For finding an optimum solution in transportation problem _____________

The use of decision models.

A. Is possible when the variables value is known

B. Reduces the scope of judgement & intuition known with certainty in decision-making

C. Require the use of computer software

D. None of the above

51. In the basic EOQ model, if the lead time increases from 2 to 4 days, the EOQ will ______________

double increase
remain constant
but not double
decrease by a factor of two

Correct answer: (B)
remain constant

52. When the sum of gains of one player is equal to the sum of losses to another player in a game, this situation is known as ______________.

two-person game
two-person zero-sum game
zero-sum game
non-zero-sum game

Correct answer: (C)
zero-sum game

53. In the network, one activity may connect any ______________ nodes

Correct answer: (B)
2

54. Graphical method is also known as ______________.

Simplex Method
Dual Simplex Method
Big-M Method
Search-Approach Method

Correct answer: (D)
Search-Approach Method

55. If the given Linear Programming Problem is in its standard form then primal-dual pair is ______________.

symmetric
un symmetric
square
triangle

Correct answer: (B)
un symmetric

56. The method used to solve Linear Programming Problem without use of the artificial variable is called ______________.

Simplex Method
Big-M Method
Dual Simplex Method
Graphical Mehtod

Correct answer: (C)
Dual Simplex Method

57. When the total demand is equal to supply then the transportation problem is said to be ______________

balanced
unbalanced
maximization
minimization

Correct answer: (A)
balanced

58. For finding an optimum solution in transportation problem ______________ method is used.

Simplex
Big-M
Modi
Hungarian

Correct answer: (C)
Modi

59. Linear Programming Problem is a technique of finding the ______________.

optimal value
approximate value
initial value
infeasible value

Correct answer: (A)
optimal value

60. Any solution to a Linear Programming Problem which also satisfies the non- negative notifications of the problem has ______________.

solution
basic solution
basic feasible solution
feasible solution

Correct answer: (D)
feasible solution

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Posted Date:-2021-12-03 17:49:00

A activity in a network diagram is said to be ______________ if the delay in its start will further delay the project completion time.
A feasible solution of an Linear Programming Problem that optimizes the objective function is called ______
A game is said to be strictly determinable if__________
A Linear Programming Problem have ______________ optimal solution
All of the following may be used to find the EOQ except ________
An Iso-profit line represents _______
An n-tuple of real numbers which satisfies the constraints of Linear Programming Problem is called ________
An _____ represent the start or completion of some activity and as such it consumes no time
Any solution to a Linear Programming Problem which also satisfies the non- negative notifications of the problem has _______
At any iteration of the usual simplex method, if there is at least one basic variable in the basis at zero level and all the index numbers are non-negative, the current solution is _______
Charnes method of penalty is called ________
Chose the correct statement: A degenerate solution is one that______
Customers arrive at a box office window, being manned ny single individual, according to Poisson input process with mean rate of 20 per hour, while the mean service time is 2 minutes. Which of the following is not true for this system?
Economic order quantity results in ________
Float analysis is useful for ________
For a salesman who has to visit n cities, following are the ways of his tour plan ______
For finding an optimum solution in transportation problem ______________ method is used.
Graphical method is also known as ______
If an artificial variable is present in the basic variable column of optimal simplex table, then the problem has ______ solution.
If one or more variable vanish then a basic solution to the system is called ____
If the given Linear Programming Problem is in its canonical form then primal-dual pair is _______
If the given Linear Programming Problem is in its standard form then primal-dual pair is ______
In a network diagram an event is denoted by the symbol _________
In a transportation table, an ordered set of______________ or more cells is said to form a loop
In Program Evaluation Review Technique the maximum time that is required to perform the activity under extremely bad conditions is known as _______
In the basic EOQ model, if the lead time increases from 2 to 4 days, the EOQ will ________
In the network, one activity may connect any ______________ nodes
Linear Programming Problem is a technique of finding the _____
Linear Programming Problem that can be solved by graphical method has ______
Operation research approach is typically based on the use of ________
The assignment problem is a special case of transportation problem in which ______
The assignment problem will have alternate solutions when the total opportunity cost matrix has _____
The average arrival rate in a single server queuing system is 10 customers per hour and average service rate is 15 customers per hour. The average time that a customer must wait before it is taken up for service shall be ______________minutes.
The cost of a slack variable is ________
The difference between free float and tail event slack is ________
The dummy source or destination in a transportation problem is added to ______
The initial event which has all outgoing arrows with no incoming arrow is numbered _______
The irreducible minimum duration of the project is called ______
The method used to solve Linear Programming Problem without use of the artificial variable is called ______
The model in which only arrivals are counted and no departure takes place are called ___________
The problem of replacement is felt when job performing units fail _______
The region common to all the constraints including the non-negativity restrictions is called the _______
The server utilization factor is also known as ________
The time between the placement of an order and its delivery is called as ______
When the sum of gains of one player is equal to the sum of losses to another player in a game, this situation is known as ____
When the total demand is equal to supply then the transportation problem is said to be ______
Which of the following methods is used to verify the optimality of the current solution of the transportation problem______
________ is used for non-repetitive jobs
________ of a queuing system is the state where the probability of the number of customers in the system depends upon time
_________ method is an alternative method of solving a Linear Programming Problem involving artificial variables

