The Solution of Homework #2(Isye 6202)

The solution of homework #2(ISyE 6202)

The Solution of Homework #2

#12.22

We have 6 product families and 40 storage bays that can be used for storing product families. For further reference, we number each location as follows:

Then the problem can be solved as follows:

Assumptions:

• We consider single-command type of operation.
• Distance is computed to/from the center of each location.
• The activity associated with each location is distributed evenly between receiving and shipping (otherwise, we would have accumulation of material in the warehouse or “systematic” shortages).
• Distance is measured according to the rectilinear metric.

Parameters:

• I = the index set of sku’s: I = {1,2, …, 6}
• J = the index set of locations: J = {1,2,…, 40}
• THi = number of units of sku i handled per unit of time
• Ni = number of storage locations allocated to sku i
• dj = expected travel distance per unit load for location j

Decision variables:

• Xij = 1 if location j is allocated to sku i; 0 otherwise for all iI and jJ.

From the problem statement, we have the following data with respect to each product family:

where, Ni =storage locations required by SKU I = Area to be occupied by SKU I / (20 ft * 20ft) and THi = throughput rate of SKU I.

Since , this problem corresponds to a balanced transportation problem and its formulation is:

Since the cost cij associated with variable xij has the product form of a term associated with SKU I (THi / Ni)and a term associated with location j (dj), We can get the optimal solution of the above problem using the following matching procedure:

• Rank all the available storage locations in increasing distance of dj
• Rank all sku’s in decreasing turns of THi/ Ni
• Move down the two lists, assigning the next most highly ranked sku i to the next Ni locations

Now we can compute the expected distance for each location as follows:

where, Lin,j and Lj,out represent the distance from the location j to the receiving and shipping dock respectively and dj=0.5*(Lin,j + Lj,out). Using the actual layout of the 40 locations, we get the following ranking profile for them:

The rank of each sku has been already computed in the first table above. So, following the matching procedure outlined above, an optimal allocation is as follows:

Notice that there are several alternative optimal solutions. The one provided above seeks also to cluster all the locations assigned to each SKU in the same area of the warehouse, since this would facilitate the better monitoring and management of the warehouse activity.

# 12.24

We have 3 products (X, Y, and Z) and 44 storage bays that can be used for storing products. Again, we number locations as follows:

Assumptions:

• We only consider single command case.
• Distance is computed to/from the center of each location considering round trip.
• In an aisle, a worker/vehicle moves along the center of the aisle.
• Travel distances are computed as shortest paths through the depicted warehouse aisle structure.

Since , this problem corresponds to a balanced transportation problem and we can use the same definitions for parameters and variables and the same formulation as in the problem #12.22. We can compute the following:

Now we can apply the same optimal solution procedure as in #12.22 and the computation of the expected distance for each location is as follows (Notice that in computing these distances, we have taken advantage of the symmetry of the warehouse layout with respect to the central horizontal axis).

The location ranking based on the expected travel distances computed above is as follows:

So an optimal solution is:

The solution of homework #2(ISyE 6202)

# 5.3

We are considering dedicated storage policy for three products (1, 2, and 3). According to the problem, the (daily) storage requirement for each product is a discrete random variable distributed according to a uniform distribution, defined by the following probability mass functions:

Let Qi denote the number of storage locations allocated to product i (i=1,2,3). Then, we can formulate the problem as follows:

Since , we can compute the feasible range of each Qi:

Then we can solve the problem using Dynamic Programming, as we did in class. The optimal solution is Q1*=100, Q2*=150, and Q3*=200. (The detailed computation can be found in the attached tables.)

1)Q1=88 2) Q1=89 3)Q1=90 4) Q1=91

5)Q1=92 6) Q1=93 7)Q1=94 8) Q1=95

9)Q1=96 10) Q1=97 11)Q1=98 12) Q1=99

13) Q1=100

# 5.10

For this problem, we have the following conditions.

Assumptions:

• No salvage value at the end of 500 periods
• We do not consider operating cost of storage after construction (C1 = $ 0)

Parameters:

• T = length of planning horizon (T=500)
• C0 = construction cost per storage position with C0 = $63
• C2 = leasing cost per storage position with C2 = $0.3
• dt = demand of period t

Variables:

• N = storage space to be constructed in order to minimize total costs over the 500 periods

Then we can formulate the problem as follows:

As it was shown in class, this is equivalent to the following optimization problem:

where

• F = the number of periods with storage demand dt > N

This last problem indicates the following solution procedure:

The optimal number of storage locations to construct for this item is equal to the first demand value for which the corresponding partial sum exceeds 210; therefore, N* = 3600.

# 2.

Let yi (i=1,2,3) denote the number of new storage locations to be allocated to each SKU i.

Obviously, 0yi5 with y1+y2+y3=5 for all i=1,2,3. Also, let Si = Ni + yi, where Ni is the current number of locations allocated to SKU i. Then, our problem is formulated as follows:

Now we can apply Dynamic Programming for solving this problem. We can get Fi(Si) from the table of Poisson distribution (c.f., pg. 264 in Francis et. al.)

1) S3=24 2) S3=25 3) S3=26

4) S3=27 5) S3=28 6) S3=29

From the above calculation we can see that the optimal solution is S1*=14, S2*=20, S3*=25 and the maximal service level is 0.746. For comparison, the original service level, with N1=12, N2=18, N3=24, was F1(12)F2(18)F3(24) = 0.7916*0.8195*0.8432 = 0.547.

APPENDIX : Solution of # 5.3 considering the storage requirement distributions as continuous random variables.

Remember that the considered optimization problem is formulated as follows:

where, Fi is the cumulative distribution function of Xi.

is equivalent to . And since , we can compute that .------(1)

But since Fi(Qi) is a monotone increasing function with respect to Qi, and the problem is the minimization problem of the sum of continuous random variables(Qi’s), the optimal solution exists on the boundary points of the feasible region. Hence, the optimal solution satisfies the constraint as equality, and the above problem is equivalent to the following:

We can prove that optimality is attained on the boundary of the feasible region as follows: Suppose that Qi*(i=1,2,3) is an optimal solution that is not on the boundary; i.e.,. Then there exists a Qi*, let’s say Q1*, such that . But this contradicts the optimality of Qi*(i=1,2,3), i.e., the sum of Qi*’s is not minimal among the triplets of Qi’s satisfying the constraint.

Now since

we can obtain an equivalent two-variable minimization problem by solving explicitly constraint (2) for one of the problem variables, e.g., Q3. Then,

(3)

and substituting this expression in the objective we have:

s.t

50  Q1  100(4)

0  Q2  200(5)

100 200(6)

Next notice that

So z is a monotone decreasing function with respect to Q1 and for any value of Q2, it is minimized at Q1=100.

When setting Q1=100, we have

So z(100,Q2) is a convex function with respect to Q2, and it is minimized at Q2=122.47, since . But from condition (6), Q1=100 => Q2150, and since z is a monotone increasing function with respect to Q2 in [122.47, 200], it follows that: Q2=150. If we substitute Q1=100 and Q2=150 to equation (3), we obtain Q3=200. So the optimal solution is Q1*=100, Q2*=150 and Q3*=200.