003 005 005 005 005 006 003 004 005 005 005 005 005 002 004 004 004 004 005 005
ID: 343787 • Letter: 0
Question
003 005 005 005 005 006 003 004 005 005 005 005 005 002 004 004 004 004 005 005 002 003 004 004 004 004 004 001 013 003 004 004 004 004 001 002 003 003 003 003 003 001 002 003 003 013 003 013 000 002 002 013 003 16 44 17 00 13 39 6789 10 1 6 7 8 9 0 1 2 6 7 8 9 0 1 2 6 7 8 9 0 1 2 6 7 8 9 0 1 2 6 7 8 9 0 1 2 6 7 8 9 0 1 2 6 7 8 9 10 5. 30 3 4 4 0 0 2 2 2 2 1. 0 0 0 0 6 1 0 0 0 0 9 1 0 0 0 0 3 1 0 0 0 0 1 2 0 0 0 0 6 2 0 0 0 0 0 2 0 0 0 0 0 3 0 0 0 0 0 789 10 5 6 7 8 9 0 5 6 7 8 9 0 5 6 7 8 9 0 5 6 7 8 9 0 5 6 7 8 9 0 5 6 7 8 9 0 1 6 7 8 9 0 1 6 7 8 9 0 1 5. 013 iza 102 ass nos 000 000 000 011 iza uzi 008 /20 iza IRA 005 125 005 012 azi 102 izz 102 012 ois nis izo izo /20 13 17 18 18 18 011 15 16 16 17 009 013 014 15 15 15 008 012 09 34 10 37 19 12 29 41 13 00200 0 45670° 4 5 6 7 8 9 4 5 6 7 8 9 4 5 6 7 8 9 4 5 6 7 8 9 4 5 6 7 8 9 5 6 7 8 9 5 6 7 8 9 56789056Explanation / Answer
Arrival rate, = 4 per day
Service rate, = 5 per day
/ = 4/5 = 0.80
Refer table 2, Look for value of Lq corresponding to / = 0.80, for M=1, Lq = 3.2
Total cost per day = M*Cs + Lq*Cw = 1*1109+3.2*271 = $ 1976.2
For M=2, Lq = 0.152
Total cost per day = 2*1109+0.152*271 = $ 2259.2
Total daily cost is lower for M=1, therefore, optimal number of docks is 1
a. Number of docks = 1
b. New value of / = 4/5.71 = 0.70, for M=1, Lq = 1.633
Total cost per day = 1*(1109+100)+1.633*271 = $ 1651
This total daily cost is lower than that in part a, therefore the answer is Yes, because the daily total cost with the new equipment is $ 1651, which is lower than without the new equipment.
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