x1,…, xn) a maximum (or minimum) then x is an optimal solution and the corresponding value of z is the optimal value.

Mathematical Formulation of Linear Programming Problems

There are mainly four steps in the mathematical formulation of linear programming problem as a mathematical model. We will discuss formulation of those problems which involve only two variables.

(1) Identify the decision variables and assign symbols x and y to them. These decision variables are those quantities whose values we wish to determine.

(2) Identify the set of constraints and express them as linear equations/inequations in terms of the decision variables. These constraints are the given conditions.

(3) Identify the objective function and express it as a linear function of decision variables. It might take the form of maximizing profit or production or minimizing cost.

(4) Add the non-negativity restrictions on the decision variables, as in the physical problems, negative values of decision variables have no valid interpretation.

There are many real life situations where an LPP may be formulated. The following examples will help to explain the mathematical formulation of an LPP.

01. A diet is to contain at least 4000 units of carbohydrates, 500 units of fat and 300 units of protein. Two foods A and B are available. Food A costs 2 dollars per unit and food B costs 4 dollars per unit. A unit of food A contains 10 units of carbohydrates, 20 units of fat and 15 units of protein. A unit of food B contains 25 units of carbohydrates, 10 units of fat and 20 units of protein. Formulate the problem as an LPP so as to find the minimum cost for a diet that consists of a mixture of these two foods and also meets the minimum requirements.

Suggested answer:

The above information can be represented as

Let the diet contain x units of A and y units of B.

    Total cost = 2x + 4y

The LPP formulated for the given diet problem is

Minimize Z = 2x + 4y

subject to the constraints

02. In the production of 2 types of toys, a factory uses 3 machines A, B and C. The time required to produce the first type of toy is 6 hours, 8 hours and 12 hours in machines A, B and C respectively. The time required to make the second type of toy is 8 hours, 4 hours and 4 hours in machines A, B and C respectively. The maximum available time (in hours) for the machines A, B, C are 380, 300 and 404 respectively. The profit on the first type of toy is 5 dollars while that on the second type of toy is 3 dollars. Find the number of toys of each type that should be produced to get maximum profit.

Suggested answer:

Mathematical Formulation

The data given in the problem can be represented in a table as follows.

Let x = number of toys of type-I to be produced

y = number of toys of the type – II to be produced

     Total profit = 5x + 3y

The LPP formulated for the given problem is:

Maximise Z = 5x + 3y subject to the constraints

Graphical Solution of Linear Programming Problem :

Let us take the following example :

minimise     Z  =  180x + 160y subject to       6x + y >= 12       3x + y >= 8       4x + 6y >= 24       x <= 5       y <= 5       x,y >= 0

Since there are only two variables in this LP problem we have the graphical representation of the LP given below with the feasible region (region of feasible solutions to the constraints associated with the LP) outlined.

To draw the diagram above we turn all inequality constraints into equalities and draw the corresponding lines on the graph (e.g. the constraint 6x + y >= 12 becomes the line 6x + y = 12 on the graph). Once a line has been drawn then it is a simple matter to work out which side of the line corresponds to all feasible solutions to the original inequality constraint (e.g. all feasible solutions to 6x + y >= 12 lie to the right of the line 6x + y = 12).

We determine the optimal solution to the LP by plotting (180x + 160y) = K (K constant) for varying K values (iso-profit lines). One such line (180x + 160y = 180) is shown dotted on the diagram. The smallest value of K (remember we are considering a minimisation problem) such that 180x + 160y = K goes through a point in the feasible region is the value of the optimal solution to the LP (and the corresponding point gives the optimal values of the variables).

Hence we can see that the optimal solution to the LP occurs at the vertex of the feasible region formed by the intersection of 3x + y = 8 and 4x + 6y = 24. Note here that it is inaccurate to attempt to read the values of x and y off the graph and instead we solve the simultaneous equations

3x + y = 8 4x + 6y = 24

to get x = 12/7 = 1.71 and y = 20/7 = 2.86 and hence the value of the objective function is given by 180x + 160y = 180(12/7) + 160(20/7) = 765.71

Hence the optimal solution has cost 765.71

It is clear that the above graphical approach to solving LP’s can be used for LP’s with two variables but most LP’s have more than two variables. This brings us to the simplex algorithm for solving LP’s.

Importance of study of Micro Economics in modern society:

It is very important to read micro economics in modern life. Because today’s main problem is found in economics. So, society alliances people faces problem economic problem, for that it has importance a lot. Below discussing the importance of study of micro economics in modern society.

1. In daily life of human being: Various economic problems have an influence in daily life of human being. It means in daily life men face limited wealth and unlimited scarcity. From the knowledge of economics as well as micro economics men find the solution of these problems.

2. Proper use of wealth: For moving the scarcity of human being how to improve the wealth and how to improve the use of wealth know from the micro economics.

3. Expand the scope of intellectuality: Reading micro economics the thinking and intellectuality level of human being can be expanded.

4. To the businessmen and industrialists: By getting the forecasting about future, businessmen and industrialists always have to produce goods. Micro economics provides those forecasting for produce goods.

5. To the social workers: To give solution about the problem of unemployment, poverty etc. social problems sufficient knowledge about micro economics needed. So, it is important for social worker to study micro economics.

6. To the leaders of laborers: To consist trade union and to be aware about the right of laborers, leaders of laborers should have the knowledge about micro economics. Because there is relation between labor work right and micro economics.

7. To the politicians: To do politics politician must have the knowledge about the micro economics. Without the knowledge of micro economics politicians can not get prosperity in politics. Because they should have the knowledge about general market, money market, principles of revenue etc.

8. Government administration: In government administration the employees should have the knowledge about micro economics. The knowledge about demand aid supply of goods and raw materials, money market, real market etc. is needed for being an employee of government administration.

9. In case of saving tendency: Micro economics does not teach for being frugal, it teaches to be a good saver of money as well as wealth.

10. For implementing plan: For nation organizational and individual life, plan is very important. How much a product will produce, how it will be produced, how it will be distributed etc. keeping in mind, plan is implemented. This knowledge can be gained only by micro economics.

So, for implementing good and effective plan study of micro economics is needed. So, from above discussion we are now able to understand that how much micro economics is needed for modern society.