There are two phases to solve the transportation problem. In the first phase, the initial basic feasible solution has to be found and the second phase involves optimization of the initial basic feasible solution that was obtained in the first phase. There are three methods for finding an initial basic feasible solution,

NorthWest Corner Method
Least Cost Cell Method
Vogel’s Approximation Method

This article will discuss how to optimize the initial basic feasible solution through an explained example. Consider the below transportation problem.

ong>Solution:

Step 1: Check whether the problem is balanced or not.
If the total sum of all the supply from sources O1, O2, and O3 is equal to the total sum of all the demands for destinations D1, D2, D3 and D4 then the transportation problem is a balanced transportation problem.

Note: If the problem is not unbalanced then the concept of a dummy row or a dummy column to transform the unbalanced problem to balanced can be followed as discussed in this article.

Step 2: Finding the initial basic feasible solution.
Any of the three aforementioned methods can be used to find the initial basic feasible solution. Here, NorthWest Corner Method will be used. And according to the NorthWest Corner Method this is the final initial basic feasible solution:

Now, the total cost of transportation will be (200 * 3) + (50 * 1) + (250 * 6) + (100 * 5) + (250 * 3) + (150 * 2) = 3700.

Step 3: U-V method to optimize the initial basic feasible solution.
The following is the initial basic feasible solution:

– For U-V method the values ui and vj have to be found for the rows and the columns respectively. As there are three rows so three ui values have to be found i.e. u1 for the first row, u2 for the second row and u3 for the third row.
Similarly, for four columns four vj values have to be found i.e. v1, v2, v3 and v4. Check the image below:

There is a separate formula to find ui and vj,
ui + vj = Cij where Cij is the cost value only for the allocated cell. Read more about it here.

Before applying the above formula we need to check whether m + n – 1 is equal to the total number of allocated cells or not where m is the total number of rows and n is the total number of columns.
In this case m = 3, n = 4 and total number of allocated cells is 6 so m + n – 1 = 6. The case when m + n – 1 is not equal to the total number of allocated cells will be discussed in the later posts.

Now to find the value for u and v we assign any of the three u or any of the four v as 0. Let we assign u1 = 0 in this case. Then using the above formula we will get v1 = 3 as u1 + v1 = 3 (i.e. C11) and v2 = 1 as u1 + v2 = 1 (i.e. C12). Similarly, we have got the value for v2 = 1 so we get the value for u2 = 5 which implies v3 = 0. From the value of v3 = 0 we get u3 = 3 which implies v4 = -1. See the image below:

Now, compute penalties using the formula Pij = ui + vj – Cij only for unallocated cells. We have two unallocated cells in the first row, two in the second row and two in the third row. Lets compute 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 = -5
For C21, P21 = 5 + 3 – 2 = 6
For C24, P24 = 5 + (-1) – 9 = -5
For C31, P31 = 3 + 3 – 8 = -2
For C32, P32 = 3 + 1 – 3 = 1

The Rule: If we get all the penalties value as zero or negative values that mean the optimality is reached and this answer is the final answer. But if we get any positive value means we need to proceed with the sum in the next step.

Now find the maximum positive penalty. Here the maximum value is 6 which corresponds to C21 cell. Now this cell is new basic cell. This cell will also be included in the solution.

The rule for drawing closed-path or loop. Starting from the new basic cell draw a closed-path in such a way that the right angle turn is done only at the allocated cell or at the new basic cell. See the below images:

Assign alternate plus-minus sign to all the cells with right angle turn (or the corner) in the loop with plus sign assigned at the new basic cell.

Consider the cells with a negative sign. Compare the allocated value (i.e. 200 and 250 in this case) and select the minimum (i.e. select 200 in this case). Now subtract 200 from the cells with a minus sign and add 200 to the cells with a plus sign. And draw a new iteration. The work of the loop is over and the new solution looks as shown below.

Check the total number of allocated cells is equal to (m + n – 1). Again find u values and v values using the formula ui + vj = Cij where Cij is the cost value only for allocated cell. Assign u1 = 0 then we get v2 = 1. Similarly, we will get following values for ui and vj.

Find the penalties for all the unallocated cells using the formula Pij = ui + vj – Cij.

For C11, P11 = 0 + (-3) – 3 = -6
For C13, P13 = 0 + 0 – 7 = -7
For C14, P14 = 0 + (-1) – 4 = -5
For C24, P24 = 5 + (-1) – 9 = -5
For C31, P31 = 0 + (-3) – 8 = -11
For C32, P32 = 3 + 1 – 3 = 1

There is one positive value i.e. 1 for C32. Now this cell becomes new basic cell.

Now draw a loop starting from the new basic cell. Assign alternate plus and minus sign with new basic cell assigned as a plus sign.

Select the minimum value from allocated values to the cell with a minus sign. Subtract this value from the cell with a minus sign and add to the cell with a plus sign. Now the solution looks as shown in the image below:

Check if the total number of allocated cells is equal to (m + n – 1). Find u and v values as above.

Now again find the penalties for the unallocated cells as above.