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Answer:

Sequoia Airlines faces the problem of planning the optimal use of its resources during the next six months. It must hire new employees and develop them through three stages of training, first as trainees, then as junior flight attendants, and finally as full flight attendants. Moreover, Sequoia must meet staff requirements for all flights during the training period. This problem is analyzed using total personnel cost as the selection criterion.

Other resources in addition to the available personnel create constraints that must be considered when hiring and training new employees. We will solve the problem by using linear programming and then follow up with an analysis of the objective function and the constraints and assumptions on which they are based. We will also offer an interpretation of the results.

Because cost is the only criterion we will use to solve this problem, the objective function is straightforward. One assumption is noteworthy: we assume that once an experienced attendant becomes an instructor, he or she continues to receive the instructor's salary even when there is a surplus of instructors who act as attendants. The objective function, therefore, becomes the sum of the number of employees in each class during each period multiplied by the appropriate wage.

Minimize:

We must also deal with several resource constraints. First, we must meet the requirement for hours of in-flight attendant service. We assume that a surplus instructor works the same number of hours as an experienced attendant does.

Next, we must satisfy the constraint that requires the total number of hours for junior attendants to be less than or equal to 25 percent of the total attendant hours in each month.

Also, there must be one instructor for every five trainees.

The remaining constraints deal with the relationships between the status of employees during the current and prior periods. In this case we make assumptions that pertain to the initial staffing levels for the positions of junior attendant, full attendant, and instructor.

We see a total of 42 constraints in the following computer model.

Min 1050J1+ 1050J2+ 1050J3+ 1050J4+ 1050J5+ 1050J6+ 1400F1+ 1400F2+ …Subject to(1) 140J1+ 125F1+ 125S1 = 14000(2) 140J2+ 125F2+ 125S2 = 16000(3) 140J3+ 125F3+ 125S3 = 13000(4) 140J4+ 125F4+ 125S4 = 12000(5) 140J5+ 125F5+ 125S5 = 18000(6) 140J6+ 125F6+ 125S6 = 20000(7) 105Jl-31.25Fl-31.25S1 = 0(8) 105J2-31.25F2-31.25S2 = 0(9) 105J3-31.25F3-31.25S3 = 0(10) 105J4-31.25F4-31.25S4 = 0(11) 105J5-31.25F5-31.25S5 = 0(12) 105J6-31.25F6-31.25S6 = 0(13) 1I1-.2T1 = 0(14) 112-.2T2 = 0(15) 113-.2T3 = 0(16) 114-.2T4 = 0(17) 115-.2T5 = 0(18) 116-.2T6 = 0(19) 1J1 = 8(20) 1F1 = 119(21) 111 = 6(22) -111+ 1Sl+.2T1 = 0(23) -112+ 1S2+.2T2 = 0(24) -113+ 1S3+.2T3 = 0(25) -114+ 1S4+.2T4 = 0(26) -115+ 1S5+.2T5 = 0(27) -116+ 1S6+.iT6 = 0(28) 1J2-.8T1 = 0(29) 1J3-.8T2 = 0(30) 1J4-.8T3 = 0(31) 1J5-.8T4 = 0(32) 1J6-.8T5 = 0(33) -.95J1-.92F1+ 1F2 = 0(34) -.95J2-.92F2+ 1F3 = 0(35) -.95J3-.92F3+ 1F4 = 0(36) -.95J4-.92F4+ 1F5 = 0(37) -.95J5-.92F5+ 1F6 = 0(38) -.005F1-.95I1+ 112 = 0(39) -.005F2-.9512+ 113 = 0(40) -.005F3-.9513+ 114 = 0(41) -.005F4-.9514+ 115 = 0(42) -.005F5-.9515+ 116 = 0

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