Solutions Assignment4

7/23/2019 Solutions Assignment4

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Assignment 4 solutions, BUS 220 – Introduction to Decision Sciences

Problem 1: (30 points)

The spreadsheet for this problem is as follows:

Selected cell formulas are as follows:Cell FormulaB9 =NORMINV(RAND(),$C$4,$C$5)F10 =COU NTIF(B9:B508,”>40000”)

a. Most simulations will provide between 105 and 130 tires exceeding 40,000miles. The percentage should be roughly 24%.

b.

Mileage In Most

Simulations Number of Tires

ApproximatePercentage

32,000 80 to 100 18%

30,000 42 to 55 10%

28,000 18 to 30 4%

c. Of mileages considered, 30,000 miles should come closest to meeting thetire guarantee mileage guideline.

Problem 2

a.(10 points)Let

x1 = number of units of product 1 produced

x2 = number of units of product 2 produced

1

2

4

5

6

8

9

10

11

12

13

14

15

17

18

19

20

22

23

A B C D E F G

Grear Ti re Co mpany

Tire MileageMean 36500Standard Deviation 5000

Simulation ResultsTire Mileage

1 38,379 Mileage Number Percent2 36,597 Exceed 40,000 118 23.6%3 28,820 Less Than 32,000 88 17.6%4 38,387 Less Than 30,000 48 9.6%5 39,638 Less Than 28,000 25 5.0%

6 34,548

Note: To reconstruct the com plete speadsheet: 1. Block rows 21 to 505 2. On the Insertmenu, clickRows 3. Copy row 14 (Tire 6) to fill rows 15 to 506. Trial 500 wil l appear in row 508 of the spreadshe et.

499 34,613500 38,730



2

Min P 1( d 1

) + P 1( d 1

) + P 1( d 2

) + P 1( d 2

) + P 2( d 3

)

s.t. 1 x1 + 1 x2 - d

1

+ d 1

= 350 Goal 1

2 x1 + 5 x2 - d 2

+ d 2

= 1000 Goal 2

4 x1 + 2 x2 - d 3

+ d 3

= 1300 Goal 3

x1, x2, d 1

, d 1

, d 2

, d 2

, d 3

, d 3

0

b. (10 points)In the graphical solution, point A provides the optimal solution. Note thatwith x1 = 250 and x2 = 100, this solution achieves goals 1 and 2, but

underachieves goal 3 (profit) by $100 since 4(250) + 2(100) = $1200.

0 100 200 300 400 500

100

200

300

400

500

600

700

o a l 1

o a l 2

A (250, 100)

B (281.25, 87.5)

o a l 3

x2

x1



3

c. (10 points)Max 4 x1 + 2 x2

s.t. 1 x1 + 1 x2 350 Dept. A

2 x2 + 5 x2 1000 Dept. B

x

1, x

2

0

The graphical solution indicates that there are four extreme points. The profitcorresponding to each extreme point is as follows:

Extreme Point Profit 1 4(0) + 2(0) = 0

2 4(350) + 2(0) = 1400

3 4(250) + 2(100) = 1200

4 4(0) + 2(250) = 500

Thus, the optimal product mix is x1 = 350 and x2 = 0 with a profit of $1400.

d. (10 points)The solution to part (a) achieves both labor goals, whereas the solution to part (b) results in using only 2(350) + 5(0) = 700 hours of labor indepartment B. Although (c) results in a $100 increase in profit, the

problems associated with underachieving the original department labor goal by 300 hours may be more significant in terms of long-term considerations.

. x2

x1

0 100 200 300 400 500

100

200

300

400

1

3

4

2

(250,100)

(0,250)

(0,0)

(350,0)

D e p a r t m e n t B

Fe asible Re gion

D e p a r t m

e n t A



4

Problem 3

a.

(10 points)

Note that getting at least 10,000 customers from group 1 is equivalent to x1 =

40,000 (25% of 40,000 = 10,000) and getting 5,000 customers is equivalent to x2 =

50,000 (10% of 50,000 = 5,000). Thus, to satisfy both goals, 40,000 + 50,000 =

90,000 letters would have to be mailed at a cost of 90,000($1) = $90,000.

Let x1 = number of letters mailed to group 1 customers

x2 = number of letters mailed to group 2 customers

d 1

= number of letters mailed to group 1 customers over the desired 40,000

d 1

= number of letters mailed to group 1 customers under the desired 40,000

d 2

= number of letters mailed to group 2 customers over the desired 50,000

d 2

= number of letters mailed to group 2 customers under the desired 50,000

d 3

= the amount by which the expenses exceeds the target value of $70,000

d 3

= the amount by which the expenses falls short of the target value of $70,000

Min P 1( d 1

) + P 1( d 2

) + P 2( d 3

)

s.t. x1 - d

1

+ d 1

= 40,000 Goal 1

x2 - 1 d 2

+ 1 d 2

= 50,000 Goal 2

1 x1 + 1 x2 - d 3

+ d 3

= 70,000 Goal 3

x1, x2, d 1

, d 1

, d 2

, d 2

, d 3

, d 3

0

b. (10 points) Optimal Solution: x1 = 40,000, x2 = 50,000

c. (10 points) Objective function becomes

min P 1( d 1

) + P 1(2 d 2

) + P 2( d 3

)

Optimal solution does not change since it is possible to achieve both goals 1 and 2in the original problem.

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Solutions Assignment4