Rowntown Cab Company has 70 drivers that it must schedule in three 8-hour shifts

Question

Rowntown Cab Company has 70 drivers that it must schedule in three 8-hour shifts. However, the demand for cabs in the metropolitan area varies dramatically according to time of the day. The slowest period is between midnight and 4:00 A.M. the dispatcher receives few calls, and the calls that are received have the smallest fares of the day. Very few people are going to the airport at that time of the night or taking other long distance trips. It is estimated that a driver will average $80 in fares during that period. The largest fares result from the airport runs in the morning. Thus, the drivers who sart their shift during the period from 4:00 A.M. to 8:00 A.M. average $500 in total fares, and drivers who start at 8:00 A.M. average $420. Drivers who start at noon average $300, and drivers who start at 4:00 P.M. average $270. Drivers who start at the beginning of the 8:00 P.M. to midnight period earn an average of $210 in fares during their 8-hour shift.
To retain customers and acquire new ones, Rowntown must maintain a high customer service level. To do so, it has determined the minimum number of drivers it needs working during every 4-hour time segment- 10 from midnight to 4:00 A.M. 12 from 4:00 to 8:00 A.M. 20 from 8:00 A.M. to noon, 25 from noon to 4:00 P.M., 32 from 4:00 to 8:00 P.M., and 18 from 8:00 P.M. to midnight.
a. Formulate and solve an integer programming model to help Rowntown Cab schedule its drivers. b. If Rowntown has a maximum of only 15 drivers who will work the late shift from midnight to 8:00 A.M., reformulate the model to reflect this complication and solve it c. All the drivers like to work the day shift from 8:00 A.M. to 4:00 P.M., so the company has decided to limit the number of drivers who work this 8-hour shift to 20. Reformulate the model in (b) to reflect this restriction and solve it

Explanation / Answer

There are 70 drivers to be engaged in three 8hours shifts but constraints about minimum number of drivers are for every 4-hour time segments and average earning is also about 4-hour time periods. The given data is represented in the following table:

Let X, Y and Z be the number of drivers allocated to 8-hours shifts numbered as 1, 2 and 3 respectively. In other words X from 8A.M to 4P.M, Y from 4P.M. to midnight and from midnight to 8A.M.

Objective function is to have maximum earnings so it is formulated as under:

Maximize X(420+300) + Y(270+210) + Z(80+500)

Now the constraints about minimum numbers of drivers for 4-hours time segments

X >= 20 and X>= 25 ; Y >=32 and Y >= 18 ; Z >=10 and Z >= 12

In other words X >= 25 ; Y >= 32 and Z >= 12

X, Y, and Z are number of drivers therfore these are positive integers

Finally constraint about total number of drivers

X + Y + Z = 70

b part states that Rowntown has maximum 15 drivers who will work from midnight to 8A.M. so additional constraint will be: Z <= 15

c part states- All drivers wants to work in the 1st shift from 8A.M, to 4P.M. so maximum limit for this shift is 20

and therefore additional constraint is X <= 20 but then minimum requirement of 25 between noon and 4P.M. needs to be removed

Now we can also think about about flexible timimgs for the 8hours shifts covering the given six 4hours time segments. In all there could be six shifts strating after every four hours.

Let x1 be the number of drivers starting at 4AM, x2 starting at 8AM and x3 starting at noon, x4 starting at 4PM, x5 starting at 8PM and x6 starting at midnight. The formulation will be as follows:

Maximize total earnings = x1(500+420) + x2(420+300) + x3(300+270) + x4(270+210) + x5(210+80) +x6(80+500)

Subject to x6+x1>=12, x1+x2>=20 , x2+x3>=25 , x3+x4>=32, x4+x5>=18 and x5+x6>=10

x1+x2+x3+x4+x5+x6 = 70 and x1,x2,x3,x4,x5 and x6 are positive integers

Solution to the first case of fixed three shifts, drivers starting only at 8AM, 4PM and midnight, requires minimum 25 in shift1, 32 in shift2 and 12 in shift3 that totals 69 less than 70. As the maximum earning is in the first shift so final solution is 26 drivers in shift1, 32 in shift2 and 12 in shift3 and total earning is

26*720 + 32*480 + 12*580 =18720+15360+6960 = 41040

Shift no. 4hour segment Minimum no. Average earning Maximum no. 1 8A.M - Noon 20 420 20 1 Noon-4P.M 25 300 20 2 4P.M. -8P.M. 32 270 2 8P.M. -Midnight 18 210 3 Midnight-4A.M. 10 80 15 3 4A.M.-8A.M. 12 500 15

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