Capps Shoes Company must meet the following demand for its two main products during the next two months.

	Month 1	Month 2
Airhead	2000	3200
Grounded	1600	2400

At the beginning of month one, 500 Airhead shoes and 400 Grounded shoes are on hand. This company has 100 workers. Workers are paid $9 per hour in month one and $9.5 per hour in month two. Each worker can work up to 160 hours in month one and 150 hours in month two before receiving overtime. Each worker can work up to 20 hours of overtime per month and is paid $13 per hour for overtime labor.

It takes four hours of labor and $15 of raw materials to produce a pair of Airhead shoes and three hours of labor and $12 of raw materials to produce a pair of Grounded shoes.

At the end of each month, a holding cost of $3 per pair of shoes left in inventory is incurred. Given the size of Cappss warehouse, a maximum of 2000 pairs of shoes can be held at the end of each month. The company owner likes to keep at least 200 pairs of each type in inventory at the end of month two as safety stock to meet unexpected demand in the future.

During these two months, buying from a subcontractor is also available. Buying a pair of Airhead shoes from the subcontractor costs $70, and its capacity is 200 shoes per month. Buying a pair of Grounded shoes from the subcontractor costs $62, and its current capacity is 150 shoes per month.

The company wants to determine its optimal production schedule to minimize its total costs and meet the demand. Let,

At: # of pairs of Airhead shoes to produce in month t, t=1, 2

Gt: # of pairs of Grounded shoes to produce in month t, t=1, 2

IAt: Inventory level of Airhead shoes at the end of month t, t=1,2

IGt : Inventory level of Grounded shoes at the end of month t, t=1,2

SAt: # of pairs of Airhead shoes to buy from the subcontractor in month t, t=1,2

SGt: # of pairs of Grounded shoes to buy from the subcontractor in month t, t=1,2

Ot: # hours of overtime used in month t , t=1,2

The linear Programming model of this problem is given in the following.

Min Z=9(4A1+3G1)+9.5(4A2+3G2)+15(A1+A2)+12(G1+G2)+13(O1+O2)+ 70 (SA1+SA2)+62(SG1+SG2)+3(IA1+IA2+IG1+IG2)

Subject to:

500 + A1 + SA1 = 2000 + IA1 (1)

IA1 + A2 + SA2 = 3200 + IA2 (2)

400 + G1 + SG1 = 1600 + IG1 (3)

IG1 + G2 + SG2 = 2400 + IG2 (4)

4A1 + 3G1 160*100 + O1 (5)

4A2 + 3G2 150*100 + O2 (6)

O1 20*100 (7)

O2 20*100 (8)

SA1200 (9)

SA2 200 (10)

SG1150 (11)

SG2 150 (12)

IG1 + IA1 2000 (13)

IG2 + IA2 2000 (14)

IA2 200 (15)

IG2 200 (16)

At, Gt, IAt,IGt, SAt, SGt, Ot0 t=1,2

Simplify the above model and write the algebraic formulation that can be solved easier in Excel. You need to have all decision variables on the left-hand side of the constraints, only one constant value on the right-hand side of the constraint and simplify the coefficients to have only one coefficient for each decision variable. In the objective function, you also need to simplify the coefficients and have only one coefficient for each decision variable.