Term Paper on Linear Programming Applications

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  • Topic: Limit of a function, Limit superior and limit inferior, Function
  • Pages : 27 (3960 words )
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  • Published : June 26, 2012
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Term Paper

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Linear Programming Applications

Submitted to : Md. Abul Kashem
Associate Professor
Department of Management Information System
Dhaka University

Submitted by : Md. Forhad hossain
ID NO: 61221-14-029
Semester: 1st , Batch : 14th
Department of management information system
Dhaka University
Date of Submission: 02/05/2012
INDEX

1. Hospital scheduling ………………………………………page-3

2. Marketing Research…………………………………… page-7

3. Blending Problem……………………………………….page-11

4. Production Scheduling………………………………….page 13

5. Product Mix………………………………………….....page 15

6. Agriculture………………………………………………page 18

7.Auditing………………………………………………….page 24

8. Assignment problem…………………………………..page 26

9. Labour Planning……………………………………….page28

10. Media selection problem……………………………..page 31

Hospital scheduling

Modern general hospital is concerned with the staffing of its emergency department. A recent analysis indicated that a typical day may be divided into four periods. In each period, a different level measured “cases ” is experienced.

|Time period |Average number of cases | |7 A.M.- 1 P.M |120 | |1 P.M.- 7 P.M. |90 | |7P.M.-1A.M. |180 | |1 A.M.-7 A.M. |60 |

Past experience indicates that one emergency room nurse can handle 2.5 cases each hour. Thus, for example between 7 a.m.- 1 p.m. there is a need for 8 nurses to handle the 20 cases per hour(120 cases per 6 hours). From 1p.m-7 a.m there is need for 6 nurses to handle 15 cases per hour(90 cases per 6 hours). From 7a.m-p.m., there is a need for 12 nurses to handle 30 cases per hour(180 cases per 6 hours). From 1a.m.-7 a.m, there is need for 4 nurses to handle 10 cases per hour(60 cases per 6 hours).

Emergency room nurses work in shift of 12 hours. The normal shifts are : 7 a.m.-7p.m; 7p.m.-7 a.m.

Find the least-schedule that will assure complete coverage of the emergency room.

Formulation

The problem is to find the minimum number of nurses, assuming that they all receive same wages. Thus, the least cost schedule means the minimum number of nurses.

Decision variables

X1 = the number of nurses starting at 7 a.m.

X2= the number students starting at 1 p.m.

X3= the mumber of students starting at 7 p.m.

X4=the number of nurses starting at 1 a.m.

The objective function

Minimize z = x1+x2+x3+x4

The constraints

To structure the constraints, it is necessary to assume first the demand in each time period is constant regardless of the hour observed. This enable the computation of the number of required nurses.

|Time period |7 A.M.- 1 P.M |1 P.M.- 7 P.M. |7P.M.-1A.M. |1 A.M.-7 A.M. | |Required nurses |8 |6 |12 |4 |

Thus the demand constraints are

1x1+1x4 ≥8

1x1+1x2 ≥6

1x2+1x3 ≥12

1x3+1x4 ≥4

Optimal solution

OPTIMAL SOLUTION

Objective Function Value = 20.000

Variable Value Reduced Costs
-------------- --------------- ------------------ X1 8.000...
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