Suppose that a developer is choosing tenants for a shopping center. There are four possible tenants: a DEPARTMENT STORE, a TOY STORE, a SHOE STORE, and a HARDWARE STORE. If each store were to be located in isolation outside a shopping center it would earn a certain level of gross profit per period (this is the level of profit before subtracting out space rent). In addition, each store requires a certain number of square feet of floor space, which is the same regardless of whether or not it locates in a shopping center. The relevant values for each store type are as follows: Store Type: Gross Profit in Isolation Required Square Footage
Department $100,000 9,000
Toy $11,200 1,000
Shoe $7,800 800
Hardware $7,000 1,100
When the stores are located together, each store earns a greater gross profit from the extra customer traffic generated by the nearby locations of other stores. The increase in gross profit for each store resulting from the presence of other store types is as follows
Affected Store Type:
Department Toy Shoe Hardware
Department -- $6000 $8000 $1000
Toy $2000 -- $600 $300
Shoe $2000 $500 -- $200
Hardware $1000 $400 $200 --
Suppose that the shopping center developer charges a rent equal to each store’s gross profit, leaving the store with a net profit of exactly zero. Total rent from the shopping center is then equal to the total gross profit of all its stores. The developer’s profit is thus equal to total gross profit of all stores in the center minus the cost of providing space to the stores (that is, total rent minus the cost of providing space). Finally, suppose that due to high-quality construction and the need to provide common space around the stores, the construction cost for shopping center space is high. In particular, suppose that the cost per period per square foot of store floor space is $10.
A) Give an intuitive explanation for the pattern of incremental profits from the presence of other stores in the second table above
B) Suppose the developer were to construct a single-store shopping center (a contradiction in terms, perhaps). There are four types of such centers (a Department Store alone, a Toy Store alone, etc.). Using the information above, compute the developer’s profit from each of these four types of centers
C) Next compute the developer’s profit from the various types of two-store shopping centers (i. e., Department Store plus Shoe Store, Department Store plus Toy Store, etc.; how many possibilities are there?)
D) Finally, compute profits from the various types of three-store centers and from the single type of four-store center
E) Comparing your answers from (b), (c), and (d), identify the optimal shopping center (the one yielding the highest profit to the developer). Explain intuitively why this particular collection of stores is optimal.

Answer is explained in the explanation section below.

Explanation:

Solution:

a) Explanation:

Closeness of different shops at one place would result in the increased Profit because of the following reasons:

1. People usually wish to save extra bucks whenever possible, so in this case, they will be saving their travel cost because, they don't need to go somewhere else for other products. As they will be getting every product they need in a single complex.  

2. Secondly, because of the greater visibility of different products people would buy things they don't even need for the time being,  resulting in the increase in profits.

b) Single Store Shopping Centre:

For Department:

Rent of the store = $ 100,000

Cost ($ 10 per sq. foot) = 9000 x 10 = $ 90,000

Profit = $ 100,000 - $ 90,000  = $ 10,000

For Toy Store:

Rent of the store = $ 11,200

Cost ($ 10 per sq. foot) = 1,000 x 10 = $ 10,000

Profit = $ 11,200  - $ 10,000 = $1,200

For Shoe Store:

Rent of the store = $ 7,800

Cost ($ 10 per sq. foot) =  800 x 10 = $8000

Profit = $ 7,800 - $8000 = -$200

For Hardware Store:

Rent of the store = $7,000

Cost ($ 10 per sq. foot) = 1,100 x 10 = $11,000

Profit =  $7,000 - $11,000 =  -$4000

c) Two-Store Shopping centers:

Department + Toy :

Toy affected Score = $6000

So,

Profit of Affected Department = $10000 + $6000 = $16000

Profit of Affected Toy Store = $2000 + $1200 = $3200

Total Profit = Profit of Affected Toy Store + Profit of Affected Department

Total Profit =  $16000 + $3200 = $19200

Similarly,

For Department + Shoes:

Shoes Affected Score = $8000

So,

Profit of Affected Department = $10000 + $8000 = $18000

Profit of Affected Shoe Store =  -$200 + $2000 = $1800

Total Profit  =  $18000 + $1800 = $19800

Similarly,

For Department + Hardware:

Hardware affected Score =  $1000

So,

Profit of Affected Department = $10000 + $1000 = $11000

Profit of Affected Hardware Store =   -$4000 + $1000 = -$3000

Total Profit  = $11000 -$3000 = $8000

Similarly,

For Toy + Shoes:

Shoes affected Score =  $600

Profit of Affected Toy  = $1,200 + $600 = $1800

Profit of Affected Shoe Store =  -$200 + $500 = $300

Total Profit  = $1800 + $300 = $2100

Similarly,

For Toy + Hardware:

Hardware affected Score =  $300

So,

Profit of Affected Toy  = $1,200 + $300 = $1500

Profit of Affected Hardware Store = -4,000 + 400 = -3600

Total Profit  =  $1500 -3600 = -2,100

Similarly,

Shoes + Hardware:

Hardware affected Score = $200

Profit of Affected Shoes  = -200 + 200 = 0

Profit of Affected Hardware Store = -4,000 + 200 = -3800

Total Profit  = 0 - 3800 = -3800

Similarly,

d) Three-store Shopping centers:

For Department + Toy + Shoes:

Department Profit = 10,000 + 6,000 + 8,000 = $24000

Toy Profit  = 1,200 + 2,000 + 600 = $3800

Shoes Profit = -200 + 2,000 + 500 = $2300

Total Profit = $24000 + $3800 + $2300 = 30,100

For Department + Toy + Hardware:

Department Profit = 10,000 + 6,000 + 1,000 = $17000

Toy Profit = 1,200 + 2,000 + 300 = $3500

Hardware Profit = -4,000 + 1,000 + 400 = -$2600

Total Profit = $17000 +  $3500 - $2600  = 17,900

Department + Shoes + Hardware:

Department Profit =  10,000 + 8,000 + 1,000

Shoes Profit = -200 + 2,000 + 200

Hardware Profit = -4,000 + 1,000 + 200

Total Profit = 18,200

Toy + Shoes + Hardware:

Toy Profit = 1,200 + 600 + 300

Shoes Profit =  -200 + 500 + 200

Hardware Profit = -4,000 + 400 + 200

Total Profit = -800

Four store Shopping Centre:

Department + Toy + Shoes + Hardware:

Department Profit =  10,000 + 6,000 + 8,000 + 1,000

Toy Profit = 1,200 + 2,000 + 600 + 300

Shoes Profit = -200 + 2,000 + 500 + 200

Hardware Profit = -4,000 + 1,000 + 400 + 200

Total Profit = 29,200

E)

Now, if you see and compare between the profits of above calculated different category stores, then maximum profit is gained by 3 store shopping center. So, the optimal store is 3 store shopping center.

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