module b solutions

MODULE B

END-OF-MODULE PROBLEMS

B.1 Let x = number of standard model to produce y = number of deluxe model to produce Maximize 40x + 60y Subject to 30 30 450

10 15 180

6

0

x y

x y

x

x y

,

B.2

Feasible corner points (x, y): (0, 3), (0, 10), (2.4, 8.8), (6.75, 3). Maximum profit is 100 at (0, 10).

Quantitative Module B: Linear Programming 1

B.3

Feasible corner points (x, y): (0, 2), (0, 10), (4, 8), (10, 2). Maximum profit is 52 at (4, 8).

B.4 (a) Corner points (0, 50), (50, 50), (0, 200), (75, 75), (50, 150). (b) Optimal solutions: (75, 75) and (50, 150). Both yield profit of $3,000.

2 Instructor’s Solutions Manual t/a Operations Management

B.5 (a) Adding a new constraint will reduce the size of the feasible region unless it is a redundant constraint. It can never make the feasible region any larger.

(b) A new constraint can only reduce the size of the feasible region; therefore the value of the objective function will either decrease or remain the same. If the original solution is still feasible, it will remain the optimal solution.

B.6 (a) Let number of liver flavored biscuits in a package number of chicken flavored biscuits in a package

Minimize

(b) Corner points are (0, 40) and (15, 25). Optimal solution is (15, 25) with cost of 65. (c) Minimum cost = 65 cents.

B.7

Let number of air conditioners to be produced number of fans to be produced

Maximize

Profit: @ a: ( , ) Obj @ b: ( , ) Obj @ c: ( , ) Obj *

@ d: ( , ) Obj

The optimal solution is to produce 40 air conditioners and 60 fans each period. Profit will be $1,900.


B.8

Let number of Model A tubs produced number of Model B tubs produced

Maximize

Profit: @ a: ( , ) Obj @ b: ( , ) Obj @ c: ( , ) Obj *

The optimal solution is to produce 200 Model A tubs, and 0 Model B tubs. Profit will be $18,000.

B.9

Let number of benches produced number of tables produced

Maximize


Profit: @ a: ( , ) Obj @ b: ( , ) Obj *

@ c: ( , ) Obj

The optimal solution is to make 262.5 benches and 25 tables per period. Profit will be $2,862.50. Because benches and tables may be matched (two benches per table), it may not be reasonable to maximize profit in this manner. Also, this problem brings up the proper interpretation of the statement that “One should make 262.5 (a fractional quantity) benches per period.”

B.10

Let number of Alpha-4 computers number of Beta-5 computers

Maximize:

Profit: @ a: ( , ) Obj *

@ b: ( , ) Obj

The optimal solution is to produce 10 Alpha-4 and 24 Beta-5 computers per period. Profit is $55,200.


B.11

Note that this problem has one constraint with a negative sign.

The optimal point, a, lies at the intersection of the constraints:

To solve these equations simultaneously, begin by writing them in the form shown below:

Multiply the first equation by 5, the second by –3, and add the two equations and solve for x2: . Given: then

and

Thus, the optimal solution is: , The profit is given by:


B.12

The optimal point, a, lies at the intersection of the constraints:

To solve these equations simultaneously, begin by writing them in the form shown below:

Multiply the second equation by –3, and add it to the first and solve for x2: . Given: then

and

Thus, the optimal solution is: , The cost is given by:

B.13 The fifth constraint is not linear because it contains the square root of x and the objective function and first constraint are not because of the term.

B.14 (a) Using software, we find that the optimal solution is:

(b) There is no unused time available on any of the three machines.


B.15 (a) An additional hour of time on the third machine would be worth $0.26. (b) Additional time on the second machine would be worth $0.786 per hour for a total of $7.86

for the additional 10 hours.

B.16 (a) Let number of students bused from sector i to school j. Objective:

subject to

(b)

Distance “student miles”

B.17 Because the decision centers about the production of the two different cabinet models, let:

number of French Provincial cabinets produced per day number of Danish Modern cabinets produced each day

The equations become: Objective (Maximize revenue)

The solution is:

, , Revenue = $3930/day

B.18 Problem Data SolutionWorkers Hire Hire Hire


Period Time Period Required Solution 1 Solution 2 Solution 31 3 AM–7 AM 3 0 3 32 7 AM–11 AM 12 16 9 143 11 AM–3 PM 16 0 7 24 3 PM–7 PM 9 9 2 75 7 PM–11 PM 11 2 9 46 11 PM–7 AM 4 3 0 0

S = 30 S = 30 S = 30

Let number of workers reporting for the start of work in period i, where , 2, 3, 4, 5, or 6. The equations become: Objective:

(Minimize staff size) Subject to:

Note that three alternate optimal solutions are provided to this problem. Either solution could be implemented using only 30 staff members.

B.19


Define the following variables:

The appropriate equations then become: Objective: (minimize handling and storage costs)

Cost: @ a: ( , ) Obj *

@ b: ( , ) Obj

The optimal solution is to produce 7500 round tables and 5000 square tables, for a cost of $115,000.

B.20 The original equations are: Objective: (maximize)

where: number of coffee tables/week number of bookcases/week

Optimal: , , Profit = $96

B.21

The original equations are: Objective: (maximize)

The optimal solution is found at the intersection of the two constraints. Solving for the values of and at the intersection, we have:


B.23 Let the number of class A containers to be used the number of class K containers to be used the number of class T containers to be used

The appropriate equations are: Maximize:

Using software we find that the optimal solution is:

, ,

or , ,

and

Profit

B.24 (a) The unit profit of the air conditioner must fall in the range $22.50–$30.00 (b) The shadow price for the wiring constraint is $5.00, and it holds within the range 210–280

hours.

B.25 Let number of newspaper ads placed number of TV spots purchased

Given that we are to minimize cost, we may develop the following set of equations: Minimize:

Note that the problem is not limited to unduplicated exposure (for example, one person seeing the Sunday newspaper three weeks in a row counts for three exposures). Solution:

ads, TV spots, cost = $18,500

B.26 (a) Minimize:


(b) Solution:

B.27

Maximize:

@ ( , ) Obj @ ( , ) Obj @ ( , ) Obj *

The optimal solution occurs at boy’s bikes, girl’s bikes, producing a profit of $21,930 However, the possible optimal solutions are so close in profit that sensitivity analysis is important here.

B.28 Let number of medical patients number of surgical patients

The appropriate equations are: Maximize


Optimal: , , Profit = $9,551,659 or: 2790 medical patients, 2104 surgical patients, Profit $9,551,659

Beds required:

Medical uses: beds

Surgical uses: beds

Here is an alternative approach that solves directly for the number of beds:

Maximize revenues

where no. of medical beds = 61.17 no. of surgical beds = 28.83 Revenue is $9,551,659, as before


module b solutions

Documents

number of benches

number of model b tubs

number of tables

maximum profit

number of alpha

number of fans

yield profit

optimal point