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Variance and Standard Daviation Part 5-Basic Mathematics-Assignment Solution, Exercises of Mathematics

This is solution to assignment of Basic Mathematics course. This was submitted to Karunashankar Sidhu at Institute of Mathematical Sciences. It includes: Optimal, Solution, Transportation, Objective, Cost, Variance, Standard, Deviation

Typology: Exercises

2011/2012

Uploaded on 08/03/2012

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1)
How is an optimal solution to a Transportation problem changed if a constant
is added
to each cost in
(i) one row say rth row
(ii) one column say kth column?
By how much is the optimal value of the objective function changed?
Solution:
Consider the standard form of the objective function;
11
mn
ij ij
ij
zcx


Now if we add
to each cost of rth row, the total cost(objective function) will become;

1( ) 1 1
1( ) 1 1 1
11
mn n
new ij ij rj rj
iirj j
mn n n
ij ij rj rj rj
iirj j j
mn
ij ij r
ij
new old r
zcxcx
cx cx x
cx a
zz a










Since
ar is constant, so if some feasible solution of the original TP minimizes zold then
the feasible solution will also minimizes the znew.
On adding
to every cost of rth row of a TP, the optimal solution will not change,
while the total transportation cost will be increased by (
ar).
Similarly, we can show that on adding constant
to every cost of kth column of a TP,
the optimal solution ‘ll remain unaltered, while the total transportation cost will be
increased by
bk.
2)
A company has three plants A,B and C, and three ware houses X, Y and Z. The number
of units available at the plants is 60, 70 and 80 and the demand at X, Y and Z are 50,80
and 80 respectively. The unit cost of the transportation is given as follow:
Find the allocation so that the total transportation cost is minimum.
Solution:
We solve this by Least Cost method.
X Y Z
A 8 7 3
B 3 8 9
C 11 3 5
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How is an optimal solution to a Transportation problem changed if a constant  is added

to each cost in (i) one row say r th row (ii) one column say k th column? By how much is the optimal value of the objective function changed?

Solution: Consider the standard form of the objective function;

1 1

m n ij ij i j

z c x  

Now if we add  to each cost of rth row, the total cost(objective function) will become;

1( ) 1 1

1( ) 1 1 1

1 1

m n n new ij ij rj rj i i r j j m n n n ij ij rj rj rj i i r j j j m n ij ij r i j new old r

z c x c x

c x c x x

c x a

z z a

   

    

 

Since  a r is constant, so if some feasible solution of the original TP minimizes z old then

the feasible solution will also minimizes the z new.

 On adding^ ^ to every cost of rth row of a TP, the optimal solution will not change,

while the total transportation cost will be increased by (  a r).

Similarly, we can show that on adding constant  to every cost of k th column of a TP,

the optimal solution ‘ll remain unaltered, while the total transportation cost will be

increased by  b k.

A company has three plants A,B and C, and three ware houses X, Y and Z. The number of units available at the plants is 60, 70 and 80 and the demand at X, Y and Z are 50, and 80 respectively. The unit cost of the transportation is given as follow:

Find the allocation so that the total transportation cost is minimum. Solution: We solve this by Least Cost method.

X Y Z

A 8 7 3

B 3 8 9

C 11 3 5

Solve the following problem whose objective is to minimize the total cost, when a manufacturing company has to face assigning 6 different machines to 5 different jobs.

Jobs/Machines 1 2 3 4 5 A 2.5 5 1 6 1 B 2 5 1.5 7 3 C^3 6.5^2 8 D 3.5 7 2 9 4. E 4 7 3 9 6 F^6 9 5 10 Solution: Here the jobs are not equal in number. So it is an unbalanced assignment problem. Therefore a dummy job is to be introduced. Further the matrix involves elements in decimal. We can make them complete numbers by multiplying the whole cost matrix by

  1. Thus we get new modified cost matrix as follows: Jobs/Machines 1 2 3 4 5 6 A^5 10 2 12 2 B 4 10 3 14 6 0 C 6 13 4 16 6 0 D 7 14 4 18 9 0 E 8 14 6 18 12 0 F 12 18 10 20 12 0

Subtracting the row minimums and column minimums, we obtain the new cost matrix as follows;