Currently, State University can store 200 files on hard disk, 100 files in computer memory, and 300...

Question

Currently, State University can store 200 files on hard disk, 100 files in computer memory, and 300 files on tape. Users want to store 300 word-processing files, 100 packaged-program files, and 100 data files. Each month a typical word-processing file is accessed eight times; a typical packaged-program file, four times; and a typical data file, two times. When a file is accessed, the time it takes for the file to be retrieved depends on the type of file and on the storage medium (see Table).

Storage Medium		Time (Minutes)
Word Processing	Packaged Program	Data
Hard disk	5	4	4
Memory	2	1	1
Tape	10	8	6

a If the goal is to minimize the total time per month that users spend accessing their files, formulate a balanced transportation problem that can be used to determine where files should be stored.

b Use the minimum cost method to find a bfs.

c Use the transportation simplex to find an optimal solution.

David · Accepted Answer

Solution:  
A. Formulation : Given 
Storage Medium 
  Time (Minutes)   
Word Processing Packaged Program Data 
Hard disk 5 4 4 
Memory 2 1 1 
Tape 10 8 6
Tabulating the given data we get:
Word Proc. Pckgd. Prog. Data
5 
 
4 
 
4 
 
2 
 
1 
 
1 
 
10 
 
8 
 
6 

300 100 100
The total capacity is 600 files and total files to be stored are only 500. Hence we need to add 
a dummy column of a requirement of 100 files.
B. Basic Feasible Solution
I. Thus the transportation initial cost matrix is
5 
 
4 
 
4 
 
0 
 
2 
 
1 
 
1 
 
0 
 
10 
 
8 
 
6 
 
0 

200
100
300 
   300  100  100             100
Hard Disk 
Memory 
Tape 
200 
100 
300 
Capacity 
Files 
II. Now C14, C24, C34 all have the same least cost (time). Arbitrarily C34 is chosen. Assign the lower 
value among the corresponding capacity and files to be stored in the cell and make suitable 
deductions.
5 
 
4 
 
4 
 
0 
 
2 
 
1 
 
1 
 
0 
 
10 
 
8 
 
6 
 
0 
100
III. Repeat the step II until all the capacity over various devices is completely utilized. 
Least Cost Element : C22 or C23 (C22 chosen arbitrarily)
5 
 
4 
 
4 
 
0 
 
2 
 
1 
100 
1 
 
0 
 
10 
 
8 
 
6 
 
0 
100
IV. Least Cost Element : C13 
5 
 
4 
 
4 
100 
0 
 
2 
 
1 
100 
1 
 
0 
 
10 
 
8 
 
6 
 
0 
100
   300  100  100             0
200
100
200 
   300  0  100             0
200
0
200 
   300  0  0             0
100
0
200 
V. Least Cost element : C11 
5 
100 
4 
 
4 
100 
0 
 
2 
 
1 
100 
1 
 
0 
 
10 
200 
8 
 
6 
 
0 
100
The number of cells allocated is 5. This is less than the number of cells according to the formula 
(m+n-1) where m=No. of rows, n-=No. of Columns 
m+n-1 = 3=4-1 = 6. Since the allocations are only 5, the solution is degenerate hence we need to 
make a dummy allocation of small quantity ‘d’ in the lowest unallocated cell i.e in this case, C24. 
Thus the final allocated table is as shown below which gives the basic feasible solution.
5 
100 
4 
 
4 
100 
0 
 
2 
 
1 
100 
1 
 
0 
d 
10 
200 
8 
 
6 
 
0 
100
C. Check for optimality (UV Method)
I. Using ‘UV’ method, the values of u1, u2, u3, v1, v2, v3, v4 are found using the cost entries of 
allocated cells. Formula :

Currently, State University can store 200 files on hard disk, 100 files in computer memory, and 300 files on tape. Users want to store 300 word-processing files, 100 packaged-program files, and 100...

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