The New Kilkenny Highway Commission is installing a camera-based highway monitor

Question

The New Kilkenny Highway Commission is installing a camera-based highway monitoring system to track the license plate numbers of cars entering and leaving the highway. They are examining two different camera systems. System A can track three cars per second, takes up 1 foot of space on the overhead beams, and costs $4,000. System B can track 13 cars per second, takes up 5 feet of space on the overhead beams, and costs $10,000. The highway commission cannot spend more than $80,000 for the system, and has 25 feet of overhead beam space available. Formulate a linear programming model to maximize the number of cars the system can track. List the extreme points and determine the solution graphically. You do not need to submit your graph.

Explanation / Answer

let the number of camera systems A that will be used be a and let the number of camera systems of B that will be used be b. Thus a, b are the decision variables.

Thus the number of cars that can be tracked per second by A will be 3a and the number of cars that can be tracked per second by B will be 13b. Total = 3a+13b. This is our objective function and has to be maximized.

Constraints are :(i) 4000a+10000b<=80,000 (the highway commission cannot spend more than $80,000 for the system)

(ii) a+5b<=25 (The highway commission has 25 feet of overhead beam space available)

(iii) a,b<=0 (it cannot be negative).

Now to plot the graph, we take the constraints and line is drawn by considering the constraint as equality

(i) 4000a+10000b=80,000. putting b as 0, 4,000a = 80,000 or a = 20. putting a as 0, 10,000b = 80,000 or b = 8. Line will be a = 20, b = 8

(ii) a+5b = 25. putting b as 0, a = 25 and putting a as 0, 5b = 25 or b = 5. line will be a = 25, b = 5

Feasible region is obtained and it is inside part of the intersection point of the two lines.

Extreme points are coordinates (a,b) = (i) 0,0 (ii) 0,5 (iii) 20,0 and (iv) 15, 2

Objective value at (i) 0,0 are 3*0+13*0 = 0

at (ii) 0,5 = 3*0+5*13 = 65

at (iii) 20,0 = 3*20+13*0 = 60

at (iv) 15,2 = 3*15+13*2 = 45+26 = 71.

Thus the maximum number of cars per second is 71 and number of systems A = 15 and systems B = 2

All constraints are satisfied.

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