Question

Question 12.14. The storage rack of a Unit Load AS/RS system is 300 ft long and 75 ft high. The I/O point is located at the lower left-hand corner of the rack. The S/R machine travels at 300 FPM in the horizontal direction and 60 FPM in the vertical direction. The time required for a pick-up or deposit operation is 0.25 minutes.

A. What is the expected time to complete a SC cycle, assuming shared (random) storage? What is the corresponding throughput per hour, measured in operations per hour?

B. What is the expected time to complete a DC cycle, assuming shared (random) storage? What is the corresponding throughput per hour, measured in operations per hour? What is the percent increase in throughput if one uses a DC cycle?

C. Assume that during peak activity 75% of the operations are retrievals, 20% of the retrievals are performed on a DC basis, and 8% of the operation time must be set aside for maintenance. How many operations per hour can an S/R machine handle during peak activity?

D. A preliminary stock assignment plan has been worked out that would concentrate approximately 80% of the activity with a rectangular sub-region A whose northeast vertex is (100, 37.5), measured in feet, and whose southwest vertex is the I/O point (0, 0). Within region A and its complement, region B, the activity is uniformly distributed. What is the expected percent increase in throughput? Assume all commands are single-cycle

Answer 1

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Convert the rack dimensions from feet to time (minutes)

So, Length wise

300 feet/300 FPM = 1 minute

Height wise

75 feet/60 FPM = 1.25 minutes

The total average time to complete an SC cycle is the sum of the average S/R travel time and 2 times pick-up/deposit operations.

E = T(1 + Q^{2}/3)

T = max (H,V) and Q = min(H,V)/T

H - Horizontal time

V - Vertical time

Therefore,

T = max (1,1.25) = 1.25

Q = min(1,1.25)/1.25 = 0.8

E = 1.25(1+(0.8)^2 / 3) = 1.25 * (1 + 0.21) = = 1.25 * 1.21 = = 1.51

The total average time to complete an SC cycle = 1.51 + 2*0.25 = 1.51 + .50 = 2.01 minutes

Throughput time = 60 mins / 2.01 minutes = 29.85 operations per hour.

DC cycle time

The total average time to complete a DC cycle = sum of the average S/R travel time + 4 * pick-up/deposit operations.

E = T/30 [ 40 + 15*(Q)^2 - (Q)^3 ] = 1.25/30 [ 40 + 15*(0.80)^2 - (0.80)^3] = 0.04 [ 40 + 9.6 - 0.51 ] = 0.04 * 49.09 = 1.96 minutes

The total average time to complete an DC cycle = 1.96 + 4*0.25 = 1.96 + 1.00 = 2.96 minutes

DC Cycles per hour = 60 mins / 2.96 mins = 20.27

Operations per hour = DC cycles per hour * 2 = 20.27 * 2 = 40.54 (Throughput time)

Here The total average time to complete an DC cycle < 2 * The total average time to complete an SC cycle

which is, 2.96 < 2*2.01

Percentage increase in throughput if one uses the DC cycle = [(40.54-29.85) / 29.85] * 100 = [10.69/29.85] * 100 = 39.54 %