SOLUTIONS:  CHAPTER 6

2.  desired output - 33 1/3 units per hour

     operating time = 60 minutes per hour

CT = operating time / desired output  or 60/33.33 or 1.8 minutes per unit

a&b) station    tasks    time

            1            a          1.4

             2            b,e      1.3

             3            d,e,f    1.8

              4           g,h       1 .5

                                      6.0

efficiency = total time/(ct* number of stations) or  6.0/7.2 or 83.3%

4.    Minimum time = 1.3 minutes

Task    following tasks

    a            4

    b            3

    c            3

    d            2

    e            3

    f            2

    g            1

    h            0

Station        Eligible    Assign    Time Left    Idle time

i                    a            A            1.1

                    b,c,e        B              .7

                                    C              .4

                                    D              .3                .3

II                    d            E               .0                 .0

III                   f,g          F                .5 

                                    G                .2                .2               

iV                    h            H               .1                 .1       

                                                                         .6

Idle percentage + the sum of idle times divided by (number of stations * ct)  

or idle % = 6/4(1.3) or 11.54%

PROBLEM #7

    cT = OT/D = 7(60)/500 or .84 minutes or 50.4 seconds

N = sum of t/CT or 193/50.4 or 3.83 which requires  4 stations

I'm not posting the tables in the interest of time but for part d the answer is :  I = 1- 193/50*5 or 22.8%

PROBLEM 11 (I am not drawing the figure, but the contents are:

    1    5    4

    3    8    7

    6    2

PROBLEM # 15

    Several answers are possible so long as departments 2, 4, & 8 are close to 3, department 4 is close to 5 and 5 is close to 1

5    

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