B.1 Let X = Number of Standard Model to Produce

MODULE BEND-OF-MODULE PROBLEMSB.1 Let x = number of standard model to produce y = number of deluxe model to produce Maximize 40x + 60y Subject to 30x  30y  450 10x 15y  180 x  6 x, y  0B.2 y 2018161412 Optimal x  0, y  10 1086420 2 4 6 8 10 12 14 16 18 20 xFeasible 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 y 201816141210 Optimal x  4, y  8 86420 2 4 6 8 10 12 14 16 18 20 xFeasible 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. x2 400350300250200150 Feasible Region 100500 50 100 150 200 250 300 350 400 450 x12 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 x1  number of liver flavored biscuits in a package x2  number of chicken flavored biscuits in a package Minimize x1  2x2 Subject to x1  x2  402x1  4x2  60 x1  15 x1, x2  0 (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 160140 b Drilling Constraint 1201008060 c40 Wiring Constraint Feasible Region 20 d 0 20 40 60 80 100 120 Number of Air Conditioners: x1Let x1  number of air conditioners to be produced x2  number of fans to be produced Maximize 25x1 15x2 Subject to 3x1  2x2  240 wiring2x1 1x2  140 drilling x1, x2  0 nonnegativity Profit: @ a: (x1  0 , x2  0) Obj  $0 @ b: (x1  0 , x2  120) Obj  25  0 15 120  $1,800 @ c: (x1  40, x2  60 ) Obj  25  40 15  60  $1,900 * @ d: (x1  70 , x2  0) Obj  25  70 15  0  $1,750 The optimal solution is to produce 40 air conditioners and 60 fans each period. Profit will be $1,900. Quantitative Module B: Linear Programming 3 B.8 300250 a 200 x2 150 b100 Feasible Region 50 c 0 50 100 150 200 250 300 x1Let x1  number of Model A tubs produced x2  number of Model B tubs produced Maximize 90x1  70x2 Subject to 125x1 100x2  25,000 steel20x1  30x2  6,000 zinc x1, x2  0 nonnegativity Profit: @ a: (x1  0 , x2  200 ) Obj  90  0  70  200  $14,000.00 @ b: (x1  85.71, x2  142.86 ) Obj  90  85.71 70 142.86  $17,714.10 @ c: (x1  200 , x2  0) Obj  90  200  70  0  $18,000.00 * The optimal solution is to produce 200 Model A tubs, and 0 Model B tubs. Profit will be $18,000. B.9 300250200 x2 150 a 100 Optimal Solution50 Feasible Region b c 0 50 100 150 200 250 300 350 x1Let x1  number of benches produced x2  number of tables produced Maximize 9x1  20x2 Subject to 4x1  6x2  1,200 hours10x1  35x2  3,500 board-feet x1, x2  0 nonnegativity4 Instructor’s Solutions Manual t/a Operations Management Profit: @ a: (x1  0 , x2  100) Obj  9  0  20 100  $2,000.00 @ b: (x1  262.5, x2  25 ) Obj  9  262.5  20  25  $2,862.50 * @ c: (x1  300 , x2  0) Obj  9  300  20  0  $2,700.00The 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 40 Optimal Solution 30 Feasible Region is Heavily Shaded Line a x2 20 b 100 10 20 30 40 50 x1Let x1  number of Alpha-4 computers x2  number of Beta-5 computers Maximize: 1200x1 1800x2 Subject to 20x1  25x2  800 total hours x1  10 Alpha-4s x2  15 Beta-5s x1, x2  0 nonnegativity Profit: @ a: (x1  10, x2  24 ) Obj  1200 10 1800  24  $55,200 * @ b: (x1  21.25 , x2  15) Obj  1200  21.25 1800 15  $52,500 The optimal solution is to produce 10 Alpha-4 and 24 Beta-5 computers per period. Profit is $55,200. Quantitative Module B: Linear Programming 5 B.11 505x1  3x2  150 4030 x1  2x2  10 x2 20 aFeasible Region 3x 5x 150 10 1  2 0 10 20 30 40 50 x1Note that this problem has one constraint with a negative sign.The optimal point, a, lies at the intersection of the constraints: 3x1  5x2  1505x1  3x2  150To solve these equations simultaneously, begin by writing them in the form shown below: 3x1  5x2  1505x1  3x2  150Multiply the first equation by 5, the second by –3, and add the