Please help me with linear programming question pls! no graph needed.?
Your company has launched the latest version of its web browser, “Visual Cruise.” As sales manager, you are planning to promote Visual Cruise 4.0 by sending sales forces to software conventions running concurrently in Saint Louis and Detroit. You have 6 representatives available at each of your Little Rock, Ark. and Urbana, Ill. branches, and you would like to send at least 5 to the Saint Louis convention and at least 4 to the Detroit convention. The Saint Louis convention will last for three days, while the Detroit convention will last for two days. Air fares (per person) and hotel accommodation costs (per person) are shown below.
Little Rock to DT Airfare is $ 200, Little Rock to St Louis is $400
Urbana to DT Airfare is $ 200, Urbana to St Louis is $ 100
Hotel Accomodation: St Louis Motel is $60/night/person
DT Grand Hotel is $100/night/person
How many representatives should you send from each branch to each convention in order to minimize the total (air travel and accommodation) cost? What will the total cost amount to?
I’m not entirely sure that you need to use linear programming for this question, based on the information you have provided. It seems fairly straightforward that as you have enough representatives available in each branch to send to the conventions, you should send them to the geographically nearest locations, which is cheaper to fly to. The staff from the Little Rock Branch should fly to Detroit and the staff from the Urbana branch should fly to St Louis. Linear programming normally focuses on making maximum use (or minimum use) of a number of limited resources, which I don’t think is really necessary in this case.
The only constraint that I can see is that you are looking for the lowest possible cost. If you wanted to save money, you send fewer representatives. There is no limiting factor in terms of the resources available in staff, time, or any other variable. Based on the information you have provided, it would be easier to just do the maths.
You would probably want to measure the relative benefits of sending representatives and you could use this as the basis for an analysis. The result would probably be that sending too many representatives would provide little additional benefit, and you normally find that these types of stand are limited to two or three people. If you want to send four representatives, the cheapest option it to send four people and to choose the cheapest location to fly from. To decide if you were going to send more, you would need to work out the value of their attendance at the event, and the value of their disengagement from other duties.
In general, I would say that if the benefit outweighed to cost, it would be worth sending them. Advertising, particularly in relation to a new product, and more so in the case of word of mouth advertising, is invaluable. The benefit of having extra staff at the event may be more difficult to quantify, and there is always a point at which the cost of advertising will outweigh the benefit you can derive from it. If you are going to use linear programming, you must thus give careful consideration to the variables that you use and the values that you attach to them.
I have included an example below, but you could amend the variable to include a projection of the profit that could be derived from the attendance of a representative at each location, machine hours would be the amount of labour hours that you could afford to devote to the activity, and raw materials per unit would be the cost of sending a representative to each location. You could also introduce a constraint on the cost of flights and accommodation that you would be willing to pay. This would give you an understanding of the maximisation of each resource available to you whilst aiming towards the highest possible level of promotion (and hopefully sales!) from the endeavour.
The procedure for linear programming is as follows,
define variables
ie selling price, maximum demand, machine hours
establish constraints (inequalities)
ie
3x + 2y _< 500 machine hours
x+y _< 200 travelling costs
2x + 3y _< raw materials per unit
100x + 200y = Profit per sales person per location
produce a graphical representation and define area of feasibility. the optimal points lie at the points on the far right of the area of feasibility above the profit line(assuming that you are looking for maximisation – if you are looking for minimisation, it goes the other way round). there are likely to be a number of different points in this area which would offer a suitable return whilst ensuring that all resources are maximised.
this can be solved using simultaneous equations as follows,
4x + 3y = 700
4x + 4y =800
y = 100
x = 100
this would indicate that you should sell 100 units of y and 100 units of x, giving you a total profits of £30,000 whilst making the best use out of the resources available to you, and by developing a strong customer base through maximising the resource of demand.
Let x and v be the number of representatives from Little Rock and Urbana respectively to be sent to the Saint Louis Convention. Similarily, let y and z be the number of representatives from Little Rock and Urbana respectively to be sent to Detroit Convention.
Then, the cost function here is the minimum total(air travel and accommodation) cost:
C = air travel cost+accommodation cost
C = [400x+200y+100v+200z]+
[(x+v)*60*3+(y+z)*100*2]
C = 400x+200y+100v+200z+(x+y)*180+(y+z)*200
C = 400x+200y+100v+200z+180x+180y+200y+200z
C = 580x+580y+100v+400z
subject to the following constriants:
Maximum number of representatives avaliable is 6 at the Little Rock branch, then we have the constriant
x+y< =6
Maximum number of representatives avaliable is 6 at the Urbana branch, then we have the constriant
v+z<=6
At least five of the representatives from each of the Little Rock and Urbana braches must be sent to Saint Louis Convention, then we have the constriant
x+v>=5
At least four of the representatives from each of the Little Rock and Urbana braches must be sent to Detroit Convention, then we have the constriant
y+z>=4
Hope this helps.