two equations and solve for x2: x2  300 16  18.75 . Given: 3x1  5x2  150 then 3x1  150  5x2  150  5 18.75 and 56.25 x   18.75 1 3 Thus, the optimal solution is: x1  18.75, x2  18.75 The profit is given by: C  4x1  4x2  4 18.75  4 18.75  $1506 Instructor’s Solutions Manual t/a Operations Management B.12 1008x1  2x2  160 x2  70 8060 3x1  2x2  120 x2 40 Feasible Region20 a x1  3x2  80 0 20 40 60 80 100 x1The optimal point, a, lies at the intersection of the constraints: 3x1  2x2  120 x1  3x2  90To solve these equations simultaneously, begin by writing them in the form shown below: 3x1  2x2  120 x1  3x2  90Multiply the second equation by –3, and add it to the first and solve for x2:x2  150 7  21.43. Given: 3x1  2x2  120 then 3x1  120  2x2  120  2  21.43 and 77.14 x   25.71 1 3Thus, the optimal solution is: x1  25.71, x2  21.43 The cost is given by: C  x1  2x2  25.71 2  21.43  $68.57B.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 x1x2 term. B.14 (a) Using software, we find that the optimal solution is: x1  7.95 x2  5.95 x3  12.60 Profit  $143.76(b) There is no unused time available on any of the three machines. Quantitative Module B: Linear Programming 7 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 Xij  number of students bused from sector i to school j. Objective: minimize total travel miles  5XAB  8XAC  6XAE  0XBB  4XBC 12XBE  4XCB  0XCC7XCE  7XDB  2XDC  5XDE 12XEB  7XEC  0XEE subject to XAB  XAC  XAE  700 number students in sector AXBB  XBC  XBE  500 number students in sector BXCB  XCC  XCE  100 number students in sector CXDB  XDC  XDE  800 number students in sector DXEB  XEC  XEE  400 number students in sector EXAB  XBB  XCB  XDB  XEB  900 school B capacityXAC  XBC  XCC  XDC  XEC  900 school C capacityXAE  XBE  XCE  XDE  XEE  900 school E capacity(b) Solution: XAB  400XAE  300XBB  500XCC  100XDC  800XEE  400 Distance  5,400 “student miles”B.17 Because the decision centers about the production of the two different cabinet models, let: x1  number of French Provincial cabinets produced per day x2  number of Danish Modern cabinets produced each day The equations become: Objective 28x1  25x2 (Maximize revenue) Subject to 3x1  2x2  360 hours, carpentry1.5x1 1x2  200 hours, painting0.75x1  0.75x2  125 hours, finishing x1  60 units, contract x2  60 units, contract x1, x2  0 nonnegativity The solution is: x1  60 , x2  90 , Revenue = $3930/day B.18 Problem Data Solution Workers Hire Hire Hire 8 Instructor’s Solutions Manual t/a Operations Management Period Time Period Required Solution 1 Solution 2 Solution 3 1 3 AM–7 AM 3 0 3 3 2 7 AM–11 AM 12 16 9 14 3 11 AM–3 PM 16 0 7 2 4 3 PM–7 PM 9 9 2 7 5 7 PM–11 PM 11 2 9 4 6 11 PM–7 AM 4 3 0 0 S = 30 S = 30 S = 30Let xi  number of workers reporting for the start of work in period i, where i  1, 2, 3, 4, 5, or 6. The equations become: Objective: x1  x2  x3  x4  x5  x6 (Minimize staff size) Subject to: x1  x2  12 x2  x3  16 x3  x4  9 x4  x5  11 x5  x6  4 x1  x6  3 x1, x2, x3, x4, x5, x6  0 Note that three alternate optimal solutions are provided to this problem. Either solution could be implemented using only 30 staff members. B.19 2015 Feasible Region x2 102x1 1x2  20 5 a x2  5 0 2 4 6 8 10 12 14 16 18 20 x x1Quantitative Module B: Linear Programming 9 Define the following variables: x1  thousands of round tables produced per month x2  thousands of square tables produced per monthThe appropriate equations then become: Objective: 10x1  8x2 (minimize handling and storage costs) Subject to x2  5 square tabletop contract2x1 1x2  20 total labor capacity x1, x2  0 nonnegativity Cost: @ a: (x1  7.5, x2  5 ) Obj  10  7.5  8  5  $115* @ b: (x1  0 , x2  20 ) Obj  10  0  8  20  $160 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: 9x1 12x2 (maximize) Subject to: x1  x2  10 gallons, varnish x1  2x2  12 lengths, redwood where: x1  number of coffee tables/week x2  number of bookcases/week Optimal: x1  8, x2  2, Profit = $96 B.21 102nd corner point 8Optimal corner 6 b c x2 4 1st corner point 24 th corner point a d 0 2 4 6 8 10 x1The original equations are: Objective: 3x1  5x2 (maximize) Subject to: x2  63x1  2x2  18 x1, x2  0 nonnegativity The optimal solution is found at the intersection of the two constraints. Solving for the values of x1 and x2 at the intersection, we have: 10 Instructor’s Solutions Manual t/a Operations Management x2  6 18  2x 18  2  6 6 x  2    2 1 3 3 3Profit  3x1  5x2  3  2  5  6  6  30  $36B.23 Let x1  the number of class A containers to be used x2  the number of class K containers to be used x3  the number of class T containers to be used The appropriate equations are: Maximize: 8x1  6x2 14x3 Subject to: 2x1  x2  3x3  120 Material2x1  6x2  4x3  240 Time x1, x2, x3  0 nonnegativity Using software we find that the optimal solution is: 1 2 x1  0 , x  17 , x  34 2 7 3 7 or x1  0 , x2  17.143 , x3  34.286 and 6 Profit  582  $582.86 7B.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 x1  number of newspaper ads placed x2  number of TV spots purchased Given that we are to minimize cost, we may develop the following set of equations: Minimize: 925x1  2000x2 Subject to: 0.04x1  0.05x2  0.40 city exposure0.03x1  0.03x2  0.60 exposure in NW suburbs 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: x1  20 ads, x2  0 TV spots, cost = $18,500 B.26 (a) Minimize: 6x1a  5x1b  3x1c  8x2a 10x2b  8x2c 11x3a 14x3b 18x3cQuantitative Module B: Linear Programming 11 Subject to: x1a  x2a  x3a  7 x1b  x2b  x3b  12 x1c  x2c  x3c  5 x1a  x1b  x1c  6 x2a  x2b  x2c  8 x3a  x3b  x3c  10 All variables  0 (b) Solution: x1b  6 x2b  3 x2c  5 x3a  7 x3b  3 Cost  $219B.27 400 Labor Hours300Optimal Solution 200 Feasible Region 100 Production Limit0 50 100 150 200 250 300 350 400 450 Boy’s BicyclesMaximize: 57x1  55x2 Subject to: x1  x2  3902.5x1  2.4x2  960@ (x1  384 , x2  0) Obj  57  384  55  0  $21,888 @ (x1  0 , x2  390 ) Obj  57  0  55  390  $21,450 @ (x1  240 , x2  150) Obj  57  240  55 150  $21,930 *The optimal solution occurs at x1  240 boy’s bikes, x2  150 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 x1  number of medical patients x2  number of surgical patients The appropriate equations are: Maximize 2280x1 1515x212 Instructor’s Solutions Manual t/a Operations Management Subject to: 8x1  5x2  32,850 patient days available3.1x1  2.6x2  15,000 lab tests1x1  2x2  7,000 X rays x2  2,800 operations x1, x2  0 nonnegativity Optimal: x1  2790 , x2  2104, Profit = $9,551,659 or: 2790 medical patients, 2104 surgical patients, Profit $9,551,659 Beds required: Use: Medical: 8  2790  22,320 Surgical: 5  2104  10,520 32,840 22,320 Medical uses:  68%  61 beds 32,840 10,520 Surgical uses:  32%  29 beds 32,840Here is an alternative approach that solves directly for the number of beds: Maximize revenues  104,025x1 110,595x2 Subject to: x1  x2  90 beds8x1  5x2  32,850 patients yr141.44x1 189.8x2  15,000 lab tests45.63x1 146x2  7,000 x-rays73x2  2,800 operations where x1  no. of medical beds = 61.17 x2  no. of surgical beds = 28.83 Revenue is $9,551,659, as before Quantitative Module B: Linear Programming 13

B.1 Let X = Number of Standard Model to Produce

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